INTRODUCTION TO
NUMBER THEORY BY
TRYGVE NAGELL Professor of Mathematics
University of Uppsata
JOHN WIL.EY & SONS, II7G. NEW YORK ALMQVIST & WIKSELL, STOCKHOLM
Printed in Sweden. IMP: GALA, 151 ALJMQVIST & WIESELLS BOKTRYCKER1 AB
PREFACE
Natural number is the original mathematical concept and the most fundamental. Speculations about the nature and properties of whole numbers doubtless constitute the oldest form of Inathematical thought. It is known that the Sumerians and Babylonians as well as the Ancient Egyptians had a fair knowledgeV of the properties of natural numbers. But first in connection with the Greeks is it
possible to speak of a proper theory of numbers. Pythagoras (circa 500 B. C.) and his pupils pursued extensive studies in the field of integers. The first systematic presentation of results in number theory with proof is to be found in Euclid's E1cviiruta (circa 300 B. C.). Among the later Greek mathematicians, Diophantos (circa A. D. 350) was of the greatest significance in the development of number theory; six of the thirteen books of his Arztlunetir have been preserved. It is also certain that number theory has a very old tradition in India. where it flourished during the period between A. D. 500 and 1200. Western Europe became acquainted with Greek mathematics mainly through the agency of the Arabs. But development was slow, and we cannot speak of an independent Western theory of numbers before the seventeenth century. The French mathematician Fermat (111011GG5) may rightly be regarded as the father of more recent number theory. Its further development before the nineteenth century was associated chiefly with the names of Euler (17071783). Lagrange (17361813), Legendre (1752 1833) and Gauss (17,4718.55). The first textbook in the theory of numbers was published in 17118 by Legendre under the title sur la thc%orie des nombres. But the really basic work is Gauss's book Di.cquisitiocaes
which appeared in 1801.
With that work number theory became a systematic science. Gauss himself considered that it was the greatest of all his works.
6
PREFACE
His opinion on the importance of number theory is expressed in his remark: "Mathematics is the queen of the sciences, and the theory of numbers is the queen of mathematics." The last hundred years have been characterized by an intensive development of number theory in many different directions. It is the aim of this book to give the reader a brief introduction to the most important results in the elementary theory of numbers. The book reproduces, in the main, lectures which I have given at the University of Uppsala. It should be possible for those with only the elementary college foundations of arithmetic and algebra
to read the greater part. Sections 27, 28 and 29 together with Chapters V and VII require a slightly wider knowledge of algebra.
In Sections 13. 16 and 17 and in Chapter VIII some simple results from analysis are used. Most of the exercises are not of a routine character but are really intended to supplement the theory with known and new results which are not otherwise included in the text. I should like to express my warmest thanks to Professor Dr.
Ernst Jacobsthal and to Dr. Sven Gellerstedt for their valu able help Uppsala, December 1950. TaYGVE NAGELL.
CONTENTS
CHAPTER I DIVISIBILITY I'agc
Section
1.
Divisors
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2. Remainders
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Primes
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12 13
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16 19
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4. The fundamental theorem . . . . . . . . . . . . . 5. Least common multiple and greatest common divisor li. Moduls, rings and fields . . . . . . . . . . . . 7. Euclid's algorithm . . . . . . . . . . . . . . . . 8 Relatively prime numbers. Euler's pfunction . . . 9. Arithmetical functions . . . . . . . . . . . . . . 10. Diophantine equations of the first degree . . . . . ii. Lattice points and point lattices . . . . . . . . 12. Irrational numbers . . . . . . . . . . . . . . . . 13;. Irrationality of the numbers a and :r . . . . . . .
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23 26 29 32 34 38 40
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J'xerei sr c (140)
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CHAPTER 11
ON THE DISTRIBUTION OF PRIMES 14. Some lemmata . . . . . . . . . . . . . 15. General remarks. The sieve of Eratosthenes 16. The function :r (.vl . . . . . . . . . . . . .
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47
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51
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17. Some elementary results on the distribution of primes 57 18. Other problems and results concerning primes . . . . 64 CHAPTER III
THEORY OF CONGRUENCES 19. Definitions and fundamental properties . . . . . 20. Residue classes and residue systems . . . . . . 21. Fermat's theorem and its generalization by Euler
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6$ 69
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71
CONTENTS
8
Page
Section
22. Algebraic congruences and functional congruences . 23. Linear congruences . . . . . . . . . . . . . . . . 24. Algebraic congruences to a prime modulus . . . . . 25. Prime divisors of integral polynomials . . . . . . . 26. Algebraic congruences to a composite modulus . . . 27. Algebraic congruences to a primepower modulus . . 28. Numerical examples of solution of algebraic congruences
73 7 (i
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83 85 510
of integral polynomials with regard to a
29. Divisibility
prime modulus
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30. Wilson's theorem and its generalization . 31. Exponent of an integer modulo u . . . 32. _Moduli having primitive roots
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33. The index calculus . . . . . . . . 34. Power residues. Binomial congruences 35. Polynomials representing integers. . . .
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93 99 102 107 111 115 120
36. Thue's remainder theorenl and its generalization by Scholz 122 (4189)
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CIIAPTER IV
THEORY OF QUADRATIC RESIDUES 37. The general quadratic congruence . . . . . . . . . 38. Euler's criterion and Legendre's symbol . . . . . . 39. On the solvability of the congruences .c 2 =  2 (mod p) 40. (,auss's lemma . . . . . . . . . . . . . . . . . . 41. The quadratic reciprocity law . . . . . . . . . 42. Jacobi's symbol and the generalization of the reciprocity law . . . . . . . . . . . . . . . . . . . 43. The prime divisors of quadratic polynomials . . . . . 44. Primes. in special arithmetical progressions . . . . .
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132 133 136 139 141 14.5
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CHAPTER V
ARITHMETICAL PROPERTIES OF THE ROOTS OF UNITY 45. The roots of unity . . . . . lS(i . . . . . . . . . .
46. The cyclotomie polynomial . . . . . . . . 47. Irreducibility of the cyclotoinic polynomial
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158 160
CONTENTS
9
nc tiu
Page
48. The prime divisors of the cyclotomic polynomial .
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49. A theorem of Bauer on the prime divisors of certain polynomials
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50. On the primes of the form u y  1
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51. Some trigonometrical products . 52. A polynomial identity of Gauss
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168 170 173 174
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53. The Gaussian suns
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177
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180
CHAPTER V1
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
54. The representation of integers as sums of integral squares
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55. Bachet's theorem . . . . . . . . . . 56. The Diophantine equation .r2  D y2 = 1 5 4. The Diophantine equation x2  D !/2    1 C' 58. The Diophantine equation if' D 59. Lattice points on conics . . . . . . . . 60. Rational points in the plane and on conics 61. The Diophantine equation a x2 + 1. r/2 ± r:2 .
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CHAPTER VII
DIOPHANTINE EQUATIONS OF HIGHER DEGREE 02. Some Diophantine equations of the fourth degree with three unknowns 227 . . . . 63. The Diophantine equation 2 ., 4  y4 = 22 . . 232 . 64. The quadratic fields K(11 11. K ('h _') and K (V) 235 .
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66. Diophantine equations of the third degree with an infinity of solutions . . . . . . . . . . . 246 . . 248 . 117. The Diophantine equation .<' + 9' ._ = 0 . . . . 251 68. Fermat's last theorem . . . . . . . . . . 69. Rational points on plane algebraic curves. Mordell's
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CONTENTS
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Page
Section
70.
Lattice points on plane algebraic curves. Theorems of 260 Thue and Siegel . . . . . . . . . . . . . . . . .
T;xercisex (123171)
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265
CHAPTER V111
THE PRIME NUMBER THEOREM 71. Lemmata on the order of magnitude of some finite suns . . . . . . . . . 275 . . . . . . . . . . . . 72. Lemmata on the Mc bius function and some related .
functions .
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278 283 286 298
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Further lemmata. Proof of Selberg's formula . . . 74. An elementary proof of the prime number theorem. 73.
Exercises (172180)
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Table of primithe roots . . . . . . . . Fundamental golutiony of equations x2  11 Yame index . . . . . . . . . . . . . Subject index . . . . . . . . . . . . .
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CHAPTER I
DIVISIBILITI
1. Divisors.  The elementary theory of numbers deals primarily with the properties of the posit/re or natural numbers
1, 2, 3, 4, 5, .. . In order to simplify reasoning and the mode of presentation, it preferable to operate with the larger system of all
is, however, integers
0, ±1, ±2, ± 3, +4....
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The sum, the difference and the product of two integers are themselves integers; but the quotient of two integers is in general
not an integer. There are only a finite number of natural numbers which are less than a given number. In a set of natural numbers there is always a least number. Either all natural numbers have a certain given property, or there is a least natural number which does not have this property. Theorems in number theory are often proved by rnathcmiatieal idle etiou. By this is understood that the proof proceeds according to the following schema:
By trial or in some other way we are led to the hypothesis: Every natural number n has the property E. 2. We show that the number n = 1 has the property E. 3. We assume that the natural number in (or, if necessary, every natural number c m) has the property E. 4. We prove by means of this assumption that the number m + 1 has the property E. 5. We are then entitled to conclude that all natural numbers have the property E, and the truth of the hypothesis in step 1.
1 is established.
CHAPTER I
12
if a, b and c are three integers such that b 5,4 0 and
a=be, we say that b is a divisor of a, or that a is dirigible by b. We say also that h divides a, or that a is a multiple of b. To ascertain whether the given b is a divisor of the given a, we have clearly only to determine if a is among the (infinitely many) numbers
0, ±G, ±2b, ±3b, ±4b, ... The multiples of the number 2 are the eren numbers; the other integers are called w1d. Every integer a, which is different from zero, has clearly only
a finite number of divisors, and these can he determined by a finite number of trials. The numbers ± I and t a. the socalled trivial divisors. always occur among the divisors. A divisor d of a is a proper divisor when 1 < d < a. All integers are divisors of the number zero. if is a rational number p" 0, a possible common divisor 1 of the integers a and b can be determined (by trial) and divided out. We can thus always assume that a given fraction
is irreducible.
2. Remainders.  Let a and b be two integers. li p4 0, and let
us consider the multiples of b which are g; a. Let bq be the greatest of these. The number r = a  b q is then clearly <; 1111 but > 0.1 Thus we have Theorem 1. If a and b are integerx and b
0, a unique integer q
exists such that
a=bq+r, where
0r
The number r uniquely determined by the division of a by b is called the least nonnegative remainder or principal remainder of a
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1,,,y
1,1
we understand the absolute value of h. In words, if G
have I h I = b ; if b <: 0, we have I b
  b.
0, we
13
DIVISIBILITY
modido b. The remainder r is equal to zero if and only if a is divisible by b.
If c is a real number ? 0, we denote by the symbol [e]
the greatest integer < c. Thus we have for example 3, [] =0, [121= 1.
Assume that the numbers a and b in Theorem I are positive. Then the quotient q in (1) satisfies the inequalities
1
a; thus we
have
In equation (1) we can also consider
q so chosen that the
remainder r fulfills the condition (3)
This is done by letting bq be the multiple of b which lies nearest a. If a lies halfway between two multiples of h, we choose the greater multiple; this case occurs only when h is an even number and a is the product of h and an odd number. The remainder which fulfills condition (3) is called the least absolute remainder of a modxlo b. It is uniquely determined by it and b.
3. Primes.  If the natural number n (> 1) has only trivial divisors, it is said to be a prune number or simply a prime. All other natural numbers > I are called composite numbers. It is clearly possible to decide whether a given number is a prime or
composite by a finite number of trials. Among the first ten natural numbers, 2, 3, ii and i are primes, and 4, 6, 8, 9, 10 composite numbers.
Every natural number n (> 1) has at least one prime divisor, i. e. a divisor which is a prime. For, the least divisor > I of n
CHAPTER I
14
must clearly be a prime q, and the number n can now be written in the form n = q )n, where nn is a natural number. The following theorem was proved in Euclid's Elementa (9th book) :
Theorem 2. There is an if/inity of pri1nes.
Proof. It is sufficient to show that, for every given prime, there exists a prime which is greater. Let us arrange the primes in order of ascending magnitude, and let us number them accordingly, so that we put pI = l. 1`2 = 3, p3 = 5, etc. If we now put P1 p2   p = F, the number P + 1 is clearly not divisible by any of the first v primes. If q denotes the least prime divisor of P + 1, then q > p,,. The theorem is thereby proved. The method of proof, the same in principle as that of Euclid, also provides a possibility for determining increasingly large primes. 4. The fundamental theorem.  We begin by proving a lemma:
Let p be a prime and a a natural number not divisible by 1,. Then only the following positive multiples of a are divisible by p:
a.2p. a31)....
(1)
Assume, in particular, that a in is the least positive multiple of a which is divisible by p; then clearly 1 < in < p. Now let a h be an arbitrary positive multiple of a divisible by p. According to Theorem 1, we may put
r=h  mq, where q and r are integers, 0:_!S r < in. Thus the number
a r = ah  amq divisible by p. But according to the definition of the number in, we must have r = 0, and h is therefore a multiple of in. Since a p is divisible by p, in must be a divisor of the prime p; and since 9n > 1, we must have in =p. From this we conclude that every positive multiple of p divisible by a is included in is
the sequence (1).
From the lemma we obtain at once
15
DIVISIBILITY
Theorem 3. If the prime p dirides the product ab of the natural numbers a and b, then it neust divide at least one of the two factors a and b. For, if the number a were not divisible by p, then, according to the lemma, the number b must be a multiple of p. Theorem 3 is also to be found in Euclid's l;lcmenta (7th book). Euclid's proof is based, however, on the algorithm named after him. His algorithm is given in Section 7.
After these preparations we now continue with the proof of the fundamental theorem of number
Theorem 4. Every natural number n (> 1) can be expressed as the product of primes (prince factors) in the form (2)
ee = PI P2 ... Pr,
(e = 1).
There is only one such e. pression as a product (decomposition into Prime factors). if the order of the factor., is zeot taken into consideration.
Proof: The first part of the theorem is proved by induction in the following way. It is valid for the number 2. Assume that it is valid for all natural numbers < n. Then it is valid for n also. For, as we have seen in Section 3, ee can be written as a product, n = pI nl, where p is the least prime divisor of n. But, according to the hypothesis, the natural number n1, since it is < ii. can be written as a product of primes in the form assuming that it is > 1. Thus expression (`3) is 7111  P2P3 ' valid for u. The number r of prime factors is of course finite. Assume now that, besides (2), we also have a decomposition of ee into prime factors as follows, n = gIg2 ... q.,
where the factors qj are primes. If we now apply Theorem 3 to the identity (3)
Pipe " ' Pr = qI q2 ... q.
CHAPTER I
16
we see that the prime 71 must divide one of the primes p;; if we take this prime to be p1, then we must have pl = q1. On dividing (3) by pl, we obtain the identity A P3 .
1), _ (12 q3 ...
Q8
By analogous reasoning we see that )P2 = q2. Continuing in this
way, we have finally that r = s and that the numbers q1, q2, .... qr coincide with the numbers P1, P2, ..., p. , disregarding the order. The second part of the theorem is thereby proved. From the first part of the proof it is easy to deduce how the prime factors of a given number can be determined and the number expressed in the form (2). p, denote all distinct prime divisors of n; we Let P1, P2, may then express it in the form u =
(4)
pr
.
i=1
where a; is a natural number which depends on p; and n. Theorems in number theory can often be proved by means of induction. By this we mean that the proof proceeds according to the following schema: 1. By trial or in some other way we are led to the hypothesis: Every natural number n (> 1) has the property E. ?. We show that all tht primes have the property E. 3. We assume that the natural number in has the property E. 4. We prove by means of this assumption that the number nip has the same property, if p is an arbitrary prime. Then, by Theorem 4 (first part) all integers > 1 have the
property E, and the truth of the hypothesis in step
I
is
established.
5. Least common multiple and greatest common divisor. If the n integers a1, a2, ..., an are all different from zero, they have an infinity of common multiples; e. g. one of these is the product a1 a2 an. Consequently there must be a lea: t positire common multiple of the n numbers; it is denoted by the symbol .
(0
{a1, a2,
.,
ar+}.
DIVISIBILITY
17
If al, a2, ..., a are n integers, not all zero, they have but a finite number of common divisors; the numbers + 1 always occur among these. There is a greatest common divisor of the n num
bers; it is denoted by the symbol (al, a2,
(2)
. .
.,
a number _! 1. We shall also speak of the greatest com
mon divisor of the numbers in an infinite set of integers. We have the following theorems: Theorem 5. The least positive common multiple of the integers al,
a2...., a,, is a divisor of all the common vlldtiples of these number
Proof. The sum or the difference of two common multiples is itself a common multiple. Let in be the least positive common multiple. If Al is an arbitrary common multiple, by Theorem 1 we can write
r=111mq, where q and r are integers and 0 < r < in. Since vt q is a common multiple, r is likewise. But r < vm; hence, from the definition of in it follows that r = 0. Therefore DI is a multiple of m. Theorem 6. If d = (al, a2, ...,
there exist n integers x1, x2, ...,
x,, such that + an x,: = d.
al xl + a2 x2 +
(3)
Every common divisor of the integers al, a2, of (1  (al, On . ., a,,).
..., a is a divisor
.
Prool: Let us consider the (infinite) set M consisting of all the integers of the form al x1 + a2 x2 
(4)
.
. + a .z.'n,
where x1, x2, ..., x.,, run through all the integral values 0, ± 1, ± 2, etc. The sum or the difference of any two numbers in M
is itself a number in M. In particular, M contains all the numbers al , ae i ., an. The numbers in M clearly have the greatest .
.
2  516670 Trygve Nagell
CHAPTER I
18
common divisor d. For d is a divisor of all these numbers, and no number d1 > r1. in M has this property. since such a number r11 would be a divisor of all the numbers u1, a2, ..., an. Let do denote the least natural number in the set M. Further, let _V be any number in M. We shall show that X is a multiple of do. By Theorem 1 we can write
r= `duq, where q and r are integers and 0 5 r < do. The number r belongs to M, being the difference of two numbers in M. But, since r < do, this is only possible for r = 0. Thus all numbers in M are multiples of clo. Hence do is the greatest common divisor of the numbers in M, and therefore do = d. Thus the first part of the theorem is proved. If dl is any common divisor of the numbers al, a2, ..., an, it follows from equation (3) that d1 must be a divisor of rl. Let c be a natural number. From Theorems 5 and G we derive the rules and {a1, a2,
.
.
., anf c = {a1 e, a2 c,
.
.
., an C).
For two integers it is easy to prove Theorem 7. If a and b are natural numbers, irr, hare
Proof. By Theorem 5, the number Vu =
ab
{a, b)
is an integer. Then, a must be a divisor of a divisor of b (aab). In consequence,
(
1) is a common multiple
of a and b. Hence, by Theorem 5, {a, b) ah
(a, b)
(a, b), and b a
«li
is a divisor of
and therefore (a, b) is a divisor of mn. On the other hand,
since the numbers
19
DIVISIBILITY
a
_ {a, b}
and
G
71!
b
=
in
(a, L} a
are integers, the number iii is a common divisor of a and L. Thus, by Theorem 6. only when in = (a, b). Exanmple.
911
is a divisor of (a, b). But this is possible E. D.
If a = 12 and b = 15, we have (a, l.) = 3 and {a, b} = 60,
and in accordance with Theorem 7
3'60=12.15 180. 6. Moduls, rings and fields.  A set of numbers is called P. modtd when it has the following properties:
1. The set contains at least one number 0. If the numbers a and b belong to the set, their difference a  b also belon(,s to the set.
2.
Each modul contains the number 0. If it modul contains the
number a, it also contains the number  a, since  a = 0  a. If a modul contains the numbers a and L. it also contains the number a + b, since a + b = a  ( b). Examples of moduls are: 1. The set of all integers. 2. The set of all even integers. 3. The set of all rational numbers. 4. The set of all real numbers. 5. The set of all complex numbers.
But, the natural numbers obviously do not form a modul. Let aI, a2, . . ., a,,, be any numbers 0. The set of all numbers of the form aI xI + a2 ;r2 *
'+
a,,, ,rm .
where xI, .r2, ..., xm are integers, forms a mogul, which we denote by M(a,, a2, ..., a,,,) or, more briefly, by (a,. a2, .... am).
The number system al, a2...., a,,, is called a ycnerali?rv syOrfx of the modul. If a modul has the generating system NI P2, ., #, of r numbers, but no generating system of ., numbers, for < r, we say that the modal has the rank r. The system #I, then forms a ba.+'is of the modul.
Ch APTER 7
20
We shall prove Theorem N. A,iii rnodrrl M of (rational) integers caisi.,d.s of all m.zdti))l,,.q q/' the lrv(Xt l,asitire timber in M.
Proof. Let r( be the least natural number in M. If a is a number in M, there exist two integers q and r such that
r=a d q; where 0 c r < d (Theorem 1). Now, it is clear that the number r belongs to M. But by the definition of rl this is possible only if r = 0. Thus a is a multiple of d. If is clear that there is no other number in M having the same property as d. The modal [1] consists of the set of all integers. If the number I belongs to the inodul M containing only integers, we obviously have M= 111.
An immediate consequence of Theorems ti and 8 is Theorem I).
I/' al, a2..... a are iutrgcrs [al, a2,
....
[(al. (12,
0, ire burr ,
rz ].
Hence, any ruodul containing only integers has the rank 1. The modul [1,11/2] has the rank 2; for the equation (a+ b V2) x = 1, where a, b and x are integers, is possible only for b = 0, since I/2 is irrational.
A inodul is called a ring/ when it has the following property: If a and li belong to the rnodul, the product a b also belongs to the ruodul.
When 1) is an integer, the modul [1, 1/J is a ring. This is apparent from the relation ((1 + b 1'D) ((. + (I 1 1))  u e + b (11) + (a rl } b e) I'D.
From Theorem 8 we obtain Theorem 10. :Irir/ modul containing only (rational) integers is a ring.
This result is not valid for moduls in general. Thus the modul [V2] is not a ring, since the product V2 Y_'22_ = 2 does not belong to the modul; in fact, the equation 2 = `? t is not. pos
sible for any integer t.
DIVISIBILITY
21
A rim' is called a field when it has the following property: If a and 1, belong to the ring, the quotient also belongs to b the ring, provided b 0. Examples of fields are: 1. The set of all complex numbers. 2. The set of all real numbers or the rral field. 3. The set of all rational numbers or the rational ,iiPlcl.
There exist rings which are not fields. Thus the set of all integers is a ring but not a field. Every field K contains all the rational numbers. For, let a be 0 in K. Then K contains the number a = 1. any number a Thus, applying addition and subtraction, we see that all integers belong to K. Finally, applyingin division, it is clear that all rational numbers belong to K. Let a be any number 5,4 0, and consider the set of all numbers of the form ao+ala+a2a2+ +ama", bo i hl a + b2 a2 + + h all b,.. ni and it are integers, m 0 and if ? 0. This set is obviously a field; we denote it by K (a). Thus K (1) is the
where ak.,
rational field. If 1) is a rational number which is not the square of a
rational number, the number VD is_irrational (for the proof see Theorem 19 in Section 12). K (VD) is said to be a quadratic Every number in such a field may be written in the form _ a + bVD where a. b, a and (l are integers. Multiplying
r+aV1)
numerator and denominator by c  cl VD, we have 1
#
2
Dd2(arb(II) ad bc)VD)= if + rl l),
where it and r are rational numbers. The field K (VI)) is rral when I) is positive; in?aginary when D is negative. 7. Euclid's algorithm.  Let a and al be natural numbers, cr > rrl. If a is not divisible by aI, the principal remainder a2
CHAPTER I
22
of a modulo al is a positive number < ol. Dividing al by a2, we get the principal remainder a3 of al modulo a2. If a3 0, we may in the same manner find a new principal remainder a4 < a3. Repeating this procedure a certain number of times, we obtain a sequence of successively decreasing integers ? 0: (11>a2>r13>a4> and we must finally arrive at a division for which the principal remainder a,+, is equal to zero. Hence we have the following system of relations: a = a1 q1 + a2 .
() < a2 < al ,
al  a2 q2 + a3 .
0
(1) 1
a,
a,l r],1
F a 0
where q1, q2,
..., q, are positive integers.
It is easy to see that the number a, is the greatest common divisor of the numbers a and al. For, from the last relation in (1) it is clear that a,, is a divisor of a,1: from the preceding it follows that a,, is a divisor of a,2. Continuing in this way, we finally see that a,. is a divisor of both al and a. On the other hand, if d is a common divisor of al and a, it is evident from the first relation in (1) that '1 is a divisor of a2; from the second relation it follows that d is a divisor of a3. Continuing this argument, we finally see that d is a divisor of (t,. Hence a,. __ (a, al).
This method for determining the greatest common divisor of two integers is called the Tmelidr'alr algorit11717. It is given in Euclid's .Lle;n('17ta. Itll book.
For a = 288 and al = 158 we have the following algorithm: Z,
2," 158 1 + 130, 158130.1+_'8, 130=
2R18.1+10,
1810.1+8, 10
8=
8. 1 +2, 2.4.
23
DIrls1B(LiTY
Hence the greatest common divisor of 288 and 158 is 2. It is easily seen that, in the algorithm (1), the principal remainders may be replaced by the least absolute remainders. In this way the algorithm may clearly be much shortened. Example. We consider once more the case a = 2S8, aI = 158. Using the least absolute remainders the algorithm takes the form 288 = 1,58 2  28, 158 = 28 6  10,
28== 10.3 2, 10= 2.5. Thus the number of divisions in the algorithm is reduced from seven to four. 8. Relatively prime numbers. Euler's q function.  Let al. a2, ... ,
a,,, he integers having no common divisor > 1. Then (al, a2, .. .. a,,,)= 1,
and we say that the numbers are relatively prime. The numbers are said to be relatively prime in pains if (at, as) = 1 for all i
and j, i
,j.
Example. The numbers S. 9. 10, 12 are relatively prime; the numbers 5. 8, 9, 11 are relatively prime in pairs. When (a, b) = 1, we say that a is prime to b and vice versa. When (a, b) =  I and (a, c) = 1, we clearly have (a, b c) = 1. For (a, li c) is a divisor of both a c and b c and therefore of (a c, b e) = r (a, b) = c; further,
(a, b c) is
a divisor of a and thus of
(a, c) = 1.
When it is a natural number, we denote by p (n) the number of natural numbers C n which are prime to it. (Lulu's q';/irnctiou or totirnt of n.) For the first fire integers we find p (2)  1, 9' (3) = 2, 9% (4) = 2, When 1, is a prime, we clearly have 9' (1) = 1,
99 (5) = 4.
9' (P) =p  1, since none of the numbers 1, 2, 3, p)  1 is divisible by p. When pa is a power of the prime p, we find .
(P (1,") = 1)"'(27  1).
CHAPTER I
24
For, between
I
and pa there are the following multiples of p: h' 1, p 2, 1)  3, . . ., p.h,:I,
and no other natural number < pll has a divisor in common with pa which is > 1. Hence we have pa  p"1 P (1)a) = We shall establish the general result:
Theorem 11. If N is a natural number with the (liihreut prime factors hi, P2, . , pr, then
(')
9°(i1I
11=(1T
\
.(11
)(1 P'2) 1'i \
\
Pr
Proof We use multiplicative induction (Section 4). Theorem 11 is
true when i is the power of a prime. Suppose that it
is
true for a certain natural number N. Then we shall show that it is also true for the number 11 r1 = \l?. where p is any prime. If a1, a2, ... , a, denote the natural numbers < N which are prime to N, then v = (p (N). It is easily seen that the p .p (1) numbers (3)
a;+h11, (i=1, 2, .., v; h = 0,
1, 3,.... p 
1),
constitute all natural numbers < \"p which are prime to N. For if c is prime to N, so is the principal remainder of c modulo X. If N is divisible by p, the numbers (3) are also prime to p. In this case we have by (2)
1 (iV )=pp(\)= `1 1 1)1
)
... ( 1\
1)
.
pr
Hence Theorem 11 is true for n 1. Suppose next that V is not divisible by p. To determine cp(\1) we have to compute how many of the numbers (3) are divisible by p. The following natural numbers < Nl = 11 'p are multiples of p : (4)
1 1), 2 .17, 3.1), ... , 1\r p.
Here (kp, N) = 1 if and only if (k, .\') = 1. Thus exactly p (V)
of the numbers (4) are prime to X. Hence, the number of
25
DIVISIBILITY
multiples of p among, the numbers (3) is also = p (N). Finally by (2) we get
9,(N1)=pp (N)rp(V)=.v(i % ,
tlIl)..
and Theorem 11 is proved by multiplicative induction. From Theorem 11 we immediately derive Theorem .12. 1/' the natural numbers :I1 and N are relatirely prime, then
q?(1IN)=q(.)q(N). A direct proof of this proposition will be given in Section 20. Finally we prove Theorem 18. If N is a natural n?nnber, then q, (d) = 1",
(5)
'r
the sun? being extended over all positive dirisors d of N.
Proof Consider the sequence
1, 2, 3,.... A'1. X.
(6)
If
rl
is any positive divisor of N, there are in this sequence
N multiples of d, namely the numbers 1
d. 2 d .....
l
d.
Which of these numbers have the greatest common divisor d with A'?! The greatest common divisor of k d and N is d, if and only if k, i1) = 1. There are cp 1 N) numbers k d with this pro17
perty. Since all numbers (6) have a greatest common divisor
with N, we have (rl
CHAPTER f
26
But, when d runs through all positive divisors of \, so does 25
Hence relation (5) is proved, even without applying Theorems 11
and 12. We shall show in Section 9 that Theorem 13 may be used for proving Theorem 11.
~
Exanmplc. Y = 60 has the positive divisors d = 1, 2, 3. 4, 5, 6. 10, 12, 15, 20, 30, 60.
Further p (l) = 1,
T. (2) == 1,
9' (3) = 2,
T. (4)  . 2,
T (5) = 4.
p (6) = 2,
9'(10)4. x(12)=4. 4T (15)=8. q(20) = 8, T (:30)cp(60) 16, and we get in accordance with Theorem 13
1+1+2+2+4+2+4!4+8+8+8+16=60. 9. Arithmetical functions.  A function f (u,) defined for all natural numbers n is called an arithnrrtical,/'iinctioo. Examples of such functions are: Euler's function q, ();) and n! = 1 2.3 ():  1) n. We have already met arithmetical functions of
several variables such as the functions (a1, a2, .... an) and
{al, a2, ..., a modulo b is an arithmetical function of two variable integers a and b. We denote by r o?) the arithmetical function which indicates the number of positive divisors of )7. If n has the distinct prime factors 171, p2, ... , pr and if 17
we clearly have (1)
z (n) = 11(1 + a;). i=1
It is easy to establish the following relation for T (n) : (2)
DIVISIBILITY
27
For, if f (n) is the sum on the right, we have [1la11
= I or 0 ,
according as h is a divisor of n or not. Hence .f (n) ./' (n  1) = i(n).
From this the relation (2) follows at once. an important arithmetical function is the Mubitt.' function It, (n), defined as follows:
It (1) = 1; p (n) = 0, if n is divisible by the square of any prime; , Pr are distinct primes. ' Pr) = ( 1)r. if Pl, P2, u (pi 1'2 Thus we have: u (2) _= It (3)
L. It (4)
0. It (G) = i 1, u (10)
+ 1. y(30)=1, etc. An integer is called a square free number if it is not divisible b} any square > 1. We prove 1.1.
(3)
For erert/ natural number ii > 1 we hare
S = ' u (d) = 0. t
the sum being c'xtendrd ores all )tuxitire diri.ors d of n.
Proof. We need only extend the sum (3) over all positive squarefree divisors d of n. We prove the theorem by multiplicative induction. It is true when it is any prime p, since S,. _= u (1) + p (p) = 0. Suppose that it is true for n = rn. Then we shall show that it is also true for it = )it)), when p is any prime. If in is divisible by p, it is easily seen that Sm), contains
the same terms as S.. Since, by hypothesis, S= 0, we also have Sm p  0. If no is not divisible by p, we clearly have
8mt,(p 6 4 it p6). the soul being extended over all positive squarefree divisors 6 of in. Since It (p 6) =  It (b), it follows that S,,,t, = 0.
28
CIIAPTCii I
We shall apply Theorem 14 for proving the llibius inversion formula. Theorem 15. If 1"(n) by an arithmetical fimetion of it and if
G (n) _ `, F' (d).
(4)
.r
then conversely (5)
the sums being extended over all positire divisors of n.
Proof. Applying the relation (4), we can write the righthand side of (5) as a double sum (6)
the outer sum being extended over all positive divisors d of n and the inner sum over all positive divisors c of d. This double suns may be written
where J runs through all positive divisors of it, and where fir (d) is a function of d to be determined presently. In the expression
c takes a certain given value J if and only if d is of the Hence we have form A 6, where 6 is any diiisor of (li)
Iff(A)=
G
,
'a1
the sum being extended over all positive divisors 6 of d
.
But
according to (3) this sum is equal to zero except when A = n. Consequently since YJ (n) = 1, the double sum (J) has the value F(n), and Theorem 15 is proved.
If we apply Theorem 15 to formula (5) in Theorem 13 we obtain the following expression for Euler's function:
29
Dl VISIBILITY
trllll \.
Inl =
rl.
or n !c<
n <, j P! Pi
the first sum being extended over all distinct prime factors of n, the second sum being extended over all pairs of distinct prime factors of n, etc This formula is clearly the same as formula (2) in Theorem 11. 10. Diophantine equations of the first degree.  From Theorem 6 follows at once Theorem 1fi. The neees8ar,f and suflirient condition for the lir ar equation a1 .x 1
(1)
c12 X2
1 rr x',. = C,
with integral eoefpri, t, Is ill, cr2, .... a, and e, to be solvable in integers .r1, x2. . , x is flint the IyreateNt common divisor dtrtde e. (al, a 2 , ,
In the case of two unknowns we further prove Theorem 17. 1f the linen; equation
axtb!, =r,
(2)
with integral eoe%lieients it. b and
x=
y = 71. the %urnculae
lrct..
the integral solution
we obtain all sulutiume in integers x and y by f
b
(3)
u
where t riots through all integer... Proof: It is easily seen that a and y given by expressions (3) satisfy equation (2), considering that it : ! bit = r. Now suppose
that x, y is an arbitrary integral solution of (2). Then ax
brt=e =a;+b,1.
CrrAPTER r
30
Hence a (.i'  $)  :  b (,q  7j).
(4)
If d = (a, b) is the areatest common divisor of a and b, it follows
from this that f is a divisor of x  5. therefore
where t is an integer. Finally we get from (4)
Thus equation (2) has an infinite number of integral solutions
when it is solvable. A solution may always be found by trial. We shall show how to find a solution by Euclid's algorithm. There is no loss of generality in supposing c = (a, b). We write the algorithm (compare Section 7) in the form
aq1b = a2, b  Q2L12=a3. (5)
av3  (jv2 (42 = ((+1 a,2  qr1 a,1 ((, _ ('.
Eliminating relation
from the two last equations, we obtuin the av2 (1 + q,.1 qv2)  a, 3 (I, 1 = C.
from this equation and the third equation from the end in (5), we obtain a relation of the form Eliminating
a,s P + a,4 Q = C,
where P and Q are integers. By continuing this procedure and by successive elimination of av_3, a,}. etc., we finally obtain a relation of the form (l A  b B = e,
31
DIVISIBILITY
where A and I3 are integers determined by Q1. r12, .... q,I. This may easily be verified by induction. In this way we have found an integral solution x = A. y = B of equation (2). Exaniple. If a = 15, b = 11, c = 1, the algorithm is
15111=4, 11 2.4 =3,
413 =1. By elimination we find
4.311.1=1, 15 3114=1 and thus we have the solution .c = 3, y
4 of the equation
The problem of solving the equation (1) in a (> 3) unknowns may be reduced to the problem of solving an analogous equation
in only it  1 unknowns. Suppose that we know a. set of solutions
='I a'2 = $2.
a'I
11
of the equation (ti)
r111i 1
d y = c,
n2 .. 2
where d  (a,,I. By the method just developed, we can determine a set of solutions of the equation
a x, = d,l. Then equation (1) clearly has the solutions x,
The general solution of (1)
(i = 1, 2, ..., 7r).
finally given by the system of
is
formulae x, = i + l/1 t1 + l12 t2 +
+ L,31
(i = 1, 2,
..
,
n)
where b1, b2i .. ,, b _I depend on the numbers a1, a2...., a,, only.
and where the parameters t1. l2.
.
.
.
i run through all integers.
CHAPTER T
32
It is easy to show this by induction from n  1 to n. starting from Theorem 17.
Equations (1) and (2) are the simplest examples of socalled Diophantine or indeterminate equations. Let f' (x, y, z, ...) be a polynomial in the variables x, if, z, etc.. with integral or rational coefficients. The problem of solving the equation (x, Y, 7, . . .)  0 (7) in integral or rational numbers x, t/, z, etc., is called a Diophantine problem; (7) is said to be a Diophantine or indeterminate equation. Simultaneous systems of such equations may also be investigated. In his work Arithmetira the Greek mathematician Diophantos examined and solved a great number of indeterminate equations
of the first four degrees.
In Chapters VI and VII we shall treat some Diophantine equations of higher degrees.
Lattice points and point lattices.  In number theory it is often advantageous to make use of geometrical ideas and 11.
interpretations. The integers may be represented as a set of equidistant points,
unit distance apart, on an infinite straight line in the plane, the socalled real number axis.
Let us consider a plane rectangular or oblique Cartesian coordinatesystem with the abscissae x and ordinates y; x and y need
not be measured by the same unit of measurement. A point y) whose coordinates x and rJ are integers is called a lattice point. The set of all lattice points is said to be a plane point lattice. Let us draw all straight lines parallel to the coordinate (x,
axes through the lattice points. The geometrical figure thus obtained is called a plane lattice. Fig. 1 represents a point lattice in ordinary rectangular coor
dinates, and fig. 2 a point lattice in oblique coordinates. The equation ax + by = c represents a straight line. The problem of solving this equation in integers x and y is equivalent to the problem of finding the lattice points situated on the straight line in question.
33
DIVISIBILITY
.
.
L
.
+/
.
.
x .
o
.
t
.
.
.
.
Fig. 1.
Fig. 2.
Example. The straight line 4 x  3 y = I passes through the lattice points (1, 1), (4, 5), ( ?,  3), ( h,  7), etc. (see fig. 3).
Fig. 3.
Suppose that the straight line has integral coefficients, or, what is obviously the same, rational coefficients. Then, by Theo
rem 17, the straight line passes through an infinity of lattice points if it passes through one lattice point. This does not hold for a straight line with irrational coefficients a, b, e, except when
there exists a positive number w such that the numbers a w, bw, ew are all rational. Thus the straight line 3 :iIWilt Tr1jgt.e \aprll
34
CHAPTER I
J=v2. r only passes through one lattice point, namely the origin. If a straight line passes through two lattice points, the coefficients of its equation are determined by the coordinates of the lattice points as rational numbers, and therefore the line passes through infinitely many lattice points. This can be generalized to three dimensions. Thus we can speak of lattice points and point lattices in space. The equation
ax  by+cz=tl represents a plane in a Cartesian coordinatesystem with the coordinates x, il and .:. The problem of solving this equation in
integers x, y, z is equivalent to the problem of determining all lattice points situated on the plane. More generally one may study the distribution of lattice points on a given curve in the plane or on a given surface in space; this leads to the problem of solving Diophantine equations in two or three unknowns. Minkowski has created a theory treating questions of the following type: Find approximations to the number of lattice points included inside a given closed curve in the plane or inside a given closed surface in space. The startingpoint of this theory is the following theorem: If a parallelogram, A whose middle point is the origin has the area 4. then besides the origin, there is at least one lattice point inside 4 or on its boundary.
This theory will, however, not be developed in the present volume.
12. Irrational numbers.  The general theory of irrational numbers belongs to analysis. But special types of irrational numbers are of great importance in elementary number theory. The irrationality of the number Y5 was doubtless discovered by some disciple of Pythagoras. Later Greek mathematicians probably proved the irrationality of other quadratic surds. But only the proof of the irrationality of 1'2 has been preserved; it
is to be found in the 10th book of Euclid's Elc ncria. The Euclidean proof is a special case of the proof of Theorem 18 given below.
35
DIVISIBILITY
In general it is very difficult to decide whether a certain given real number is rational or irrational; the number may for example
be given as the sum of an infinite series. According to school arithmetic, the necessary and sufficient condition for a real number to be rational is that its decimal expansion is finite or periodic. But, the real difficulty consists in deciding whether the decimal fraction is periodic or not. By the following theorem we can construct special irrational numbers.
Theorem M. If f(x) = x" 4 it, x"1 + c tcitlt integral co(ji icnt,, al. az, of
the equation f (.x) = U, tltcu
$
.
a is a poltluontial in a,,, and if is it root is either an integer or an 
,
irrational number.
Proof. We may suppose a,,  0. Putting E = t where r and .c ,
(
0) are integers having no common divisor >1, we have rI ; al)111.9 + ..  a fi"  0.
If s were divisible by the prime p, r would also be divisible by p. But, since this is contrary to hypothesis, we must have s = 1, and the theorem is proved. In particular, we have
Theorem ill. If a and N are natural ntnubcrx, aril if l7 is not the nth power of a natural number, then
i. an irrational number.
The irrational numbers constructed by means of Theorems IS and 19 belong to the extensive class of numbers which are called algebraic numbers. When f(w) is a polynomial in x with rational coefficients, any root of the equation f(x) = 0 is called an algebraic ututtber. In particular, the rational numbers are algebraic. Any number which is not algebraic is said to be transcendental. The existence of transcendental numbers was proved in 1851 by Liouville.
CHAPTER I
36
Let a be any real number. We divide the interval 01 in t equal parts; in each subinterval the left end point only is included. In fig. 4 t = 8 t
i Fig. 4.
Now let g be an integer ? 0, and let x be the least integer > a y. Then
0<xay< 1. Putting y = 0, 1, 2, ... , t, we get t + 1 numbers x  a y, which all belong to the interval 01. Since the number of subintervals is t, there is at least one subinterval containing at least two of
the numbers x  ay. Thus there exist a natural number h < t and two pairs of integers x', y' and x", y", such that h
t
1<x'ay'<
h
I <:e" ay,,< t 1
Putting we find (1)
1.eayI <j
If we suppose g" > y', then y is one of the numbers 1, 2, ..., t. Hence we have proved: There corresponds to every real number
a and to every uatural number t at least one pair of integers x and y such that 1 < y _S t and such that inequality (1) is satisfied. From this we also obtain the inequality (2)
I
In (1) as well as in (2) we can obviously suppose that x and y are relatively prime. If a is rational and =1 where (a, b) = I
and b > 0, then inequality (2) has only a finite number of solu
DIVISIBILITY
37
tions in relatively prime numbers x and y. For, if a y > 0, we get _ Ir, n 1hxayl> 1 .r.
y
al
and
=1)/I
'J
,!t
t,
Hence, if ('l) is satisfied, y < b.
On the other hand we have Theorem 20. If a is a real irrational numlwr, inequality (2) has
an infinity of solutions in relatirely prime integers x and y.
Proof. Let /1 be a natural number. Applying the result just obtained, we then determine a pair of relatively prime integers x1 and ill such that rI
711
where 1
1
u < "1 tl
f' 1
rpl s t1 . Since rc is irrational, 9h 54 0. Then we choose
a natural number /.2 > 1 and determine the relatively prime in'11
tegers T2 and y2 such that x2 '12
Y2
1 .
a<
y2 t2
<1 l2
c 11
with 1 < Y2 < t2. Repeating this procedure we obtain an infinite sequence of successively decreasing positive numbers )h>172>?1a> . . >71r> ....
where the number .r;
satisfies the inequality
This proves Theorem 20 The procedure just developed readily gives an infinite sequence of successively better approximations
38
CHAPTER 1
to the real irrational number a by means of the rational numbers x'
!/c
in the proof of inequality (1) we applied the socalled Dirichlet box principle: If more than t objects are distributed in f boxes,
at least one of the boxes must contain two or more objects. This extremely simple principle has nevertheless been very effective
in many mathematical proofs. 13. Irrationality of the numbers e and r  Let e = 2.7182R .. .
be the base of the system of natural logarithms, and let z = 3,14159 ... be the length of the circumference of a circle with the radius If. We then prove the following theorems. Theorem 21. The n umber e is irrational.
Proof. In the introduction to analysis it is proved that. when
it
+
+
II +...+
0
la
1
1
n!
are satisfied for all natural numbers n. Suppose that e is rational
and =
b,
where a and b are natural numbers; we can clearly
suppose b > 1. Now, choosing the number n in (1) so large that h divides n!, we obtain from (2)
0
But this is impossible. since both a
and n! s are integers.
Thus the number e is irrational. This theorem was first established by Euler.
DIVISIBILITY
39
Theorem 22. Thr number r is irrational. a
Proof Suppose that 3= b , where a and b are natural numbers. For any natural number n we define the two polynomials J.

l)x)n
Tn (a
(x) 
rr i
and I)n
f (x)
(x),
where fI"') (x) is the mth derivate of [(.r). has integral coefficients and no terms The polynomial in x of degree less than the nth. It then follows that f (.r) and all its derivates take integral values for x 0; and, since
f(r)=f(ax) f(.rx), this also applies for . = ;r. By computation we find that rlGr [I' (x) sin x  F(x) cos 4 _ [I,'" (x) + 1'(r)] sins = f'(x) sin .r. Hence a
(3)
n
f f (x) sin x ax = [I," (,r) sin x  I'(x) cos x] = F(a) + F (0). 0
0
Here F(s) + I'(O) is an integer, since f'n') (n) and f(m) (0) are integers. But. for 0 <x < r, we obviously have
Since the infinite power series z
s
is convergent for all x, we can choose the number n so large that
(:za)n<1 !r
n!
CHAPTER I
40
Hence, according to a mean value theorem in elementary integral calculus, the integral (3) must have a positive value < 1. On the
other hand, we have just shown that the righthand side in (3) is an integer. Thus the assumption sc = b leads to a contradiction, and Theorem 22 is proved. This result was first established in 1761 by Lambert. The proof reproduced above is due to Niven and was published in 1947.
In reality, the numbers e and ;r are transcendental numbers. This was proved for e by Hermite (1873) and for r by F. Lindemann (1882). Exercises 1.
If n is an integer, show that the product
(n 1)(2)1 1) is divisible by 6.
2.
If n2 is the square of an integer which is divisible neither by 2 nor by 3, show that the number lea + 23 is divisible by 24.
3. Show that the number 3),'  I can never be a square, when n is an integer. 4. If p denotes the nth prime in the sequence of natural numbers, prove the inequality
Suggestion : Apply the proof of Theorem 2. 5. Show that the least absolute remainder of a niodulo 7) is
abI27a]+b 6.
[],
where a and b are natural numbers. If g is a natural number > 1, show that any natural number IV (? g) can be represented uniquely in the form ,l" = eO + ('1 9 + C2 y" +
+ c,,, gna,
41
DIVISIBILITY
where the integral coefficients c; satisfy the conditions
0cct
(z=0,1,2, ., in1)
0
9 4+g3+g2+g+1 is never a square, when the natural number g is The number
3.
g4r2gs I2g2+29+.5
8.
is a square only when the natural number g is equal to 2. Prove this by means of the result in the preceding exercise. If n is a natural number and a > 0, prove the formula ni rr
ll 12
A=0
9.
If n is a natural number, and if v denotes the number of positive divisors of n, show that the product of all these divisors is equal to OT.
10. Let pI, p2.... , pr be the distinct prime factors of the natural number n and suppose
Show that the sum of all positive divisors of is is equal to TlPij+1 i=1 11.
pr  1
Suppose that pl, 1'2 .. , ltr are the distinct prime factors of the product a b, and suppose that .
r
a = ]Jptt (at > 0), i=1
r
b=
11 1),& t=1
(Pt > 0).
CHAPTER i
42
Further, denote by vg the least of the two exponents aj and /3; and by ,uj the largest of the same exponents. Prove the formulae
r
{a, b} =111)=
(n, b) f=1
.
i1
12. Prove the relations (a1, a2,
.
., an)
(a1, a2.
.
{a1, (72. .. ., an} _ al, a2. .
.
., ak .
J
. .
ak}.
.
.
.. (am. .
.
.
., {am .
.
.. a
),
.. an}}.
13. Prove the relations (a, { b, c})
(a, b). (a. c) },
{a, (b, r), = ({a, b}. {a, c}). 14.
Prove the relation
((a, b}, {a, (}. {b, c}) =
{(a,
b), (a, r), (b, c)}.
15. Show that the number of irreducible fractions between 0 and 1 whose denominators do not exceed the natural number
n is 71
I'P (m) M=1 16. Let N, in and n be natural numbers. Find, by means of the Euclidean algorithm, the greatest common divisor of the numbers
_l'n  1 and X",  1. N. Find all natural numbers 12 such that cP (u) _ 24.
18. Find all natural numbers 9)i < 100 such that the equation T (n) = in
has no solution. 19. If 12
is a natural number > 1. show that
a=g129(n),
DIVISIBILITY
43
the sum being extended over all natural numbers a which are prime to a and < n. Find all the natural numbers n satisfying the inequality 7: T (7:) < n.
Let F (n), G ()i) and H (n) be three arithmetical functions which satisfy the conditions G O2) = L H (d), r!
the sum being extended over all positive divisors d of n, and fl
F(n)
G (G). Lam]
Show that F(n)
r=
[t] H
What formula do we obtain by putting in the preceding exercise H(n) = 1 for all n? What formula do we obtain by putting in Exercise 21 G (n) = log n and applying Theorem 15? Prove the formula, n
n,:1
valid for all natural numbers n. Determine the arithmetical function W (n) defined by the relations F(d),
(1) = 1 and 1u (n) a
the sum being extended over all positive divisors d of n:. Suggestion: Apply the inversion formula of Mobius.
Show that there are infinitely many primes of the form 4n + 3, i. e. leaving the remainder 3 on division by 4.
Suggestion: Apply the same method as in the proof of Theorem 2.
44
CHAPTER I
27. Show that there are infinitely many primes of the form 6 n + 5,
i. e. leaving the remainder i on division by 6.
Suggestion: Apply the same method as in the proof of Theorem 2. 28.
If the number
2r1
is a prime, then 1) is a. prime. (11ersenne primes.) 29. If the number
2"+1
is a prime, then n is a power of 2. (Format primes.) 30. Show that every odd number can be written as the difference of two integral squares. In how many ways is this possible? 31. Solve completely the Diophantine equation
119x29y=8 in integers x and y. 32. Solve completely the Diophautine system
2x+ 5,, 11:=1, x  12y + 7 = 2 in integers r, y and z. 33. A natural number that is equal to half the sum of its positive divisors is said to be a perfect number. The least perfect number is clearly 6. Prove the following theorem of Euclid (Elen?e)ita, 9th book):
If 21" I is a prime, then 2"'1(2m  1)
is a perfect number. This is the case for nn == 2. 3. 5,
7,
13, 17, 19, 31, 61, 89.
107 and 127. No other perfect numbers are known than these twelve.
34. Prove the following theorem of Euler: Every even perfect number must be of the form just indicated
in the theorem of Euclid. No odd perfect numbers are known.
45
DIVISIBILITY
3:5.
Let it be a natural number having the distinct prime factors , pr. Prove the formula
14, I'2,
V (12 =' q (11) 112 i ( Or I; 9' 0r)AP2 ... 11,,,
the sun, being extended over all natural numbers a prime to it and < it. Suggestion : Start from the formula 12+1'2+.. I(/r1)2=1, 1,(,11)(21?1) and apply the inversion theorem of M bius.
36. Let n be an odd natural number having the distinct prime Pr. Prove the formula factors PI, P2,
,cl=II cp
(//1l)(p211 ... (Pr1),
1)r
the sum being extended over all natural numbers a prime to n and < s n. Suggestion: Start from the formula
1 +::+3+
r(1r1)=,A,(),2 1)
and apply the inversion theorem of Mobius. 37. Let n = 2 h + 1 be an odd natural number having exactly tc distinct prime factors of the form 4 t + 1 and exactly v distinct prime factors of the form 4 t + 3. Further, for / = 1, 2, 3, 4, let A, denote the number of integers prime to n in the interval I (r  1) it  I rn. Finally, for / = 0, 1, 2. 3, let B, denote the number of positive integers < it which are prime to it and of the form 4 t + r. Prove for u > 0 that
and Bu  = B1
B.,  B;;
Prove for p = 0 that l
=14+(1)'2'
:I. = 43 =
99()1)(1)1,2s,
46
CHAPTER I
and Bo
(n).,'s, B3
Bs__.j
9i
38. The Farey series of order n is the ascending sequence of irreducible fractions  satisfying the following conditions:
(a,b)=1 and 0a
i'
U
1
6'
1 .
1
1
I'
:i,
2
1
'5
2
:1
3'
2 :1'
3
I
4' 5'
1
I
Prove the following theorems a 1. If and c are two consecutive fractions in a Farey series,
< l , then
bead=1.
b 2.
If
a
are three consecutive fractions in a Farey
d and y series, a < d then b
J
39. Show that cos 1 and sin 1 are irrational nunibers. Suggestion: Like the proof of Theorem 21. 40. Show that the number e is not a root of a quadratic equation with rational coefficients.
Suggestion: Prove by a slight alteration of the proof of Theorem 21.
CHAPTER II
ON THE DISTRIBUTION OF PRIMES
14. Some lemmata.  We first prove Theorem 25.
If x is ? 0, and if' a is a natural nzunber, then a
Proof. Since
a
[[] is the greatest integer C [x] [x] a
_
where 0:5: n < 1, we have [x] a
[a,
+r a
0 and _ a  1. Putting x = [.r] +
where r is an integer x
[x]
we can write
+
(i
it
 FrIl ; N + I a
ca
From this the theorem follows at once, since
IIT ` Theorem V.I. D4 a be a natural number, and let 1) be a prime. Then the exponent of the highest I)owrer of p which divides )1! 123. n is equal to (2)
V  I P" I + L,?" I +
[?''s J
t
Proof The series (2) continues so long as the power of p is 5 n. If h, denotes the number of terms in the sequence 1, 2, 3, . ., Which are divisible by p', the required exponent Y is 11
CHAPTER II
48
obviously equal to hl + h2 1 /13

.
which are divisible by 1)'' are
The natural numbers < n
r
1
P
Thus we have h = [], and the theorem is proved. By Theorem 23 we have It.
/411 = I p
Hence, we can determine the numbers h, successively by divisions
by p instead of by powers of p. This gives a quicker method for determining N.
Example. If n = 3000 and p = 7, we get 3000
_
t; l
128+el.
1e2, =8+f?3 and
428 7
8 7
where el, e2, e3 and e4 are positive numbers < 1. In this case it follows that X = 428 + 6 1
8 i
i
I = 498.
By means of formula (2) we prove Theorem 25. If n, III. )12, n3, ete., are natural numbers x1rch that
N=n1 + n2
1
n3 +
,
the quotient _
121
(3) 1111
)121
1131 ...
is an integer.
Proof. Let m be a natural number. From the relation 11
= ))1 + 1I2
11
we obtain the inequality [in
7,1.
I))
+
))3
in
{
...
ON THE DISTRIBUTION OF PRIMES
49
Let p be a prime factor of n!, and put into this inequality successively 7n = p, p2, p3, etc. By addition we then obtain the inequality
[l]
[n] (4)
N1 [1,2J
P'
t=1
:J i P
According to Theorem 24, the sum on the lefthand side is the exponent of the highest power of p which divides 0, and likewise, the All sum on the righthand side is the exponent of the highest power of p which divides n,! (v = 1, 2, 3, etc.). Thus it follows from the inequality (4) that the highest power of 11 which divides the denominator of the number (3) also divides the numerator. Hence the number (3) is an integer. Q. E. D.
Theorem 25 may also be proved by combinatorial reasoning. In fact, the number (3) is a socalled polynomial coefficient; more
precisely, it is the coefficient of the power product .1.n, xn . xn, R
1
1
in the development of (x1 + x2 + xg +
By their manner of formation, the polynomial coefficients are necessarily integers. In particular, when n = nI + )?2, the quotient
_r(n1)
n!
(n 111+1) n
1 23 ... it,
771! 7?21
is a binomial coefficient.
A set of nz objects chosen from a given set of n objects, without regard to order, is called a combination of n things taken 7n at a time. It is easy to show that the number of these combinations is
n(n= 1) 1
(n)11+ 1)(u/ :1. 3
in
M
We have the following corollary to Theorem 25: The product of )2 consecutive integers is dirisible by n!
This is obviously true also when the integers are negative. 4 516570 Trygue \ugell
CHAPTER 11
50
The following theorem, which is due to Sylvester (1883), is of a very general character. Theorem 26. Let us consider \r objects and a number of mathema
tical properties El. E2. Es, etc. Let Ni denote the number of objects hating the propertq E let \;,.; denote the number of objects having the two distinct properties E; and Ej, let V,j,J, denote the number of objects having the three distinct properties E;, E, and Ek7 etc. Then the number of objects baring now, of
the properties El, E2, Es, etc., is equal to
the sums being extended orer all values of the indices i, j, k, etc. satisti/ing the conditions: i = 1, 2. 3, etc., in the firxt suns, i > j ? 1 in the second stmt, i > j > k ? 1 in the third sutm, etc.
Proof. Let A be an object having exactly r of the properties El, E2, E2, etc. Then A contributes 1 to the number N. If r>0, A contributes 1 to r of the numbers Ni. If r> 1, A contributes
to (2) = a r (r  1) of the numbers \;, j . If r > 2, A contributes I to 1
r ( 3J
_r(r1)(r2) 1 2.3
of the numbers 1,j,x, and so on. We finally see that the object A contributes
1(I)
+
(')(3)

=l1
1)r=0
to the sum (5), if r > 0. On the other hand, if r = 0, it contributes 1. Consequently the number of objects having none of the properties is given by (5).
Suppose now that the N objects in Theorem 26 are the natural numbers < x, and thus N = [x]. Suppose further that E; denotes divisibility by the natural number a;. We then obtain the result: Theorem 27. Let a1i a2...., am be natural nsnnbers such that a,) = 1 if 1 J. Then the number of natural 'n embers < x
ON THE DISTRIBUTION OF PRIMES
are not is equal to
51
b!/ any one o/'the nrnnbers «I, a2, ..., a,,,
'
[ t ]  L [rr+ i
t
lT
r
l
ae ax akJ
the sums being exlended over all q/ the indices i, j, k, etc. ; the eondclrou:: i = 1. . 3, Pfc., in the /first I in the seeond wrnr, i > j > k =' I in the third .sunr, etc. i>
For the number of natural numbers < x which are divisible by every on(, of the numbers U. u etc., is obviously x 1 cc; crJ n,,
If, in Theorem 27, the numbers aI, a2, a3, etc.. denote the distinct prime factors of [x], we obtain a new proof of the
formula for Euler's pfunction in Theorem 11. 15. General remarks. The sieve of Eratosthenes.  It is theoretically possible to decide 'whether or not a given natural number n is
a prime by trying to divide it by every smaller natural
number. For, if n is not a prime, it must have a positive divisor > 1 and < n. This method does not presuppose any prime to be previously known; it is, however, inapplicable for large values of n. If the primes < Yn are already known, the question can be decided in a much shorter time by trying whether or not n is divisible by any one of these primes. Provided that it is not too large, the question can be solved by means of a factor table or a prime table. The largest prime table yet published was worked out by D. N. it gives the primes up to 10006721. By inspecting a prime table one observes that the prime numbers gradually become more scarce the farther one goes on in the sequence of natural numbers. In the ten intervals 1100, 100200, primes:
...,
900luau there are the following numbers of 25, 21, 16, 16, 17, 14. 16, 14, 15, 14.
CHAPTER 11
52
In the ten intervals each of one hundred numbers between 10000000 and 10001000 the corresponding numbers are 2, (i, 6. 6, 5, 4, 7, 10, 9, 6.
The largest number known at present to be a prime is 21
1 = 17014118'0'46046923173168730371,)884105727.
this was shown by Lucas.
The distribution of the primes in detail is most irregular. In an interval of relatively many primes, there may occur long sequences of consecutive composite numbers. Thus, there are no
primes between 1327 and 1361. A gap of this length does not reoccur until between the primes 8467 and 8501. When n is any positive integer, it is easy to construct sequences of a consecutive composite numbers; for instance, the numbers
(u F1)!+2,(n! 1)!+3... ,(1:±1)!+n+1 are all composite. On the other ]land, pairs of primes which have the difference 2, socalled prime twine, occur relatively often;
we have the following eight pairs of prime twins less than 100: 3, 5; 5, 7; 11, 13; 17, 19; 29, 31; 41, 43; 59, 61; 71, 73.
Among the first hundred primes after the number 100000000 there are ten pairs of prime twins. There are probably an infinity of pairs of prime twins; but the proof of this conjecture is at present beyond the resources of mathematics. When the primes = tax are known, the primes < x may be found in the following way. We write up the sequence of all integers s 2 and < .': in their natural succession. We first strike out all numbers divisible by then all numbers divisible by 3, further all numbers divisible by 5, etc., and finally all numbers divisible by q, where q denotes the greatest prime  J 'x. The remaining numbers obviously consist of all the primes that are > Vx and < x. For such a number cannot have any prime
factor < Vx, and it cannot be the product of two numbers > 1 `a.
This simple but effective method is known as Eratosthe
nes's siere method.
ON THE DISTRIBUTION OF PRIMES
53
Eramlde I. We consider the case x = 26 and apply the sieve method. The prime numbers < 1126 are 2, 3 and 5. We write down the integers from 2 to 26; we.first mark by a bar every second number counting from 2, then every third number counting
from 3 and finally every fifth number counting from 5. Then the sequence looks like this: 2. 3. 4, 5. 6, 7, 8, 9, 10, 11, 12, 13, 14, 1 n, 1 G,
17, 1 h, 19, 20, 211, 22, 23, 24, 25, 26;
The numbers not barred 7, 11, 13, 17. 119 and 23
are the six primes
 V26 and < 26.
L.rannple 2. If we take .r  300, the primes
1 300 are
2, 3. 5. 7, 11, 13 and 17.
Applying the sieve method we find the following 55 primes 1 300 and < 300: 19, 23, 29, 31, 37. 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89. 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 103, 167. 173. 179, 181, 191, 193. 197, 199, 211. 223. 227, 229. 233, 239. 241. 251, 257, 263, 2091 271, 277, 281, 283. 293.
By means of the sieve method we can also calculate the nionbcr
of primes which do not exceed a given limit This number is usually denoted by fr (x). For instance we have 7c (10) = 4, rc (Y300) = 7, c(300)=62. After the application of the sieve method to the sequence of integers _> 2 and < r, there are left exactly z (x)  :c (V x) integers. It is, however, possible to deduce
another expression for the number of integers remaining. For if we replace aj by pt in Theorem 27, and suppose that pI, ps, ..., l)m are all the primes < Yx, we find the following expression for the number in question
1 +[x
]
[Pil+[x_]

CHAPTER 11
54
% Te thus obtain the formula
it (rl) IJ
:r (.r)  :r (lr') _  1 +
(1)
the sum being extended over all positive divisors of the product t'1 P2 .. PM
It is, however. possible to improve this result considerably, as
was shown by Meissel. The formula estahlislled br him gives the best method up to now for numerical calculation of n (x). The following table an idea of the wav in which the function .T(.r.) increases. q
11
100
25
4000
5511
300
46
500(1
000
300
6000
783
44110
82 78
700))
900
500
05
811011
1007
Goo
109
:1000
1117
700 800
1_'5
111111111
1220
139
1000(10
0502
0(10
15a
100000))
78498
1000
11;8
1000000(1
0111.579
2(100
303 430
100000000
5761455
1000000000
50847478
:3000
The value of :r (1011) was calculated by Bertelsen from the formula
of Meissel; v (x) has not been calculated for values of .r larger than 1011.
16. The function c (x).  Legendre and Gauss occupied themselves with the problem of finding simple functions which give good approximations to 7r (x) for large values of x. Thus, in his book Thcorie de uo;ijbr (179); Legendre states that the function log x  1.08011
gives a good approximation to n (x).1 By means of prime tables ' here and in the following, log denotes the natural logarithm.
ON TAE DISTRIBUTION OF PRIMES
55
Gauss discovered that :r (x) may be very well approximated by each of the functions x log .L and
du log u
(1) 2
But he gave no proof of it. The function (1) is the socalled integral logarithm of X. The first demonstrated results are. however, due to Tchebychef,
who (1850), among other things, proved that the inequalities x 8 log .r. i (2)
< z (x) < _
log a
8
are valid for all sufficiently large values of x. He also showed
that the quotient of the numbers :a (x) and
log x
has the limit
I for increasing :r, provided that the limit exists. In 18911 Hadamard and Vallee Poussin, independently of each other, proved the existence of this limit and thus the relation lim
(3)
CC
n (X)
:r/lo:C
= 1.
Their proof of this theorem, the socalled prime nunmber theorem, is based on the theory of Riemann's zeta function (,) defined
by the infinite series (4)
(,) = 1
1
1
T
.$
31;
1La
for all complex values of s = a A i t when a > 1. Subsequently, by analytic continuation, (s) can be defined for all s I. The connection of the zeta function with the primes is obvious from A'nlcr's identzly
(
5 (.) _
=8 = n=1
II r
1
' ,=8 .
(a > 1)
CHAPTER II
56
the infinite product being extended over all primes p. To prove this identity we first verify that the infinite product is convergent the and different from zero for a> 1. For, since series Y, px I,
is absolutely convergent for a ; 1.
1
Since
+ I)R + p
+
and, therefore, because of the absolute convergence,
where p runs through all primes .r, and where n runs through all positive integers which have no prime factor >.r. Hence
Here the absolute value of the right hand side is obviously less than
Hence we have
x 1111,^x rim
1
l
1) = 0, .J tx
which proves the truth of identity (5). It may be observed that, in this proof, we make use of the fundamental theorem of number theory (Theorem 4). Riemann, perceiving the fundamental importance of the zeta function for the study of the distribution of primes, developed the elements of a theory for this function. He also formulated six hypotheses which lie could not prove. Especially the position of the imaginary zeros of the function appeared to be of great
57
ON THE DISTRIBUTION OF PH13MI,
importance for the applications to prime number theory. According to Rielnanu's famous but still unproved hypothesis, all All the other the imaginary zeros have the real part f7 hypotheses of Riemann have been proved by later investigators. Important contributions to the theory have also been made by Mangoldt. Landau, Bohr. Hardy. Littlewood and title Selberg. These results belong, however. to the higher analysis and will
not be developed in this book. We shall only mention the following result of Titchlnarsh
:rLr) Li(.r)
((1)
01.a..r
:...
a (log .Ail it is valid for all sufficiently large values where a. of r; E is a positive number. k and a are certain positive con
stants, and 0 denotes a function of x which varies between the limits  1 and 1. This formula, which was proved in 1935. expresses the best result up to now for the function T(x). It is easily seen from the formula that r (.r) is approximated by Li(.r) with Great accuracy. This is verified by numerical examples. For instance, if x = 1000000000. we have. apart from the decimals, 17.5i..
this difference is less than
L,000
.
of the value 7 (10").
It was shown by Littlewood that the difference :r (x)  Li (.r) assumes both positive and negative values infinitely often. There was a sensation when recently an elementary proof of the prime number theorem was given by Atle Selberg (1948). The proof is elementary in the sense that it uses practically no
analysis, except the simplest properties of the logarithm. We shall give this proof in Chapter VIII. 17. Some elementary results on the distribution of primes.  Let
us put, for x ? 2,
P =ll psx where the product extends over all primes have
1,
Then we
CHAPTER II
58
pr
I
I
P
P
'.
r
[rl .}.1
dit
I
=1n
f du It
it 1
1
or
Pa>logx.
(1)
Hence lint
(2)
TW
Thus, corresponding to every d > 0 there is a natural number I such that
11<
(3)
ICY
where p1 i p., ..., pr are the first r primes. If B (x) denotes the
number of natural numbers c x which are not divisible by any one of the primes p1, p$, ..., pr, we have by Theorem 27 B
(4)
(d) rl
l]
the sum being extended over all positive divisors of the product p, .V . Pr. Since clearly 7r(x)
it follows from (4) that /
+a(d)X
and, since the number of divisors of the product p, P3 we have r
r + 2r
7t
Then, by (3). we obtain (5)
1
.ri II 1 1  1 .
t(x) < r + 2r + jdx.
pr is 2r,
ON THE DISTRIBUTION OF PRIMES
59
Let us choose the number 5 greater than the rth prime and such that
r
;
21' < L. td
t
it follows that
Then, for all x
?L: (x)
We can thus state Theorem 28. link
7r)= 0.
.c r
When P, has the above significance, we get
log(l +1
Pr= nSz
P
+
lV+1y+ P
Since, for any positive r, log (1 + x) < .i . we obtain Since, lo" 71X< ,, r
...1=1+L 1) 2+1)1 ,,gx13
7,ax`17
As before the snnnnmations extend over all primes p c x. The
value of the last sum on the righthand side is obviously less than
Hence we have
log P. < , + 1. P3x
Combining this result with inequality (1), we can state Theorem 2.9.
The summation being extended ores all primes < x,
we hare the ineq.Lalilq 4J
> log loI  1.
As a consequence we see that the infinite series S t over all primes is divergent.
CHAPTER II
60
Instead of the function z (x) it is often more convenient to consider the function z9 (x) =2 log p, 1Isa
the sum extending over all princes p5; x. We shall prove the following theorem of Tchebychef: Theorem
There exist two po.,eitise conxtants c awl el such that
30.
(6)
for all x > 2.
Progf. Let n be an integer 2: 2. If pm is the highest power of the prime 1) which divides the binomial coefficient _(2n)!
Cn
(n}=n!77!' C
(7)
then, by Theorem 25, we have 2
2
777
where r is the highest integral exponent which satisfies the inequality pr C 2 71,
(8)
and thus
rlo~2n1 [.fore JJ
The difference 2
LP' has either the value zero or the value 1, and therefore m S r.
(9)
Now we have 2n 7I
(11+1)(n+2)

)1
277
n
1
77+11
h
ON TIIE DISTRIBUTION OF PRIMER
and
61
on the other hand, by (9) and (8),
21, < (:2) _ II pm < II pr p,2n
)z
p 2n
the products extending over all primes p < 2 n. Therefore, by taking the logarithm, .
No 2<
L.J
p 2n
r lo agp=
2 lo[L]iogp.
every p we have
12L? n log p
and for all p
< lo;L2 i,
lob p '
l1? n
lo ?)a log])
Thus we obtain for s = N2 n
j2
n log 2< v llo
b
log p + ,,
log p
p.;. n
or
,log2
V2 n log2 i + 79(3n).
Hence, for all sufficiently large integers n, z`} (2)1)> n log 2V
log222> 12 ()2 + 1)log2.
If 2 n :9.t < 2 n + 2, we obtain
0(x)_>_49(2n)(n+ 1) log 2 >
.
log2
for all sufficiently large x. This proves the first inequality in Theorem 30.
The number (7) is clearly divisible by all primes p which are a and _ 2 n. Therefore we have P,
2st I 0
(22n)> (2:1)> 11v 1,
CHAPTER 11
62
and, by taking the logarithm, 2nlog 2>d (2
id.
If x = 2" (h integer > 1), it follows that i9(x)=(0 2",d,2"') + (c9 2""1 0(2",2) +
<(2''+2'''
!
+ 2)log;2<21, +1log2=2.elob;2.
Further, if 211 < x < _', we get, for all .,; 'l, 79 (.e) X 0 (2"") < 21 +1 log 2 < 4x log 2,
which proves the second inequality in Theorem 30. From Theorem 30 we easily deduce Theorem 30 a. There exist t,ro positive can tants P_, and c;, such that (10)
C2
too,./
< ; (x) < C3
log x
for all e> 2. In fact we have r(x)log.r. Pax
and so by (6)
(x)
.T.
log x
log x
On the other hand, when we put q = Vx, pgx
6 (x) > Ylo).P R: ]oga; [,r
r (1
)].
pr
and since 7L( C <
)2z9(.v)y
log x
log .e
Q. E. D
Starting from the relation
log /?! =, (["] 4 Pan
1
2] +
I
P
) loelcr p.
ON THE DISTRIBUTION OF PRIMES
63
we shall prove Theorem 30 b. The sum being extended orer all primes :5; x, ice
hare the formula
/there 0 is a frutctio)t of .r such that I0! is less than a po.itire constant.
We have 2
± in G 1)
pl;
I +
PL'.31
+
<),=` (nz+ P
'It
100, nP> P
loge=n
1ogp ;,.p (1)  1)
>(n).
)
ran
1
P
Further, we see that )z
s+..
I1
Consequently, applying Theorem 30, we obtain (12)
11ogn!
Y,logp=a, Pign p
where a is a function of it such that
Ia
is less than a positive
constant. For every integer h ? 2 we have
logh hlogh(h1)log,(h1)(h1)log(l + k]
/
where the last term is less than 1. Hence
)i logn  )t  I <
log h < it log it.
Combining these inequalities with (12) we obtain the formula (11)
for all .r ? 2.
Finally we shall prove that the prime number theorem is equivalent to the theorem: (13)
limn .r 00
( =1. x
CHAPTER II
64
We have
0(4
T(x)logx
and, for 3
?I`
log J
log y
Hence
_ _ :i (x) log' :c
z9 (x) ;l
y log x
L
x
./'
lo, log q
0 (.Y.)
x
Choosing y = xa, where d = I  logllog x we obtain
Since, according to Theorew
U<
311..9 (.r) >
rx
t x.. we have
logtog 21
Here the righthand side tends to zero for increasing a'. Hence lien
oo
:v (x) log 0 (x)
= 1,
which proves the theorem.
The proofs of Theorem 30 and of the following results in this section give an idea of the elementary methods applied by Tellebychef. It is apparent how the constants e, e1, r2 and cc3 may be
determined numerically. With our simplified method it is not possible to obtain values of the constants as good as those found by Tchebychef.
18. Other problems and results concerning prunes.  A polynomial in .c'. (x) = ao 1 a1 .r i u2 .12 i
+ a,, xn
where the coefficients cco, al.. .. a,, are integers, represents integers for all integral values of the variable a'. We prove
65
ON THE DISTRIBUTION OF PRIMES
Theorem 31. No polyinozziial f (x) with integral coefficients, which is
not a constant, can represent only primes for all integral x ? xo .
Proof. Let x0 be an integer such that f (xo) = p is a prime. We consider the identity a= [(xo
(xo + p y) .f (xro)
p ii)'  x;,]
1=o n ki)Xi
i=u
aI
(')i
ry
(lp p)2 + .. .
If y is an integer, the last sum is an integer which is divisible by p. The integer f (.ro + p y) is therefore divisible by p. From algebra we know that the three algebraic equations +1)y) = 0 and .f (.c'o + p y) _ ±p
have at most 3n solutions y, if J '(.v) is of degree n. Thus, for sufficiently large integral values of y the number f (xo + p y) is an integer which is different from zero and different from ± p. Since the number f'(.ro + p y) is divisible by p, it cannot be a prime. Thus the theorem is proved. There exist, however, special classes of polynomials which produce primes for a long sequence of consecutive integral values of the variable. A remarkable example is the polynomial .,c + 41,
which produces primes for the following 80 consecutive values:
x=0, ± 1, ±2....,±39 and 40. Recently (1947) W. H. Mills succeeded in proving the existence
of a positive constant A such that the expression
[All
yields only primes for all integral values of x. The proof is, however, based on the most advanced results in prime number theory and will not be reproduced in this book. 5516670 Trygve Nagell
CHAPTER. 11
66
Let u be a natural number % 3, and let r be one of the numbers 1, 2, 3, ..., n 1. A necessary condition for the existence of infinitely many primes in the infinite arithmetical progression
r,r+n,r+2n,r+3n,....r+xn.... is obviously that (). u) = 1. According to a famous theorem of Dirichlet this condition is also sufficient. Dirichlet's proof of this theorem requires methods from higher analysis. Atle Selberg showed (1949) that the elementary method which he developed for proving the prime number theorem, could also be applied to prove Dirichlet's theorem. In the same Year another elementary proof of Dirichlet's theorem was published by H. Zassenhaus; it depends on the theory of algebraic numbers. None of these three proofs will be given in this volume. However, in Chapters IV and V we shall prove the theorem in some special cases.
The question whether a polynomial of second degree in one variable will represent an infinity of primes for integral values of the variable has not yet been solved, not even in any special case.
There are many other solved and unsolved problems concerning primes. We mention the following conjectures of Goldbach (1742): 1.
Every even number AV = ti is the sum of two odd primes;
2. Every odd number X
9 is the sum of three odd primes.
If the first conjecture is true, the second one is also true. By means of methods from higher analysis Vinogradov proved (1937) the truth of Goldbach's second conjecture for all sufficiently
large values of N. The first conjecture has not yet been proved. But N. Pipping has verified that it is true for all N = 100000. Viggo Brun has generalized the sieve method in different ways
in order to make it more effective and applicable to other sequences than that of all natural numbers. By means of his new sieve method he showed (1919) that every positive even integer N can be written as the suns of two positive odd integers Qi and (9g, v = 01 + Q2,
ON THE DISTRIBUTION OF P1I1MES
67
where l21 and 02 are the products of at most nine prime factors. It has since become possible to replace nine by four in this result.
Brun also applied his method to the study of the frequency of prime twins. He showed in 1921 that, if T (x) denotes the number of prime twins which are < x, then 100x (log x)=
From this it follows that the series (:l
is
I)+(111
r
11i)+(117+19)+( 9
convergent or perhaps finite. (Compare the remark after
Theorem 29.)
For proving a theorem in the theory of groups Bertrand conjectured (1845) that there is at least one prime >x and <2x2, when x % 4. It is easily seen that the truth of this conjecture follows from the inequalities (2) of Tchebychef in Section 16. From this the problem arises of finding still smaller intervals which contain at least one prime. The best answer to it was given in 1937 by Ingham, who proved the result: There exists a positive constant k such that there is at least one prime >.c and < x + k x5'
CHAPTER III
THEORY OF CONGRUENCES
19. Definitions and fundamental properties.  Let n be an integer 0. The integers it and h are said to be congruent modulo n, or for the modulu,e n, when their difference a  b is divi
sible by n. To express this fact we write a = b (mod n),
where the symbol  is to be read "is congruent to." This relation is called a congruence niodudo n. The number n is the modalus
of the congruence. When the difference a  b is not divisible by n. we say that a and b are incongruent nwodulo n and write a
b (mod n).
These concepts and notations are due to Gauss; he introduced them in his Dixquisitiones arithnmetiear.
From the definitions we easily get the following rules for operations with congruences: I. 1/' a = b (mod n) and b = e (mod n), then it = e (mod n). For, since a  h = h it and b  cs = hl n, by addition we have
ac=(h+h1)n.
II. If a = b (mod n) and e m d (mod n)4 then a ± c = b ± d (mod a). For, since a  b = h n and r  d = hl n, we have a ± e
.(b±(1)_(h±hl)n.
III. If a = b (mod n) and c = d (mod u), then a c = h d (mod n). For, since a  l) = It n and e  d = hi n. we have a c  b d _ (b hl
d h  h hl n) n.
In particular, by putting c =d we get IV. if a = b (mod u). Then a r = be (mod n).
By repeated use of Rule III with c = it and d  h, we find
THEORY OF COIVGRUENCES
69
V. If a = b (mod )i), then a"'= b" (mod n) for any positive exponent m.
By using successively Rules V, IV and II, we get the following more general result:
VI. If f(x) is a polynomial in x with integral coefficients and if a = b (mod n), then f (a) = J'(15) (mod )?).
According to Rule IV a congruence may be multiplied by an arbitrary integer. But, in general, it is not permitted to divide a congruence by an integer, even if the quotients are integers. This is illustrated by the following example. We have 5 9 = 5.3 (mod 10), but since the difference 93 is not divisible by 10, the common factor 5 cannot be cancelled. We have, however, the following rule
VII. If m a  m t (mod n) and/ if d is the greatest common rliri,or \ oof m and it, then a = t (mod d)
Hence, in particular,
VIII. If m a = m. t (mod it), and if m and n are relatively prime, then a = b (mod )i).
The Rules IVI are quite analogous to those valid for equations in ordinary algebra. Suppose that a and b are congruent modulo it. Then it is evident that they have the same principal remainder modulo n,
and vice versa. Either of the numbers a and b is said to be a residue of the other modulo it. It is clear that (a. re) = (b. n). In the sequel the modulus is always supposed to be positive. 20. Residue classes and residue systems.  The integers a and t are said to belong to the same residue class modidu n when they have the same principal remainder modulo nn. It is evident that
there are in all
n:
classes modulo n, corresponding to the it
possible values of the principal remainder
0,1,2,...,?21.
CHAPTER III
70
The necessary and sufficient condition for the integers a and b to belong to the same residue class modulo ii is obviously that a = b (mod )2).
Any set of it integers al, a2... , a representing all the residue classes modulo
is called a roniplete residue system inodulo n. The simplest system of this kind is 0. 1, 2, . ., ii  1. it
Theorem 32. !/' the natural numbers in and it are rclalire7y prime
and r is an i;tiegrr, the it
ntnnber..
r. it/ ' r, `? in +
(1)
(it
1) in
: r
ftnvit a rntndtlrto res?Ilur ..gslrnr iiiodulu it. Proof. It is sufficient to show that the numbers (1) are in
congruent modulo it. If we suppose h )n y r
( nzod n)
with ii then (li.  k) in = 0 (mod n) and, since (in. u) = 1, h =1r (mzod n), which is contrary to hypothesis. Moreover we can prove Theorem ..i. if the natural number in and it are relatively prime, if .c ruffs through a complete residue sztstenz inodulo 72. and a'1'
y rims through a rrunplete residue system inodulo in. then the in n numbers
ntx + gill
(2)
form a complete residue system moduclo in n.
It
sufficient to show that the numbers (2) are incongruent modulo ann. Suppose that Proof.
is
mx + it y = mx1 + nnyfi (mod in n).
Then inx  in.xl (mod n) and ny =T nY1 (nmod in), and, since (in, )i) == 1.
't' = x1 (mod n) and
ii
y1 (Itlod n,).
THEORY OF CONGRUENCES
71
If a residue class modulo n contains a number prime to n, all the numbers in the class are prime to n. A residue class modulo n with this property is said to be prime to n. It follows from Section 8 that there are q' (n) residue classes prime to rr. Any system of T(n) integers representing all the residue classes prime to n is called a reduced residue system nrorlrrlo n. We conclude by proving Theorem 34. If the natural nunthers in and n are relatively/ prime, if' it runs through a reduced residue, syaeni modulo n, and if r rims through a reduced residue system modulo in. then the 4p (in) rp (n) integers
mu + nV
(3)
form a reduced residue system tnodulo rn rr.
Proof. Since (in. n) _ (u. n) = (r. m) = 1, we have (nr if + n r. rrr) _ (n r, rrr) = 1,
(in it + n r, n) _ (nr it. n) = 1, and thus
(nru + 12 r, nru)=1.
The number m.c + n y is prime to in a only when x is prime to rr and y is prime to rrr. Since, by Theorem 33, the numbers (2) form a complete residue system modulo inn, the numbers (3) form a reduced residue system modulo tan. In particular it follows that op (m u) = qr (in) qr (n),
when ()n. n) = 1. giving a new proof of Theorem 12. 21. Fermat's theorem and its generalization by Euler.  Fermat stated the following theorem without proof in 1640.
Theorem 35. If p is a prinzc and a an integer not divisible by p, then the difference
aP1 1 is divisible by p.
CHAPTER III
72
Proof It is sufficient to show that the congruence xn = x (mod p)
(1)
is satisfied for any integer x. It follows from Rule VI that we
only need consider the values x= 0, 1, 2.. . p  1. We use mathematical induction. The congruence (1) holds for x = 0. Suppose that it holds for x = a. Then it may be shown that it also holds for .r  a + 1. The binomial coefficient _P(1,1)
1'
k)
.(hk+1) k!
is obviously divisible by the prime p when k = 1, 2, 3..... p  1. Hence pI
(a + 1)P = 0 + 1 +
p
I (k
ak = ar + 1 = a! I (mod N).
Thus Theorem 35 holds for any natural number and therefore
for any integer. The first proof of it was given by Euler in 1736. He established later (1760) the more general result: Theorem 36. I/' n is a natural number and a i.q prince to n, then
a'p:";  1 (mod 7t).
(2)
Proof. Put q, = q? (n). Let a,,. a2..... a,r be a reduced residue system modulo n. Then the numbers
al a, a2 a, .... a,, a are evidently incongruent modulo n and prime to n. Hence they also form a reduced residue system modulo n. Therefore. taking the product, we have al a a2 a

a,r a =alas
a,r (mod n),
and, if we divide each side by the product of the numbers at, we conclude that a''''") = 1 (mod n).
Remark. It often occurs that we already have (3)
crr m 1 (mod n)
T}ICORY OF CONGRUrl'CES
73
for a positive exponent f  rp (n). This is illustrated by the example n = 12. q' (12) = 4 and
12=52= i2=112= 1
(mod12).
We shall treat the question of the minimum value of the exponent f in (3) in Section 31. 22. Algebraic congruences and functional congruences.  A poly
no}vial in x ./,(X) :  a0', .ll

n1
X,,,1
L
am
is called an integral polynomial when the coeff}cienis no a1,
.
,
a,,,
are integers. An integral polynomial represents integers for all integral values of the variable .r.. The integral polynomial f (x) is said to be prunitire when the greatest common divisor of the coefficients ao, al. , an, is equal to 1. Let f (x) be an integral polynomial and n a natural number. If c is an integer such that J '(r) is divisible by n, we say that r is a root or solution of the algebraic co)'gr'uenre (1)
f (.r)  0 (mod
n),
and also that c is a root of f(x) modirlo n. When r is a solution of the congruence (1), all values x for which .r == r (mod n) are also solutions. All the solutions belonging to the same residue class modulo n as c are considered as a single solution. Therefore, to determine all the solutions of the congruence (1), we need only try the values x = 0, 1, 2, .... r:  I (or other representatives of a complete residue system mnodulo n).
If bis the first number in the sequence of integers
br,br}....,bm,....b1,bo which is not divisible by n, we say that the congruence bo + 7i1 x + lie x2 + ... + b,,, x", + ... + br xr °'0 (mod )i)
is of degree m. We speak of linear, quadratic, cubic and biquadratic congruences according as the degree is 1, 2, 3 or 4.
74
CHAPTER III
The congruence 18x3 + x2  3.r. + 2 = 0 (mod 6)
is quadratic and equivalent to the congruence
x23x h 2=0 (mod 6). We find by trial that the congruence
x5+2x'+x3+2x22xI3=0 (mod 7) has only the solutions (mod 7). The congruence
x2 + 2  0 (mod 5)
has no solution at all. The problem of solving the congruence (1) is obviously equivalent to the problem of solving the Diophantine equation ,f (x) = n y
in integers x and y. Let f (x) and g (x) denote two integral polynomials and put + r,. ,!r (x)  g (x) = co .r" + PI x'  1 +
If all the coefficients co, cl, . ., c,, are divisible by the natural number n, we say that f (.r) and g (x) are identically congruent .
naodulo n, and write (2)
f (x) = g (x) (mod n).
This relation is a functional congruence or an identical congruence.
When f (x) and g (x) are constants, this new notion coincides with our earlier notion of congruence. If all the coefficients in f(x) are divisible by n, f(x) is identically congruent to zero modulo n. The rules for operations with functional congruences are analogous to those for ordinary congruences. If f(x), fI (x), g (x) and gI (x) are integral polynomials, we have as a direct consequence of the definition the following rules:
THEORY OF CONCRUENCES
75
If .f (x) = .fl (x) and f1 (x) = g (x), then
1.
.f (x) = y (x)
If f (.r) =,/i (x) and g (.r) = gl (.r), then
11.
f(x) ± g (x) = f 1(x) ± 91(x)
and
.f (x) 9 (x) _ A (x) gi (4.
The modulus
)i
is here a natural number, the same in all the
congruences.
The identical congruence (2) holds for every x, and therefore
;.r may be chosen as an arbitrary integer. On the other hand, if we have
f
g
(mod u)
for all integral values of x0, we cannot conclude that the polynomials f (x) and y (x) are identically congruent modulo n. This is evident from the following example. The congruence
(x+1)(x+2) (xr n)=2x(x1) (x)i +1)(mod n) is satisfied by all integral values of x, because each side of it is divisible by n ! and therefore by n (Corollary to Theorem 25).
But it is obvious that this congruence is not identical for n > 1.
If the integral polynomials
g (x) and h (x) satisfy the
identical congruence f (x) = g (c) h (r) (mod )i).
we say that f (r) is divisible by g (x) modulo n, or that g (x) is a divisor of f (x) modulo u. f (x) is a multiple of g (x) modudo n. Theorem 87. A neeessarg and sufficient condition for a to be a root of the. integral polynomial f(s) modulo n is that ,f(x) be divisible h x  c niodulo n.
Proof: It is evident that the condition is sufficient. If f(e) is divisible by n, we have the identical congruence f (x)
f'(x)  f (c) (mod n).
CHAPTER III
76
Suppose that the degree of ft r) is BIZ and that
Hence
,f (x)  f(r) .r
x  cr 
=Y, any_,.i=1
c
_
x
e
a, _1.(T,1 + Cx' 2 +
C2Xv3
4
+
c''1)  g (x),
,=1
where g (.r) is an integral polynomial in x of degree nn  1. Thus we have the identical congruence f (x) = (x   c) g (x) (mod n),
which shows that the condition is also necessary. Renz.ark. One may even consider algebraic congruences with several unknowns x, y, z, etc., of the type .f (x, y, ', ...)  0 (mod n),
where f (x, y, z, ...) is a polynomial in x, z/, z, etc., with integral coefficients.
In a following section we shall consider certain types of nonalgebraic congruences.
23. Linear congruences.  In a congruence of the first degree a x i b  0 (mod n)
the coefficient a is (by definition) not divisible by n. If a and Ia are relatively prime, the numbers
b, a + b, 2a + b,
.
.
.,
(nl)a+b
form a complete residue system modulo n (Theorem 32). Hence just one of these numbers is = 0 (mod n). and we have proved Theorem 38. If (a, n) = 1, then the linear congruencr
ax +b=0 (modn) has exactly one $olntion modulo n.
77
THEORY OF CONGRIJENCES
By means of Fermat's theorem (Theorem 35) the solution may be written out explicitly by the formula x =  b a`c
("'1 (mod n).
Theorem 38 is contained in the more general Theorem 39. The congruence
ax+b=0 (mode)
(1)
has no solutions or exactly d= (a, n) solutions, according a. b is not, or is, a multiple of d.
Proof When the congruence (1) has a solution, it is clear that d = (a, n) is a divisor of b. If we assume that b is divisible by d, the congruence a
h
x+
= 0 imod
(Theorem 38). Then the
has exactly one solution :ro modulo
congruence (1) is satisfied by all the numbers x0 + y a, where y
is an arbitrary integer, and only by these numbers. Hence the incongruent solutions of (1) modulo ua may be represented by the d numbers ao, xo 4
..v0
+ (cl  1)
i
cl
Finally we prove the following theorem on sets of simultaneous linear congruences.
Theorem 40. If no two of the natural numbers nl, n2.... has a common divisor > 1, the congruences (:')
x = al (mod n1), x = a2 (mod 122).
.
.
.
`?)
, x = ar (mod nr)
always hare common solutions x. These solutions are f/icru b!r the nacmters in a certain residue cla.vs modulo V  "0 2 nr
CILtPTER 111
78
If we put 1V = rrj v; (i = 1, 2, . , )), then clearly , 1'r) = 1. Now, according to Theorem 6 (Chapter 1) the equation Proof:
(1'1i v2,
.

.
+'Vrf/r=I
M1'11/1 + 1'21/2 +
has integral solutions y , 1/Z... r), we have
.... Putting; 1'; //j = yj (i = 1, 2,
,
zl + ?2 + ...
where z; = 0 (mod u;), if system (2) is satisfied by
j
x = al ?1 } a2 ?. +
(3)
I.
1
?r = 1,
and z; = I (mod rrj). Thus the 
 ar 7r (mod.,'V).
If further xo is an arbitrary integer satisfying the system (2) and if x is defined by (3), the difference x  is obviously
... , and thus also by their product 711 »2 ??I.= N. Hence xo belongs to the same residue class modulo V as x. This theorem was known by the Chinese mathematician Sun divisible by all the numbers rr; (i = 1, 2,
Tse (about A. D. 250). Examples. 1.
In the linear congruence
6x=3 (mod 15)
a = ti, b =  3, n = 15 and thus 'i = (a, ,e) = 3. Using the same method as in the proof of Theorem 39 we find .'0  3 (mod 5), and hence .1, = 3. 8. 13 (mod 15). 2.
In the simultaneous system x = 1 (mod 4), .' = 2 (mod 3), x = 3 (mod 5),
a1 = 1, a2=2, U3=3, nl = 4, n12 = 3, n3=5; thus V=60 and 7'1 = 15, 1'2 = 20, v3 = 12. The equation I by, + 20 12 + 12 i/3 = 1
has the solution Jr = Y2 = 1, !/3 = 3, hence ?1 = 15, ?2 =  20, z3 = 36, and according to formula (3) x = 53 (mod 60).
THEORY OF COP GR[TENCES
79
24. Algebraic congruences to a prime modulus.  In the following Theorems 4143, f'(x) denotes an integral polynomial such that the algebraic congruence Ax) ° 0 (mod p), where p is a given prime, is of degree rn. Thus, when ao denotes the coefficients of x'n' in f ao is not divisible by p. Theorem 41.
1I7en the algebraic congruence of degree 211
f (x) = 0 (mod p),
where p is a prince, has S incolilgruent Solution r1, C2, ..., C, 7izodulo p, the following identical congruence holds:
f (,r) _ (x  r1) (.r  e2)
(1)
(x  e,) g
(mod p).
Here we hare g (x) = ao xm8 + h (x), where h (x) is an integral polynomial of degree < 111  s  1. In particular, if ?n = s, we hare g (x)  t1
Proof. We use mathematical induction. The theorem is true for s = 1 (according to the proof of Theorem 37). Suppose that 1. it is true when the number of incongruent solutions is We shall establish its truth when this number is s. By hypothesis we have the identical congruence (2)
.f'('V) _ (.!  c1) (x  c2)
(x  c81)fi (a') (mod p),
where fl(x)aox'n11 is an integral polynomial of degree Since cR is a root of f (x) = 0 (mod p). it is clear that (e8  e1) (e8  e2) ... (c8  r81)fi (c8) = 0 (mod p).
Hence ft (e8) = 0 (mod p).
According to Theorem 37 we must have the identical congruence fi (x) _ (x  cr) g (x) (modp),
where g (x)  aox'n8 is an integral polynomial of degree
CHAPTER III
80
Theorem 4k'. The algebraic congruence of degree m
f (x) = 0 (inod p),
(3)
where p is a prime, has at most in
incongruent solution.:
modulo p.
Proof. Let us suppose that the congruence (3) has the incongruent solutions eI, r2, .. , cIuodulo p. From Theorem 41 we conclude that .f (x) = ao (x  (I) (.c
 r._)

 (x 
(mod 1)).
If there were another solution y incongruent to c, (i = 1, 2. we would have ao (7  el)
c2)
('  c,,,) = 0 (mod p).
This is impossible since none of the numbers ao and y  c; is divisible by p.
The theorem is not true when the modulus is a composite number. Thus the congruence xz
 1 = 0 (mod 12)
has the four incongruent solutions .L _ ± 1. ± 5 (mod 12). Theo,rn, 43. Suppo.c that the algebraic conyrucuec of degree n, ./'(x) = 0 (mod ),),
(4)
cohere p is a prince, has in incongruent solutions modulo p. Suppose also that g (.e) is a divisor of J '(.c) nuudulo p, and that the congruence
g (x) = 0 (mod p)
(5)
is of degree
It.
Then the last congruence has exactly It in
cw,yruent solutions modulo p.
P,v,of. By hypothesis we have the identical congruence (6)
J '(x) = y (x) h (x) (mod p),
where h (x) is an integral polynomial. Suppose that the congruence
81
THEORY OF COI'CRUENCES
It (.L) = 0 (mod 1))
(7)
is of degree v. Then we get from (6) u r = in. Further. suppose that the congruence (5) has exactly It, incongruent solutions modulo p and that the congruence (7) has exactly vi incongruent solutions modulo 1). Then it follows from (6) that every root c of f (x) modulo p is a root either of g (x) or of It (.,) modulo p. For if , f (c) is divisible by p, so is g (c) h (c). Hence we have 1Ii + vi nt. It follows from Theorem 42 that p1 <,u and ri < v,
and thus ui + vi  in. Hence we have It,
'
ri = )it and pi = u.
vi = v. This proves Theorem 43.
Finally we prove Theorem 44. Thr product of ttco prindticr polynomials is also a priori ti re pol ynomtmial.
Proof. Let g (a) and It (x) be two primitive integral polynomials.
Suppose that the product g (x) h (x) is not primitive. Then there exists a prime p such that the identical congruence (8)
g x) h (x) = 0 (mod p)
holds. We now form gi (x) from g (x) and hi (x) from It (x) by rejecting all terms of g (.r.) and h (x) whose coefficients are divis
ible by p. The polynomials gi (x) and hi (x) have the following properties. They are both different from zero. The coefficients of the highest powers of x are not divisible by p. They satisfy the identical congruences rh (x)  g (.x) (mod p) and hl (x) = h (x) (mod p). Hence the identical congruence gi (x) hi (.1) = 0 (mod
1))
holds. This is impossible, however, since the coefficient of the
highest power of .r in the polynomial gi (x) Its (x) is not divisible by p. Hence the product g (x) h (x) must be primitive. 25. Prime divisors of integral polynomials.  Let J '(.;r) denote an integral polynomial which is not a constant. If c is an integer such that f (c) is divisible by the prime p, we say that p is a prime divisor of the polynomial f (x). We prove 6  516670 Trygre i\ agell
CHAPTER III
82
Theorem 45. Every integral polynomial ,f (x) which is not a constant
has an infinity of prime divisors.
Proof. We consider the integral polynomial f (x) = ao x'n + ct,, x  I + ... + a.,
where in ? 1. When am = 0, all the primes are prime divisors of J '(x). Thus we may suppose a,,, = 0. Now assume that f(.r) has the prime divisors PI, P2, . , 1), and no others. If 11 112 ... P, a,,, ,
we get where
9(y)=1
4Ii/+_12t/2+
±:lmya,
is an integral polynomial in y. The coefficients Al are all divisible by the product. PI Ps . ' ' Pr.. Hence none of the primes , Pv can be a prime divisor of the polynomial q (y). Pi, P2, Every prime divisor of g (y) is a prime divisor of j'(x). We must' consequently have g (y) = ± 1 for every integral value of y. The algebraic equations g (y) _ ± 1 have, however, at most '? n: roots. Thus it is possible to find an integral value of y such that g (y) is divisible by a prime different from the primes Pi,1'2, Pr Hence Theorem 45 is true. It may also be formulated in the following manner: 117hen f(x) is an integral polynomial which is not a constant. the congruence
f (x) = 0 (mod 1))
is solvable ,fur an injmitq of printer p.
It is easily seen that all primes are prime divisors of the linear polynomial ax + b, except the primes which divide a and not L. For the linear congruence
ax + b = 0 (mod p)
is always solvable when the prime p is not a divisor of a (Theorem 38).
THEORY OF COAGRUENCES
83
In Chapter IV we shall determine all the prime divisors of any polynomial of the second degree, and in Chapter V those of some other types of polynomials of higher degree. :26. Algebraic congruences to a composite modulus.  Let f (x) be
an integral polynomial in x, and let II be a composite integer such that 7l =pi ps' ... par, (a,> 1), where 1l1, p2, ..., p,. are distinct primes. Then it is evident that the problem of solving the algebraic congruence
f(x) = 0 (clod )r)
(1)
is
equivalent to the problem of solving the system of algebraic
congruences
f'(x) = 0 (mod I)i'), 0 (nlod p"2),
0 (mod jlar).
If congruence (1) is solvable, every congruence in this system is
solvable and vice versa. Suppose that the first congruence in the system has the solution e1, the second congruence the solution e2, etc., and finally the All congruence the solution (,,.. Ac cording to Theorem 40 (Section 23), the system of congruences (Inod pi'), (2)
,r. = r2 (mod 1y2*),
.c = er (mod ))ar)
representing all the numbers in a has exactly one solution certain residue class modulo u. Thus the number $ is a solution of congruence (1), and we obtain all the incongruent solutions modulo n of this congruence when in system (2) the number c, (i = 1 , 2, ... , r) runs through all the incongruent solutions modulo par of the congruence (3)
f (x)  0 (Inod ]in,).
CHAPTER III
84
In particular we state Theorem 46. If congruence (3), where i = 1, 2, . . ., r, has vi incongruent solutions modalo pat, then congruence (1) has VI V ... l,r
incongruent solution: nuodalo n.
From the preceding results it is evident that we need only consider congruences whose moduli are powers of primes. Applying Theorem 46 we prove Theorem 47. Let n be a natural number such that )! = 23 pl' `)Q, ... 1)ar r3
where pi, P2, ... , Pr are distinct odd primes, and denote by `' the number of incongruent solutions modulo n of the congruence
r 2  I  0 (mod n).
(4)
Then X = 2r ,Jor P = 0 and 1, 1' = 21'+I for P = 2, and X = 2r+2 for # ? 3. This is also true for r = 0, i. e. when n is a power of 2.
Proof. As just proved above, we need only consider the congz uences (5)
X2  1 = 0 (mod pa'), (i = 1, 2,
.
,
)),
and (6)
x2  1 = 0 (mod 2,3).
It is evident that the congruence (5) has only the two solutions x = ± 1 (mod pal). The congruence (6) has for P = 0 and 1 only one solution, which may be represented by the number 1. For ft = 2 it has the two solutions ± 1. For P = 3 it has the four solutions ± 1, ± 3. Suppose next P ? 3. Then only one of the numbers x  1 and x + 1 is divisible by 4. In this case a necessary and sufficient condition for x to be a solution of the congruence (6) is that x  ± I (mod 2,3I); thus we have the four solutions
85
THEORY OF CONGRUENCES
x
± 1, + 1 + 231 (mod 2,3).
Hence, applying Theorem 46, the theorem follows as stated above.
27. Algebraic congruences to a primepower modulus.  Consider a primitive integral polynomial f (.r) = ap x1n +
al.1.ne1 + ...
+ a,1 x +
an,,
whose discriminant D is different from zero. When we have by Taylor's theorem (1)
T ('c  2
.f(x) =M) + (x
is arbitrary
+
+ (x
)m
fim)
)
717!
In this identity the coefficients
l / r ()
977
ao
n _r
a,
N7  1
mr1
4 (10,
are obviously integers if is an integer. Suppose that p is a prime and y a root of the congruence (2)
.1(x) = 0 (mod p).
Then f (.,c) has the divisor x  modulo 17 (Theorem 37). If f (x) also has the divisor (x  )2 modulo p, we say that is a multiple root of the congruence (2). Otherwise ; is a simple root of this congruence. If is a multiple root of (2), we obtain by (1) the identical congruence (3)
l (r) =
)2
i/ (a') (mod p),
where g (x) is an integral polynomial in x of degree in  2. Con versely. when this congruence holds, it follows from (1) that f is divisible by p. When and ,f'( ) are both divisible by p, congruence (3) holds. Hence a necessary and sufficient condition for the root to be a multiple root is that f" be divisible by p.
CHAPTER III
86
From algebra we know that the discriminant of a polynomial .f (x) is an integral polynomial in the coefficients of f (x.). Hence
it follows from (3) that the discriminant 1) of the polynomial f (x) is congruent modulo pp to the discriminant D1, of the poly
nomial (x  5)2 g (x). Since 1JI = 0 and DI  0 (mod p), we also
have D = 0 (mod p) and we state Theorem 48. If congruence (2) has a multiple root, the discriminant
1) is dirisible by the prime p.
By a similar argument it is easy to establish the more general result
Theorem 49. Let p be a prime, and let f (x), g (x) and h
be
primitire integral polynoanialc satisfying the identical congruence
f
(g ,a' )2 li
(mod p).
If g (x) i.. not identically congruent to a constant modulo the discriminant D of f (.r) is diriciblr by p.
F,
If we know the solutions of the congruence f(x) = 0 (mod p)6),
(4)
(a a 1)
it is possible to deduce the solutions of the congruence = 0 (mod pa+I).
(5)
It is clear that every root of (5) is of the form i t p', where is a root of (4) and t an integer. We seek a value of 1 such that + tpa will be a root of t5). By Taylor's theorem we have .f
tpa) =.f (5) + f (5) /I)"
a)2 +
or
.f( +
+f'(.)tpa (modpft+I).
Hence we have to solve the congruence (6)
f" (
) t =  f (e) (mod p).
THEORY OF CONGRUE\CES
UI
is not divisible by p, that is, if is a simple If the number root of congruence (2), congruence (6) has exactly one solution 1 modulo p (Theorem 38 in Section 23). If f is divisible by p, congruence (6) has either 1) solutions modulo p or none at all. Hence we have proved Theorem G0. If 5 is a simplr root of congruence (2), congruence (4) ha.> r.ractly one root modulo p" congruent to modulo p.
If i., a multiple root (2). congruence (4) has at most p'11 root. modulo p" congruent to nzodulo p.
Theorem 51.
Combining Theorems 48 and 50 we have Theorem 52. Suppose that the diserimminant D of the primitive integral polynomial f (x) is not diri..ible by the prime p. Denote the number of incongruent solutions cf' the congruence by .f (x) = 0 (mod p).
(2)
Then the number of incongruent solutions of the congruence .1(x) = 0 (mod p")
(4)
is al o N. is
Suppose next that the prime p is a divisor of D and that I) divisible by p" and not by 1pi4 1. If congruence (2) has no
multiple root, Theorem 52 is still true. Suppose further that
is a root of the congruence
.1(x) = 0 (mod p""
1)
and a multiple root of congruence (2). Then the number /(5) is at most divisible by p«. For if this number were divisible by p"+1, it would follow from identity (1) that an identical congruence
.f (x) = (x  $01 g (x) (mod p!,+')
must hold, where g (x) would be an integral polynomial in x. By the property of the discriminant D just mentioned we would have
D = 0 (mode"+l), which is contrary to hypothesis.
CHAPTER III
88
is divisible by p'3 and not by ph+1, We now suppose that ,u. Also the number ,/"( + t p't+1) where t is an arbitrary integer, is clearly divisible by pis and not by p3+'. Suppose further that s1 is a root of the congruence
hence f
(x) ° 0 (mod p"),
(7)
where a ? 2 it a 1, and yI = 5 (mod p" t 1) Then all the 1).3 numbers
E1+it p"3
(8)
(u=0,1,21'.......1)
are incongruent roots of congruence (7). For since./"'
is divis
ible by p' and since 2 a  2 # ? 2 a  `l ,u ? a r 1, it follows that the number ./*(,&I
a
=,f'i',l)
3./x (51) 'r z (1, p"
r r11)
3)Zf (e1)
is divisible by p"
Hence we see that the incongruent roots of congruence (7), provided that they are multiple roots of (2), form a certain number of systems of the type (8); and every system contains exactly incongruent roots. Let us determine the roots of the congruence .f (x) = 0 (modp0l+1)
(9)
corresponding to the roots the form
(8)
of congruence (7). They are of
1 + rrp"15 d
r?p",
where it and r are integers. We have l)"13 + e p") = t'(1 + ul)r:+1) + 21)".f'
(511 + u q),_;3) +
and. since f' (1 + 16p"') is divisible by p, ,t'(,", + 112)43 + 1, pry) B
(el y up"_h) (mod pr:+1)
Since 2a2,62a'I1, we have ,f(E1 +
uprr3)=,f(1) + up"i3/''(51) (modp"+1).
Here the lefthand side is divisible by p"11 if and only if p"+1),
THEORY OF CONGRUENCES
89
where it is determined by the congruence
f ll
it
(mod p). Pa
Since J" (yl) is not divisible by p6+1, this congruence has exactly one solution uo modulo p (Theorem 38). Hence. corresponding to the Iii roots (8) of congruence (7) we have derived the following ]);I roots of congruence (9) i + (u + hp) p01+3 + l,p
(10)
where h= 0, 1. 2, and 10, 0, 1, 2, ..p1. The numbers (10) are incongruent modulo Every root of (9) congruent to 1 modulo must be congruent to some of the p.I numbers (10) modulo p"+1. If we put i + rtop"11= Zs, we see that the system (10) modulo p"+1 can be replaced by the p"1.
system 2+tp"
+1,
where t = 0, 1, 2, ... , p'3  1. This system is analogous to the system (8).
Finally we conclude that the number of roots of congruence (9) is exactly the same as that of congruence (7). This is also true for the roots which are simple roots of (2) according to Theorem 50. Hence Theorem 53. Let p be a prime and .,mppo.e that the discrilrrirraart D of the prinr/tire intcyral polynomial f(x) is divisible by p, and not by p'1+'.: their the con fJruenee
f (x) = 0 (mod p"),
where a> 2,u a 1, has exactly the same number of roots as the congruence (11)
.f (x) = 0 (mod e
Of course, roots always means roots incongruent for the modulus of the congruence.
If congruence (2) has exactly a simple roots and exactly A,
CHAPTER III
90
multiple roots, it is evident by Theorems 50. 51, 53 and 42 that the number of roots of congruence (1 1) is at Most + 7.1 172" < m D2.
Hence Theorem 54. Let f(x) be a prinritire inhpral polynomial of degree
in and irith a di..eriminant D different from zero; then the congruence
>'(.,) = 0 (Inod 1;=)
has at most in D2 rorfs.
This result was found in 1921 independently by Ore and the author.
28. Numerical examples of solution of algebraic congruences. We shall illustrate flie theory developed in the preceding section by some examples. 1.
If 1) is a prime, we find by trial that the congruence x3 _
.,,2  2 x + 1 = 0 (mod 1))
is solvable for p _ 7, 13, 29. 41, 43, 71, 83 and 97 and for no other prime < 100. If p 7. the congruence has three roots. For p = 7 it has the triple root x 2 (niod 7), as we see from "3
 x2  2 x r 1 = (.r + 2)3 (mod 7).
The discriminant of the polynomial on the lefthand side is = 49. It is possible to show that the congruence is solvable for p = 7 and for all primes of either of the forms 7 t  1 and 7 t  I and for no other primes. 2.
If p is a prime, we find by trial that the congruence :,.3 _x r 5 = 0 (mod p)
solvable for 1) = 5, 11, 13, 19, 23. 29, 31, 37, 41, 43 and is not solvable for ]p = 2. 3, 7, 17. For p) = 5, 29 and 43 the congruence has three roots and for p = 11 two roots, one simple is
and one double. For the other values of p there is only one root, which is simple. The discriminant of the polynomial on the lefthand side is =  671 =  11 61.
91
THEORY OF CONGRUENC S
3. The discriminant of the polynomial Y '(x) _ .rs + 3 x + 9
is I)_ .3s5. 17.
Since .1(x) = .!.s (mod 3),
the congruence
f (x) = 0 (mod 3)
has only the triple root r, = 0 (mod 3). By the method developed in Section 27 we find that the congruence J '(x) = 0 (mod 32)
has the roots x = 0, 3, 6 (mod 9), and further, that the congruence f (x) = 0 (mod 32)
has the roots x = 6, 15, 24 (mod 27). which are all congruent to 6 modulo 9. Finally, we consider the congruence 0 (mod 3"),
(1)
where a = 3. If this congruence has the root $1. clearly it also has the three roots .) * , yl . 2 3x1. yel i
Suppose that congruence (1) has only these three roots. Then
we shall show that the congruence f (a.) = 0 (mod 3rr+1)
(2)
also leas exactly three roots. Every root of (2) must be of the form Jtj + it 3"1 + r.. 31%
and v are integers. The number f' (x) = 3 (x2 + 1) is divisible by 3 and not by 9. We have where
it
1(sl
1( 3.:1
and, since 2 a  2 ? a
r  3") .r(51 1 it  31) (mod 1.
3r:+I),
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92
it . 3"'/" (51) =0 (mod 3c+1) or
X1)
U = .
(mod 3).
f .3.
This congruence has exactly one root it. Hence, it is shown that
congruence (1) has exactly three roots for all a ? 3. For a= 4 we find the roots x = 6. 33, 60 (mod 34) and for a = 5 the roots x  6, 87, 168 (mod 35).
From this example we see that it is advantageous in special cases to modify the method.
4. We choose the same f (x) as in the third example and consider the congruence f (x) = 0 (mod 5). It is obvious from
A r)  (r + 1) (.r + 2)2 (mod 5)
that the congruence has the double root .r =  2 (nod 5) and the simple root x =  1 (mod 6). Now the congruence
/ ( 2)  5
2) = 0 (mod 52)
or
5
has no solution
75 ,t _ 0 (mod 25)
Hence, it follows that all the solutions of
u.
the congruence f (x)  0 (mod 511)
(3)
are =  1 (mod 5) %vhen a only one solution. 5. The polynomial fl.,)
2.
This congruence has therefore
,.4
 .r2 + I has the discriminant
I)=24.32. We find J'('') =
(mod 3).
Thus the congruence /'(x) = 0 (mod 3)
has no solution at all.
THEORY OF CONGRUENCES
6. The polynomial f (x) = x= + ..
7
D= 27. We find
93
has the discriminant
f(x) = (x  1)2 (niod 3).
Thus the congruence f (.r) = 0 (mod 3)
has the double root ,c = 1 (mod 3). The congruence 0 (mod 3")
has for a = 2 the roots x = 1, 4, 7 (mod 9), and for a = 3 the roots x = 4, 13, 22 (mod 27); for a ? 4 it has no solution. 7. If we choose ./'(X) _ .t 5 + 4 .4  5,t.3 1 2 .r2 + x + 1
and 1) = 7, then we have identically f (X)
= (x + 1)
1)2 (mod 7).
Hence the congruence f (ax)  0 (mod 7)
has the unique root x   1 (mod 7). Thus the congruence f (x) = 0 (mod 7") has exactly one root for every a.
29. Divisibility of integral polynomials with regard to a prime modulus.  In this section polynomial means integral polynomial
in x. Let p be a given prime. Every polynomial f (x) which is not identically congruent to zero modulo 1) may be written in the form (1)
f(x) = ao X", i ai xm1 + ... +
p g (x),
where n (x) is a polynomial, where the coefficients ao, al, ... , am
are integers and where ao is not divisible by p. Then f (x) is said to be of degree vi modulo p. If f (x) is a constant not divisible by 1), then f (x) is of degree zero modulo 1).
CHAPTER 111
94
The product of two polynomials of degrees a and v is obviously
of degree It + v niodulo p. Hence the product of two or more polynomials is identically congruent to zero nlodulo p if and only if at least one of the polynomials is congruent to zero modulo p. (Compare Theorem 44 in Section 24.) Two polynomials which are identically congruent modulo p are said to belong to the same residue elasx ,uudzzlo 1). The number of polynomials of degree n incongruent modulo p is obviously (p  1))),". The polynomial (1) is said to be primary i,wodulo p when ao = 1. The number of primary polynomials of degree m incongruent inodulo p is In every residue class modulo p there is exactly one polynomial
the coefficients of which are ? 0 and < p  1. This polynomial is called the nzo iwl pot?pzzozzzial moclrrlo p in the class. The residue
class which consists of all the polynomials identically congruent
to zero modulo p has the normal polynomial 0 (the constant zero).
When the polynomials
y
and h (x) satisfy the identical
congruence (2)
.f (x) = y (x) h (.,) (mod
/'(x) is divisible by q (x) nlodulo p, q (,,) is a divisor of f(x) mo
dulo ,p and /'(x is a multiple of r 6r) nlodulo p (definitions in Section 22). We say also that,/'(x) is the product modulo p of g (x) and h (x); the polynomials y (x) and 7i(x) are the factors nlodulo p of the product.
Suppose now that f(.r) is not identically congruent to zero modulo p). Neither g (x) nor h (x) is then identically c .ngruent
to zero nlodulo p. If we replace the three polynomials in (2) by the corresponding normal polynomials we obtain: If f (x), and h (x) have the degrees jn, It and v modulo p respectively, then zn = tz + r. Thus, the degree modulo p of a divisor g (x) of ,1(,r) cannot be greater than the degree modulo p of 9(.r,)
The polynomial f(,r) has the following tririal divisors inodulo p: 1. Every integer c not divisible by p. 2. Every polynomial
!1(x) = c f (x), where c is an integer not divisible by p. In fact. e,f (.0 and r c = 1 (mod p), we have f (x) = a' g (.r) (mod p) if g and J '(x) = c d f (x) (mod p,).
THEORY OF CONGRUENCE.
95
p A polynomial of degree > 1 is called a p are trivial. It is also said to be irrcducille uwdulo p. Every linear polynomial which is not identically congruent to zero modulo p is a prime function modulo p.
Let '(x) and (1) be two polynomials which are not identically congruent to zero modulo p. ti e consider the set M of all the polynomials G (.r) which satisfy the congruence (3)
G (rc) m f (x) h (x) + 9.(; )h,, (x) (ulod 1,),
where ii (x) and h,(e) run through all polynomials in x. The sum. the difference and the product of two polynomials in M are also polynomials in M. By putting h (e) = I and hl (x) = 0 in (3) we
see that f(X) is an element of M; similarly it follows that q(x) belongs to M. There exist in M certain polynomials which are not identically congruent to zero modulo p and which have the least possible degree. Let d (;r) denote one of these polynomials;
we can suppose that d (x) is a primary normal polynomial modulo p. Then d (x) is uniquely determined. In fact, let us suppose that dl (.c) is another primary normal polynomial of the same degree modulo 1) as d (x). Then the difference d (x)  d1 (x) is a
polynomial in M; its degree modulo p is less than that of d (x), and it is not identically congruent to zero modulo p. But this contradicts the definition of d (x). Hence there are two polynomials ,i (x) and V1(x) which satisfy the identical congruence (4)
cl (x) =
y (x) f q (.z) V1 (,a:) (mod p).
Algebraic division of an arbitrary polynomial G (x) in M by d (x)
leads to an algebraic identity of the form G (x) = cl (x) 91
r (4),
where ql (j) and r (x) are polynomials and where the degree of r (x) is less than that of d (x). Hence (a (x) = d
!11 (4 ! I. (A (mod p).
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96
The polynomial r (x) belongs to M, and in consequence of the definition of d (x) it must be identically congruent to zero modulo p. Hence G (y) = d (x) !h (x) (mod p).
Thus all polynomials in M are multiples of d (x) modulo p. In particular, d (x) is a common divisor modulo p of f'(x) and g (x). On the other hand, it follows from (4) that every common divisor modulo p of f (x) and g (x) is also a divisor modulo 1) of d (x). All multiples modulo p of d (x) belong to M. The polynomial d (x), which is uniquely determined, is called the greatest common divisor modulo p of f (x) and g (x). If d (x) =1,
we say that the polynomials f (x) and g (x) are relatively prime modido p.
E,camples. By trial it is easy to show that the polynomials
xsx+3 and x  1 modulo 3. The polynomials x + 1 and a:2 + I are relatively prime modulo 5.
Finally we prove the following two theorems: Theorem 55. Let g (x) be a prince function nmodulo p. Then, if g (x)
is a divisor modulo p of the product f(x) h (x) but not of the polynomial f (x), g (x) must be a divisor modulo p of the polynomial h (x). Proof'. Since f (x) and g (x) are relatively prime modulo p, ve have by (4)
1 = f (x) y' (x) + g (x) V, (x) (mod p),
and, multiplying by h (x), we get h (x) =.f (x) Y' (x) h (x) + g (a') 'I (x) h (x) (mod p).
Since f (x) h (x) is divisible by y (x) modulo p, it follows that ff(x) h (x) ° 9 (4'P2 (x) (mod p),
where '2 (x) is a polynomial. Hence
THEORY OF CONGRCTE\CES
h (.,') = g (.r' [Y' (x)
'2
F y'I (.r,') h
97
(mod p),
which proves tile theorem. Theorem 56. 1. Every poll/nwnial 1'(x) of degree in ? 1 iliodlllo 1) can be written iii the fore) (5)
q, (x) (mod f,)
91 (.4 q2
ae a product )/' prime functio)w q) (,r) neodido 1) and a natural number C. ql
(x) is (f degree v) ),odttlo p (i = 1, 2,
Tl+1'2±
then
.  )'p=,n.
.. If 0 < e < ))
ctn( if the pri»u fu),ctio)o, ql (x) are chosen ll.N priuiarl/ noruwl politnomials. then the deco)upositiun (5) is tu)ique. apart ,fro))) the order of the prime f ilnction, .
I'rool'. Among the divisors niodulo p, of f'(x) there are certain polynomials which are of the least positive degree. Such a divisor q1(.)') must be a prime function modulo 1). For, if a divisor t (x) of ql (a') modulo p were nontrivial, it would be of lower degree than q1 (x) modulo p. Since t (e) is also a divisor of f'(x), this is only possible when t (.,') is trivial. We prove the first part of the theorem b1' induction. The assertion is true for in == 1. Sup
pose that a decomposition of the type (5) exists for all polynomials of degree < He inodulo p. Then we can prove that it also exists for a polynomial of degree In > 1. As was shown above, (x) has at least one prime divisor q1(x) modulo p. Hence = q, (x) g (x) (mod p).
(f')
where g (r) is a polynomial of degree in  I Inodulo p at most, provided that f (.c) is not a prime function. Therefore, by hypothesis, Ax) can be written in the form g (x)  c q2 (x) q3 (x)
' qa (x) (mod p)
as a product of prime functions and a constant. Combining this congruence with congruence (ti), we get congruence (a). The second part of the theorem follows directly from a remark in the beginning of this section. 7  5 1ti67U Tryyve \agell
CHAPTER III
98
It remains to prove the third part of the theorem. We may suppose that the prime functions qi (r) in (5) are primary and normal polynomials modulo 1), and that 0 c < p. Now suppose that there exists a second decomposition f (a,) of the same kind (5) (7) (7)
i'u (x) (mod p),
AX) = r1 I'1 (X)'.2
where rr(x) are primary and normal prime functions modulo p and 0 < c1 < p. Then clearly we have c1 = c and (8)
fl qr (x) i=1
(x) (mod p)
11
i=I
Using Theorem 55 we see then, that the prime function q, (r) must be a divisor modulo p of at least one of the prime functions r; (x), say of r1(.r). Since q, (x) and r1are both primary and normal, it is evident that q, (.v) = r1 (x.). In the identical congruence (8) we can divide both the members by the polynomial q1 (x) = P1. (x), since it is not identically congruent to zero modulo p
Thus we get the congruence p
11 q: (x) 
i=2
12
ri (x) (mod p).
By the same argument we prove that q2 (,r) is equal to some ri (x), say to r2 (x). Evidently this process may be continued, and we find finally that qi (x) =;j (x) for all i and that p = a.
Thus the proof of Theorem 56 is complete.
Example. Let us take p = 7 and f(x) =x8 + .[? + 2x"  2,x5 + 4x1x2
3x+ 3.
By trial we find the following result f (X) _ (x + 1) (x2 + 1)2 (x3 + 3) (mod 7),
where x + 1, x2 + 1 and xs + 3 are primary and normal prime functions modulo 7.
Remark. It is not possible to develop an analogous theory in the case when the modulus is a composite number. Theorems 55
and 56 are not valid in general.
THEORY OF CONGRUENCES
99
30. Wilson's theorem and its generalization.  When p is a prime it follows from Theorem 3:1 that the congruence xzi1  I = 0 (mod ii)
has the roots x = 1, 2, 3, .... p  1. By Theorem 41 we have then identically
01  1 = (x  1) (x  2)  (.r  p + 1) (mod j)). 
Put .r. =1) in this relation, and it follows that
(p  1)! _  1
(1)
(mod p).
This result is called Wilson's Theorem. after the discoverer. The first proof of it was, however, given by Lagran;e in 1770. The theorem may be extended to yield a criterion for primes: Theorem 57. A necessary and sufficient condition that an integer zz (> 1) is a prime is that (n  1)! + 1 be divisible bq n.
Proof: It remains only to show that the number (n  1)! + 1 is not divisible by n, when n is a composite number. If n is composite and q is a prime divisor of n, then q < zz, and therefore (n  1) ! is divisible by q. Hence the number ()z  1) ! + 1 is not divisible by n. It is, however, obvious that the test furnished by Wilson's Theorem is useless for large numbers n, since ()2  1) ! increases too quickly with 12.
Now suppose that the prime p is odd and put } (p  1)
q.
On the lefthand side of (1) replace every factor h + q (for h = 1,
2, .... q) by the congruent h  q  1, and multiply both sides by
1),z; then we obtain (q!)" =  ( 1)Q (mod p).
(2)
If the prime 1) is of the form 41z + 1, the righthand side of (2)
is  1. Hence we obtain the first part of Theorem
S.
If' p is a prime = I (mod 4), the congruence .)
CHAPTER III
100
has the ttro solutions
x+ The
i'uen
1)
(1)
!
3
(mod ji).
(3) ha., no .solution
when )p is a lmiulu
3
(mod 4).
We prove the second part of the theorem indirectly. If (3) were solvable for p = 4 n + 3, we would find by raising both sides in (3) to the qth power 1 (mod p).
xN1 = ( 1)9
But this is impossible. since by Ferlnat's Theorem (Theorem 35) :Cti1 = 1
(11104 11)).
If the prime p is of the form 4 n + 3, the righthand side of (3) is + 1. Hence one of the numbers q'  1 or q! is divisible by 1)
Hence eye have (7) )
(4)
1)!
±I
(niod1i).
For p == 3 and 23 the righthand side is ; 1; for j. = 7, 11 and
19 the righthand side is 1. In Chapter IV we shall give a rule for determining the righthand side of (4) for any prime.
Wilson's Theorem is a special case of the following more general result due to Gauss. Theorem u3.
Let a be a natural number > 2, and let N denote
the number of incongruent solutions of the congruence .42 = 1 (mod n).
(5)
Then, i f al, a2i
... ,
sy..teur modulu n,
f''` r; r
a,, are representatives of a reduced residue hare ao
%'33%5
(laodl n).
THEORY OF CONCRUENCES
101
Proof p means p (u). To every a prime to n there corresponds a unique a', also prime to it, such that aa' = I (mod n).
(i;)
Hence the numbers al. (12, .... aq can be divided into pairs a, a' whose product is congruent to I mnodulo 1). We have a = a' (mod a) only when a is a root of the congruence (5). Denote by Q the product of all the N incongruent roots of (5). If a is a root of (5), so is  a; since n > 2, the roots a and  a are incongruent modulo n. We have
a(a)=a21 (mod n).
(7)
Hence 1)',v (mod ),).
(8)
Now let Ql denote the product of all incongruent numbers a; (i = 1. ?, .... 9;) niodulo n which are not roots of (b), if there are and numbers of this kind, otherwise put Q1 = 1. By the congruence (6) it follows that loll = 1 (mod u),
and by (8)
P= QQ1.=(
1)3'
(mod a).
Q. E. D.
Applying the results of Theorem 47 to the number 1\, we find
that the product P is congruent to  1 modulo i, in the following cases: When n = 4; when a is a power of an odd prime; when it is twice the power of an odd prime. In all other cases P is congruent to + 1 modulo n.
Example. For u = 00 we have cp (n) = 16 and N = 8. The congruence
x2 = 1 (mod 60)
has the roots ± 1, ± 11. ± 19. t 29. How the residue classes modulo n which are prime to n may be divided into pairs is apparent from the following congruences 7 ( 17)
1 (mod 60)
and 1
(1)=11
(mod 60).
CHAPTER III
102
31. Exponent of an integer modulo n.  Let n be a natural number > 1 and a an integer prime to n. In the infinite sequence
a. a2, a3, a4...
(1)
there are numbers = 1 (mod n), since by Theorem 36 (2)
a`r. ('
1 (mod 72).
Suppose that ad is the first number in the sequence (1) which is 1 (mod n). Then a is said to belong to the exponent 6 modulo n. 6 is the order of a inodulo n. Congruent numbers modulo a have the same order modulo n.
Theorem. (O. Let a be a natural number % 1 and a an integer prime to n. 1. If a belongs to the exponent 6 modulo n, then the numbers (1, a2
ad
are incongruent modulo n. 2.
Further, if am = 1 (mod n),
theta m is divisible by 8. In particular, 6 is a divisor of 4P (n).
Proof. Suppose 6
k>h
1.
If we had
aR = ah (mod n).
then axh_ 1 (mod 12).
But, since 0 < k  h 6, this is contrary to the definition of 6. To establish the second part of the theorem put ))z = 8q + r, where q and r are integers, 0 < r < 6. Then we have a"
ad q+. = am = 1 (mod n),
and thus, recalling the definition of 6, r = 0. Hence 6 is a divisor of m. By (2) it follows that op (n) is divisible by 6.
Example. The following table for n = 55 gives the order 6 modulo 55 of all the positive integers prime to 55 and < 55.
THEORY OF CONGRUENCES
a
Numbers of order 6 modulo 55
I
1
1;
2
221, 34, 54;
4
12, 23, 32. 43;
5
16, 26, 31. :3'i: no
8 211
4, 6. 9, U. 19, 24, 20, 341, 41, 411, 49, 51: 2, 3, 7, 8, 13, 17, 18, 27, 2'5, 37, 3'i, 42, 47,
40
,
lu
103
48,
.5 2,
53;
i,u,nbp, s.
j
We next prove
Theorem 61.
Let n be a natural number > I and a an integer
prune to n. If a belonga to the exponent 6 modulo n and if nt is a natural number such that (m, b) =;u, then am belong.'r to the exponent
6
P
nzodido n.
Proof Suppose that the number am belongs to the exponent n modulo n. Then r is the least positive exponent such that (am)' = 1 (mod n).
(3)
Applying the second part of Theorem 60 we conclude from this congruence that my is divisible by 6. Thus, since (in, 6) = ct, we must have
where h is a positive integer. On the other hand we have m
d
(am)a = (ad)I' = I (mod n).
since a belongs to 6. Hence we conclude that h
I and v = 8 Q. E. D.
Exanmple. As in the above example we take n = 55. From the table we see that the number 2 is of order 20. Then, by Theorem 61, the numbers 24s 26, 212, 216
are all of order 5. This is verified by the table since they are
m 16, 36, 26, 31 (mod 55).
CHAPTER III
104
Let 6 be a positive divisor of
(n). If a is of order 6 modulo n, root of the
we say that a is a
x6 = 1 (mod n).
(4)
Theorem 62. Let ra be a nataral number > 1, let b be a positire divisor of (p (n) and let it he a root of ihr congruence (4). Then we hart,
1. A necessary and sufficient condition for a rout it of (4) to be a piinritire root of (4) i.5 that the nzrrnbe,s a, az,
(5)
(1rr
be incongruent modulo n. 2.
Let a be a primitive root uf' (4). A necessary and sufjricient condition for rt"` to he a rout ol'(4) is that (nr, a)  1.
.i. I/ the currgre(encr
xn'(") = 1 (mod n)
has a primitive root, it has (p (p )?,) primitive roots incongruent modulo U. P r o o f.
1 f a is a root of (4). all the numbers (5) are roots of (4).
The truth of the first part of the theorem is then ail immediate consequence of the definition of primitive root and of Theorem 30 (first part).
To prove the second and third parts we only have to apply Theorem 61 with tr = I and observe that the numbers rr, a`
. .
. aT ('
form a reduced residue system modulo n when a is of order 9;(n).
We next consider the special case when the modulus is a power of 2. Theorem 63. 1. Every odd integer x .crtisfies the congruence x2,32=
i)
1 (mod 2i')
when 13 ? 3. 2.
The number 5 is it primitive root of (6) when f ? 3.
105
THEORY OF CONGRUENCE' .9.
The nun?Lrr8 + 52:3
2
,form a reduced residue syste»z 'uodulo 2,3 irheu
Proof. The first part of the theorem is true for
3.
= 3, since
12 = 3= = a2 = 72 = 1 (mod 8).
If (6) holds, then '32
=1+2;t.
where t is an integer. By squaring we get x2' 31 +2.j1t+22,312 and 3
2'
1
= I (mod
23+').
We thus conclude by induction that the congruence (6) is true for all fi ? 3. To prove the second part of the theorem we suppose that the number 5 is of order 6 modulo 2.3. According to (6) and to Theorem 60 the exponent 6 is a divisor of 2+32. If 6 < 2.12, then 6 would be a divisor of 2.3s and b`
= I (mod M.
We can, however, show that for 213 _=
(7)
=3
1 + 23' +
2,3 T,
where 7' is an integer. This is true for P = 3. If (7) is true for a given value of the exponent #, it is also true when fl is replaced by fi + 1. For. by squaring both sides of (7), we have :)
24
=1+
2,1 + 2s+1 (T +
Va + 2,11 7' + 2;31 T$).
Thus we conclude by induction that the relation (7) is valid for : 3. Hence we cannot have 6 < 2,31, and it is clear every that 6 = V2.
CHAPTER III
106
The truth of the third part of the theorem follows from the second part and from the fact that the congruence  5k (mod 2i')
5/°
is not satisfied for any fi
2.
Now we introduce a new arithmetical function ,p(n) defined in the following way: yi (n) = 'P 07)
for n = 1, 2, 4 and n =pa, when p is an odd prime; 2.
zp (n)
for n = 211, when 3.
; 'P (n)
3.
i, (n)=.
iV'(Ni')
for any n having at least two different prime factors. Here where PI, P2, etc., are the different prime factors of n. {a, b, ...) denotes as in Section 5 the least common multiple of a, h, etc. From this definition follows
Theorem 04. If n is an integer > I and if a is prince to n. then a'°i") = 1 (mod )?).
For, by Theorem 311 this congruence is satisfied for to = 1, 2, 4 and n = tpa, where 17 is an odd prime, and by (6) also for n 213. Hence, using the definition, we see that the congruence is satisfied for any integer n. Theorem 64 has the corollary : Except for the cases n = 1, 2, 4, pa and ?pa, ithere p is an odd prime, we hare ay `f"') = I (mod ii).
In fact, the number y (n) is a divisor of a T(n), apart from the exceptions mentioned.
THEORY OF CONGRUETINCES
107
32. Moduli having primitive roots.  If n is a natural number > 1, and if et belongs to the exponent T (n) modulo n, a is said to be a primitive root niodido n or of the number n. We will now determine all moduli which have primitive roots. The number 1 is a primitive root modulo 2. The number 3 is a primitive root modulo 4. From the corollary to Theorem 64 it follows that every integer n which has a primitive root and is different from '? and 4 is either the power of an odd prime or twice such a power. We shall prove Theorem U.. 1. The natural number n > 1 has primitive roots if
n has one of the rabies
n= 2, 4. p° and 2p, inhere p is an odd prime, and in no other ca.,(,. 2. The number of incongrucnt primitive roots modulo n i+ then 9 9(99 n').
.i.
If d is a priniitire root of the odd prime p, and if the number 1 is not divisible by pp2, then g is a primitive root of Jf, for anp/ positive exponent a.
Proof The theorem is true for n = 2 and n = 4. We now have to distinguish three cases. First case: n = the odd prime p. Let 6 be a positive divisor of 1p  1, and denote by Z (6) the number of incongruent integers modulo p which belong to the exponent 6 modulo p. Then clearly (1)
1Y.(6)= p 1, d
where the summation extends over all positive divisors 6 of p  1. We have x1'`1  1 = (xd  1) h (x), where h (x) is an integral polynomial. By Theorem 35 the congruence xi'I
 1 = 0 (mod p)
108
CHAPTER III
iiicongrueut roots 1. 2, ... , p  I inodulo p. Apwe plying Theorem 43 with f'(x) = x) i  I and q (.r) see that the congruence
has the p
1
,x.1 1 = 0 (niod p)
(2)
has exactly b incongruent roots inodulo 1). If this congruence has a primitive root a, the numbers a, n2, a3.... , a') are the complete set of incongruent roots modulo p of (2) (Theorem (i2. first part). Among these roots exactly q.,(b) are primitive roots of (2) (Theorem 62, second part). Therefore it follows that congruence (2) has either q; (b) incongruent primitive roots Inodulo 1) or none at all. Hence we have either x (b) = 0 or x (b) = q (b).
It follows from Theorem 13 that (3)
where the summation extends over all positive divisors b of p  1. Comparing (3) and (1), we see that we never can have y (b) = 0; thus x (b) = (p (6) for any 6. Hence it is proved that every odd
prime has primitive roots. Second raxr: )) = p', 1) odd prime, a? 2.
Let q be a primitive root of p. If the number (/P1  I is divisible by p2. there exists another primitive root yI = h' + 1) of p such that !lii  (!I + p))' _ !!i'1 + (1)  1) q)'21) = 1  p qP2 (niod p2),
 1 is not divisible by 2. Thus we can choose the primitive root (1 modulo 1) such that qPi 1 is not divisible by 1)2. This condition satisfied, g is also a primiIt is clear that the number
yii
tive root of pa for any a. To show it we begin with the proof of the following lemma: The number is divisible by
pai anal not by pa.
By hypothesis this is true for a= 2. Suppose that (4)
Pa'2,P1)  1 4 r
pKi,
109
THEORY OF CONGRUENCES
where the integer c is not divisible by p. We raise both sides in (4) to the pth power and expand the righthand side by the
binomial theorem to obtain
fPa1(111:(1Tcjt'')J'=1
rj)a
is an integer. Since 2u it follows that 1 where
J,
p(p1)p'("1)rbjr"11.
(.2.1,

I
all,.,
3'1  3 are
a+
where the integer cl is not divisible by p. Hence the lemma is proved by induction. Suppose next that q belongs to the exponent 6 modulo p It follows from Theorem (S(( that 6 is a divisor of
Since q is a primitive root of p, the number p  is a divisor of S by Theorem 00. Thus 6 = p (p  1), where 0 < P C a  1 If 6 11"1 (p  1). then pu2 (1)  1) would be divisible by 6. and I
.
we would have 1 (mod 1)a),
which is contrary to (4). Hence 6 =p 1 (p  1). and (i is a priori. tive root of p". Third case: (i = 211" p odd prime. Among the primitive roots of p," there are also odd numbers. For, if p is even, then q + pa is odd. Every odd primitive root y of p" is a primitive root of 21,u. For, if y belongs to the exponent r) modulo 2p", then 6 is a divisor of T, (2 pc) = , (pa); further, since q belongs to the exponent (p (pa) modulo p", 6 ? p(i)a)
Hence 6 = op (pa) = (p (2p"). Thus the proof of the first and the
third part of the theorem is complete. The second part is a direct consequence of Theorem 62 (third part).
The primitive roots of a given modulus may be determined by trial. At the end of the book we give a table of the least primitive root of the first 150 primes. Examples. The prime 7 has the T. (G) = 2 incongruent primitive roots 3 and :1.
Since the number 36  1 = 7  13 is not divisible by 72, the number 3 is a primitive root of any power of 7.
CIIAPTER III
110
The prime 13 has the rp (12)  4 incongruent primitive roots 2, 6, 7 and 11.
If the rratual number rr (." 1) has and if d is a positive divisor of q;()?), there the
Theorem Ills.
r'oot.+,
x" = 1 (mod n)
(5)
has exactly d
roots 'modulo it.
Proof: By Theorem 62 this is true when d = be a primitive root of n. Then the number
Now let ry
r' (n)
d
belongs to the exponent d modulo n, and the numbers t2.
(6)
are incongruent modulo n (Theorems li0 and 61); hence, these numbers are roots of (5). Now let 6 be an arbitrary root of (5). Then it follows from Theorem 60 that 72 = gr` (mod )?),
where h is a multiple of
Therefore we have
21  ' (mod n),
where k is an integer ? 0, i. e. the number iy is congruent to one of the numbers (6) modulo n. Hence, there are no other solutions of (5) than those given by (0)). This proves Theorem 66.
A supplement to Theorem 35 is Theorem 67. Let n be a natural uctmber ;> 1. 1. There alrrays rxist integers which belong to the exponent ?p(n) modrtlo n.
2. Every integer prince to n belongs modulo n to an exponent which is a divisor of zV (n). .i. At least T(6) integers incongruent modulo n belong to a gir(ii poNitire divisor 6 of p(n) nrorlido n.
111
THEORY OF CONCRUE\CES
Proof. Suppose that n is divisible by p'i and bv no higher power of the prince p,. Lot p; denote a primitive root of p i, when p1 is odd; for p; _ 2 and ai 2, 9t denotes the number 5, and for pi = 2 and ai °= 2, !l, denotes the number 3. Now, aloply ina Theorem 40, we determine the common solution of the simultaneous congruences x 91 (mod pi'), x 9a (mod p 2 , .. , x = g, (mod p r).
where pl. p2..... )r are all the different prime divisors of n. Let the common solution be
x = (mod n). If belongs to the exponent 6 modulo n, then 6 is a divisor of yi(n). (Theorems 64 and 60.) On the other hand, $ belongs to the exponent T(1)1i) = y (prci) modulo ),'zt, when p, is odd, and when ),t = 2 and a; = 1 or 2; if p2 = 2 and a, > 3, then, by Theorem 63, 5 belongs to the exponent I q' (2'i) == y , ( 2 '  r ) modulo 2 "i.
Hence 6 is a common multiple of all the numbers y} (pAi) for i = 1, 2.... , t (Theorem 60). According, to the definition of y, (n) we have then 6 = y' (n). Thus the first part of the theorem is proved.
The second part of the theorem is a direct consequence of Theorems 64 and 110. Suppose that belongs to the exponent y, (u) modulo n. Let 6
be a positive divisor of y' (,t) and put
y b7a)
= q. By Theorem 6 1
it is then clear that the number ri  
belongs to the exponent 6 modulo n: and any number iol", where It is prime to 6. belongs to the same exponent 6 modulo n. This completes the proof of Theorem 67. In the example given in Section 31 the modulus is it  55, thus (,t) = 20. From the table we see that sixteen numbers
belong to the exponent 20 modulo 55, twelve numbers to the exponent 10 modulo 55, etc. Since the numbers 8 and 40 are not divisors of y)(55)=20, no numbers belong to the exponents 8 and 40. 33. The index calculus.  Let a be a natural number having primitive roots. If y is a primitive root of n, the numbers
CHAPTER III
112
(1)
1. //, /f2,
..
,
fT(x
I
form a reduced residue system modulo it (Theorem 60). In the set (1) there are T,(qr (n)) primitive roots, and these are the numbers q`, where c is prime to T (n). If ca is an arbitrary integer (n)  1 prime to it, there exists among the numbers 0, 1. 2.... exactly one number it such that a = rj!' (mod n).
The number It is called the index of the nnnlbrr a acith respect to the base g rnodaalo j?. and we write I
or. shorter,
= ind, it
l  ind a,
when no misunderstanding is possible. Example. The number 7 is the least positive primitive root of n = 41. Since 1:5 = i 3 (mod 41), the number 1.5 has the index 3 with respect to the base 7 miodulo 41.
We readily verify the followincr rules for the index calculus. 1.
Lt. 111.
IV. V.
ind (a b' ind a i ind b (mod (ra)). ind ((a'a) _ q ind a (mod 97 ()? ), when q is it natural number.
ind I = 0, independently of the choice of the primitive root. ind q= 1. when II is the primitive root chosen for base. Ind ( 1)  r (n), if it > 2.
The correctness of the last rule follows immediately from the congruence
gm("'  1 = (q `r(")  1) (q 'a (n` + 1) _ 0 (mod )1).
For, since y is a primitive root, we must have q1 `a i"i =  1 (miod )?).
The first four rules valid for the index calculus show an obvious analogy to the rules valid for logarithms. Many types of congruence problems may be solved more easily
by means of the index calculus. The condition for this is, of course. that index tables have been computed for all possible
THEORY OF CONGRUENCES
113
moduli up to a certain limit. Gauss at the end of his Disgttisitiones gives tables of indices for moduli up to 100. The Canon arithmeticlis of Jacobi contains tables of indices for all prime power moduli < 1000.
In the following example of an index table, the modulus is n = 19 with the primitive root 2. 'Number.
'
1
Index ......I 0 I
+ 3
4
° I 13 1
7
16
141 6
11 11? 11
I111
3 18
17
1?113
1
3
5
14
16
7
II"PI 111
4
17
lU
18 9
Since we have
ind (n  a) = ind ( a) = s' (n) + ind it (mod en)),
the latter half of the table may be omitted. If the number n has primitive roots, the linear congruence a x = h (mod rr),
where (a, n) _ (b, n) = 1, can be solved by use of index theory. In fact, this congruence is equivalent to ind a + ind x m ind b (mod
;n'
and therefore x is uniquely determined by the congruence ind x = ind b  ind a (mod
ri ).
I: sample. Let us consider the congruence
II H 9 x = 7 (mod 13).
The prime 13 has the primitive root 2, and we obtain the following index table. Xumbr'r
1
Index
0
2314
5
6
7
89
1
11
12
1141'3
9
J
11
31 8
10
I
6
H
Then we get
ind x = ind 7  ind 9 = 11 8=3 (mod 12) and
x = 8 (mod 13). 8  516670 Tr ,gre \ agell
114
CHAPTER III
The general binomial congruence a x'n = b (mod n)
may be treated in the same manner, as will be shown in the next section. By using index theory it is also possible to solve the exponential congruence
ax = b (mod n),
where (a, n) = (b, n) = 1. In fact, if n has a primitive root, this congruence implies x ind a = ind b (mod op ;)2 ).
Thus it is evident that the number (970?, ind a) must be a divisor of ind b. Hence, in this case, there are just (op(n., ind a) incongruent solutions modulo p oz). Example. Find the solutions of 7x = 5 (mod 17).
The prime 17 has the primitive root 3, and we obtain the following index table. Numbei Index
3
1 . ...
0
14
4
F+ I
12
5
7
.5
I1
10
HH
3
11
12
13
14
7
13
4
9
16
c,
II 8
Then we get 11 x = 5 (mod 16), and
x = 15 (mod 16).
Finally we shall show how it is possible, by use of index theory,
to determine the order f modulo n of a given integer a. The number f is, by definition, the least positive exponent that satisfies the congruence at = 1 (mod n).
If the modulus n has primitive roots, we have (modp (n)).
THEORY OF CONCRUENCES
115
Hence, putting If _ (g:
it. ind a),
we clearly have
34. Power residues. Binomial congruences.  Let it be an in0, and let a be an integer prime to n. If q is a natural number j 2 such that the congruence teger
x'c = a (mod n) is solvable, we say that the number a is a qth porter residcumodcdo n. In particular: the number a is a quadratic, cubic or
biquadratic residue modulo n according as q = 2, 3 or 4. Let p br an odd prime, let a be an integer not dirisible bq 1), and let n = p" and S = (q, pin)). Thrn the con
Theorems 66(,?.
gruence
x'r = u (mod n)
(1)
ha.v exactly S incongruent Nulutions modcclo it. if furl a is ditni 6. Otheruirr it has no solution. Proof: If we choose a primitive root modulo it. it follows from
(1) that q
ind x = hid a (niod ip (n)).
This is a linear congruence in the unknown ind x. Hence, applying Theorem 39, Theorem 68 follows. Exampl es. 1.
Let us consider the congruence xs = 3 (mod 13).
Here 6 = (8, cp .13))  4. IVe may take y = 2. Then ind 3 = 4. 8
ind x = 4 (mod 12). thus ind r = 2 (mod 3), and ind x = 2. 5, 8, 11 (mod 12),
and finally
x = 4, 6, i, 9 (mod 13).
CHAPTER III
116
2. Let us consider the congruence
x12
= 13 (mod 17).
Here 6 = (12, pC1 7)) = 4. We may take g = 3. Then ind 13 = 4, 12 ind:r,  4 (mod 16), and ind x = 3, 7, 11, 15 (mod 16), and finally
x = 6, 7, 10, 11 (mod 17).
3. Let us consider the congruence
a" = 4 (mod 29).
Here 6  (7,
29.) = 7. We may take g  2. Then ind 4 = 2.
But the congruence 7
ind x = 2 (mod 28)
has no solution. Hence the number 4 is not a 7th power residue modulo 29. 4.
Let us consider the congruence
x$ = a (mod p),
where p is a prime ? 5, and where a is not divisible by 1). If p = 6 nt  1, then 6 = (3, 6 mn  2) = 1. In this case the congruence has exactly one solution. If p = 6 m + 1, then 6 = (3, 6 m) = 3. In this case there are either no or three incongruent solutions. An example of the first category is the congruence
r,g = 2 (mod 7),
which has no solution. An example of the second category is the congruence X'= 6 (mod 7),
which has the solutions x m 3, 5, 6 (mod 7). According to Theorem 68 the congruence (1) is solvable if and only if a =.V'" (mod n),
THEORY OF CONGRUENCES
117
where h is an integer ? 0. Hence q'(") d
a
= 1 (mod .n).
Conversely, if this congruence is satisfied, and if a = #7 (mod n), 0, we have
where y is an integer
7(n)
d = 1 (mod n).
g7
Since g is a primitive root, the exponent y
8") is a multiple of
(n), and therefore b is a divisor of y. Hence we have proved Theorem 69. Let p be an odd prime, let a be an integer not divisible by p, and let n =1)" and b = (q, (p (W). The necessary and condition for the congruence (1) to be solvable is that the congruence m{ni
a
d
= 1 (mod n)
hold.
A supplement to this result is Theorem 70. Let p be an odd prince, and let a be an integer not divisible by p. Further, suppose that q is a natural number 2 not divisible by p. If the congruence x4 = a (mod pa)
is solvable for a 1, it is also solvable for all ('integral) exponents a> 1. Proof. If we put n  pa and 8 = (q, op (n,), then 8 = (q, p  I). If the congruence x7=a (mod P")
is solvable, we have by Theorem 69 r(a)
a 4 =1 (mod pa).
CHAPTER III
118
Hence q'(n1
rr'p
 =I Fp't.
where t is an integer. If we raise both sides of this equation
to the pth power, it follows that PO))
'Y Im(ne) J =(1 +P' i)n
n
=1
J
(I')p' t +
where ti is an integer since 2a a
m(pn) a
(1)
p=': 0 
= 1 + p,:TI t1'
a + 1. Therefore we have
I (mod
Hence, from Theorem 69 it follows that the congruence xQ  n (mod p"±I)
is solvable, and Theorem 7() is proved by induction.
Further, we can prove Theorem 71. If p is an odd prinir, and if n = p" and 6 = (q, 9P (M), there are
9)(n)
qth power residues incongruent mnodulo it.
Proof. By Theorem 69 the number required is equal to the number of incongruent solutions of the congruence T (")
,ca=1 (mod n). By Theorem 66 this congruence has exactly a°7) incongruent solu
tions. Hence the theorem.
Example. If n  17. there are four biquadratic residues in the interval 0  it, namely 1, 4, 13 and 16. We next consider the congruence (2)
xs = n (mod _'").
where a is odd, and prove
THEORY OF CONGRUENCES
119
Theorem 72. 1. If q and a are odd numbers. the congruence (2) has exactly one solution. 2.
Let a be an odd number and q = 2 m, where in is odd. Let the exponent a be ? 3. Then the congruence (2) has four incongruent solutions if a = 1 (mod 8); otherwise it has no solution.
.3.
Let a be an odd inunber and q = 2 in, where ni is odd. Then the congruence x9 = a (mod 4) has two incongruent solutions if a = I (mod 4); otherwise it has no solution.
Proof. If a ? 3, we have by Theorem 63 (3)
a = ( 1)h h" (mod 2"),
(4)
x  ( 1)" .5y (mod 2"),
where h, k, u and y are integers ? 0. Now suppose that q is odd. By introducing (3) and (4) in (2) we get
(1),i.59y=( I)h
5k
(mod 2a).
Hence tC = h (mod 2) and by Theorem 63 q y = k (mod 2a2).
This linear congruence has exactly one solution y. Therefore, the congruence (2) has exactly one solution x. In the proof we have supposed a ? 3. but the result is clearly valid also for
a=1 and a=2.
Suppose next that q = 2 in, m odd and a ? 3. By introducing (3) and (4) in (2) we get 52my _ (1)h . 51 (mod ?a).
Hence the number h is even, and thus a = I (mod 4). Therefore 2 m y = k (mod
2a2)
This implies k = 0 (mod 2) and a = I (mod 8). When this condition is fulfilled, there are two incongruent solutions y modulo 2a2, and consequently four incongruent solutions x modulo 2a.
Finally, it is evident that the congruence x2m=a (mod 4)
CHAPTER III
120
is solvable if and only if a = 1 (mod 4). When a = 1 (mod 4) it has the two solutions x = ± 1 (mod 4). Hence the proof of Theorem 72 is complete.
The theory developed in this section may also be used for solving the general binomial congruence a xm = b (mod n).
According to the results in Section 26, the problem can be reduced to the case where the modulus is a primepower. Example. We consider the congruence 11 xs = 17 (mod 56).
(5)
The number 3 is a primitive root of the prime 7. From the congruence
11 xs = 17 (mod 7) we conclude
ind x = 1 (mod 6)
4 I 3
and
ind x = 1, 3, 5 (mod 6). Hence
x = 3, 5, 6 (mod 7).
(6)
From the congruence 11.r
17 (mod 8)
we conclude 3 x m 1 (mod 8) and (7)
x= 3 (mod 8).
Combining (6) and (7) we finally get the following solutions of (5):
x = 3, 19, 27 (mod 56). 35. Polynomials representing integers.  An integral polynomial
f (x) represents integers for all integral values of There exist, however, other polynomials with the same property. An example is the polynomial of degree n
THEORY OF CO\GRtENCES
121
r(xl) (xn1) n!
12
which, by Corollary to Theorem 25, takes integral values for all integral values of x. When a polynomial represents integers for all integral values of the variables, we shall call it, for the sake of brevity, an i. r. polynomial (i. r. = integer representing). For such polynomials in one variable we prove Theorem 73. Erery i. r. polynomial J '(x) of degree n in the variable
x may be written in the form
f(x)=A0+Al(1) + A2 (2X) +
(1)
where the coPfffcPents :10.A1i .
.
+ A, (x),
., A. are
integers.
Proof Every polynomial f (x) of degree ii may be written in the form f(,r) = co + c1 (X1) +
(2)
r2 (2"')
+ ... + r" C.) , x
where the numbers co, r1, ..., c,, are uniquely determined. This assumption is true for polynomials of degree zero. Suppose that it is true for all polynomials of degree < n  1. Then it is also true for the polynomial f (x) of degree n. For, `if the coefficient of x" is ao, the polynomial g (x) _, f (.c)  ao n ! (x) is at most
of degree n  1. Hence the assumption is true for g (x), and by induction for all f (X).
Now suppose that f (x) is an i. r. polynomial expressed in the form (2). Since ,f (0)  co. the coefficient ro is an integer. Suppose
that the coefficients ro, c1, ..., cri are all integers. Then the coefficient
Cr
is also an integer. For by putting x = r in (2),
we have .f fr) = Co + r1 ( 1)
P2
( y) +
+ rr_ 1 1'
l + rr .
Since f (r) is an integer, we see that Cr is also an integer.
CHAPTER III
122
Hence, by induction, Theorem 73 is proved.
In particular it follows: If f (x) is an i. r. polynomial of degree n, the polynomial n! f(x) is an integral polynomial. If an i. r. polynomial for all integral values of the variables represents integers, which are all divisible by the same integer d, we say that the polynomial has the same fixed divisor d. For such polynomials in one variable we prove Theorem 74. Erery i. r. polynomial ,f (x) of degree n in the variable x, which ha, the fixed divisor d, may be writtenn in the ,form
f(x)= 40+A1(i)+
(3)
where the coefficients 10, A1,
..., A are integers divisible bi/ d.
P r o o f. The integer rlo is divisible by d, since f(0) _ AO Suppose that the coefficients AO, A , ,..., A,I are all divisible by d. Then the coefficient Jr is also divisible by d. For, by putting x = r in (3), we have
A0 + A1I 1J rt
_12
() + J
F
IrI
r r1) + ,.
Hence, Theorem 74 is proved by induction.
In particular it follows: If a primitive integral polynomial g(x) of degree n has the fixed divisor d, then d is a divisor of n!. For i. r. polynomials in several variables there are results analogous to Theorems 73 and 74. 36. Thue's remainder theorem and its generalization by Scholz. 
The following result due to Axel Thue is very useful for many questions in number theory. Theorem 7:1. Let n he a natural number > 1, and let c denote the
least integer > l'n. Then for any integer a prime to n, there exist two natural numbers x and y not exceeding e  1 such that (1)
a i/ = ± x (mod )i).
Proof. We consider all numbers of the form a y + x, where x and y are numbers in the set 0, 1 , 2, ... , e  1. Since there
THEORY OF CONGRUENCES
123
are in all e2 > )r such numbers, at least two of them must have the same principal remainder modulo n. (Dirichlet's box principle, see Section 12.) If we suppose a!/I + J'1 = e+!/2
`
.''2 (mod n).
we can write ''2  .r1 (mo(l )r).
a (!/1  !/2) 
(2)
Here
0
0
.'2I
 r2I<e1.
For, if one of the differences "2 and J1  //2 were = 0, the other one would also be = U. By putting f1  !/2 = / and .r2  x1 .r in (2), we obtain a congruence of the type (1), and the theorem is proved. However, it is easy to see that by a slight alteration of this proof we can obtain the following more general result (Arnold Scholz) :
Theorem i'/;. Let n he a natural number denote two natural i2unzbers such that
of> n, e > 1,
a
> 1, and let e and f
f
Then for anti integer a primp to n, there exist two natural numbers x and p/ such that
aid
ail
± x (mod n)
0
If n is a prime, we can clearly suppose (x, y) = 1 in these theorems. Then it is convenient to write congruence (1) in the form (3)
a_t
(mod n). 1
1:'xaur1)le 1. Putting n = 7, in Theorem 75, we have e = 3 and obtain the following six representations of a in the form (3):
1=1. 2=2, 3='_ 4
5=;,6 =L
In all these congruences the modulus is 7.
Cn APTER III
124
Example 2. Putting n = 11 in Theorem 75, we have c = 4 and obtain the following fourteen representations of a in the form (3):
1=1, 2, 3=1 8 ,'
6 12, 71= = _ 1 = ::
4J
5
1=82 +91i 1 = __1
_R
P
1
In all these congruences the modulus is 11. In Example 1 there is only one representation of every a; in Example 2 there are two representations of the numbers 3, 4. 7 and 8. (Compare Exercise 89.) We shall use Theorem 75 for proving theorems on the representation of natural numbers as sums of squares (see Chapter VI). Exercises
41. Find the highest power of 12 contained in 120!. 42. With how many zeros does the number 2000! end? 43. Let p be a prime. Show that there exist exponents in with the following property: There is no natural number n such that pm is the highest power of l) contained in n!. Develop a method for determining the numbers na. 44. Find the greatest common divisor of the binomial coefficients
11, ( , ... , (n
n l
.
45. Show that the fraction
(m+n1)! M! 0 is an integer, when the natural numbers ?it and n are relatively prime.
46. Let m and n be natural numbers. Show that the number (m n) !
is divisible by the number n! (m!)".
47. Show that the fraction (2 m) ! (2)a)!
m! n! (m+v)! is an integer when in and n are natural numbers. (Catalan.)
125
THEORY OF CONGRUENCES
48. Let n be a natural number and put In = 22' + 1.
Show that the numbers I;, and F,,, are relatively prime when n 74 m. 49. Show that the congruence
22x + 1 = 0 (mod p)
has solutions x for an infinite number of primes p. 50. Find all natural numbers x such that the product x(x + 1)(x + 2) has no other prime divisors than 2, 3 and 5. 51. Let in and n be two natural numbers and n odd. Show that the sum 1+
m m+n 1
+
1
m+22n
+...+
1
nt+xn
is an integer only for .c = 0 and to = 1. This result also holds for an even n; the proof is, however, more difficult in this case, but may be accomplished by applying Bertrand's theorem: There is at least one prime between x and 2 x  2 when x z 2. 52. Find all positive rational solutions x and y of the equation XI/ = yX
and in particular the integral solutions. 53. Let f (x) be an integral polynomial in x, and let n and in be natural numbers. Show that there exists an integer x0 such that each of the numbers f(xo), .f(xo + 1), .....f(xo n  1) has at least m different prime divisors. 54. Let v be = 0 or = 1 according as )i is divisible by the square of a prime or not. Show that the infinite decimal fraction
6=0.vlv2VS 3'4 ..
is an irrational number.
CHAPTER III
126
55. Let J '(x) be an integral polynomial in x which takes positive values for all x 1. Show that the infinite decimal fraction n = U..f (1) J'(2) .f (3) .f (4) 
. .
is an irrational number. If, for example, f (x) =.c 2, the decimal fraction is 71 = 0.119162536446,181100 .. . 56. Prove Hermite's formula
i2 1fij I~i[x] =2 =
[;] [l
Q
As before [z] denotes the greatest integer < z. 57. Let r(k) denote the number of positive divisors of the natural number k. By means of the formula in Exercise 56 prove the following relation: [.1
.'Y z(k)=xlogx+(2y1)x+b
k=1
where y is Euler's constant = lim (1 +
+ +
= 0.6772 ... , and where b is a function of is less than a positive constant for all x.

log n
such that 161
58. Show that the number of lattice points inside and on a circle having centre at the origin and the radius 1'X_ is
where 6 is a function of .,c such that 161 is less than a positive constant for all x. Suggestion: Show first that _t (x) = V (l i 2 the sum being extended over all integers it such that I it < lax.
The sum can be evaluated approximately by considering the integral 1Z
f1 0
d it
THEORY OF CONGRUENCES
127
We define the functions O(x) and tp(x) for any x by the formula
0(x)logp, pSZ
the sum being extended over all primes p <x, and
'l'(x)=il(x) + 0(x3) + 0 (x)+..., where the series breaks off when x11'll < 2. Prove the following relations:
k
'0
x P;9X
GPX 1
P
s
log 1, = log [XI!
and [x]
/,
7 ( l)k1 O I k=1
`
= log [x]
2 log [ x] !.
11
As usual, [z] denotes the greatest integer < z. By means of Stirling's formula, known from analysis, prove the following inequalities:
log[x]!<xlog.rrx+ Ilogx+log V:9n +;' and
log [.r] ! > x log x  x  z log x + log Y 27r,
and thus, for x ? 4, log [x]!  2 log [I x] ! < x log 2 +
log x  log Y8 a +
Use the latter inequality to prove the relation t' (x)<2x, where v(x) is the function defined in Exercise 59. By means of the results in Exercises 59 and 60 prove the inequalities
2
the sum being extended to all primes p < x. (Mertens.)
CHAPTER I11
128
62. Let f
ax2 + bx + c be a primitive irreducible integral polynomial, and let p be a prime dividing D = b2  4 a c. Show that the number of incongruent solutions of the congruence f (x) = 0 (mod pm)
is < 2 1 D I for every exponent nt. 63. Show that the number
n!+1 has at least two different prime factors for infinitely many positive integers n. Prove the same proposition for the number Suggestion: Use Wilson's theorem.
64. Prove the following theorem of Wolstenholme: If p is a prime > 3, then the numerator of the fraction 1+
+
+ 1 + 3
3
1
p1
is divisible by p2. 65. Let
):
be a natural number > 1. If a belongs to the ex
ponent e and b to the exponent f modulo n, and if (e, f) = 1, show that ab belongs to the exponent of modulo n. 66. Prove the following theorem: If a belongs to the exponent 4 S (8
integer) for the prime modulus p, then a belongs
to the same exponent modulo p. 67. If a is an integer, and if ax  1 is divisible by p for x = p 1 but for no positive value x < p  1, then p is a prime. (Lucas.) 68. Let t and n be relatively prime natural numbers and let 5113
.
)ij,
where (10, nI) = 1. If y denotes the greater of the integers
a and P, prove the following theorem: In the decimal expansion of
t , the period begins after y terms; if the number
12
THEORY OF CO1\GRLJENCES
129
10 belongs to the exponent f modulo n1i the number of terms in a period is .1. 69. Find the numbers which belong to the exponent 21 modulo 2" for I < t:!5; a  2. In particular, if a ? 4, show that the numbers belonging to the exponent 22 are the numbers ± 3 (mod 8).
70. Show that a primitive root of p", where p is an odd prime, is also a primitive root of p. 71. Show that if g is a primitive root of the odd prime 1) and if gP1  I is divisible by p2, then g is not a primitive root of p" for a 2. 72. Show that if g is a primitive root of the prime p = 4 n I 1, then  y is also a primitive root of p. 73. Show that if g is a primitive root of the prime p = 4v ; 3, then  g belongs to the exponent (p  1) modulo p. 74. Show that if a belongs to the exponent (p  1) for the prime modulus p = 4)1 + 3, then  a is a primitive root of p. 75. Show that if p = 8)2 + 3 and q = (p  1) = 4 ?1 1 are both primes, then :3 is a primitive root of 1). 16. Show that if p = 8 n  1 and q = j(p  1) =4)i 1 are both primes, then  2 is a primitive root of p. 77. Show that if 1, = 4 n + 3 and q  2 n + 1 ()i > 1) are both primes, then  3 is a primitive root of 1p. 78. Show that 3 is a primitive root of any prime of the form 2n+ 1 (l1> 1). 79. Show that if S denotes the sum of all positive primitive roots < p of the prime p, then S
u (1,  1) (mod 1)),
where p is the 3 bius function. 80. Show that if a belongs to an odd exponent for the prime modulus p, then the congruence
a''+ 1 =0 (mode) has no solutions x. 9  516670 Trygve Nagell
130
CHAPTER III
81. Determine the solutions x of the congruence 2 (mod 11).
.Y = 3x
82. For what values of a is the congruence
10'=a (mod 41)
solvable?
83. Show that the congruence a x,2
!, t/2 = c (mod p)
is always solvable in integers x and y when the number a b c is not divisible by the prime p. 84. Show that the congruence aa3 + b r/3 = e (mod p)
is always solvable in integers x and y when the number 7 a b c
is not divisible by the prime p. 85. In Section 35 the notion of fixed was also defined for polynomials in several variables. Find the possible fixed divisors of the homogeneous integral polynomial
ail
1
3
b.r3g
ey,4
which is supposed primitive. What conditions must the coefficients fulfill, if the polynomial shall have the fixed divisor 6 ?
86. Prove the identity n"
5 (S)'
valid for all s = a + it, when a> 1. 87. Let Q (x) denote the number of positive squarefree integers c. Prove the relation
where 8 is a function of x such that 161 is less than a positive constant for all x.
THEORY OF CONCRUENCES
131
88. Put U
`'('r)=> L=1 and prove the relation 3 x 2 + b t. log x ,
cry
where b is a function of x such that I b I is less than a positive constant for all x. 89. According to Theorem 75 (Thue's theorem) there exists for each integer it not divisible by the prime p at least one representation of the following type: it = +
(mod f)l,
where x and ;q are natural numbers < V Show that there are, for given numbers it and p, at most two such representations with (x, y) = 1. If there are two representations, the righthand side in the congruence is .
!J
in the one and  ' in the other. ?/
3. 7, 23 and 41 there is for each integer a only a single representation. Show by applying the result in the For
preceding exercise that no other primes p have this property.
CHAPTER IV
THEORY OF QUADRATIC RESIDUES
37. The general quadratic congruence.  Let n be an integer 0 and let D be an integer prime to n. According to the definition in Section 34 the number D is a quadratic residue modulo n or of the number n, if the congruence
x^D (mod n)
(1)
is solvable. If this congruence has no solution, I) is said to be a quadratic nonresidue modulo n or of the number n. In this chapter we shall treat the two following main problems:
Find the quadratic residues of a given modulus n. Find all moduli of which a given integer is a quadratic residue. It is easy to show that the solution of the general quadratic congruence
ax 2 + bx + c = 0 (mod n), (n > 1), where a, b and e are integers and a not divisible by n, can be
(2)
reduced to the solution of a binomial congruence of the type (1). For, if I) = 1,2  4 a c, the congruence (2) is equivalent to
4a2x2+4abx+4ac=(2ax +b)2 Il= 0 (mod 4an); by setting 2 ax + b =y, this gives 9,2
= D (mod 4 a n)
with the condition y = h (mod 2 a).
Consider next the case in which the numbers D and n in congruence (1) are not relatively prime. We shall prove Theorem 77. Suppose (D. n) = d, d = e2 f . D = dal and n = d nl.
Here e, f, al and nl are integer.., and f is square free. Then the necexsary and xu%licient condition for the congruence (1) to be
THEORY OF QUADRATIC RESIDUES
133
solvable is that (f, n1) = 1, and that f a1 be a quadratic residue of n1.
Proof: Both D and n are divisible by e2. Therefore, if the congruence (1) is satisfied, x must be divisible by ef. Thus, putting x = efy, we must have fy2 = al (mod ni). Hence (f )?1) must be a divisor of a1. This implies that (f, n1) = 1, and we conclude from (f y)2 = f ai (mod n1) that f a1 is a quadratic residue of ni. On the other hand, suppose that (f, n1) = 1, and that f'a1 is a quadratic residue of n1. Then there exists an integer y such that y2  fai (mod n1). If we determine an integer x such that f x = !/ (mod n1), it follows that ,1x2 = a1 (mod n1). Hence, after multiplication by e 2f = d, c2 f2 x2 = dal (mod d n1).
Putting e f x = a , we get 72
= D (mod n), and Theorem 77 is proved. Thus it is sufficient to examine the binomial congruence x2 = D (mod n),
where the integer D is prime to the modulus n. According to what was shown in the beginning of Section 26, it is sufficient to consider the case in which n is a prime power. If we put q = 2 and a = D in Theorems 70 and 72, we obtain the following
result: A necessary and sufficient condition for the integer D to be a quadratic residue modulo n is that D be a quadratic residue of all prime factors of n, furthermore that D = I (mod 4), when
n:
is
divisible by 4, and that D = 1 (mod 8), when n is
divisible by 8.
38. Euler's criterion and Legendre's symbol.  In Theorem 69
put q=2, a = D and n = p = 2 h i I = an odd prime. Then 6 = (2, N  1) _ :3 and
rs) = It.
It follows from the congruence
DPI  1 = (D''  1) (D" + 1)
0 (mod p)
that D' is either congruent to + I or to  1 Inodulo p. Thus we can state
CHAPTER IV
134
Theorem 78. Zf p = 2 h rirrarlratic rr.,irtue H Dh i.+
I
I
is an odd prime, the i'rteyrr D i.+ a
or a quadratie to
.} 1 or to

of p, I
rr(0(I1110 p.
This theorem, known as Eider's criterio)?, may also be proved
directly in the following way without the use of the theory of primitive roots. Every integral square which is not divisible by p is clearly congruent to some of the h (p  1) squares 12. 2. 3
.
hs.
It is easily seen that these numbers are incongruent modulo p.
Hence, any quadratic residue of p is congruent modulo p to exactly one of these squares. But it follows from D = :( 2 (mod p) that
D'
xr'I  I (mod p).
Suppose next that 1) is a quadratic nonresidue of p. and consider the congruence .r. y  D (mod p).
(1)
To every integral value of c in the interval 0  p there cor responds exactly one value of ?/ in the same interval; and we never can have cx = y (mod p). Thus we can write down a (p  1) =h collgruences (1) in which ..c runs through one half of the numbers
1. 2, 3... .])1. while y runs through the other half.
Multi
plying together all these congruences we have
1.23
(p
1) !  Dh (mod )r).
and by Wilson's theorem D' =_ 
I
(mod p).
From this proof we also obtain Theorem ill. If' p i. all odd ))rime, there are just (ix )??am quadratic
as nunresidues modulo p, i. e.. there are each kind.
(p  1) of
This result is, however, a corollary to Theorem (i8. as is easily seen by putting q = 2, a = D, n == p and 6 = 2.
THEORY OF QUADRATIC RESIDUES
135
For an odd prime p, Legendre introduced a symbol defined in the following manner:
('P) l (D)
+ 1, when D is a quadratic residue of p;
 1, when D is a quadratic nonresidue of p. p The following relations are valid for this symbol:
(D) 1)
(D) 1)
and further
DiPIi (mod ? ),
(W), when D  D' (mod p), P
`DD'\  (D) (D)
From the last relation we conclude: The product of two quadratic residues or of two quadratic nonresidues is a quadratic residue. The product of a quadratic residue and a quadratic nonresidue is a quadratic nonresidue. If g is a primitive root of p, the incongruent quadratic residues are represented by the even powers 1, g2, g4
gps
and the incongruent quadratic nonresidues by the odd powers 9, 93, g5, ... , gt'2.
By putting D =  1 in Theorem 78, we obtain Theorem 80. The number  1 is a quadratic residue of all primes of the form 4)1 + 1, and a quadratic ttonresidue of all primes tre have of the form 4 n + 3. Using Legendre's P
It follows from Theorem 80 that the odd prime divisors Of the polynomial x2 + 1 are the primes of the form 4 n + 1. This result is already known from Theorem 58 in Section 30.
CHAPTER IV
136
According to relation (4) in Section 30 we have (3)
(p
11 y
J
I = + I (mod p),
when p is a prime of the form 4n + 3. Let m denote the number of quadratic nonresidues of 1p in the interval 0  i p. As a consequence of our statements just made about the product of residues and nonresidues, we can conclude: On the righthand side of the congruence (3) + I or  I is to be taken according as the number in is even or odd. For, the number  1 is a nonresidue modulo p. If p = 7, there is one single positive nonresidue < 11), 2 In namely 3. If p = 19, there are three positive nonresidues < p, namely 2, 3 and 8. If p = 23, there are four positive nonresidues < 11), namely 5, 7, 10 and 11.
Suppose now that p is a prime of the form 4n + 1. Then the number of quadratic residues in the interval 0  p is equal to (p  1). For if a is a residue in this interval, then p  a is a residue in the interval 1 p  p. Thus, there are the same number of residues in both intervals. Also the number of nonresidues in the interval 0  j1p is equal to (p 1). 39. On the solvability of the congruences x2 = + 2 (mod p). We shall prove the following theorem: Theorem 81. The number 2 is a quadratic residue of all primes of either of the ,furm. 8)2 + I and 8n + 7 ; it i.. a quadratics nonresidue of all primex of either of the,frirmx 8n + 3 and 8 n + 5.
Proof. Suppose first that the prime 1) is of the form 8 n + 1; then we have xp1  I = (xs  1) 9 ('x)

(x4
+ 0.171 (x),
where g (.r) and g, (x) are integral polynomials in x. Therefore, according to Theorem 43, the congruence .r4 + I _ 0 (mod p)
has exactly four incongruent roots x. Determining the integer p such that
THEORY OF QUADRATIC RESIDUES
137
aj.r = xz + 1 (mod p),
it follows that /2 = 2 (mod p).
Hence, the number 2 is a quadratic residue of any prime p (mod 8).
Next we shall show that the number 2 is a quadratic nonresidue of all primes p = ± 3 (mod 8). This is correct for 1p = 3. Suppose
there exist primes  + 3 (mod 8) of which the number 2 is a quadratic residue, and suppose q is the least of these primes. Then the congruence 2 (mod q)
would be solvable. We can suppose that the solution x is positive, odd and < q, since  x is also a solution. Then we should have x2  2 = q f where the integer f satisfies the following inequalities: z (1)
Q <.t' < q
< q.
Thus, the number 2 would be a quadratic residue of any prime factor of the odd integer f Hence, by (1) and by the hypothesis
on q, every prime factor of f would be either of the form 8 in + 1 or of the form 8 m  1. Therefore we should have f = ± 1 (mod 8). and x.2  _) = q,1'= ± 3 (mod 8). But this conx2
 2 _  1 (mod 8). Consequently we conclude that there is no prime q, and that gruence is impossible, since
1;1
=1
for all primes p = ± 3 (mod 8). Finally we show that the number 2 is a quadratic residue of all primes of the form 8 17  1. By Theorem 80 it is sufficient to show that the number  2 is a quadratic nonresidue of these primes. Suppose there exist primes =  I (mod 8) of which the number  2 is a quadratic residue, and suppose q is the least of these primes. Then the congruence x3 =  2 (mod q)
CHAPTER IV
138
would be solvable. We can suppose that the solution x is positive, odd and < q, since .x is also a solution. Then we should have x' + 2 = qf, where the integer f satisfies the following inequalities:
(1
(2)
< q.
q
Thus, the number  2 would be a quadratic residue of any prime
factor n of the odd integer f Hence, by (2) and by the hypothesis on q the prime n could not be of the form 8 in  1. It could not be of the form 8 m  3 either; for, if n = 8 m  3, we should have, according to the result just proved,
8ni 3 and
= 1
8m3) (?
1
1'
Thus, n would be either = + I or = + 3 (mod 8) and the same would hold for f. Therefore we should have 2 = q.1' =  I ors  3 (mod 8). But this is impossible, since x$ + 2 = 3 (mod 8).
Hence, we conclude that there is no prime q, and that
or
for all primes p of the form 8 11  1. Thus the proof of the entire theorem is complete. Using Legendre's symbol the result may be written in the form (3)
Combining Theorems 80 and 81 we get
THEORY OF QUADRATIC RESIDUES
139
The urnnbr;  2 ix a quadratic residue of all prime. o/' either of the,/bnns 8n 1 awl 8 n + 3; it is a quadratic noureciclue of all primes of either of the forma 8 u 5 and
Theorrvn
8n+7.
We shall also establish some results on biquadratic residues. By means of the identity
x4 + 4 =((x + 1'2+ 1)(x 12+ 1) and Theorem 80, we deduce immediately Theorem R3. The numlcr  4 is a biquadratic residue of all prime. of the ,/born 4n + 1 and of no other primes.
Further we have Theorem 84. The number  1 i.y a biquadratic residue of all primes of the form 8 n T 1 and of no other prime > 2. Proof. From Theorem 80 we know that the congruence
r4 =  I (mod p)
(4)
is solvable only if 1, = 1 (mod 4). From the proof of Theorem 81 (first part) we see that congruence (4) always has solutions if 1) = 1 (mod 8).
Now suppose that p = S (mod 8). If (4) were solvable, we should have from this congruence nI
xjiz
pI
(.r.4) 4 _ ( 1) 4
1 (mod p).
But this is contradictory to Fermat's theorem. 40. Gauss's lemma.  We shall establish the following useful lemma due to Gauss: Theorem 8;. Let p be an odd prime and D an integer not dirisible by p. If p denotes the number of integers in the sequence (1)
1 .A 2 D,
.(1)
1)D,
140
CHAPTER IV
p are > } p, then we have
whose principal remainders
(D1
= ( 1)
Proof. The numbers (1) are clearly incongruent modulo p. For, the congruence h D = k 1) (mod p) only holds for h = k (mod p). Let al, a2, .a u be those of the principal remainders modulo p) of the numbers ( 1 ) which are > z p, and NI, P 2 , f. those
which are < 11). Then A + u = t (p  1). The numbers p  aI,
p  a2, ... , p  a,, are all in the interval 0  Ip. None of these numbers is congruent to any one of the numbers flp modulo p. For, p  a = flj (mod p) and a, = rD, flj D (mod p) implies r + s = 0 (mod p); but this is impossible, since r and s belong to the sequence 1, 2, ..., j (p 1) and have a positive sum < p. Thus, the a (p  1) numbers Y1+N2,....flx,pa1,pa2,....pa
(2)
are all the natural numbers < Y (p  1). Hence, forming the product of all the numbers (2), we get #1 #2 ... YZ . (p  a1) (p  a2)
(p
(1)  a,,)
y 1) ! = ( 1)" . (P. 2)! W (P1) (mod p).
Since
it follows that
Di(n1)  (n) (mod p), p
I I=(l)', and Gauss's lemma is proved.
Now suppose that D is a positive number. For k = 1, 2, .. (p  1) we put
kD= p('hD]
,
+ rk,
where rk is the principal remainder of kD modulo p. Then, taking
the sum over all k, and recalling the identity
141
THEORY OF QUADRATIC RESIDUES
we get the relation k«1)
s(1)21)D=p
(3)
x _1
Ir D
p
+A+Ii trill+A+B,
where A is the sum of the numbers a1 , a2 ...
, a and B the
sum of the numbers #1 i j2.... , f .. Further we have A
(Tr
L
1+2T
ai) T 2#,
'+
11(p1)=n(121)
or
,upA+B=''(p21).
(4)
Eliminating B between (3) and (4) we obtain
I(p21)(II1) (DIcc)p+2d. Hence (5)
,u= 1II +(p2 1)(D 1) (mod 2).
When D = 2, then DI = 0, and thus p ='1(1)2  1) (mod 2).
Hence we have a new proof of formula (3) in Section 39. 41. The quadratic reciprocity law.  We first prove a theorem of Eisenstein: Theorem S(i. Let a and h be two odd integers > 3. li' (a, b) = 1,
and a'=v(a 1), b'=1(I,1), we have 11 n=1
aL. cr
a=1
G
Proof. We consider the a' b' integers (1)
buar
for n = 1, 2. ... , a' and r = 1, 2.... , V. None of these numbers is equal to zero. For, since a and 1 are relatively prime, the equation b u = a u implies n a t, v= b t. Exactly
CHAPTER IV"
142
I
a'
[hit]
of the numbers (1) are positive. For, if u is fixed, then ba> a ' b +r for r = 1. 2. , Further, exactly .
.
1r, ar
(G
of the numbers (1) are negative. For, if a is fixed, then b it < a r
for a = 1, 2.... ,
[J.
Thus the theorem is proved.
The proof may be interpreted geometrically as follows: In a twodimensional rectangular coordinatesystem with the abscissae x and the ordinates y we draw the straight line L from the origin to the point (a, b). In the first quadrant we mark the lattice points (x, y) which satisfy the conditions
l <x
Fig. 5.
THEORY OF QUADRATIC RESIDUES
143
It is readily seen that these lattice points are distributed in the following manner: On the straight line L there are no lattice points, since (a, b) = 1. Between L and the xaxis there are Sl lattice points; between L and the yaxis there are S2 lattice points. (In fig. 5 it is a = 15 and b = 11.) Now we will apply this result to prove the quadratic reciproeihj lair: Theorriii 87. relation
I,/' p awl q are distinct odd prirues, ice hair the (Pq)
where
= ( 1)",
h=a(1) 1)
(q
1).
Proof Putting D = q in relation (5) in Section 40, we
get
y = 11 (mod 2) and
P
where 1P,11
31=
q ar
J
C,
Interchanging p and q, we get the analogous relation
where
e fvI'
r=I
Cr

G ) (P)
9`
Finally, by putting in Theorem 86 a = 1), 1, = q. it follows that 111 + \' = a' b' = k (p  1) This proves the reciprocity law.
(q
CHAPTER IV
144
It is readily seen that this theorem may also be formulated as follows: If at least one of the primes p and q is = 1 (mod 4), then p is a residue or a nonresidue of q according as q is a residue or a nonresidue of p. If p and q are both   1
(mod 4), then p is a residue or a nonresidue of q according as q is a nonresidue or a residue of p. The quadratic reciprocity law was stated by Euler (1783) without
proof, but was first proved by Gauss (1796), who gave not less than eight different demonstrations of it. How great the interest in this theorem has been among mathematicians is apparent from the fact that about forty other proofs, more or less different, have been published since Gauss. Among the authors of these proofs we mention Cauchy. Jacobi, Kronecker. The reciprocity law facilitates considerably the determining of Legendre's symbol. This is illustrated by the following example: (59)
59 131)
(5913)
(591) ___
(I)=  (13) = _ 13
1
7
=+
Let us use the reciprocity law for determining the primes 1) of which the number  3 is a quadratic residue. We get (
13)
p1)(1)(11)(3)(1)1(Pi)_
31.
Thus we can state Theorem A3, The nia ber  3 is a quadratic re,idue of'all priiiies of all primes of the form 6)1 1, and a quadratic of the form 6 u  1. Similarly, from the relation
we deduce
The ii vib(r 5 i., it quadratic residue of all prime.. of either if' the forms 10 u ± 1, acid a quadratic nonresidue of
Theorem 181.
all primes of either ul' the farmx 10 n ± 7.
THEORY OF QUADRATIC RESIDUES
145
As a consequence of Theorems 88 and 89 we see that the polynomial
x2+3 has the following prime divisors : 2, 3 and all the primes of the form G n ± 1; and, that the polynomial
l. 2a has the following prime divisors: either of the forms 10 n + 1.
5 and all the primes of
42. Jacobi's symbol and the generalization of the reciprocity law.
n)
Legendre's symbol
is defined for primes p only. Jacobi
introduced a more general symbol
1n),
defined for all odd natu
ral numbers P in the following way: When P =P1 P2
1/m
is a product of primes Pi P2, , pm, distinct or not, and when D is an integer prime to P, then (1)
D
`I'/ 
(D 1 1121
(D_PM
)'
where the factors on the righthand side are Legendre symbols. For P = 1 the Jacobi symbol is defined by the relation (2)
(n = * 1. 11
If D is a quadratic residue of P, then (D) = + 1. since all factors on the righthand side of (1) have ``the value i 1. On the other hand, if D is a quadratic nonresidue of P, it is not always true that (P) _  1. In fact, when an even number of factors on the righthand side of (1) have the value  1, the product has the value + 1. 10518870 Tryyre vagell
CHAPTER JV
146
From the definitions we easily get the following rules for operations with the Jacobi symbol: 15
I.
I/' the integees D and D' are prince to the odd positire integer P, then we hare the relation
(D) (D) = (DD').
(3)
To prove it we have only to apply the corresponding rules for the Legendre symbol.
II. if the integer., 1) and D' are prince to the odd po.itire integer P, and i f D = D' (mod P). then ire hare the relation
(D).
(D)
il
(1)
T1
For the corresponding rule holds for the Legendre symbol. III. If the integer D i.. prince to the odd po.4tire integers P and Q. then we hare the relation
(D) (D) = ( D ).
(5)
The proof follows from the definition, if we write P and Q as products of prime factors. The following rule is analogous to the rule given by formula (2) in Section 38: IV.
any odd positive P ice hare the relation

1
(6)
1
F
For, writing P as a product of prime factors in the form m
P=
f=1
+pl),
we get m
P=1+
(pi  1) (mod 4), f=i
THEORY OF QUADRATIC RESIDUES
147
or
1(P  1) =
(p%  1) (mod 2). i=1
Since
fl ( pill
(1,1)
,
we obtain the relation (6) by using formula (2) in Section 38. Furthermore, we prove the following rule analogous to the rule given by formula (3) in Section 39: V. For any odd positive P we have the relation
1') = (
(7)
1)1u1,.
For, writing P as a product of prime factors in the form P= P1 P2 '
' pm, we get P2=II(1+p1).
i=1
Since every number 1)2 1 1 is divisible by 8 and every product of two of them by 64, we obviously have m
P2  1 = 2: (l ii  1) (mod 64), 14
i=1
or n'
(2p)w
I (P21 (1 1) (mod 2). i=1
Since
(2)
obtain the relation (7) by using formula (3) in Section 39.
Finally we shall prove the following generalization of the reciprocity law:
Theorem 90. If P and 9 are two positive, odd, and relatively prime inteyere. then we have the relation (P)(Q)=(1),,
CHAPTER IV
148 Iohere
h=z Proof. Suppose P = P1 P2
Pr,
Q = qI q2
q8,
where Pl, P2, . . Pr, q1i q2, ... , q8 are primes. Then
(P) = \P1/
.LP2/
... (PQr)
and
 (q1) (q8) ... ( q8)l (4)()_ 11(11)l \q) (
Thus
Q)
where, by the reciprocity law (Theorem 87), t has the value
t=12 r
8
r
j(pi1)'(gt1)=1.j(pi1).
[=1j=1
i=1
}(qj1). j1
In the proof of formula (6) we showed that
Q(pi1)=}(P1) (mod 2).
ial
Then we also have the analogous congruence
(qj 1)=iI(Q 1) (mod 2). j=1
Hence
t=.I(P 1)
21(Q 1) (mod 2),
which proves Theorem 90.
Example. The number 2137 is a prime. Decide whether or not the number 666 is a quadratic residue of this prime. We have
149
THEORY OF QUADRATIC RESIDUES
_ 664 2137 
2
_
333
(`213 7) 2137)
r333 _ 55 `139  (139
k291
\'19 /

333 \12137]

_ 139 333(333 )
(2137)
(139)
(29)
55)
(55}
29
(293
l 3/
(5b 29)
\ 31l
+1
43. The prime divisors of quadratic polynomials.  We have already determined the prime divisors of the polynomial x2 + 1 (Theorem 58 in Section 30). Further, from Theorems 81 and 82 (Section 39) it is apparent what primes are the prime divisors of the polynomials a2  3 and x2 + 2. Now we shall generalize these results and show how to determine the prime divisors of all quadratic polynomials. We showed in Section 37 that it is sufficient to consider polynomials of the form x2  D. (1)
If .D is a perfect square (2, this polynomial is the product of the linear factors x  C and x + C; in this case any prime is a prime divisor of the polynomial. Suppose next that D is not a perfect square, and let D = C2 Di where Dl is a squarefree number
1.
Putting x = Cy, the polynomial (1) becomes C2 (,2  Dl).
This polynomial obviously has the same prime divisors as (1). Therefore, it is sufficient to consider the polynomials (1) in which
IJ is a squarefree integer We shall first prove
1.
Theorenn Iii. If P is a square free., old integer > 1, then /m (2)
=0,
where the vum is extended over all nuniber.c m in a reduced residue s?/xteni modulo P.
CHAPTER IV
150
Proof. There always exists an integer b such that
(7;)=
(3)
 1.
P For, let p be a prime factor of P, put T" = , and denote by i a quadratic nonresidue of p. Then we can determine an integer b satisfying the congruences b = j9 (mod 1)),
b = I (mod P');
this is possible by Theorem 40, since (p, P') = 1. This number b satisfies the relation (3), since b
j)  \p11P'l  \pl lF'l
 1.
If the number nt runs through a reduced residue system modulo P,
so does the number nib, since b is prime to P. We therefore obtain
(MP)=G).(y)S.
S= in
Q. E. D.
Hence S = 0.
Let ,u denote the number of incongruent numbers a modulo P
such that
\'1
= + 1, and let v denote the number of incongruent
numbers b modulo P such that (P) states that
1. Then Theorem 91
it = v = lop (P).
(4)
Now we pass to the determination of the prime divisors of the polynomial (1), where 1) is a squarefree integer 71 1. It is
evident that the prime
`?
and every prime factor of I) are
prime divisors of the polynomial. Therefore, apart from these primes, the problem is to determine the odd primes p for which
\)+1. `
It is convenient to distinguish four different cases.
THEORY OF QUADRATIC RESIDUES
151
Case 1. D == ± P  1 (mod 4); P > 0. Let al, a2...., a,. denote the z op (P) odd integers in the interval
0  3 P for which
(a1) _ + 1, and let b1, b2.. . .. b, denote the
p(11) odd integers in the same interval for which (11)
Then, the necessary and sufficient condition for the prime p
(which is not a divisor of 2 D) to be a prime divisor of the polynomial (1) is that
pmas (mod2P),
(i=1,2. ...v).
For, it follows from Theorem 90 that \pl\1'/=\1'1=+1.
On the other hand, the primes q which are not prime divisors of the polynomial (1) are characterized by the congruence conditions
gmbt (mod2P),
(z=1,
...,v).
.F'xa»zple 1. If 1) = 21, we find that the prime divisors of the polynomial x2  21 are, apart from 2, 3 and 7, the primes p satisfying any one of the congruences
p m 1, 5, 17. 25, 37, 41 (mod 42).
Example 2. If D =  15, we find that the prime divisors of
the polynomial x2 + 15 are, apart from 2, 3 and 5, the primes p satisfying any one of the congruences p m 1, 17, 19, 23 (mod 30).
CaNe 11. D=±Pm3 (mod4); P>0. Let a1, a2...., a, denote the . ip (P) integers in the interval 0  l P which are =1 (mod 4) and for which (a) _ + 1. Let bl, b2, ... , b, denote the 99 (P) integers in the interval 0  4 P z which are 3 (mod 4) and for which C.:)  1. 16
=
CHAPTER IV
152
Then, the necessary and sufficient condition for the prince p
(which is not a divisor of 2 D) to be a prime divisor of the polynomial (1) is that either 1' = a; (mod 4P),
(i = 1, 2..
p = br (mod 4 P).
(.j = 1, 2,
or .
v).
For, it follows from Theorem 90 and formula (6) in Section 42 that.
P)
=
(LP
Example 3. If I) = 15. we find that the prime divisors of the polynomial x2  15 are. apart from 2, 3 and 5, the primes p
satisfying any one of the congruences p = 1. 7, 11, 17, 43, 49, 53, 59 (mod 60). ("asp 111. D = ± 2 P = 2 (mod 8); P > 0.
Let aI, a2.... , ay, denote the T (P) integers in the interval 0  8 P which are .
.
are
I (mod 8) and for which (
1. Let bl, b2
., li,r denote the T (P) integers in the interval P)
3 (mod 8) and for which (rP
0  8 P which
1.
Then, the necessary and sufficient condition for the prime p (which is not a divisor of D) to be a prime divisor of the polynomial (1) is that either p = a; (mod 8P),
(i=1,2.....4),
l  l)f (mod 8 P).
(.i =1. `_', ... ,
or
)
For, it follows from Theorem 90 and formula (7) in Section 42 that (D) =(1)a(p2] (t) =\a'/+ 1.
Example 4. If D =  6, we find that the prime divisors of the polynomial x2 r G are, apart from 2 and 3, the primes p
satisfying any one of the congruences 1)
1. 5, 7, 11 (mod 24).
THEORY OF QUADRATIC RESIDUES
Cage IV. D = Let aI, a2i
... ,
153
22 P _ 6 (mod 8).
a,r denote the (p (P) integers in the interval 0 8 P
which are either = 1 or = 3 (mod 8) and for which Let bI, b2,
... ,
b,, denote the go (P) integers in the interval 0  b P
which are either = 5 or = i (mod 8) and for which Then, the necessary and sufficient condition for the prime p (which is not a divisor of I)) to be a prime divisor of the polynomial (1) is that either
/  ar
(mod 8 P),
(i = 1. 2.....
l) m bi (mod 8 P),
(J = 1, 2, ... ,
01,
)
For. it follows from Theorem 90 and from formulae (6) and (7) in Section 42 that 11;i
1?
\
l = Gl
Y/ =
+ 1.
Example :i. If D = 6, we find that the prime divisors of the polynomial x2  6 are, apart from 2 and 3, the primes p satisfsina any one of the congruences p = 1, 5, 19, 23 (mod 24).
The results obtained may be expressed, less precisely, in the following manner: Let D be a square free integer
1. J iewn 1 the. 92 (41 D I) integerk
prime to 4 I I) I in the interval U  4 11) 1. there are, ,u = z q: (4 1
rl,
the

1)
property: Ever// prime
of the polynomial ors  D is congruent to any one of the lrrc»rbers )  1 ')  2 ,.. rr, morlrrlo 4 I D 1, or it is a divi,or of 2 D. 44. Primes in special arithmetical progressions.  In Section 18 we mentioned the following theorem of Dirichlet: If r and n are relatively prime natural numbers, then there are an infinity of primes = r (mod rr). By applying the results of the preceding section we shall prove this theorem in some special cases. It
CHAPTER IV
154
follows from Theorem 58 that the odd prime divisors of the polynomial 2'2 + I are the primes of the form 4 n + 1. Now. according to Theorem 45 every integral polynomial which is not
a constant has an infinity of prime divisors. Thus, there are infinitely many primes of the form 4n + 1. It follows from Theorem 88 that the prime divisors (different from 2 and 3) of the polynomial x2 + 3 are the princes of the form 6 n + 1. Thus. there are infinitely many primes of the form 6 n + 1. More generally we have Theorem 12. There are infinitely many primes of each of the forms
4n + 1, 6n + 1, 8n3, 8n1, 8n + 3. 12n1, 12n + 5.
12n5.
Proof. We consider the following six polynomials in x:
Jt
1.2 (2.r.. + 1)2 + 4,
js (.r) = P= (2,r +
1)2
+ 2,
f:4(x)°12P2x21, f(x) = P2 (ti.r. + 1)2 + 4, .16 (x) = 3 P2 (2.,. + 1)2 + 4,
where P is an odd integer. We have, from the results in Section 43:
1. The prime divisors of the polynomial ji (x) are the primes of either of the forms 8 n + 1 and 8 ?a  3, with the exception of the primes dividing P. are the primes of 2. The prime divisors of the polynomial f2 either of the forms g n + 1 and 8n  1, with the exception of the primes dividing P. 3. The prime divisors of the polynomial J; (x) are the primes of either of the forms 8 n + 1 and R n + 3, with the exception of the primes dividing P. 4. The prime divisors of the polynomial f4(x) are the primes of either of the forms 12 n + 1 and 12n  1, with the exception of the primes dividing P.
THEORY OF QUADRATIC RESIDUES
155
5. The prime divisors of the polynomial fs (x) are the primes of either of the forms 12 n + 1 and 12 n + 5, with the exception of the primes dividing P. 6. The prime divisors of the polynomial f6 (x) are the primes of either of the forms 12)? + 1 and 1271  5, with the exception of the primes dividing P. Let ,f; (.r) be any one of the six polynomials just defined. For i = 1, 2, 3, let m = 8; for i = 4, 5, 6, let in = 12. Then, the prime divisors of f: (:t4) are the primes p (not dividing P) which
are either = 1 or = r (mod nz), where r is a certain number prime to in and not = 1 (mod m). Now assume that there are only a finite number of primes r (m.od m), and denote by P the product of these primes. If P has this value, the number f; (x) cannot, for an integral value of r , be divisible by any prime = r (mod ni). For f (x) is congruent to one of the numbers  1, 2 or 4 ulodulo P. Therefore, as a consequence of the properties of the prime divisors of f (x) just mentioned, we see that the number f; (x) is the product of primes ° I (mod en). But, this is impossible. since such a product is itself = 1 (mod ni). It is, however, easy to verify that f; (x) = r (mod nt),
for all i. Hence, the hypothesis that the number of primes = r (mod nn) is finite is false, and Theorem 92 is proved. We finish by proving Throrew 98. There are infinite y many primes of the form 8n + I. Proof. It follows from Theorem 84 that the odd prime divisors
of the polynomial x4 + 1 are the primes of the form 8 n + 1. Assume that there are only a finite number of primes = 1 (mod 8),
and denote by P the product of these primes. Then, the number (2 P!/)' + 1 would not be divisible by any prime = 1 (niod 8).
But, this contradicts the fact that every prime factor of this number must be = 1 (mod 8).
CHAPTER V
ARITHMETICAL PROPERTIES OF THE ROOTS OF UNITY
45. The roots of unity.  According to the rules valid for complex numbers we have (cos T, + i sin T)" = cos nz c' i i sin n i'
for all integers n. (Moivre's formula.) Hence, we conclude that the algebraic equation has the roots (1)
e,,,  = cos
2±rm )[
+ i sin
:3"n I!
On=0,1,2,....ii  1).
It is apparent from their position in the complex plane that the numbers (1) are all distinct. For, if C is the circle with radius 1 and centre at the origin, the numbers (1) form the vertices of a regular polygon with a sides inscribed in (' so that one vertex lies on the positive real axis. The n numbers (1) are called the nth roots of unity. The number cos
+ i sin
2z >n
does not change if n is replaced by in + n t, where t is any integer.
The number + 1 is always among the roots (1), the number  1. however, only if is is even. The product of two nth roots of unity is itself an nth root of unity.
THE ROOTS OF UNITY
157
If sm denotes one of the numbers (1) which has the property that all the numbers i,1
9
0 En=, 1 Eni, Eon,
(2)
ni
are distinct, we say that sm is a primitive nth root of unity. Then the numbers (2) represent all the nth roots of unity. We now prove the following theorem: A necessary and .aflicient condition ,for sm to be a primitive nth root of unity is that the integer m be prime to n.
Proof. Suppose that m and n have the common divisor d > 1. Then not all the numbers (2) can be distinct; for by (1) we have 11
s'1 = 1 = s0 M
.
On the other hand, suppose that (m, )i) = 1. Then the numbers (2) are distinct; for if r
we should have
cos
2 a7n(rs) + i sin 2Z,n(r+) = 1. n
is
But r  s is no multiple of n, since I r  s I < a. Thus the number of primitive nth roots of unity is equal to the number of positive integers < zz and prime to n, and consequently equal to q, (n). The number sI is a primitive nth root of unity. When ii is a prime, each nth root of unity is primitive, except eu = 1. From the preceding result follows at once:
If s
is a primitive nth root of unity which satisfies the algebraic
equation
Z  1 = 0,
the positive integer N must be a multiple of it.
For n = 2 the roots of unity are + 1 and  1, of which the latter is primitive. For n = 3 there are two primitive roots, namely
CHAPTER V
158
el=(1+i13), and e2e=Y(1i13), which are the roots of the equation e2 + e + 1 = 0. For n = 4 there are two primitive roots, namely ± i. 46. The cyclotomic polynomial.  The polynomial of degree T ()i)
F. (x) = ll (x  ea),
(1)
0
the product extending over all primitive nth roots of unity, is called the cpclotomic polynio#nia7 of index n.
Let pl, P2, ... , pr denote the distinct prime factors of ii; further, put 110
and for I
v
(2)
=x" I
r
JJ _ JJ(xPIPl.,...P1,  1)
the product extending over all the v indices ix which satisfy the conditions
1 = it
Then we have the identity 110 112 (3)
rn(x)
where all the II, with an even index v occur in the numerator and all with an odd index in the denominator. In fact, let eL = cos
tab + i sin tab !1
)1
be an arbitrary nth root of unity, and suppose (b, )i) = d. A necessary and sufficient condition for the polynomial n
a P,,Pj,...Pi,,
1
to be divisible by x  eb is, in consequence of the results in Section 45, that the degree of the polynomial be divisible by
THE ROOTS OF UNITY )t
d
;
159
this im plies that d must be divisible b y every one of the
primes pi,, pi.,
..
pi,..
Denote by u the number of distinct prime factors of d. If 1z > 0, clearly the product IL is divisible by )(',)
(:r  sb
,
and by no higher power of a  sb; for
(is the number of
combinations of fi elements taken v at a time. IIo is divisible by x  sb, and not by the square of this linear function. Consequently, the numerator in (3) is divisible by a power of x  sb, whose exponent is equal to 1+
and the denominator in
(3)
is divisible by a power of x  eb,
whose exponent is equal to
(1)
+(3)+
Now we have
'  (")
+
("Z)
 (P) +
Hence. if d > 1 and IA > 0, the righthand side of (3) is not divisible by x  sb. On the other hand, if d = I and u = 0. the righthand side of (3) is divisible by x  8b and by no higher power of this linear function. Since, in this case, Sb is a primitive 7th root of unity, we have established the identity (3).
From (3) it follows that 1 (x) is an integral polynomial in x. For both the numerator Ho 112.. and the denominator 111113 are integral polynomials, in which the highest power of ..v has the coefficient 1. Carrying out the division in the usual manner, we obtain a quotient which is an integral polynomial in x. From (3) we easily deduce the identity
CHAPTER V
160
(4)
1 n (),P)
F
I ,n P (x)
provided that p is a prime which does not divide n. On the other hand, if p divides n, we clearly have F" 1, (x) =
(J)
1''v, (xi).
Applying the formulae (3), (4). (5) we calculate the following special cyclotomic polynomials
1'2(x)_.; + 1, F3(:) ..,2 FS (.T) = x4 + C3 + x2 + ., T 1. 11'9
=x6 *
.X'3 + 1,
1,16 (x)
1,
=xx
=x4 x3  x2 .r
.1 10
x
r
1,
'.'4 (X) = X'
1,1 (x) = x4 t 1.
11.2 (x)=x4.r2 + 1.
h20(x)_a'1.1
F' 21 (x) =x'12 211
x9  x61 6 __
rx x
1,
x3  x t 1 .
If p is a prime, we have F1,(x).,r
1
(6)
For n > 1 the constant term in F (x) is equal to 1. To prove it we have only to put .r = 0 in (3). By putting x = 1 in (3), we get the following result for n > 1: (7)
P. (1)
 lIp, when n is a power of the prime p,
1, when n has at least two distinct prime factors.
47. Irreducibility of the cyclotomic polynomial.  A polynomial i'(x) in x with rational coefficients is said to be reducible when there exist two polynomials in x, not constants. with rational coefficients, such that f (x) = g (x) h (x).
Otherwise the polynomial f (x) is said to be irreducible. We prove the following lemma: Lemma 1. Let f (x) and g (x) be two polynomials with rational coefficients. If g (x) is irreducible, and if f (x) and g (x) have a common zero, then f (x) is divisible by g (x).
THE ROOTS OF UNITY
161
Proof. Let a be the common zero. The greatest common divisor d (x) of f and g (x) cannot be a constant, since it has the factor x  a. Since g (x) is irreducible, it has no other divisors than constants and divisors of the form ay (x), where a is a 0. Hence d (x) = a g (x) and therefore f'(x) is rational number divisible by g (x).
A consequence of this result is that an irreducible polynomial can never have any zero in common with a polynomial of lower degree; here the coefficients are supposed to be rational. We next prove
Lemma 2. If the integral polynomial f (x) = x'" + 01 X!,1 +
+ cf,
is divisible by the polynomial with rational coef g (x) = xm + b,,
x'°I + ... + b,,,,
these coefficients are necessarily integers.
Proof. We may suppose that f (x) = g (x) 11 W,
where the polynomial h (x) has rational coefficients. Let 111 be the least natural number such that Mg (x) is an integral polynomial, and let N be the least natural number such that \'li (i) is an integral polynomial. The polynomials .11g (v) and V h (x) are then primitive polynomials. Hence, according to Theorem 44, the product 111Vg (x) h (x) is also a primitive polynomial. But, since 111 Kg (x) h (x) =11 Nf (x), we must have 11= N = 1. Thus Lemma 2 is proved. Lemma 3. Let g (x) = x'R + aI
xm1 + ... + am
be an integral polynomial acith the zero.q xI, x.z, ...,
G(.r.)=xm
+.91xm1 + ... + A,,,
be the polynomial whose zeros are the n limbers 11516670 Tr;/g ie Nagell
and let
CHAPTER. V
162
.>'i, X.P. .. .,
where p I's a prime. Thc)i the tartfi(Kllt: X11. 12, ..., A. are iuteger.y, and all the
.11a1, _12(r2...., A.,,a,,, are diri..itle tg p.
Proof. By the main theorem on symmetric functions we know that every symmetric integral polynomial ill X17 x2...., xm is an integer. We now apply the polynomial theorem for calculating the expression )n
n
where the sum extends over all indices i satisfying the following conditions: I < it < i2 < < i, < in. Obviously every polynomial coefficient pl
k1! k2!
,!
.k,
+ k, = p, is divisible by p, if it is > 1. Hence we obtain an equation of the form where k1 + k2 +
( 1 '.
11,)1.
_ ( 1)' _l, + I'S (1'1, .7'2, ... ,
is a symmetric integral polynomial of the numbers xl, x2, ..., x,,, and consequently an integer. Since, by Theorem 35 aP = a, (lnod p), it follows that a,. = A,. (mod p) Q. E. D. for all v. where S (el, x2i
We shall prove the following theorem: The cyclotomic polynomial is
it indirectly. and suppose that the decomposition (1)
V?, (x) =ffi
is reducible having .t, (x),
where fi (x), .2 (r).... , (x) are irreducible distinct polynomials with rational coefficients. in which the highest power of x has the coefficient 1. By Lelnma 2 the polynomials are integral.
THE ROOTS OF UNITY
163
We first show that these polynomials are all of the same degree. Let q be a root of the equation fi (.<) = 0. In consequence of the properties of primitive roots of unity (Section 45) there exists a natural number k such that qA is a root of the equation f2 (i) = 0. The number k is clearly prime to n. We now form the polynomial g (x). whose zeros are the kth powers of the zeros of f (x). Since the equation g (x) = 0 has a root in common with the equais irreducible. it tion f2 (x) = 0, and since the polynomial f2 follows from Lemma 1 that every root of jZ (x) = 0 is a root of 0. Therefore the degree nI of fi (x) cannot be less than the degree n2 of f2 (x). In just the same way we prove that )12 ? nI. Hence, we have n1 = n2. and all the irreducible factors f;(.r) on the righthand side of the identity (1) have the same degree. The roots of f,(x)= 0 are the kth powers of the roots y
of J1 (X) = 0.
Let 11 be a number greater than n and greater than the absolute value of any of the coefficients in the polynomial differences
it
f (x)
for all i and j, i F&J. If T denotes the product of all primes
11, except the princes which divide k, we put Q=Tn + 1,.
Then the Qth powers of the roots of f1 (x) = 0 are obviously the roots of .J2 (x) = 0. Let us put Q = q1 02 ... 9S,
where the numbers qi are primes which, by our hypothesis on Q,
must be >0l. Let h1(x) be a polynomial whose zeros are the q1th powers of the zeros of fi(x). Starting from hI (x) we form a new polynomial h2 (x) whose zeros are the 92th powers of the zeros of hI (x). Continuing in this way, we obtain a sequence of polynomials ('')
h1 (xL h2
h3 (,c'),
..,
hR (x),
in which the highest power of x is supposed to have the coefficient 1. All polynomials are of the same degree as f i (x), and
CHAPTER V
164
we have h, (x) = f2 (x). They are all irreducible, since f2 (x) is irreducible. Hence, every polynomial h, (x) coincides with some of the polynomials J; (x). In the sequence (2) not all the polynoinials can coincide with fi(x), since J , (x) =J2 (x) fl (x). Let
be the first polynomial in the sequence which is t) different from ,li (x). Then the zeros off (x) are the firth powers of the zeros of fl (x). Hence, by Lemma 3, all the coefficients in the polynomial difference ,
Jl (4 J% (x)
must be divisible by the prime
But, since q,. > .M'. this is contrary to our hypothesis on the number 11I. Consequently, the polynomial cannot be reducible. Q. E. D.
48. The prime divisors of the cyclotomic polynomial.  The cyclotomic polynomials have the property in common with the polynomials of the second degree that their prime divisors are characterized by certain congruence conditions. W' a shall first establish the following main result: Theorem 94. If' q is n prime uhirh does not divide n, we have: 1.
The necessary awl sclf ticient ewidition for the co)rgruence
0 (mod q)
(1)
to be
i.,e that q = 1 (mod n).
If q = 1 (mod n), the solatioj,s of congruence (1) are the number, uhirh belun i to the exponent n modulo q. Thu., the numbe, of incongruent ..olutions niudcclo q is T(n).
If x
a .solution of congruence (1), the number F (x) is dirisible by e.i'uctllt the sane po#rer of q as x"  I Pro f: Since F (0) = 1. a solution .r of congruence (1) cannot
be  0 (niod q). If F. (x) is divisible by g, at least one of the factors in the numerator on the righthand side of relation (3) in Section 46 is divisible by q. Hence the number x"  1 is divis
ible by q. If we suppose that the solution x belongs to the exponent p modulo q, the number ,u must be a divisor of n.
THE ROOTS OF UNITY
165
Further suppose f2 > 1, and denote by PI, P2.... 'P111 the dis
tinct prime factors of ' . If q divides the number n
:r`'  1,
(2)
where d is a product of different prime factors of ii, the number
1 must be a multiple of uu and thus a multiple of d. Hence. every prime divisor of cd must belong to the set of primes 171, P2, ..., J)m
Now suppose that the number a!' no higher power of q. Thus .c" = 1
+
1
is divisible by qF and by
t q ",
where t is not divisible by q. Raising each side of this equation to the kth power, we have
xxI'=1 +ktgs
g2"f1=1 +f2q$,
where fl and t2 are integers. If k is not divisible by q, neither
is the number t2 divisible by q. Thus the number x"'  1
is
divisible exactly by the same power of q as the number x<<1. In the number (2) we have
= k,i, where k is all integer not
d divide ii. divisible by q, since q does not Finally we can conclude: The product 11o 112II4 the numerator on the righthand side of relation (3) iii Section 46) is divisible by a power of q whose exponent is equal to .
y
(4121)+
i (1 + (2') and the product II1 H3 (the denominator on the righthand side of relation (3) in Section 46) is divisible by a power of q whose exponent is equal to
W) t 113
1.
166
CHAPTER V
But we have (in) +
Hence, if 14
()_() ± ...=(11)»:=0.
> 1, the number F (x) is not divisible by q. If
,u = ii, the number F. (.r) is divisible by the sane power of q as x"  1. Thus Theorem 94 is proved. It also holds for n = 1. We further establish the supplementary result: JG.
Suppose that
q
is a prime factor of n, and put
u = q" iii, inhere nl i not divisible by q. Then ire hare: 1.
The necessary and siifjieieiit condition for the congriienre
F (x) = 0 (mod q) to be solvable is that q = 1 (mod n1). If q = 1 (mod n1), the ,olutions of congruence (3) are the numberx irhich belong to the exponent nl q. Tliits the number of incongruent solutions niodulo q is p(n1). I!' .r is a solution of congruence (3). the number F (a') is diri, ible by q and not by q2, provided that n > 2.
Proof. Suppose first that n is not a power of 2. Then for q = 2 we must have n1 > 2. Applying formulae (4) and (5) in Section 46, we get (4)
F. (x) =
F'u, x qu)
F,,,
(.x,7"1)
If F,,(x.) is divisible by q, the number (n')
F,,,
clearly divisible by q. Then by Theorem 94 we see that the number x9" must belong to the exponent n1 niodulo q. Hence :r also belongs to the exponent n1 modulo q. For q = 2 this implies iil = 1. Since by hypothesis ii is not a power of 2, q must be odd. If ;c = 1, we have n1= 1 and by (7) in Section 46. (1) = q. is
If ;c=1, we have nl=2 and
1;,(1)=Fy(1)=q.
THE ROOTS OF UNITY
167
Suppose next that a 34 ± 1. If x belongs to the exponent n1, the number (5) is divisible by q and, according to Theorem 94 (last part), by the same power of q as the number
W"')"'  1 = X.  I, which is 74 0. In that case the number F,,,
(6)
n1
is also divisible by q and by the same power of q as the number n
which is
0. Suppose that the last number is divisible by q*
and not by q"+ 1, or n
x`! = 1 + tq".
where t is not divisible by q. Raising each side of this equation to the qth power, we have :Yn=1
gtg8+
\q/12g2. +
... =1 +t1q"`1,
where t1 is not divisible by q since q > 2. Hence we conclude: If the number (li) is divisible exactly by q", the number (5) is divisible exactly by From (4) it then follows that the number F,, (x) is divisible by q and not by q2. Finally, when n (> 2) is a power of 2, we have F (x) = a2n + where m = Hence T (x) is never divisible by 4. 4 This proves Theorem 95 for all valves of u. q"+1.
In particular, we see that congruence (3) is solvable only if q
is the greatest prime factor of n. For, according to the first part of Theorem 95, we must have q > u1. Example 1. For n = 20 we have
F20 (.r)
= 0 (niod 4 1)
CHAPTER V
168
has the eight roots
x = ± 2; ± 5, ± 8, ± 20 (mod 41), since 41 = 1 (mod 20). the congruence Since 5 = 1 (mod 0 (mod 5)
F20 (x)
has the solutions .r = ± 2 (mod 5). Example 2. It follows from Theorem 95 that none of the congruences F15 (x) = 0 (mod 5) or F15 (r) = 0 (mod 3) is solvable.
From Theorem 94 we obtain
Theorem 11. If n is a natural nzunmber, there are infinitely many primes which are = 1 (mod n). Proof. Suppose that there are only a finite number of primes I (mod n) and denote by P the product of these prunes. Let us put x = n P y into F Cr), y being an integer. Then, by Theorem 94, every prime factor of F,, ()a P y) must be = 1 (mod n). But this is impossible, since F,, ()? Py)=F,, (0)= 1 (mod nP).
Thus Theorem 96 is established, by indirect proof. 49. A theorem of Bauer on the prime divisors of certain polynomials. Theorem .97.
Let m be a natural number ? 3, and let n
./ (1") _
(1)
at.
xnk
k=0
be an integral polynomial which has at least one real zero. Then f(x) ha infinitely many prime divisors which are not 1 (mod )n).
169
THE ROOTS OF LTNITY
Proof. We can suppose that f (x) has no multiple zero. Further we suppose that ao is positive. There are real values of x such that f (.x) is negative. In fact, let o be a real root of the equation f (x) = 0. If ,J'(x) 0 for all real x, the function J '(x) has a minimum value for x = o. Thus we have ,f'(o) = 0. But this is impossible since f'(.r). by hypothesis, has no multiple zero. There
fore there exists a fraction
,
,
where t and T are integers (T posi
tive) such that f Gt) < 0. Then the polynomial g (:r) = 1'n 1 (
l)=
`,a,. ' 1 xnt:
is negative for x = f. Let us put lt(x)r/
t .+; + f)
g (t)
_1
t'
x`J
l
it)  ' / (t) .,
9
It
+
i (
J t))n1 a0 x
In this integral polynomial the coefficient of x" is positive. Therefore, It (x) is positive for all values of x exceeding a certain
value ro. Since the constant term is  1, the polynomial h (x) has at least one prime divisor which is not = I (mod )12). Suppose now that h (x) has only a finite number of prime divisors which are not = 1 (mod tit), and denote by P the product of these primes. Then, for all to Px > x0 the polynomial h (ii? Px) has only prime divisors = 1 (mod i)i). Thus, we must have h (t)z Px) = 1 (mo(l nt). This is impossible, however, since from the expression for h (x) we see that h (ttt Px) =  I (mod tn). Hence, it follows that h (x) has infinitely many prime divisors which are not = 1 (mod iii). But the polynomials f (x), g (x) and h (x) obviously have the same prime divisors, possibly with the exception of the finite set of primes which divide the numbers g (t) and T. This proves Theorem 97. By means of ideal theory it is possible to establish the following result: If in is an integer ? 3, every integral polynomial (which is not a constant) has infinitely many prime divisors which are = 1 (mod in).
CHAPTER V
170
50. On the primes of the form ny 1.  For any natural number
re
we define the two polynomials Un (.c) and 1',,(x) by
the equation
(x + ,)n = t' (x) + 11'. (x),
where i is the imaginary unit. Thus we have l n (x) _ (.r + $)n + (x  j)n (1)
9
(x f r )n  (x  n In i
positive divisor of n, the polynomial 1' (x) is divisible by the polynomial F, (x). For, putting r: =It v, we have U,, (;X') + i 1n (x) = (x + i v)" _ (UT, (x) + i V,
.x.')",
or
(3)
(x)),,l  3) (U,.
(U,
1. (x) = I ,. (x)
(x))"s (y. (x))2
+
I.
Lemma 1. A. prime q which iv =  I (mod 4) cannot, for the .cane
integral value of
,
dirirle both C. (x) and V, (x).
For if q divides both numbers, it also divides the number (Um (x))2 + (Jr. (x))2 = (x2 + 1)n,
But the number x2 + 1 is not divisible by any prime   1 mod 4).
Lemma 2. Let q be a prime =  I (mod 4) which is riot a dirisor of n. If v is a g positive divisor of u, the numbers i
(4)
(
x ) an d
T ;, (x)
T". (x)
cannot be divisible by q for the sanir integral ralue of x.
In fact, put n=trv, and suppose that q is a divisor of both numbers (4). It follows from (3) that T'n(.X')
_Y(U.,(x)"'I
(mod T,. r).
171
THE ROOTS OF UNITY
Thus the righthand side of this congruence must be divisible by q. But by Lemma 1, q does not divide U, (x), and since q is not a divisor of n, u cannot be divisible by q. Lenrrrra 3. If q i.+ a prime =  I (mod 4), the number I y+l (x) is divisible by q for all integral values of x.
In fact we have I,,+I
(x)=q
)x+....I(
I
1
1`
where all the binomial coefficients, except the first and the last, are divisible by q. Hence by Fermat's theorem I',,+ 1 (.r) _ (q + 1) .x a  (q + 1) x = xQ  x = 0 (mod q).
Lemma I. Let q be a prime =  I (mod 4) with the following propertti: For a certain integral ralue of x the prime q dirides 1',,(x) but none of the numbers V. (x), where v is any positive dirisor rl" n less than u. Then we have
q _  1 (mod n). Proof. Put (n, q + 1) = v. There exist two positive integers u
and r such that it (q +
This readily follows froln Theorem 17 (Section 10). Now we have i V, (x)] [UL (x) i i T',. (x)] _ (.e + i).+r u(q+i; _ or (5)
(x) + i ,. (x) U'." (X) = Vu9+n (x).
By Lemma 3 the number V'9+1(x) is divisible by q. Therefore, since 1',,+l (.r) divides V,,(q+I) (x), the latter number is also divisible by q. Since by our hypothesis I ;, (x) is divisible by q, 17, (x) is
also divisible by q. Then it follows from (5) that the number V, (x)
(.r) is divisible by q. Since, by Lemma 1. L
(x) cannot
CHAPTER V
172
be divisible by q, we see that q is a divisor of Y., (x). But, by our hypothesis, this is possible only when v = n. Hence we have
l =  1 (mod
n).
For m > I the equation Vin (.r) _
(x + 2)m  (x  L)m
=0
has the roots
r=
+
where a is an mth root of unity 79 1. These roots are real, since they are also given by
x=cot z
(k=1.2,..
For n, > 2 let us lout 0 (x) = (x  l)`Pfm) F
1).
xt
.x
This polynomial is integral since, for in > 1. Fm
1
= I'm (z)
It is of degree q ()n), and its zeros are obviously the numbers 23
where a runs through a reduced residue system modulo in. If n and v are natural numbers. and v a positive divisor of n which is < n, the polynomials Y,, (x) and o (x) have no common zeros. Both polynomials are divisors of the polynomial T 'n (x), and therefore we have the equation (6)
(x)
11' (x) = c .
T',: (x)
where c is an integer 7tf 0, and where lj' (x) is an integral polynomial in x. Since 0,, (.r) has real zeros, it follows from Theorem 97 for in = 4, that this polynomial has infinitely many prime divisors q =  1 (mod 4). Let q be one of these prime divisors
of C (x); we further suppose that q is a divisor neither of n nor of the constant e. Let .r be an integer such that 0n (x) is
173
THE ROOTS OF i NNITV
i'n (x)
divisible by q. Then, by (6), the number i
is also divisible
v (x)
by q. Therefore, by Lemma 2, the number i'y(x) cannot be divis
ible by q, and finally, by Lemma 4, we must have q =  1 (mod n). Thus we can state Theorem .98. If n is a natural nzunber, there are infinitely many
primes which are = 1 (mod n). 51. Some trigonometrical products.  If n is a natural number > 1, it is easily shown that 2 sin
(1)
i, 2
x=1
tl.
'a
For, if we put x = 1, it follows from the identity n1
x"
(2)
x 1
cos k
I
21.:T It
 i sin 21n
that
n=[J I2isin___) (cos + isin  1=1 `
Thus n1
n(cos2 rn1, isin Inn1)fJBill n1 k?L =II2sinkI k=1
fa
Since sin
('z  k) = sin n
n
it follows from (1) for any odd n that
J ;n1) (3)
U 2 sin1'rc _°7t.
AI
By putting x =  1 into (2) we obtain analogously the formula v n1f
:.Icos
(4)
k1
kn tl
=1.
CHAPTER V
174
We shall further prove the following formula } (n1)
n
(5)
It
A=1
Since the numbers 2, 6, 10, ..., 4 k  2, ..., 2 ri  4 are incongruent modulo tt, the absolute value of the product in (5) is equal to the value of the product in (3), i. e.. equal to Vn. 1
Further, since sin is negative for all k > 4 4 2 and )? <Jn, the number of negative factors on the lefthand side of 4/<
.
(5) is
By considering the number n modulo 1 it is easily seen that 1 = [ 4]. Thus formula (5) is valid for all odd n. By analogous reasoning we also deduce the formula 2
(6)
k
1tt.
72
x=1
For sin (8  4)
sin(8k4)7r=(1)'
is negative when k lies in one of the intervals
tI
I (n + 4)  } (n + 2) and a (3)1 + 4)  j n. Thus the number of negative factors on the lefthand side of (6) is rt1
4
2 _ 11t + 4j +
18 41
It is easily shown that r is even when it = 1 (mod 4), and that r is odd when n = 3 (mod 4). 52. A polynomial identity of Gauss.  Let us put (1)
F (x, lt, k) =
(1 A (1  x11) .. (1  xhk+1)
(1x)(1x2)...(1x')
THE ROOTS OF UNITY
where
Is
175
and k are natural numbers and k :s: h. In particular
we have
F ( x,
1 : 1x
)
and F (x, h, 1i) = 1.
if k > e Is and < is, it follows from the identity
F(x,h,hk) ( 1  x'') ... ( 1  xk 1) (1  xk) (1  x11) ... (1  x.hk+I) (1 X (1  .r.) ... (1  xhk1 (1  x^k+11 (1  xhk+2)
F (r, h, h  k) = F
h, k).
F (.r, h. 0) = F (.r, is, h) = 1,
the relation ('') is also valid for is = We have
F(x,h,k+1)=
X,
1 1
x 1  a.hk1  xh + Xhk1 hk.1
1x
 F (x, h  1, k + l)
=F(r..h1,k+ 1) xhk1
1  xk+l .. (1  xhk1)   hk1 (1  xh1) x
(1 x)... (1 xk11)
Hence (3)
F(x, is, k t 1) =F(x, h  1,k + 1) + xhk1F(x,h  1,k).
This recursive formula shows that F (.r, Is, k) is an integral poly
nomial in x of degree (h  k) k. From formula (1) we see that the highest power of x has the coefficient 1. Let us define a new polynomial f f (x, h) by the equation h
(4)
.f (.r, h) = I ( 1)k F kO
h, k).
176
CHAPTER V
Then we obtain by (3) f(x,h)=1+(1)'
'+1
h  1, k1)]
+ I ( 1)k[F(.c,h  1,k) + k=1
and
11 1)11(1
J '(x, h)
 .>!,k) F (.r, It  1.h  1).
k=1
By (1) we have x,'L.
1
1  .c
'
F(x, h 1. k  1) = F (x, h2, k  1),
thus
f(.r h)
111
lk1F x.h LI
 k1.
Finally, by (4) we obtain the recursive formula (5)
.f (:c, h) = (1  x''1) f (x, h  2).
Since f (x, 1) = 0, it follows for every odd h that J'(x, h) = 0.
On the other hand, if h is even, we have .f(x,h)_(1(1x''3)
Now
.. (12). 1=1x.
Hence, for all even h, we have established the polynomial identity (6)
,j'(x, h) _ (1  x) (1  x3) ... (1  a.r,1),
or
x''1) 11 x + (1(1 X) (1.c2)
(1  x'').(1 x''1) (1x''2) ... (1x)(1.r2)(I.Y3) + _ (1  x) (1  x3)
(1  xh1)
177
THE ROOTS OF UNITY
53. The Gaussian sums.  In his investigations on the construction of regular polygons Gauss was led to the problem of determining the sums of the following type: n1/ (in, n) _
(1)
1 cos
2nnts2 n
80
+ i sin
2mmns2 n
where in and n are integers, n > 0. After much effort he at last established the following result: Theorem 99. If n is a natural naa)nber, we have (1 + 1)1/n for n = 0 (mod 4), l/n for n = 1 (mod 4), 0 for n = 2 (mod 4), i 1 n for n= 3 (mod l)_
Proof. Let us put E = cos )a
7
 + i sin
22
For n = 2 (mod 4) the theorem is trivial, since 4
E($+* n)s = Ess+su+ 4
=  Ess.
Thus, one half of the terms in the sum (1) are cancelled out by the other half. Suppose next that n is odd, and put 'I (n  1) = v. Let in be an integer prime to n, and put Em =,q. In the polynomial identity (6) in Section 52 we then put h = n  1 and x = 112. Since 1  Elkt"
 EI'
1  E2k
for every integer t, we obtain the following relation I +922+918+7112+ ... + ,1n{nli=(1922)(1
118)...
(1
or (2)
n1 y 71kik+2) = k0
12  518870 Trygve Nagell
%1
11135 ., .  (2 R (1)2k}1 
2k1). ?1
k=0
71211+4)
CHAPTER V
178
Since
we have I.
,1k(k+1) = 1 + 9 + 274 + 279 + ... + y7rs.
77,2
k=0
Further ='1k2+,2+ (n_l) k = fvtk)s
11(nk) in L. r 1 f 12 _
22
and thus v
n,2 }1 1Ink)lnkTl) = I
r
,r V
k1
s
s+
k=1
Since
1+3+5+
. + (n  2) = r2,
it follows from (?) that 1 + 11 + 114 + Ir9 + .... 
n1,2
(113  )13) ...
+2 ).
Here the lefthand side is by definition equal to T(?)i, )I), and therefore we have } (n1) (3)
(in, n) = J1 2 d sin
(4 k  2) m n
For in = 1 this product has, by formula (5) in Section 51, the value n
Hence we see that ip (1, n) has the value Vn for n = 1 (mod 4) and the value i l i for iI  3 (mod 4). Only the case n = U (mod 4) remains. When in and n are relatively prime natural numbers and h an integer, we shall prove the leinnla: (4)
op (k in, n)
T (h n, nz) _ (k, in n).
In fact, by putting E(x) = cos 2;Zx+ i sin 2nx,
179
THE ROOTS OF UNITY
we have 97 (h in, ii) . 97 (k )7. lit) =
E(h7Ns" + hu
r
,7
an
(h (1n s + n t)`)
t
m?I 2I; (//2)
= (. (!t. m n);
L0
for by Theorem 33 the numbers ))is n t run through a complete residue system modulo )j i)2 when and t run through a complete residue system modulo in and modulo n respectively. From (4) we obtain for li = 1 and in = '? j, if ii is otld : (5)
(2 .1,
n) = 9' (n, 21)
97 (1,
)r).
If fi is even, we clearly get 7.t
t:  L
2
(V)
If 1
(7)
11
0
2., k) ¢ ('' n) _ A=O L (1Y_)
u
(2i'+1 =rp('.it) n1
By formula (6) in Section 51 we get (8)
Further
({)).4)=2(1
(9)
r'' I.
and
(10)
g2 (a, 8) = l (cos 47r
+ i sin fir! _
(1 r i) ii
Finally, for in = 2 and f = 4, we have =;:;E(Ii,2),
k =0
`
97d
kU
`
Sit
J
CHAPTER V
180
In the first sum on the righthand side the numbers (2 k + 1)2 are = 1 (mod 8), and, if t=o
t
the value of this sum is obviously `'1 t_o
an
`F
since p > 1. From this we conclude that 2
(1
j E\ 71,
when n is odd, and in is a power of 2 which is > 8. Finally, by repeated use of formulae (5). (6), (11) and (9) we obtain, if fi is even and > 2: i
x(1,2 n)=9, (1,4)],`=4._(1+1) and by repeated use of formulae (5), (7), (8), (11) and (10), if
is odd and ? 3: (1, 2i,) =9,(?, n)9'(n, 8)}i
=(1 +x')1& .
Thus Theorem 99 is completely proved. Exercises
90. Let n be a natural number, let r denote the number of distinct odd prime factors of n, and let fi be the exponent of the highest power of 2 which divides n. If a is an integer prime to n, and if \ denotes the number of incongruent roots of the congruence a2 = a (mod n), prove that
1. \ = ?r for fi = U or = 1; 2. X = 2"+1 for fl _ 2; 3. V
for 1i > 3.
This result is also true it'
)I
Suggestion: Use Theorem 47.
is a power of `? and r = 0.
THE ROOTS OF UNITY
181
91. Let n, r and fi have the same significance as in Exercise 90. If , denotes the number of incongruent quadratic residues modulo n, prove that 1. An=2'2'x2)
for fl=0 or = 1;
2.
A.n=2r() for 1S=2;
3.
f3.
This result is also true if n is a power of `? and r = 0. 4. The number n: (> 2) has a primitive root if and only if the number of quadratic residues is equal to the number of quadratic nonresidues. In all other cases the latter number is at least thrice the first number. 92. Prove Theorem 88 by direct application of Gauss's lemma (Theorem 85).
93. When p is an odd prime, determine the number of quadratic
residues r in the interval 0  1, which have the property that r + I is also a. quadratic residue. 94. Prove the relation
xI [h b]
=
(a  1) (b  1) + z (d 1),
where a and b are natural numbers and d = (a, b). 95. Prove the relation '2
[]
b(
+
l
[ a] _
[a] [
[d]
`U +
where a and b are natural numbers and d = (a, b). 96. Let vi be a natural number, and let a be a positive number such that none of the numbers ka (k = 1, 2, ..., mn) are integers. If is = [m a], prove the relation
[k a] + v [a] = nt n. k=1
x=1
182
97. Show that the polynomial x4 + 1 is never a prime function to any prime modulus. (Compare Section 29.) 98. When p is an odd prime, we define the Legendre symbol also in the case in which the numerator t is divisible by p by putting ( 1 0. )
If a and b are integers. and if a is not divisible by p, prove that
'1;(ax+L)_0 =o`
90. Let ), be an odd prime, and let f(.1.) = ax' + bx + c be an integral polynomial of the second degree, where the coefficient a is not divisible by p. Put A = L2  4 a c. As in the preceding exercise we put Prove that
(t) = 0 if I is divisible by p.
if J is not divisible by p, and further that
(a) if A is divisible by p. (Jacobsthal.) 100. Let p be an odd prime, and denote by m the number of quadratic nonresidues modulo p in the interval 0  a p. Show that
If p is of the form 4)1 + 1, we have already (in Section 38) shown that vi = 3 ()?  1). 101. Let p be a prime of the form 4 a + 3. How many quadratic
residues modulo p in the interval 0  p are even? Express this number as a function of m defined in Exercise 100.
THE ROOTS OF UNITY
183
102. How many of the quadratic residues modulo p in the in
terval 0  p are even, when p is a prime of the form
4n+1?
103. If p is an odd prime, prove the formula I:
2'4 p(1)E1)p },. x=I
2
p
where the first sum extends over all quadratic residues r modulo p in the interval 0  p. Show that this sum has the value p (p  1). if p is of the form 4 n + 1. 104. If p is a prime of the form 4)1 + 3, prove the formulae H2 sin '_r
r
p
2 sin
S
p
= lp,
where the first product extends over all quadratic residues
modulo p in the interval 0  p and the second product over all quadratic nonresidues in the same interval. Find also the value of the product lipI1
11 2 sin k2ta x=I
p
where t is an integer which is not divisible by p. The product depends on the number m in Exercise 100. 105. Prove the following theorem: If p is a prime of the form 8 11 + 1, there is in the interval 0  I/ at least one prime q which is a quadratic nonresidue of 1,. Suggestion: Use Thue's theorem. 106. Prove the following theorem: If p is a prime of the form at least one odd 8 n + 5, there is in the interval 0 prime q which is a quadratic nonresidue of p. Suggestion: Suppose it is true that every prime = 1 (mod 4) may be written as the sum of two integral squares. (Compare Section 54.) 107. Prove the following theorem: If p is a prime > 3 of the form 4 n + 3, there is in the interval 0  (2 1 p + 1) at least
CHAPTER V
184
one odd prime q which is a quadratic nonresidue of p and of the form 4 m + 3. _ Suggestion: Put a = [Vp] and consider one of the numbers p  a2, (a + 1)2  p or (a + 2)2 
108. Prove the following theorem: If p is a prime > 17 of the form 4 it + 1, there is in the interval 0  Vp at least one odd prime which is a quadratic residue of p. Suggestion: Suppose it is true that every prime = I (mod 4) may be written as the sum of two integral squares. (Compare Section 54.)
109. Prove the following theorem: If p is a prime of the form
8)1 + 7, there is in the interval 0  (2 Vp  1) at least one odd prime q which is a quadratic residue of p. Suggestion: Consider the numbers p + uo, where uo is a root of the congruence it2 = p (mod 2h), and 11 = log
Lg
+ 1.
4j
110. Let P and Q be two odd and relatively prime integers > 1, and let u denote the number of integers in the sequence
1' Q, 2 Q, 3 Q, ..., 4. (P 1) Q,
whose principal remainders modulo P are > P. Show that for Jacobi's symbol we have the following relation
=(*
(9
This result is a generalization of Gauss's lemma. Suggestion: Put B (x) = x  [x + J], and let sign x denote + 1 or 1 according as x is positive or negative. Begin
by proving the relation (P1)
(P
sign 11 R h Q` h=1
1.
THE ROOTS OF UNITY
185
111. If a and b are two natural numbers, b odd, show that for Jacobi's symbol we have the following rules:
(2a'
) b
= (it) if a = 0 or =1 (mod 4), b
and
(2a°
b
)_(a) if a=2 or =3 (mod 4). b
112. Let a, b and c be natural numbers and (a, b) = 1; suppose that b is odd and < 4 a c. Show that for Jacobi's symbol we have the following rule:
(lax
b)
(b).
113. If x and y are integers and ys > 1, show that none of the following four quotients is an integer:
4x2+1 ys+2
4x2+l y32
xe2 x2+2 2y$+3' 3y$+4
114. If p is a prime, show that the solutions of the congruence Fp_1(x) = 0 (mod p)
are the primitive roots of p. F. (x) is the cyclotomic polynomial of index n. 115. Show that the sum of the q (n) primitive nta roots of unity is equal to u (n) (MObius's function). 116. If the natural number n: has at most two distinct odd prime factors, show that the coefficients of the cyclotomic polynomial F,, (x) cannot have other values than 0, + 1 and  1.
117. Put
Fn(x,y)=ll(xey), the product extending over all the primitive nth roots of unity. For what values of n is the equation
186
ChAPTRR V
F. (x, y) = p
solvable in integers x and y, if p is a prime factor of 17? Find all the solutions x and y in these cases. 118. Let 711 and 11 be integers; suppose n > 2 and (m, n) = 1; put
11 =112 sin"' 11
the product extending, over all integers a in the interval
0  a 11 which are prime to n. Prove the following propositions:
fl = 1, if n is neither of the form p" nor of the form 2p", where p is a prime. 111
(2P )m1.
_ ( 1)1
1 rl), if n is a power of the odd prime p.
(711),
if
n
is twice a power of the odd
prime p. 7171
1/2, if
/
17
is a power of 2.
119. Let in and n be integers; suppose ii > 2 and (m, )i) = 1. Prove that
ll 2 sin kzm =
1 1 n] 1'7Z k=1
11
if n is odd, 1)tnim1i, if it is even.
120. Let y be an arbitrary positive number. Show that there are infinitely many primes p such that the least positive primitive root of p is > y. Suggestion : Use the theorem that for any natural number n there are infinitely many primes = 1 (mod !1). 15
121. Show that there are infinitely many primes p such that the exponential congruence
THE ROOTS OF UNITY
187
2Q = 1 (mod p)
has a solution q which is a prime. 122. Let n be a positive odd integer, and let m be an integer prime to na. If q, (m, n) denotes the function defined in Sec
tion 53, prove the formula q, (nz, u) =
na
(??
where r = (u  1).
, i' j n,
CHAPTER V1
D[OPHANTINE EQUATIONS OF THE SECOND DEGREE
54. The representation of integers as sums of integral squares. We shall use Thue's theorem (Section 36) for proving prime p which ix = 1 (mod 4) can be exwhere x and y are natural pressed in the form 17 =.r2 hare this property. uuinbere. No other odd 2. l:rerr/ prune p which is = 1 (mod 6) can be expressed in the form p = x2 + 3 y2, where .r and y are natural nunmbers. No other primes hare this pruperty. L'rrry prinzr p :chick is = 1 or = 3 (uiod G) can be expressed in the form p = x2 + 211, where .r and y are natural numbers. No other primes have this property. 1. Ererr/ prime p which is = 1, = if (jr = 11 (mod 14) can be
Theorem 100. 1.
j.
5.
y2,
expressed in the forma p =x 2  i a/2. where x and y are natural numbers. No other primes hare this property. Erery prime p which ix =5 or = 11 (mod 24) can be expressed in the , f bran p = 2 x2 + 3 y2, where .r and y are natural nuintbers. No other primes hare this property.
A supplement to this result is Theorem 101. If c and d are given natural there ix at most one representation of the prime p in the form p = ex2 + d y2, where x and y are natural numbers.
Proof Let us consider the congruence (1)
r2 + d m 0 (mod p).
where d = 1, 2, 3 or 7, and where p is an odd prime. From the
results in Chapter IV we have: For d = I congruence (1)
is
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
189
solvable if and only if p = 1 (mod 4); for d = 2 it is solvable if and only if p = 1 or = 3 (mod 8); for d = 3 it is solvable if and only if p = 1 (mod ti) apart from p = 3; for d = 7 it is solvable if and only if p = 1, = 9 or = 11 (mod 14) apart from p = 7. If z is a solution of congruence (1), and if the modulus is a prime p, we have by Thue's theorem (mod p),
%
where x and ij are natural numbers < 1'p: we can suppose that (x, y) = 1. Congruence (1) becomes x2
and therefore
d y2 = 0 (mod p),
22 +(Ig2=mp,
where m is a natural number < d. Hence for d = I we net ni = 1 and
.22+y2=p. For d = 2 we get m = 1 or m = 2, thus either x2 + 2 y2 =j) or x2 + 2 y2 = 2 p. By putting x = 2 x1 in the latter equation we obtain
2.0 + y2 = p.
+ 3 y2 = x2 + 3 y2 = 2 1) or x2 + 3 y2 = 3 p. The second of these equations is clearly impossible modulo 4, since 1) ; 2. By putting x = 3 r1
For d = 3 we get )n = 1, 2 or 3, thus either
x2
in the last equation we get 3.2+ y2=p.
For d = 7 we get 7n = 1, 2, 3. 4. 5, 6 or If7.m is even, both x and y must be odd, and therefore the number .r2 + 7 y2 is divisible by 8. Hence the equation .e2 + 7 y2 = in p is impossible for m = 2, 4 and 6. Since  7 is a quadratic nonresidue of the primes 3 and 5, the values m = 3 and in = 5 are also impossible. If n? = 7 we get, by putting x = 7x1, 7x4 + y2=p.
Thus the first four parts of Theorem 100 are proved.
190
CHAPTER VI
Consider next the congruence
2 a2 + 3 m 0 (mod p),
(2)
where p is a prime > 3. It is easily seen that this congruence is solvable if and only if p = 1, 5, 7 or 11 (mod 24). If z is a solution of congruence (2), we have by Thue's theorem
z = ± J (mod p),
where x and y are natural numbers < V p; we can suppose that (x, y) = 1. Congruence (2) becomes 2 x2 + 3 y2 = 0 (mod p) and therefore
2x2+3y'=mp, where m = 1, 2, 3 or 4. If m = 2, x is odd and y even and y=2yli thus x2+Gyl=1).
But this equation is possible only for p x is divisible by 3 and .r = 3.71.; thus
64 +
I (mod 8). If ))t = 3,
y2 =1),
which implies that p = ± 1 (mod 8). If m = 4, both x and y were even; but (x, y) = 1. Hence we have 2x 2 + 3!/ 2 = f),
and this equation is possible if and only if p = ± 3 (mod 8). This proves the last part of Theorem 100. We now proceed to the proof of Theorem 101. Suppose that we have the two representations of the prime p (3)
p = ex2 7 d l/2
and (4)
p = c zit + d r2,
191
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
where x. y, it and v are natural numbers. Eliminating d from these equations, we have 1) (7/2  t 2) = P (1(2 y2 c
< p, uy
(5)
rx (mod 1)).
Multiplying together equations (3) and (4), we get 2 P = (c x u ± d t/ r)2 + c cl (u y T r
(6)
where the upper or lower sign may be chosen arbitrarily. If we
suppose that n y = rx, we must have u = x and v = y, since (.c, y) = (u, r) = 1. If it r/ 79 rx, it follows from (5) and (6) that
1u+/+rxl=p, c=d=1 and exu±clyv=0; this
is possible
only for x = r and y = it. Thus Theorem 101 is proved. It is easy to verify the identity ([f2 + (1 f)t) (a2 + (l192) _ (a a  d b /3)2 + d (a i + b a)2.
By means of this and the first three parts of Theorem 100 we obtain the following results: 1.
Every integer which is the product of primes = 1 (mod 4) or twice such a product can be expressed as the sum of two integral squares.
2. Every integer which is the product of primes = 1 (mod 6) can be expressed as the sum of an integral square and thrice an integral square.
3. Every integer which is the product of primes = I or = 3 (mod 8) can be expressed as the sum of an integral square and twice an integral square. These three results were stated by Fermat; but the proof was given by Euler.
55. Bachet's theorem.  The following identity of Euler is easily verified:
192
CHAPTER VI
(a2 + b2 + C2 + (12) (a2 + #2 + y2 + 62)
=(aa+ bfl+ cy+ d6)2+ (a# ba  eb + d y)2 + (ay + bbca(1fl)2 + (ab  by + cflda)2.
(1)
We shall use it for proving Bachet's theorem: Theorem 102. Erery natural number can be cxpre,.:ed as the sum of four integral squares.
Proof. In consequence of Euler's identity (1) it is sufficient to
prove the theorem for primes. The following proof is due to Lagrange_
Lemma 1. If p is an odd prime, there exist four integers x1, x2, x3 and x4 such that
xi+xQ+x3+.x¢=reel), where m is a natural number
(2)
1?  1))
clearly are incongruent modulo p. This is also true for the numbers (3)
1g2
The two sets (2) and (3) consist of p + I numbers. Thus, at least one of the numbers (2) must be congruent to one of the numbers (3) modulo p, and therefore
a'2+y2+ 1=pm, where m is a natural number. Finally we have a
2
1)
s
< j) ( 4
+ IL2 4
<
which proves the lemma.
Lemma 2. Every prime p can be expressed as the sum of four integral squares.
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
193
Proof. The lemma is true for p = 2. Suppose that p is odd, and that 1)1 is the least natural number such that [21 i x22 
(4)
2 + r24 = ))1p :i
where x1, x2i x3 and x4 are integers. According; to Lemma 1 we
have in < 1). Lemma 2 is proved if we can show that in = 1. Equation (4) may be written x2)2+ . (5)
a'1
1\(X1
, .r2)2+ (.t'3 , x;)2+ (.v3
.T'q)2
nl P.
If m is even, the following three cases will be distinguished: 1. The numbers a'1, x2, x3 and x4 are all even; 2. the numbers
xl, x2i x3 and x4 are all odd; 3. two of the numbers, say x1 and x2, are even and both the others are odd. In all these cases the numbers .1 (x1 + x2), (xl  a2), a (xg + xq) and z (a'3  ,r4)
are integers. But it then follows from equation (5) that the number in does not satisfy the prescribed minimum condition. Hence m is odd. Now suppose in > 3, and for every Xk (k = 1, 2, 3, 4) in (4) let us choose an integer !/A such that //A = ..CA (mod u))
and 1yA < Y In. Then we get
Ai+y;+1/3+i/4x +az + a3T r}=0 (mod m) and therefore (6)
If r = 0, we would have 1/1 = 92 = J3 =Y4 = 0; the numbers xl, :r.2, x3 and xa would all be divisible by in, and we would have 2
2
2
2
But in is not a divisor of 1), since I < m < p. Hence the integer is positive. Since I t/A I < ' 111, we have from (6) ja)12f 111112I 7112+4,1112>ulr. and thus
I.
CHAPTER VI
194
If we multiply together equations (4) and (6) member by member and use the identity (1), we obtain (7)
122 r . 9221) _ (x1 yl + x2 y2 + x3 y3 + x4 y4)2
+ (xl y2  x2 yl + x3 y4  x4 y3)2 + (x1 y3  x3 y1 + x4 y2  x2 y4)2
+ (x1 y4  x4 y1 + x2 y3  x3 y2)2.
All the four squares on the righthand side are divisible by m2; for we have 4
4
xoI.
Xkyk = k=1
= 0 (mod 122)
k=1
and further x'k y!  x! yk = xk xt  A X1. = 0 (mod 122).
Hence, dividing both sides in (7) by m2, we deduce
1'p=e2+2 +z2+ 2, where z1, 172, z3 and z4 are integers. But, since 0 < r < m, this is contrary to our definition of 922. Hence, the only remaining possibility is that m = 1, and Lemma 2 is proved. Thus, Bachet's theorem is true for all primes. Finally, using
Euler's identity we see that this theorem is true for all natural numbers.
Lemma 2 also gives an algorithm for determining a representa
tion of a given prime as the sum of four integral squares. There exist natural numbers n which cannot be expressed in the form
22=xi+x2+x '2
when x1, x2 and x3 are integers. It is easily seen that all integers of the form 4k (8 n2 + 7),
where k and in are integers ? 0, have this property. Conversely, it may be shown that no other positive integers have this property; but the proof, which depends on the theory of ternary quadratic forms, is rather difficult. It follows from Lebesgue's identity
195
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
(a2 + b2 + (.2 + d2)2 = (a2 t b2  r2  d2)2
+ (2ac + 2bd)2 + (2ael2bc)2
that every integral square may be written as the sum of three integral squares.
56. The Diophantine equation x2  Dy2 = 1.  We first prove the following Lemma. If D is any natural ,somber which is not a pe?fect square, there are infinitely many/ pairs of natural numbers x and y which satisfy the inequality I x2  D y21 < 1 + 211).
(1)
Proof. The number 11D is irrational (Theorem 19), and by Theorem 20 there are infinitely many pairs of positive integers
x and p such that
We have further
,1/DI
Thus
Ix2Dy2I=I.ryID1 Ix+ p11)1
and y.
Theorem 103. If D is a natural number which is not a perfect square, there ie at least one hair of natural numbers x and y which satisfy the Diophantine equation (2)
.r2Dy2=1.
Proof. It follows from the lemma that there exists at least one integer k, different from zero, such that
.C2Dy2=k
CHAPTER VI
196
for infinitely many pairs of integers x and y. Among these pairs x, y there must exist at least two pairs !/1 and x2, y2 which satisfy the congruence conditions a'1 = '2 (mod I k j) and ?/1 = Y2 (mod I k 1).
(3)
In fact, the remainders modulo I k I of the four numbers x1, x2, 4/1 and t/2 may be combined in a finite number (=A") of ways.
Hence, we can suppose that
., Dyi=.r D!/
(4)
where xj, a/1, x2 and !/2 satisfy the conditions (3). Now we have y1
VD)
i
y2 l 1)) = x1
!/1 !/2 D + 1l1 y2  x'2 YO V D.
By (3) and (4) we get
.rla'2!/iy2Dx'i, D=U (modI1I) and
.x1112 t'2 J1xI!11x1y1 Therefore
(mod IkI).
xla'2lhN2D=kit
and
xl 112  .r2 t/1 = k i',
where it and r are integers. Hence and
(X1 A VD) (?'2 + 2 I'D) = k (it + r 1 D) y11/) (.c2  y2 Vv) = k (it  r1' D).
Multiplying together the two equations member by member we get (.1i  D 1r') (.i 2  1) 1/2) = k2
(u2  D r2).
Hence, we have
rte  D t2 = 1. Here v ,E 0. For, if v = 0, we would have x1!/2 = x2!/1, If and (., I  r/1 V I)) I.r'2
92 1
1.)) 42  !/2 VD) = ± k (.r2  92 1 '/)l.
±I
DIOPHANTINE EQUATIONS OF THE SECOND DECREE
197
Theretore after division by k x1  y1 V D = ± (x2  y2 YX
Which implies x1 = ± x2 and 111 = ± r12. But we can choose x11 74 x21. Thus Theorem 103 is proved.
The theorem was stated in 1657 by Format without proof. The first complete proof was given by Lagrange in 1768. About
50 years later it was discovered that Indian mathematicians even before 600 A. D. possessed an algorithm for solving equation (2); but they had no proof that their method always gives
a solution of the problem. The proof just given is due to Dirichlet. Commonly equation (2) is called Pell',c equation: but this is unjustified, since Pell did not make any independent contribution
to the theory of this equation. Let D and k be two integers. If ..c = it and 11= r are integers which satisfy the Diophantine equation
x2Dy2=k,
(5)
we say, for simplicity, that the number if +
is a .solution of equation (5). The two solutions it + v l D and rr' r r' 1 1) are equal if u = it'
and r = r'. The first solution is greater than the second if
n+rVD>u'+r'VD.
Let us consider all the solutions x + y VI) of the equation
x2Dy2=1
(6)
with positive and 11. Among these there is a least solution x, F y, lam, in which .r1 and all have their least (positive) values. The number Xi t y1 VD is called the fundamental solution of the .
equation (6).
A complement to Theorem 103 is Theorem 10/. If I) ix a natural ninuher which IN not a perfect xquare, the Diophantine equation (6) has infinitely many ,olu
198
CHAPTER VI
tions x + y1`D. All solutions with positive x and y are obtained by the formula x, + Y. 1'11 =
(7)
i.v the fundamental solution of (6), where n runs through all natural nwnbers, and where
where x1 + i/1 IUD
( II
rn =
)1
l
.JGk k=1
xn2k 112' Dk, 1
1
(8)
` +, 2 kT 1 Jl k1
(2 k
I.1
//
Proof Clearly it follows from (7) that
x  /n
11D
_ (.r1  YI
Then, multiplying together the corresponding members of this equation and equation (7), we have x;;  D y2 = 1.
Hence xn + yn IT) is a solution of (6). Suppose now that it + v YD were a solution with positive u
and c which is not obtainable by formula (7). Then a natural number 13 would exist such that (x1x1 + gl 1 'L)" < it  r V D < (x1 +
))n' 1
J1 1
and thus xn + yn Y D < it + v
(.i,n + Jn
i
Hence, multiplying by the positive number xn  Jn Y D, we would have (9)
1 < (u + 1, YD) (x,n  ynVL) < x1 + ?/11/D.
If we put (it + v 1rD) (x  y VD) = x t
where x = u.r  r y,, D and 1/ =
1'D,
n y, we would also have
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
199
(u  v ID) (x + ynl'D)=xyl D and, multiplying together the last two equations,
1 =(:!2Dvv2)(x2Dy2)=x2Dy2. ?1
11
Hence the number x + y 1'D would be a solution of equation (6). Then, by (9), we would have
x+y1D>1, and, on the other hand,
0<x1VD=
1
.r + y1 I)
<1.
It is obvious from these three inequalities that x and y must be positive. Now it follows from (9) that
x+yiD<x1+yil'D. But this inequality is impossible, since xi + yi 1'D is the fundamental
solution. Thus Theorem 104 is proved. When D is given, the fundamental solution xi + yi VD may be
found by trial. In the expression I + D y' we put successively y = 1, 2, 3, 4, etc., until it becomes a perfect square. However, the practical utility of this method is quite limited, as is seen from the following example: For D = 94 the fundamental solution is 2543295 + 210641/94.
It was shown by Euler how it is possible to determine the fundamental solution by means of the expansion of VT into a simple continued fraction; this method is much quicker than the method by trial. For certain values of D the fundamental solution may be indicated immediately. For instance, the equation
x2(t[21)y2=1, where it is an integer > 1. The fundamental solution is u + u$ 1; for here we have yl = 1.
Zoo
CHAPTER V"I
More generally we have Theorem 10;;. Let D be a natural number which is not a perfect square.
and q are natural numbers satigf!ling the inequality
> z112 1.
(10)
and if a = 5
t11 1) is a solution of the equation
x2D1/2= 1, then the number a is the fundamental solution of this equation.
Proof The theorem is obvious for it = 1. Suppose now that >) > 1, and that xI + y1VD is the fundamental solution of (11). Suppose further that 1<_ yI < tl. Then we would have xI 
Ji an d
1
$2 5
1
l
Hence
xlrl +
o'iil Jib=d2,
where dl and d2 are natural numbers such that dI d2 = d. Thus 5
dld2 2 y]
2 !/I
 r12t/ilc.1  t122 lh 2
1.
But, since this is contrary to (10). we must have rJ = yj as asserted in Theorem 105. We have the following corollary to this theorem: Let th and u be natural ntimbrrs. If we pint
1)=It (u!/2+2). the number l
u y + r/1 VD
is the fundamental ohdion of equation (11).
By letting ii vary, we obtain an infinity of values D for which yi has the same value.
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
201
57. The Diophantine equation x2  Dy2 = 1.  While the equation .e2  Dye = 1
(1)
is always solvable if the natural number D is not a perfect square, the equation '2  1)11'2 =  1
(2)
is solvable only for certain values of D. A necessary condition
for the solvability of this equation in integers and r is obviously that all odd prime factors of I) be of the form 4 n + 1; furthermore, if D is even, it cannot be divisible by 4. However, these conditions are not sufficient, as will he shown below by an example (D = 34). If equation (2) is solvable for a given integer I) and if 1 + r, I'D is the least solution with positive 5 and 91, we say that i + 211 YD is the funclainental Qolution of the equation. The square of any solution of (2) is obviously a solution of (1). We prove the following theorem:
Let I) be a natural cumber which ie not a prefect square. Suppose that equation (2) is solvable, and that el + ?111'D solution. Then the nnn?ber i.,
Theorem 1011.
()
XI+)tIVIJ=(i+r1'VI))2=;1 FI)r1+2S1ip1VI) is the fundamental ohition of equation (1). Further, if ere pail
4Fn+rp1L( l+niVIA,
(4)
is th I
en 1 + L=1
(qi1
1/1.
`'1
i
I)1:
(5)
ipn =
,/'orntula (4) fires:
1
2
I
kI I)kI
CHAPTER V'I
202
1. All the solutions with positive E and 71 of equation (2) when n
rims through all poeitire odd integers. 2.
All the solutions with positive x = En and y = nn of equation (1) when n runs through all positive even integers.
Proof. We get from (4)
and multiplying this equation by equation (4),
E2D9i2=(1),: Hence, En + 91,E l/D is a solution of (2) or of (1) according as the
exponent n is odd or even.
The number  51 + n, 1/b is clearly positive and < 1; it is the largest of the solutions of (2) which have a negative $ and a positive 71. Suppose now that the fundamental solutions of equations (1) and (2) are not related by the formula (3). Then we must have
1 <) + th 1 .D <
)]1
1'D)$.
Hence on multiplication by the positive number  E1 + ill l1fD__ (6)
E1+71iVD_<Eo F no 1t/<El +q11D,
where the numbers Eo =  E1 x1 + 911 ?h D,
ti1o = ill xl  E1 ?11
satisfy equation (2). It is apparent from the properties of the solutions E1 + 71, IT and  E1 + 9]11 ) that we can have none of the following cases: 1. $0 > 0 and no > 0; 2. Eo < 0 and 710 > 0; 3. Eo < 0 and 910 < 0. If 50 > 0 and no < 0, we would have 7)I y1 I) > bbl x1 and y, )/1 > ill x1. whence 77I E1 ill D >'11 E1 a
,
which
is impossible, since yl D _'7.2 =  1. The numbers Eo and no cannot be = 0 since they satisfy equation (2). Thus relation (3) is true. As a consequence of this relation and of Theorem 104 we obtain immediately the proof of the last part of Theorem 106.
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
203
It remains to prove the second part of our theorem. Suppose that a + v11D were a solution of (2) with positive it and v which is not obtainable by formula (4). Then a natural number nz would exist such that I+rli1VD)2m1
(
Multiplying this relation by the positive number 1r/Il/D)2m=$2mr/2mJ D
(
we would have )7i1'D<so + 9701 D
(7)
)]11'D,
where the numbers SO = )L 62 m  t' )72 m D,
t70  )' 2 n:  4l rf2 m
satisfy equation (2). But we just proved that the inequalities (7) are possible only for no = 0; this implies it = 2m and v = 212.The numbers 2m, r72., however, satisfy equation (1) and not equation (2). Thus Theorem 106 is completely proved. Equation (2) is clearly solvable for D = 2 with the solutions = r)1= 1. More generally we have Theorem 107'. If p is a prime =1 (mod 4), the Diophantine equation
2  p272 =  1
(8)
is solvable in integers
and st.
Proof. Let xI + ?/115 be the fundamental solution of the equation
x2p/2= 1.
Then (9)
xi1=pyi.
Here xI cannot be even, for in this case we should have  1 m p (mod 4).
If xl is odd, the numbers xI  1 and xI + I have the
greatest common divisor 2. Therefore it follows from (9) that
xl ± 1 = 2 2, xl + I = 2 1»72,
CHAPTER VI
204
and 71 are natural numbers and 11 = 2 ?j. By elimination of xl we get + 1 = 2 prj2. where
Since it < !/I, we cannot have the upper sign. Thus the lower
sign must be taken, and Theorem 107 is proved. According to Theorem IOG, 5 a rl Vp is the fundamental solution of equation (8).
We shall show that the e(jnation (ill)
y2
:i4112 =  1
has no solution. The fundamental solution of the equatio,%
.,234i/2 is, as is easily shown, 35 + tiV'84. If equation (10) were solvable and had the fundamental solution si + all V'34, we would have. by Theorem lOG,
35=
+ 34x1',
G=2Siill.
But this system of equations has no integral solutions y``1 and ill, and thus equation (10) is not, solvable.
58. The Diophantine equation u2  D v2 = C.  Let II be a natural number which is not a perfect square. and consider the Diophantine equation Z42  I/ 1.1 = (',
(1)
where C is an integer 0. Suppose that the equation is solvable, and that it + e l 1) is a solution of it. If r ± q Il is any solution of the equation (2)
.
2  I) y2 = 1.
the number (u.
I
r 111) (x + yV 11) = rt.,: + r yD
(+t Y + ex) l 1)
also a solution of (1). This solution is said to be associated The set of all solutions associated with the solution it + with each other forms a rla , q/' ,:olutionas of (1). By Theorem 104 every class contains an infinity of solutions. is
DIOPIIANTENE EQUATIONS OF THE SECOND DECREE
205
It is possible to decide whether the two given solutions it + r IT) and u' + v' l 1) belong to the same class or not. In fact, it is easy to see that the necessary and sufficient condition for these two solutions to be associated with each other is that the two numbers r(u.'VC'1) L,
and l'
be integers.
If K is the class consisting of the solutions u;+
2, 3....
, it is evident that the solutions u;  r,1 i1, i = 1, 2, 3, ... , also constitute a class, which may be denoted by K. The classes
K and K are said to be conjugate,: of each other. Conjugate classes are in general distinct, but may sometimes coincide; in the latter case we speak of «iiibiguou.. classes. Among all the solutions it + r VI) in a given class K we now
choose a solution u + r" I I ) in the following way: Let v* be the least nonnegative value of v which occurs in K. If K is not ambiguous, then the number V is also uniquely determined; for the solution  u + v' 1/D belongs to the conjugate class K. If K is ambiguous, we get a uniquely determined u" by prescribing that i ? 0. The solution u* * r* Yl) defined in this way is said to be the fundanieutal solution of the rlas.N.
In the fundamental solution the number 1ug I has the least value which is possible for j u ! when if + rVI) belongs to K. The case u" = 0 can occur only when the class is ambiguous, and similarly for the case r* = 0. If C = ± 1, clearly there is only one class, and then it is ambiguous.
Suppose now that the number C in (1) is positive, and put C = N. We prove . If' if, + e 1 i) i the frurdaniented ,eolation of the rlas. K of the equation
Theorem 1O
(3)
it21)1.2.\, and il' .rl I y11'1) is the fruedauientat .crdutir,n of equation (2), ire (rare the inequeditie
206
CHAPTER VI
0
(4)
yi
/Wr
V2 (x1 + 1)
0<<11 (x1+1)N.
(5)
Proof. If inequalities (4) and (5) are true for a class K, they are also true for the conjugate class K. Thus we can suppose that u is positive.
It is plain that
it r1Dva,1=uxlV(zc2>0.
(6)
Consider the solution
(u + t'VD)(x1;r,1 L)= zcxlDry1 + (x1VY111))", which belongs to the same class as u + v 1IDR Since u + v VD is the fundamental solution of the class, and since by (6) u xt  D r yl is positive, we must have (7)
Dvy1 >_ it.
From this inequality it follows that u2 (x1 1)2 z D21wy2 _ (u2  N) (x2  1) or
x1+1
and finally
1
z It
n2 <  (x1 + 1) N.
This proves inequality (5); and it is easily seen that (5) implies (4).
Suppose next that the number C in (1) is negative, and put
C =  N. We
prove
Theorem 108 a. If u + r b'D is the fundamental solution of the class K of the equation (8)
it2I)v2= N, and if xl + ylYD is the funclanzental solution of equation (2), we have the inequalities
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
(9)
0
(10)
0<1 u1 <j(A" (XI1) X.
?11
(xl  1)
207
1',
Proof. If inequalities (9) and (10) are true for a class K, they are also true for the conjugate class K. Thus we can suppose that it ? 0.
We clearly have
(xlt)2=(
+
D)(112
I
>rt2,
or
>0.
(11)
Consider the solution
(it+vl`D)(x1.!111"D)=u.x,Dtyi +(.' which belongs to the same class as u + v Since it + v 1'D is the fundamental solution of the class, and since by (11) x1 v  l/l u is positive, we must have x1 u  y1 It Z V.
(12)
From this inequality it follows that Dv2(x.1 1)1 D!/2 112, or 1
+
\T
x
:i'i + 1
xl1
and finally 1114 C 4 (XI  1) `'.
This proves inequality (10), and it is easily seen that (10) implies (9).
From Theorems 108 and 108 a we deduce at once
Theorem 109. If D and 11' are natural numbers, and i f D is not a perfect square, the Diophantinc equations (3) and (8) hate a finite number q J' classes of solutions. The f ndamerntal solutions
208
CHAPTER VI
n% all the classes can be found after a ,finite rrwiber of trials by means of the inequalities in Theorem.. 10R and 108 a. If u* + c* 1 L is the fundamental solution ref the cla,ks K, we obtain all the solutions v + r V of K by the formula if + I.1 1) = (u$ + r$ J) (x + g 1 1)),
where x + yYD runs through all the .solutions of equation (2), including ± 1. The Diophantine equation (3), or (8), has no solutions at all when it has no solution sati.yfying the incqualitie.. (4) and (5), or (9) and (10),
A supplement to Theorems 108 and 108 a is given by Theorem 11O_ If p is a 1u imc, the Diophanline equation a2  D
(13)
lr
has at snort one volydion it + r 1 I) in which if and r satisfy the inequalities (4) and (5), or (9) and (10), respeetirely, pro
rided it ? 0. If equation (13) is mlrable. it has one or two classes of solution.,, according as the prune p dirides 2 D or not.
Proof. Suppose that if + r: 1 1) and uI + rI 1 1) are two solutions of (13) which satisfy the conditions in the first part of Theorem 110. Thus the numbers it, v, it, and rI are nonnegative. Eliminating D between the equations (14)
It'  D r2 = ± p, rri  1)1.2 = + p,
we get
u2 rl  nl r.2 _ ± 1r (r  t
Thus
it rI 
(15)
it, v (mod 1))
for the upper or for the lower sign. Further, on multiplying together equations (14) member by member we have (urrl
(11 uII y rrlr)'41r2.
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
209
In the equation (16)
2T)(zrr.I+v v\2=1
I l P ` P let us choose the sign so that the congruence (15) is satisfied.
Then the two squares on the lefthand side in (16) are integers. If a t'1 + it, r 34 0, we conclude from. (16) that (17)
1 rtrl 711111
JI P.
On the other hand, applying inequalities (4) and (5), or (9) and (10), respectively. we obtain
I t(n'I T uir
which is obviously possible only for it = it, and r = rI. Thus the first part of Theorem 110 is proved. Consequently, there are at most two classes of solutions. Suppose that u + v 11; and If  c 11) are two solutions which satisfy inequalities (4) and (i) ), or (9) and (10), respectively. These
solutions are associated if and only if j) divides the two numbers 2 it v and u2 + f) r2 = '? I) r2 + P. Since v cannot be divisible
by p, the numbers 2 it and 2 D are divisible by V. But, if 2 I) is divisible by p, so is 2 it. Thus, the necessary and sufficient condition for it + r ICI) and it  r 1 1) to belong to the same class is that 2 1) be a multiple of /). This proves the second part of the theorem Example 1. Let the equation be (18) u221.2= 119. The fundamental solution of the equation (19)
is 3 + 2 1 2. The following solutions of (19) satisfy inequalities (4) and (5):
11+1'2, 11+12, 13751'2. 13+51/2 It is easy to show that these numbers are all fundamental solutions in different classes. Thus the number of classes is four. 14515670 Trig
\'aydr
CHAPTER V
210
Example 2. The equation 1t2Gt'2= ` 39
(20)
belongs to the second category. The fundamental solution of the equation
x2G? = 1
is 5 + 2 YG. From inequalities (9) and (10) we get the fundamental solutions
+3116,Z +31"6. Example 3. The equation
it 2Gr2= 22
(21)
has only one fundamental solution 2+ 1
G.
For the solutions 2 + 16 and  2 + VG belong to the same ambiguous class.
Example 1. Let the equation be
(2)
it2  82 1.2 = `?3.
The fundamental solution of the equation
x28?1t2= 1 is 163 + 18
From (4) we get the inequality t <
9 r41 < :
But equation (22) has no solution with r = 1, 2, 3 or 4. Thus it has no solutions at all.
We finish the theory of equation (1) with the proofs of the following special theorems:
Throrem 111. If j) is a prune = 1 or =  I (mod 8), there exist tieo natural nui nbtrw a awl r vitc h that and (24)
it<1
,
r<1 V.
211
D1OPHA TINE EQUATIONS OF THE SECOND DEGREE
Further. there
t,rro
uattval ,au,i,brrs ,i awl r such that
it2.'r2= 1)
(2J) Ctnrl
< 11,, c < I l,.
(2 6)
If p IN it prime (mod 12), tlu, a exist trru ,ratioal ntnnbers ,t crud r sitch that
Throrem 11*?.
rr'  Ucs =p
(27)
rt <
(28)
1,. v < V p.
If p iq it, pri,nr = 1 (mod 12), there exist twu bers if awl r
mrut
that
n:3r'=1,
(29) mud
(30)
it <
,
< I/
1,.
Pruo/' of Theorru, M. For p _ + 1 (mod S) the congruence (lnod p)
is solvable. If a is a solution, we have by
theorem
(mod 1i)
where x and r/ are natural numbers < Yp. Heuce From this equation it follows that (x + 2y)22(.r + u)2 p. Therefore equations (23) and (25) are both solvable. If we observe
that the equation
x22,2= 1 has the fundamental solution 3 + 2 V, and if we apply Theorems 108 and 108a, we find that equations (23) and (2t) have solutions which satisfy inequalities (24) and (26).
Q. E. D.
212
CHAPTER VI
Proof of Theorem 11:2. For p
1 (mod 12) the congruence
z2 = 3 (mod 1))
is solvable. If z is a solution, we have by Thue's theorem
z m ± 'I (mode), r
where x and y are natural numbers < I/ P. Hence where in = I or 2. From the equation
.r23i,2=21), where .e and # must be odd, it follows that !x + 3 \ 2_ (x + ry)' _ 2
3
? (x + 3 r,) and x (.r + y) are integers. Therefore equation (27) is solvable for 1,  1 (mod 12), and equation (29) is solvable
where
for 1,
1 (mod 12). If we observe that the equation
x:23y2 1 has the fundamental solution 2 + V3, and if we apply Theorems 108 and 108 a, we find that equations (27) and (29) have solutions which satisfy inequalities (28) and (3o).
Q. E. D.
59. Lattice points on conics.  The results obtained in Sections 54, 56. 57 and 58 may be interpreted as theorems on the distribution of lattice points on special conies. We shall now examine the general case; let us consider the conic represented by the equation .f'(x, ) = 0,
(1)
where f (x, y) is an integral polynomial of the second degree in a: and y (Cartesian coordinates).
If equation (1) represents a the form
it can be written in
DIOPIIANTINE EQUATIONS OF THE SECOND DEGREE
213
(a.r+bl/)2+cx+dr/+e=0,
(2)
where a, b, c, d, e are integers such that A = a d  b c 0. If we put a.r. + by = 1l, we get cx + d!/ + e =  u2. Hence
x= J(bu2+du+be)
(3)
and
(a 112 + (' it t a P).
Thus, there are lattice points on the parabola (2) if and only if the following two congruences are satisfied for the same value of ll :
C1 ll2 + en + ac= 0 (mod I J
(5)
and
bit 2+dit +be=0 (mod I A
(6)
If these congruences are satisfied by the integer 111, we get the
corresponding values of x and y by putting into (3) and (4) 1l = it, + d t, where t is any integer. Thus we obtain all the lattice points on the parabola (2) by putting t = 0, ± 1, ± 2, etc., into a finite number (= r) of formulae of the type x = g. (t), y = hl (t),
(7)
(i = 1, 2,
.
,
r),
where yl (t) and hi (t) are integral polynomials of the first or sec
ond degree; at least one of the polynomials is of the second degree. The number r is the number of incongruent solutions of the congruence system (5), (6). Thus, there are either no lattice points or infinitely many lattice points on a parabola.
Example 1. The lattice points on the parabola
x22xy+y2x2y=0 are clearly determined by the two systems of formulae and
x=312+2t, y=3t2t :f =3t2+4t+ 1, y=312 + t.
214
CIIAFTER VI
I:'.rannplr !. There are no lattice points on the parabola
2x23y1=0. since the congruence
2.,),' = I (mod 3)
has no solution.
On an (or circle) there are but a finite number of lattice points, which may easily be determined by trial. Consider next the case of a hyperbola. The problem arises how to decide whether a given hyperbola passes through any lattice point or not. Another problem is to find a. method for determining all the lattice points on a given hyperbola of which we already know that it passes through lattice points. It is easily seen that the equation of the hyperbola can he transformed by means of linear transformations with integral coefficients into an equation of the type K,
(8)
where I) and \ are natural numbers. Our problem then reduces to the problem of finding all the integral solutions it and r of (8) which satisfy certain congruences (9)
it = It (mod 6), v = r (mod 6).
where It, v and 6 are integers which depend on the coefficients of the original equation (1). If I) is a perfect square, equation (8) has only a finite number of solutions, which are easily determined. In Sections 56 and 57
we have already treated the case in which I) is not a perfect square. In this case there are either no solutions or infinitely many. If equation (8) is solvable, its complete solution is given
by means of the fundamental solutions (Theorem 109). Finally we have to determine which of its solutions satisfy congruence conditions (9). Example (10)
Let the equation of the hyperbola be
5.
14x1/+7//2=
1.
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
we get
By putting if = 5 x  7 .+l and r rr2 
(11)
215
141.2
where the solutions if and r must satisfy the additional condition (12)
it =5xi7/=2r(mod 5).
Equation (11) has the fundamental solutions ± 3 + 1 14; its complete solution is given by the formulae
)(15+41'14)"
(13)
and
V14 = ±(31 14)(15+4V
(14)
where )r = 0, + 1, + _>, ! 3, etc. If lows from (13) that
if =(1)"'.3 (mod 5),
rr
is even and = 2m, it fol(mod 5),
and from. (14) that
if = ( 1)'" 3 (plod 5), v = ( 1)"' (mod 5).
If n is odd and = 2)n + 1, it follows from (13) that
it = ( 1) (mod 5). r = ( 1)"' 2 (mod 5), and from (14) that if = ( 1)m (mod 5), r = ( 1)"' 3 (mod 5).
Thus we see that condition (12) is satisfied by taking formula (13), and not by taking formula (14). Consequently we obtain the whole set of solutions of (10) by the relations C' _
(rE + 7
where it and r' are determined by formula (13).
In the theory of binary quadratic forms. i. e., homogeneous polynomials of the forum axe + bx// +cy2, the questions examined in Sections 54, 55, 56, 57 and 58 are more thoroughly.discussed.
CHAPTER Vf
216
60. Rational points in the plane and on conies.  Let K be a given field. In this section any number belonging to K is said to be a rational number. Any point (x, !!) in the plane with rational coordinates (Cartesian) x and y is said to be a rational point or a point in K. The straight line
a.r+by +r=0 is called rational if there exists a positive number co such that the numbers a o.), b o.) and c ow are all rational.
Now let (1)
F(.r, r!)=a.r2+bx!l+cy2+d.r+e!l+f=0
be the equation of a conic with rational coefficients. Suppose that 71) is a rational point on this conic, and cut the conic with the straight line
!!2/=t(x ), where t is a rational number. Then the second point of intersection is also rational, and it has the coordinates
_ dal:brl(2ci7 + e)t + ct2 X= a + bt + ct2
(2)
(l)t(bZ+ c)1 + e)t2 +t
a+bt+ct2
Thus we obtain all the rational points on the conic (1) when in formulae (2) the number t runs through all rational numbers; further if c 3,4 0. the value t = oo gives the rational point x =1",
y =  1c (b
+ c sl + r). There are infinitely many rational points
on the curve if there is one. If the conic is a parabola, there are always rational points on it; for a parabola with rational coefficients may be transformed by linear transformations with rational coefficients into the form !!2 = x. When the conic is an ellipse or a hyperbola, we have no general method of deciding whether or not the curve has rational points belonging to a given field K. Only for special fields (as for instance the ordinary
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
217
rational field K (1) are we able to decide this question. (See the following section.)
Consider, in particular, the unit circle having the equation x2 + its  1 = 0.
(3)
A rational point on this curve is ( 1. 0). Putting =  1 and '1 = 0 in (2), we obtain the whole set of rational points on the curve by the formulae (4)
'Z
_ I  l2 I+t2'
1I _
`2 t
1+t2
On the circumference of the circle (5)
a_2+ '/2=3
there are clearly no rational points belonging to the ordinary rational field K (1). For, since the number  I is a quadratic nonresidue of 3, the equation a2 1 b2 = 3 c2 is satisfied only by
a = b = c = 0. On the other hand, equation (5) is solvable in every one of the quadratic fields K(l'). K (1/2) and K(V 1); in the first field a solution is .e = 1/3, y = 0, in the second a solu
tion is x = tip, y = 1, and in the third a solution is x = 2, Just as in analytic geometry, it is convenient to pass over to homogeneous coordinates and operate with a more general concept
of point. Let X, F and Z be the homogeneous coordinates for a point in the plane. The point (X, F, Z) is said to be a rational point in K if there exists a positive number w such that the three numbers X co, Yer and /, co belong to K. The points (X. Y, 0)
are the points at infinite in the (.r, y)plane. Let F be the same polynomial as in (1). and put /,21' (71, 7) = CT(\. 1, Z).
Then consider the homogeneous equation of the second degree
in X, Y and Z: (6)
G (X, ). Z) = 0,
CHAPTER yi
218
and suppose that its coefficients are rational numbers in the ordinary rational field K(1). To every solution of equation (6) in integers X. Y and /, there corresponds a solution of equation
(1) in rational numbers .r = j , g =
j, if /.
0. Conversely, if
x, / is a solution in rational numbers of (1), and if n is a natural number such that the numbers n.x and n'j are integers, equation 6) has the integral solution X = n x, 3' = n l/,/, = n. 61. The Diophantine equation a x2 + by' + cz2 = 0.  Suppose that the field K in Section 60 is the ordinary rational field K (1), and consider once more the conic represented by equation (1). We shall show how it can be decided whether or not there are rational points on the curve when this is an ellipse or hyperbola. By means of linear transformations with rational coefficients the equation of the conic, in this case, can be transformed into the homogeneous form
e.2=0. where the coefficients a, b and c are integers 7( 0. Then our problem is to find the conditions for the solvability in integers axe i l)!/2
(1)
x, y and z (not all zero) of the Diophantine equation (1). A neces
sary condition is clearly that the coefficients a, b and c be not all
positive and not all negative. We can suppose that the
greatest common divisor of the numbers a, b and c is = 1, and that these numbers are all squarefree. We can further prescribe that (a. b)= (a, e)=(1.,
For, if we put (a, b) = rl, (a, r) = r and (b, c) =,/: we have (d, e) _ (d. f) _ (e. f') = 1. and therefore it follows from (1) that : is divisible by d, !1 by r and x by f. Then, by putting
a=deal. b=df11, c=r j'r1. ,r=frt. 11=r'!h, '=riz1, we obtain the equation
a1J,r 
r'1d£2 = 0,
where
("tf.
1;3 e) = (al.f cl d) _ (l,l e, cl d) = 1.
DIOPIIANTINE EQUATIONS OF THE SECOND DEGREE
219
If the integers x
q = 77, which are not all zero, satisfy equation (1). we say that il, is a solution of (1). If the numbers , 97 and 5 have no common divisor > 1, [;, 71. C] is said to be a proper solution. Obviously, it is sufficient to take into consideration only the proper solutions of (1). If the numbers a, b and e are squarefree, and if [x, y. z] is a proper solution, we have (x, q) = (q, ) = 1. For, if x and q had the common prime factor p, it would follow from (1) that cz2 was divisible by p2; but z cannot be divisible by p, since the greatest common divisor of x, g and z is 1; and c cannot be divisible by p2, being a squarefree number . Le ;endre and proved the following criterion for the solvability of equation (1). Theorem 11, Let a, b and r be three inteyerx sati'fit the following They are all = 0, not all positire and not all negative. Tina are all .quarc free, and (a, b) = (a, c) _ (b, c) = 1. Then the necesxarrt and sirftiei(,,t conditions for .
the sulrabilitq of the 1)iop/iantiue equation (1) in integer' x, (t.z.
not all = 0. are that be be a quadratic residue of a, that  ac be a quadratic residue of t, and that  al, be a quadratic recidl(e of C.
Proof. The three conditions are clearly necessary for the solvability. We shall show that they are also sufficient. There is no loss of generality in supposing (2)
jai < JbI
Iet.
Then we have (3)
I a b I
The number I = la c is said to be the index of equation (1). If we suppose I a e = 1, we obtain from (3) la b = 1, and therefore I a _ I b e = 1. Thus we see that only the equation
.r.2I,i2.:2=0 and the equations which are obtained from it by permuting x, y and z. have an index = 1. Theorem 113 is true for all equations with an index = 1. Suppose now that the theorem is
CHAPTER VI
220
true for all equations with an index < I. We shall then prove that it is also true for all equations with the index = 1. Suppose that equation (1) has the index I> 2. If I II = I r1 1 and we would have I b I = I r I = 1 and therefore also I a 1= 1. Thus we have Icr1
Iub1
If a b is a quadratic residue of r, there exist two integers and r such that
(4)
and I
It rz <,'1,
i b = r Q
Hence
I :I < I.
(5)
If Q = 0, we have b =  a r2, and since b is squarefree and (a, b) = 1, it follows that I =  a = ± I and r = ± 1. Then equation (1) has the solution [1, 1, 0], and the theorem is proved. Suppose next that Q 0, and let A be the greatest common divisor of the numbers art, b and c Q. Then by (4) we have A _ (a r', b. e Q)  (a r2, l,) = (a r2, a 9) = (b, r i1). 1. Thus A Since A is a divisor of b, we have (_A, a) = (_l. is a divisor of both r,z and Q. Since b is squarefree, so is A. Thus A is a divisor of r, and we can put
(6)
)_la. b=Jfl. Q1= Aq=ACys,
where (" is a squarefree number. It follows from (4) that n .I Z
.11i = cACys
and (7)
((.1
(In

if = r C y2'
where it is clear that a2.
Putting it fi = B, wry get 1 I; = a b.
r(';2) = 1.
DIOPIIANTI\E EQrAT10AS OF THE SECOND DECREE
221
Now it is possible to show that the equation (8)
_1 X2 1 B )12 i C''2 = ()
has the following properties: 1. The numbers A, B and C are all 0. They are not all positive and not all negative. 2. They are squarefree. 3. We have (_1. B) = (A, C) _ (B, C) = 1. 4. The
number  B C is a quadratic residue of A, the number  A C is a quadratic residue of B, and the number A B is a quadratic residue of C.
Proof We already know that A and C are squarefree. It follows from A B = u b that (4, B) = I and that B is squarefree. Further (C, aAj3) _ (C. AB) = 1; hence (A, C) = (B, C) = 1. If it b = A B is negative. it is clear that A, B and (' are not all positive or all negative. If a b is positive, the numbers a e and be, by the hypotheses on it, b and r, must be negative, and then it follows from (4). when we multiply by (,, that a C), = 1,
. = ('` (t) = (.2. 1 (' ),
Hence AC is negative. A B is a quadratic Further, it follows from t7) that  a .I residue of C. From the same equation we see that fir C is a
quadratic residue of A. Now  a c is a quadratic residue of b and therefore also of A which is a divisor of b. Hence the product
(  ac)(ficC)=  a/,rC= BC is it quadratic residue of A.
From (7) it also follows that ail r (' is a quadratic residue of P.
Now  a r is a quadratic residue of b and therefore also of fi which is a divisor of b. Hence ( a c) (a A c C') _ 11 2 (.2_1 C is a quadratic residue of /i, and thus  .1 (' is a quadratic residue of P. Finally it follows from (7) that /3 r C is a quadratic residue of a. Therefore, since  br is a quadratic residue of a. the product
( b() (fc C) =  bir2 (' _  _IC(fic)2
also. Thus  A C is a quadratic residue of it, and the congruence x2 =  A C (mod al is solvable. We have just proved is
CHAPTER V1
222
that  A C is a quadratic residue of P. Therefore the congruence a,2 _  A C (nmod fl) is also solvable. Since (a, /3) = I and a f3 = B,
it follows from this that the congruence x2 =  :1 C (niod B) is
solvable. Hence  .1 C is a quadratic residue of B. Thus we have proved that the coefficients A, B and ' in equation (8) have all the four properties indicated above. Obviously we have the inequalities
IABI=Iabi
If we put
.c;=Aa1ii Y, ,j=1+aal', z=CyZ,
(9)
we get by (ti) and (7). since B = a(3, a.r'a
b 1/2 +
2 = c (.'1'2 (_1 X2 + B 1,2 + CZ2) = U.
Hence, [x, y, z] is a solution of (1). We cannot have x = 1/ = e = Il, For then we should have from (9) Z = 0, and by elimination of X
0=(t3+ Aaa2)3,=c(,y21,=0, and since the numbers r, C and y are 0, 1 = 0 and finally X=0. Thus Theorem 113 is proved by induction. The proof also provides a method for determining a solution of equation (1) when it is solvable.
By modifying slightly the proof of Theorem 113 it is possible to obtain the equivalent result: Theorem 11.i a. Let a, b and r be three integers srtch that abc is square free. Then the J)iophantine equation (1) is solvable in integers .r, 1/,  not all = l if and onh, if the following four
conditions are satisfied:  b c is a quadratic residue of a,  a c is a. quadratic rt.,eidue of b,  a h i a gccadrettic residue rf e, and the a r2 + h y2 ; e ;2 = 0 (niod 8) is solvable in integers
y, 7 not all even.
223
DIOPHANTINE EQUATIONS OF THE SECOND DEGREE
Here the condition in Theorem 113: "not all the coefficients a, b, c have the same sign," has been replaced by a congruence condition modulo S. It is evident that the congruence = 0 (mod 8)
x2 '. !J' r
has no integral solutions x, .y. r not all even. Hence Theorem 113 a. holds for all equations (1) which have an index = 1. Thus to prove Theorem 113a it suffices to add the following result to the proof of Theorem 113: Lenvna. !t the c'o,,gruf uec (10)
« a2 + bg2 1
j...volrable in interer8 x, (11)
>J,
1) (mod 8)
not all et()?. thr congruence
A1'' 1;)'22 .0(I11od8) ig al.co soleuble in intrgcrs X. Y, % aol all even.
Proof From the proof of Theorem 1 13 we have the following relations between the coefficients a. b. c and A, B, C: (12)
b=_1ji, aj3=B, ab=AB.
(13)
a_la2 : j3=rCy'.
Furthermore, it is easy to see that the identity (14)
r(''2(a.L2+bg2
c:21=_l(ua.t.Tffy)2+I;(y_1aa.)2 i
C(eyr)2
holds for all ;r, y and z. We have to distinguish four cases. First case. a and b are odd. r is even. Then the solutions x and y of congruence (10) are odd. Hence (lv)
ct 
1, =  C:2 (Inod 8).
It follows from (12) and (13) that A, B, a and j3 are odd. Hence, multiplying both sides in (13) by a. we have (1G)
A+B= acC'y' (mod 8).
CHAPTER V!
224
Multiplying both sides in (15) by a A , we have (17) A  B =  a c.A z,2 (mod 8).
If y is even, we get from (16) A + B = 0 (mod 8). Hence congruence (11) has the solution X = Y= 1, Z = 0. If y is odd, we get from (16) and (17): C+.4 z2=0 Iu,od 4). Since Cis square
free, this implies that C and z are odd. Hence c(Cy2+.122)=2(C'+A)0 (mod 8). If C + A = 0 (mod 8), congruence (11) has the solution X = 1, Y = 0, Z = 1. If C I _1= 4 (mod 8). con ;ruence (11) has the solution X= 1, 1 = 2, Z= 1.
b and c are odd, a is even Then the solutions y and z of congruence (10) are odd It follows from (12) and (13) that f; and y are odd. Thus the a is it solution of (10), we number c y z is odd. Hence, if Second ca.,:e.
obtain a solution X. Y. % of (11) by means of formula (14). h urd cane. ft and c are odd, b is even.
Then x and y in congruence (10) are odd. If A is even, the number yA a  x is odd, and we obtain a solution X, Y, % of (11) by means of formula (14). If .1 is odd, (3 is even, and it follows from (13) that y is odd. Then the number c y : is odd, and we obtain a solution of (11) by means of formula (14). ca.ye. a, b and c are all odd. Then _1, B and P are odd. If y is even, x is odd and ty_i aais odd. If .c is even, y is odd and y A a  x is odd. If . is even, and if a is even, then y 4 a  x is odd. In these three cases we obtain as above a solution of (11). Finally suppose z even and a odd. Their, multiplying both sides of (13) by a, we have
_1 + B = a c C7,2 (mod 8).
(18)
From (10) we have
a+b=c:2 (111od6), and multiplying by A a. _1 + 133
( (A z2 (mod 8).
DIOPHANTINE EQLATIONS OF THE SECOND DEGREE
225
Hence A + Ii is divisible by 4. If A + B = U (mod 8), congruence
(11) has the solution 1= Y= 1, Z = 0. If A + B = d (mod 8), it follows from (18) that C is odd. In this case congruence (11) has the solution X = Y = 1, Z = 2.
Hence our proof of the lemma and of Theorem 113a is complete.
A more elegant formulation of the result is given by Let a, b and c be three integers such that a b e is square free. Then the I)iophantiue equation (1) is solvable in integers x, !/, z not all = 0 if and only if the congruence
Theorei n 11.1 b.
(19)
a a,2 + b y2 + C.Z2 = 0 (mod _V)
is solvable for all integral modidi X in iutet/ers x, y, z such that (x, y, z, N) = 1. Proof It is evident that the condition is necessary. Suppose next that congruence (19) is solvable for all X. Let p be a prime factor of c. Put V=p', and consider a solution x, y, z of (19).
If y is divisible by p. it follows from (19) that x is also since (a, p) = 1. Since e is squarefree, this implies that z is divisible
by p. But this is impossible since (x, ,y, z, p) = 1. Thus y cannot be divisible by p, and similarly we prove that x is not divisible by p. Therefore it follows from the congruence a x2 + by2 = 0 (mod p) that  a b is a quadratic residue of 11. Hence  a b is a quadratic residue of e. Similarly we find that  a c is a quadratic residue of b, and that  b c is a quadratic residue of a. Furthermore, by hypothesis congruence (19) is solvable for N = 8. Thus all the conditions of Theorem 113 a are satisfied and equation (1) is solvable.
Suppose that [x1, JI, ZI] is a proper solution of (1). Applying
the results in Section 60 we find that the whole set of proper solutions of (1) is given by the formulae Jx.
(20)
by,ae'+ b,r1 1.2.
.Jy=a//I a2'?a,riueb /i J7 = r1.210 I
1551U670 Trygee Say.!!
2,
CHAPTER VI
226
where br and x are relatively prime integers, and where d denotes the greatest common divisor of the three righthand sides. If the equation is 12 .;2 = 0,
(21)
we can take x1=  1, y1 = 0, and zl = 1. Then formulae (20) give
dx=if
dy=2ur, ± Az=u2+ r2,
where d = 1 or 2 according as the product if r is even or odd.
In the latter case, by putting u1= (r( + v) and r1= (u  r), we get
x=2ulr1i ?/=a ri, Consequently, we obtain the whole set of proper solutions of (21) with odd x and even g by means of the formulae (22)
x=tr2t:2. I/=2ity,
± 7 = 262 + v2,
it and r are relatively prime integers, the one odd and the other even. Conversely, every triplet of positive x, y and z uniquely corresponds to a pair of positive al and r. The result expressed by (22) is found in Euclid's Elementa.
where
CHAPTER VII
DIOPH ANTINE EQUATIONS OF HIGHER DEGREE
62. Some Diophantine equations of the fourth degree with three
unknowns.  In the margin of his copy of Bachet's edition of Diophantos's Arithmetira Fermat stated and proved the following theorem: The area of a right triangle with rational sides cannot
be the square of a rational number. We formulate this result in the language of number theory as follows: Theorem 11.1. The Diophantine equation
ft'4
(1)
14=z2
hax no solutions in natural wonber., x, y and z.
Proof. We suppose that (x. Vii) = cl,
x d = .e1,
a!
cl
= 91
:r,
and
c(2
y and z are positive. If we put
= j, equation (1) becomes xl_y1= %1)
where 1 is an integer. Since (.rl. 111) = 1, it follows from this relation that (x,, (!/1, zl) = 1. Thus it is sufficient to prove the theorem when x, !! and z are relatively prince in pairs. Since !/4 ± 2 cannot be divisible by 4, x is odd. Now suppose that equation (1) has the solution [.r, y, z], where x is the least (positive) integer for which (1) is satisfied. First we consider the case of an even ,y. The numbers d'2 y and x2  z then have the greatest common divisor 2. Since the f
product of these numbers is divisible by 16, one of them is divisible just by 2 and the other by 8. Therefore the numbers (x.3 Y
± r) and n (r2 :
CHAPTER 1 11
228
are relatively prime, and since their product is a biquadrate, they must themselves be hiquadrates. Hence .1:2 ±
2 a4,
T2 + z = 8 b4,
where a and b are relatively prime natural numbers, such that 1 = 2 a b; a is odd. By eliminating z, we get the equation l b4.
2 = a4
which inuv be written
=4f4
(x ;
Since the two factors on the lefthand side have the greatest common divisor 2, it follows that .('+
((2=.)('4,
xa2=2(14.
where e and it are relatively prime natural numbers such that b = c(1 On subtraction we obtain the equation t4_ (a4 = ((2
Starting from a solution U. ?] of (1) with an even ij, we have thus deduced a new solution it, a]. But we have
<x, which is contrary to our hypothesis on Suppose neat that , is odd. Then , is even. Since the numbers e2  r;2 and x2 + (12 have the greatest common divisor 2, we deduce from (1) that Y _ :.' i/'=.
/.2  ,,2 = 2 1,2,
where a and b are relatively prime natural numbers. such that 2 z = a b. Hence .+''  (12
. 1,2,
((2  b=
and by multiplication (,
?/)2
= a4  h4.
DIOPHANTINE EQLTATIONS OF HIGHER DEGREE
229
Starting from a solution [x. g.:] of (1) with an odd rl, we have thus found a new solution [a, h, x e]. But we have a < 1/a2 t h2 =
which is contrary to our hypothesis on x. This proves the insolvability of (1) when z 11.
Our method of proof is an illustration of Fermat's famous method of infinite descent. In general, it may be characterized as follows: One assumes that any natural number x has a certain property E. This assumption implies the existence of some smaller natural number a'1 which also has the property E. But this leads to a. contradiction since there must be a smallest positive integer
which has the property E. Hence we conclude that the assumption is false for every natural number .r. This method has been used with ;Treat success by Feru at and later investigators for solving Diopllantine problems. It is, however, important to notice that it sometimes occurs that certain values of ,r. do not imply the existence of any xl less than x. In these cases the assumption may be true for a finite or perhaps infinite number of positive integers x. By an appropriate modification of the reasoning it is then often possible to determine all the positive integers x which have the property E. In the next section we shall give an example of this generalized method of infinite descent From Theorem 114 it follows at once that the equation (2)
`2
7.4
has no other solutions in relatively prime natural numbers than .r = g = y = 1. For after squaring it may be written 4

(.r' 111
4
_
(';.4 _ ti4)2
J
'1W'e use Ferlnat's method for l,roving the following three propositions: .Theorem, 175.
The I)iophautine equation
p has nil
in natural r,un,hrr.; .r. ,/ owl .
CHAPTER All
230
Theorem IN. The Diophantine equation
has no solutions in natural numbers x, !/ and 2. Theorem 117. The Diophantine equation IT4 (5)
 I/4 = p ?2,
where p is a prime = 3 (mod 8), has no solutions in natural numbers x, y and z. In equations (3), (4) and (5) Ave can obviously suppose that the
natural numbers x, y and z are relatively prime in lairs.
Proof of Theorem 115. In equation (3) the number z must clearly be odd; one of the numbers x and y is even, say 1/. Sup
pose that (3) has the solution
[x, 1/, :;],
where z has its least
positive value. Since the greatest common divisor of the factors z + x2 and z  x2 is 2, we obtain from (z + x2) (2  .1.2) = 32  ,7.4 = (f4
the s` stem ±:r2=2a4,
rlx2=8114, q=tab.
where a and b are relatively prime positive integers, and a is odd. Eliminating z we obtain ± .r2 = a4  4 b4,
where the upper sign must be chosen since the lower sign does not hold modulo 4. Since the greatest common divisor of the factors a2 4 x and a2  x is 2. we get from this equation the system
a2.r,= 2(14, b =ed,
a2 +,r. = 2e4.
where a and d are relatively prime positive integers. Hence, by eliminating r, a`
(4 d (14.
Thus the assumption that [x, //, :] is a solution of (3) leads to the existence of a new solution [e. d. a]. But we have z = a4 + 40 > (14 j a2.
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
231
Since this is contrary to our hypothesis on z, we have thus proved
that the sum of two biquadrates cannot be a square. Proof of Theorem 116. In equation (4) both .r and y must be odd; z is even. Since the greatest common divisor of the numbers x2 + r/2, r + y and x  y is 2, we deduce from (4) that
x2+ y2=2a2, x+ y=2t2, xy= 2c2,
(6)
where a, b and c are natural numbers relatively prime in pairs, and z = 2 a b c. From the last two equations in (G) we get
X=0 +C2, y=h2C2. Introducing these values of .c and g into the first equation in (J) we obtain [l2 = b4  '4.
But according to Theorem 115 this equation is not. solvable in natural numbers. From this we conclude that equation (4) has no integral solution for z 34 0 either. Proof of Theorem 11;'. Suppose first that z is odd. Then either
x or
even. In the first case we obtain from (5)  1 = p (mod 8). and in the second case I =1t (mod 8) ; but this is imr/
is
possible since p 3 (mod 8). Thus z must be even. where Now suppose that equation (5) has the solution [x, x has its least positive value. : is necessarily divisible by 4.
The greatest common divisor of x  +/..x + y and a:2 + ,2 is 2. We then have either the system
± r/=2rr2, xT g=4Jrr
(7)
x2+y2=2u2.
or the system
x ± y= 2pit2.
(8)
r T //=
.>2
.. y2='1 rr2,
where it, r and w are positive integers relatively prime in pairs; if
and w are odd. By elimination of x and y we get from (7)
(9)
It4 + l 1,2 t'4 =
and similarly from (8) (10)
102 u + 4 r4 = tr2.
CHAPTER VII
232
From the latter equation it follows that if, +ptl2= 2a4, rrpit 2= 2b4,
where a and b are relatively prime natural numbers. Hence
pll2=a4b4. This Diophantine equation is of the same type as (5). But here it is odd, and we have just shown that z must be even in (5). On the other hand, since is i 2 p I2 and it  2 p r2 have the greatest common divisor 1, it follows from (9) that lr + 2pr2=a4,
ur  2pe2=b4,
where a and b are relatively prime positive integers. Hence, on subtraction
a4b4=p (2r)2.
Thus the assumption that [x, //, _] is a. solution of (5) leads to the existence of a new solution [a, b, 2 r]. But we have a
a2 b2
which is contrary to our hypothesis on x. Hence Theorem 117 is proved.
Remark. In Theorem 117 the restriction on p to be = 3 (niod 8) is essential. For in each of the cases p = 1, 5 or 7 (niod 8) equation (5) may be solvable for certain values of the prime p. For instance, when p = 41, it has the solution [5, 4, 3]. When p = 5, it has the solution [3, 1, 4]. When p = 7, it has the solution [4, 3, 5].
Applying similar methods, Billing, Lind and other investigators
have shown the impossibility in positive integers x, y and z of a great number of Diophantine equations of the type a x, a b !14 = c z2,
where a, b and c are integers. 63. The Diophantine equation 2 x4  y4 = z1.  The equations investigated in the preceding section had no solutions in natural
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
233
numbers, except equation (2) which had only the solution x = y = z = 1 in relatively prime positive integers. There exist, how
ever, equations of a similar type which have infinitely many solutions in relatively prime integers. As an example of this category we choose the equation 2 x4 ry4=22.
(1)
We denote a solution of this equation in integers r. 1/ and z by [x, r/, z]. We consider only solutions with positive x. y and z. Such a solution is said to be a positive solution. We can obviously suppose that x, y and z are relatively prime in pairs. It is easily seen that they are all odd. Equation (1) may be written /+/2
2
z2 )

Applying equations (22) in Section 61, we obtain the following system, provided that r > 1 :
3:2=a2+b2,
r12 122) a1),
where a and h are relatively prince integers (I. There is no loss of generality in supposing a + h > U. It follows from the first equation of this system that a b is even. Then we conclude from its second equation that b must be even. From the first equation of the system we obtain in the same way as before .r=r2+ a2, a=r2r12, 7) =2rr1,
where c and el are relatively prime integers < U. Introducing these values of a and b into the expression for 112. we have 2 = r 4 t 4,.3 r1  (i c2 r1'  4 (.(13 +
and (C'2 + 2 a rl  (12)2  ,/2 = s; c2 (12.
2 c. r1  ,2 is positive,
Hence, since a I b = F
r'`
_' rd
r12 + 1/ = 2,l'2,
rrl r12r1/
4f)
r14
CHAPTER VII
234
where f and g are relatively prime integers By summation we get
0 such that fg = cd.
c2+2cdd2=J2+2g2.
(2)
Since fg = c d, we can put
f=5r,
c=sir, d=>1r,
where , ,, it and t' are integers 34 0. Introducing these values into equation (2) we obtain $2(1ter2)+2 211rv=121.2+2112u (3)
and thus
Yluy
(4)
V2914V4 lr 2
The (5) (6)
+ 1'
numbers x and r, are expressed by the formulae X=`2+ (12=$2112 + 2121.2,
±/
f2g22t'2:.0)22.
Consequently, starting from a positive solution [x, rI, ] of (1), we have by this procedure deduced a new positive solution [ 11r I, I r 11 1 200] 'v4] of (1), where
1x1=11YX n21'x <x. But we had supposed x > 1; for .T = 1 the reasoning is not valid.
Conversely, starting from a positive solution [u, v, w] with u > 1, we can by means of formulae (4), (5) and (6) deduce two new, positive solutions; and 27 are determined by (4) as relatively prime integers. For it = r = 1 we get from (4) either
n= O or t = 3;9 since only the latter value is applicable, we obtain 77 = 2 and s = 3 and thus x = 13, y = I and = 239. Starting from the solution [13, 1, 239] we get by (4) 6
3351
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
235
whence the two sets of values: 1=  3, ill = 2 and $2 = 113, 212 = 84. This gives two new positive solutions of (1), of which the least is [1525, 1343, 2750257]. By repeating this procedure we obtain in succession all positive solutions of (1). Obviously the method applied here can be regarded as a generalization of the Fermatian method of infinite descent. It was first discovered by Lagrange that Fermat's method can also be used for solving completely Diophantine equations which have an infinity of solutions. Mordell has shown that the Diophantine equations of the type (7)
f (x. i/) = a .G4 + b x3 7f + C .tie !)2 + d.7!/' + e y4 = k 22,
where the coefficients are integers such that the equationf(r.,1)=0
has no multiple root, can be solved by the same method. If equation (7) is solvable in integers r, y and z, there exist a finite number of initial solutiow with the following property. Starting from these solutions the total set of solutions can be found by means of a system of formulae analogous to (1), (5) and (li). In our example the initial solution was [1, 1, 1]. But equations are known which have more than one initial solution. (Compare Section 68.)

64. The quadratic fields K (Y=1), K (1/ 2) and K (Y3). It was shown in Section 6 that the quadratic field K(VD) consists of all numbers of the form a = a + b 1 D, where a and b are (ordinary) rational numbers. Since D is rational, a and b are uniquely determined when a is given. The number a' = a  bl'D is the Conjugate of a. _ By the i,orm N (a) of a = a + b 1''D we understand the product
of a and a'; thus (a) = a a' = (a + b l I)) (a  b 1 D) = a2  D b2.
If we put f = e + rl I'D with rational c and d, we obtain from the identity (a2  I) b2) (c2  f) (/2) = (a c + b d D)2  D (a d + b c)2
the following rule for the norms: N (a(3) = N (a) x"(13)
CHAPTER VII
236
In this section number means a cumber in the field K (co), where
w denotes one of the numbers 1/ 1, V'2 or ( l +
The number a + 1) co is said to be an integer in K (co) if a and h are ordinary inte(rers. in the following, intecr means an integer in K(w). The ordinary integers are called rational integers. The norm of a= a + G co is equal to a2 + G2 in K (1  1), equal to a2 + 2 L2 in K (Y 2) and equal to I 2+3 =u2 ct/+b2 a1,'
G2
in K (l ( I + l/3)) = K (1"). It follows From this that 1"(a) is always positive in K (co) except when a= 0.
If a is an integer, N (a) is a natural number. It is clear that only a finite number of integers can have the same norm. There
are positive rational integers n such that no integer has the norm a. Thus the number 23 cannot be the norm of any integer in any of the fields K (w).
An integer with the norm I is called a u;cit. Recalling what was said above about the norms, we easily prove: In the field
and +i =±T/I. In K (1/i) there are the four units the field K(V2) there are only tho two units = 1. In1/_the field K (1" 3) there are the six units + 1, + _ ± ( 1 + 3) 1
and ± t,2 = + If . 71 and x are three integers in K((,j) such that (1)
(77
0)
we say that t1 is a divisor of y, or that y is divisible by 1, or that r1 divides $. Each integer E (O) has only a finite number of divisors; for, by (1), we have Y (5) =
(71) V (x)
Thus the natural number _V(,1) is a divisor of the natural number Consequently, there is only a finite number of possible values for V(71) and therefore also for 71. Every integer (different from zero) is a divisor of its norm. A unit may also be defined as a divisor of the nnrnber 1, since W = I when a is a unit.
DIOPHANTTNE EQT ATTONS OF HIGHER DEGREE
237
Two integers are said to be associated with each other when their quotient is a unit. The units and the integers associated with a given integer a are called the trivial divisors of a. All the divisors of a unit are themselves units. An integer which has only trivial divisors is called a prime when it is not a unit. An integer ;, whose norm is a rational prime p, is an irrational prime. For it follows from ; _ q that X (s) = 1 (Ij) \'(,') =p.
Therefore, one of the numbers Ni) (?and \'(;) is equal to p and the other to I ; hence one of the uurubers 71 ands is a unit. Consequently all the divisors of ; are trivial, and E must be prime. In the field K (I  1). 1 + 21  1 isa prime; in K (V J),
a
2 + 1 3 is a prime; in .+11 2), 1 + l 2 is a prime. Lemma 1. If s aj,,l ij are )aro integers, and ij integers % mul A :ueh that tp !
Proof'.
Put
h c).
aI
7j
rrith
0. there exi.4 taro
i 2 < 1' (rj).
where a and b are rational numbers.
Then there exist two rational integers .c. and q such that
Ia ,t I
(2)
Putting
,
IbyI C T
.r+ qce=x and
y rji=i,
we have Y) = tj
Cb  g Col.
17
If ro = 1  1
or l'
X (2) = `; (,j) . [(a
If 0) = ( I \r (A) 
V()j)
it follows from (2) that ae)2  rot (1i  J)2] < \' (71) (] + i) < 1 (Tj)
3 ), it follows that
[(ca  ;i)2  (a x) (G  y)
F (h  at)2],
and from (2) N(A) = 1'
+ a + .'1) < V (,j).
CHAPTER VII
238
This proves the lemma. We next prove Lemma 2. Every prince z which dirides the product 'q of the two integers and 21 is a divisor of either or 71.
Proof. Suppose that is not divisible by n. Then we shall prove that i is divisible by z. In the set of numbers (3)
a and fi are arbitrary integers in K (co), there exist certain numbers ,E 0 which have the least possible norm, say N. Then N < 1(n). For according to Lemma 1 we can choose the integer 9 such that
1('+#a)< N(70+ P n 34 0, since it not divisible by an. Let y = a$ + fl n be an integer such that ` (y) =N. Then, according to Lemma 1, there exist two integers x and ). such that
Here
n=xy+A with N (A) < N (y). Further we have
A=nmy
pn)=(xa) + (1
The number A therefore belongs to the set of numbers (3), and its norm is < \'(y) = `T. By the definition of iV this is possible
only if A= 0. Thus we have n = my. Here y cannot be associated with n since \' (y) = X< \'(n). Hence y is a unit. Multiplying the equation
a$+(in=y by the integer 27y1 we obtain
ae7yyl+Pn)lyI?/. Since n is a divisor of e27, the lefthand side of this equation is divisible by on, and therefore also the righthand side = tl is divisible by or.
An immediate consequence of Lemma 2 is
Q. F. D.
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
239
Theorem 11 ti. In the fields K (V 1), K (V  `?) and K (1/ 3) every
integer whose norw is > 1 can he uuiiguely e.cpressed as the product of a fib,ite Homelier of primes, apart from the order of prime facturs, and prowled that associated primes are not considered as distinct.
The proof is analogous to the proof of Theorem 4 in Section 4. We first prove that every integer with a norm > 1 has at least one prime factor. Among the divisors of which are not units there are certain divisors whose norms are the least possible. Let tl denote one of these divisors. Then, apart from the units, no divisor of has a norm < N(?J). Now suppose
where
if
and i are integers which are also divisors of . Hence,
is not a unit, we must have 7V (y)
N(21). But since
N (zl) = X (S) X (y),
\' (21), and i is a unit. we also have N(5) = \' (71). Thus \' According to the definition of a prune, 21 is then a prime. Consequently 1: has at least one divisor which is a prime. By induction it is now easy to show that can be written
as a product = a, ?2 ... 7t,.
of a finite number of primes ire. Finally, the uniqueness of this decomposition follows by the
same argument which was used for the rational field K(1) in Section 4. Combining this result with Theorem 100 (first three parts) we obtain Theorem 119. 1. In the field K(V1) there are the following
primes: The number I + 111 and its associates: the rational primes = 3 (mod 4) and their associates: the integers x + yV1 whose norms are prinzes  1 (mod 4). . In the field K(Y2) there are the following prunes: The numherx ± 1` 2: the numbers ± p where p is a rational prime = 5 or
CHAPTER 1II
240
,'.
7 (mod 8); the integers x + rl 1' 2 whose norms are primes I or = 3 (mod 8). In the field K(1r3) there are the following princes: The num
ber V3 and its associates: the rational prinrcx and their a,q.goeiatr,: the infrgers a' + lit 3 primes = 1 (mod 3).
(mod 3)
nouns are
It easily follows from Lemma 1 that if a and /3 are two integers, not both = 0, they have certain common divisors 6 with the following properties: The numbers a and have no common prime divisor; the numbers 6 are associated with each other; all the common divisors of a and (3 are divisors of 6 These numbers 6 are said to be the greatest co?nmon dirisnrs of a and P. If the numbers 8 consist of units only, we say that a and (3 are relatirelry prince and write (a, P) = 1. The integers al, a_, .... an, are prince in pairs when (ai, a;) = I for all distinct
i and j. The concepts of congruent nu,nbers, congruence, residue,e and residue classes can be defined in the same way as in the rational number theory. If the difference a  P is divisible by 5, we say
that a is congruent to # modulo $ and write a=fl (mod e).
The numbers a and j3 are then said to belong to the same residue
class modulo . Finally we prove some lemmata on numbers in the field K(1"3).
As above we put e = ( 1 The number 7. = 1  e is a prime, since its norm is 3. It is associated with the prime 1' 3. The integers in K (1  3) fall into three residue classes modulo A
which may be represented by the numbers 0, 1 and  1. In fact
et +bo=a+b(1I)=a+b (mod A), and the number a + b has one of the residues 0, 1 or  1 modulo 3.
241
DIOPHANTINE EQL?4,TIONS OF HIGHER DEGREE
Lemma :1.
If a is an integer in K(1'3) which is not divisible
by A, we have
a3 = ± 1 (mod ),4).
Proof. Since a is not divisible by A, we have a = ± 1 (mod ).),
and therefore
a3=(±1+Af)3±1+32 4322fl +A3f3, where P is an integer. Hence a3 + 1 = A3 (as
e2 N)
A3 (j3  j) (111od A41.
But it is easily seen that the number f3  p is always divisible by A regardless of the residue of (3 modulo A. Thus a3 is Q. E. D.
divisible by A4. Lemma .1.
If E is a unit, and if ; is an integer in K(13), the
congruence
3 = e (mod 2)
holds onli/ for E = ± 1.
For 2 is a prime, and the residue classes modulo 2 are obviously represented by the four numbers 0, 1, a and e2. A complete theory of the field K (V1) was given by Gauss; it is therefore called the Gaussian field. The field K(Y3) was treated by Jacobi and Eisenstein. 65. The Diophantine equation ;,3 + I13 + 53 = 0 and analogous
equations.  Suppose that , rJ and 5 are integers in K (V  3), different from zero, which satisfy the equation
3+73+;3=0.
(1)
A prime which divides both 5 and n must, by Lemma 2 in Sec
tion 64, be a divisor of 5. Thus we can suppose
rl)
=(il.;)=1.
If none of the numbers ;, 21 and 5 is divisible by A = I we derive from (1) by Lemma 3 in Section 64 that
±1±1±1=0 (mod;.'). 16  516670 Trygre X aycrr
CHAPTER VII
242
But clearly this congruence is impossible for all eight combinations of the signs. Thus we can suppose that one of the integers and 5 , say 5 . is divisible by A. Putting
= where n
ny,
1, and where y is not divisible by A, equation
(1)
becomes y3 l
113
I
)111, 3=0.
We consider, however. the More general equation t3 + 773 + E A3' y3 = Q,
(2)
where E is an arbitrary unit in K (V3). and where (E, 27) y) A) = (11, A) _ (y, A) = 1. Further we suppose that the = (21, y) = exponent it has the least positive value such that equation (2) has integral solutions. Then n 2. For, applying Lemma 3 in Section 64, it follows from (2) that ± I ± I ± E A37, = 0 (mod A4),
whence
A3 = 0 (mod )4)
and therefore nt 2. The three numbers
a1=e+ )], a2=5+p'1, are all divisible by A. For according to (2) the product al a2 a3 is divisible by A, and further al = a2 = a3 (mod A). The three integers
P1 = a' P2 j' are relatively prime in pairs. In fact, if we suppose that #I and #2 are divisible by the same prime :c, the numbers #1 #2 = 11 and 0  #1 + /32 = & would also be divisible by z: but this is contrary to hypothesis. Thus we have (j31. I32` _ 1: in the same way it may be shown that ((31, a3) = (,l2, Al) = 1.
DIOPIIANTINE EQUATIONS OF HIGHER DEGREE
243
From (2) it is obvious that one of the numbers /91, 192, 93 must be divisible by A3113. There is no loss of generality in supposing
that this number is Pl. For equation (2) does not change if y is replaced by y e or y e2.
If we write equation (2) in the form 1F'2N3=E;3n3(_',)3
it follows that C32  F2 7/31
R3 ,
= E3
3
,
where £1i E2 and E3 are units, and where yI, 71, and 51 are in
tegers relatively prime in pairs which are not divisible by A. Now we have and thus e2 e3 51  e E2 )1 + El A3 i, 3 13
= U.
L
Hence
F E,1 7l1 i e5 A3 a 3 y1 3 = U,
(3) 1
where e4 and £s are units. Since n ? 2, and since the product 5171i is not divisible by A, it follows from this equation that ± I ± E4 = 0 (mod A3).
Hence E4 = T 1. Then equation (3) can be written = 0. 63 + (+ 711)3 + e5 A3 I I Ss 1
(4)
1
This equation is of the same type as (2), and it is solvable if (2) is solvable. But since the exponent of A is less in equation (4) than in equation (2), this is a contradiction of our hypothesis on the number n. Consequently we conclude that equation (2) is not solvable and state Theorem 120. The Diophantine eq anion (1) has no solutions in ilitegcrs $, 7l and 5 in the field K(V3) if X77; = 0.
By a simple modification of the proof of this theorem we easily establish
244
CHAPTER V11
Thro'enn 121. The Diophantine equation S3
 9]3 + 3 y3 = 0
has no golutiwis iii integers
:
71 and 5 in the field K(1'3)
0.
Proof Evidently it is sufficient to show that there is no solution of the equation y3 + 1] 3
(5)
where e is an arbitrary unit in K (N ), and where
il)
y)
We suppose that has the least positive value such that equation (5) has integral solutions. Applying Lemma 3 in the preceding section we see that n > 1. By the same argument as in the proof of Theorem 120 it follows from (5) that 11
r y]
=£
7
3 '13 ?41 S1 .

2 1]
LtLo 11  E2 1/7 3 A
E3  £3 51
Finally we get the equation (+ 9]i I? + £5 AU1
7
i = 0,
which is of the same type as (5) and is solvable if (5) is solvable.
But this is contrary to our hypothesis on the number n. Hence (5) cannot be solvable. Theorem 122. The Diuplul,dine equation 3
y3 { 1]3
(t;)
has no volution8 in integer:y y e 3 = 7]3 = S3
if
r
1] and ; in K (1  3) other than
0.
Proof. We consider the more general equation (7)
&3
1]3 t:1 ;3=11,
where r: is an arbitrary unit in Kif equation (7), we say that norm N($??,') of the product
e]
and e satisfy
is a solution of (7). The
1] 4 is said to be the height of the
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
245
solution [5, yj, 5, e]. The height is a natural number. We obtain all the solutions with height 1 when t3 = 2)3 = + 1, 3 = + e1= + 1.
Now suppose that [$. 12, 5, e] is a solution which has the least possible height > 1. Then the numbers E. )) and 5 must be relatively prime in pairs. Let us denote by a, fl and y the numbers 2
2
in this or some other order. Then
a r)+Y=ll, al;y=2ev.
(8)
If S is a divisor of a and l1 such that
fa
n')
P) ,
t
= 1, it is clear that
= 1. Then it follows from the last equation in
6) (8) that
(d.
(9)
J = FI :3
1j
F" )jl'
Y
_ .1 e3
where el, FQ and e3 are units. and where $1, qI and SI are integers
relatively prime in pairs. Adding the first three of these equations and recalling that a + + y = 0. we obtain after multiplication by el I (10)
ti rF))
F3 ,
j
=ll,
where s4 and es are units. If we consider this equation modulo :3 and apply Lemma 4 in Section 64, we see that e4 =  1. Thus we have deduced a new solution [;1 )I. 51 e5] of equation (i).
This solution has height N( I nI .i) = N(). Thus according to our hypothesis we must have either N(81 _ \ I51)4)
(11)
or (12)
,'I a)=
Since 6 divides a  E? fi and a  ?y it also divides (1  t,) and ))) = 1, 6 is either a unit or associated
(1  e) )). Hence, since
with A=1n.
CHAPTER VII
246
if relation (12) holds, ; is associated with 6. From (7) and from Lemma 3 in Section 64 it is evident that. is divisible by ?,z if it is divisible by 2. Thus 6 ands must be units. It then follows from (9) that a and fi are units. Hence A5 and A71 are both the sum of two units. It is easy to see that this is possible only when and i1 are themselves units. Hence we should have \'( g i_') = 1 contrary to our hypothesis. From (11) we obtain \ (y ,1 b) 1, and thus the numbers , i1 and 6 are units. From (7) it then follows that 5 is also a unit. Hence we should have N(e it ) = 1 contrary to our hypothesis. In particular, Theorems 120, 121 and 122 also hold in the rational field K(1). Theorem 120 was stated for rational numbers by Fermat (see Section 68); the proof was given by Euler. Theorems 121 and 122 were proved by Legendre for rational numbers. 66. Diophantine equations of the third degree with an infinity of solutions.  To Theorems 120, 121 and 122 in the preceding section we add the following result: Theorem 128. If a is an integer > 2 which is not divisible by the cube of any prime, the ILioj7hantine equation (1)
.1.3
} y3 == a z3
has either no solution or infinitely many solutions in relatively prime integers y and y, with z 74 0.
Proof. Integer means here ordinary rational integer. As usual, denote by [x, g, z] a solution of (1). As in the similar cases in Section 65 we can suppose ((, y) z) z) = 1. From the solution [x, r/, z] we obtain a second solution [x1, y1i of (1) by the formulae 3 + 2 g31,
(2) (3)
6i/1 y (2 .i + )13),
(4)
b
1 =
A.
where b denotes the greatest common divisor of the three numbers on the righthand side. This is easily verified by introducing
DIOPIIANTINE EQUATIONS OF HIGHER DEGREE
247
the values of x1, y1 and zl given by (2); (3) and (4) in equation (1).
The geometrical interpretation of this is as follows: If we draw the tangent to the curve (1) at the point P(x, y, z) (homogeneous coordinates), PI (x1, y1, z1) is the point in which the tangent cuts
the curve again. P is called the tangential point of P. PI coincides with P only when P is a point of inflection. We have .c: i/; for otherwise x = g = 1 and a = 2 contrary to hypothesis. Therefore it follows from (4) that z1 74 0. From the equation .7.1 + I/i=a2i we see that (r1, 1/1) =
z1) _ (?/1, z1) = 1.
Let q be a prime factor of 6. If .r is divisible by q, it follows from (3) that +f4 is divisible by q; this is impossible, since (x, i/) = 1. Similarly we show that neither t/ nor z is divisible by q. Thus (6, x) _ (6, y) = (6, ) = 1, and the number 6 is a divisor of the three numbers .13 + 2 i/3,
2.r3 + rya,
.t'3  J3.
6 is therefore a divisor of 3x3. Hence we have either 6 = 1 or
6=3.
To prove Theorem 123 it is obviously sufficient to prove that (5)
ItI
Since y
dlz(x3_y3)I.
0 and x 74 !/, we have
.?'2+x!+!/2=4(_.r+ r/)2 + q> and since x3  93 is divisible by 6, so is ,r  y; thus I x  J I ? 6 and
I.r,3  !/3I > b.
From this it follows that I,:1 I > I I, and Theorem 123 is proved.
Each of the equations c'3 r q3 = li z3. C3 + 7/3 = 7 r3,
,c'3 + y3 = 9 23
248
CHAPTER V11
has infinitely many solutions. For the first one has the solution [17, 37, 21], the second one has the solution [2, 1, 1] and the third one has the solution [2, 1. 1]. More generally, the equation
x3+y3=(c3+ 1),3 has an infinity of solutions in relatively prime integers if c is an integer P4 0,
 I and
1. In fact, this equation has the
solution x = c, y = 1, z = 1. Further, the equation c3 + I = 2 b3
has no solutions in integers c and b, if c  I and 1 (Theorem 122), and finally the equation C3 + I = b3 has no solutions in integers c and b if c 0 and  I (Theorem 120).
67. The Diophantine equation x7 + y7 + z7 = 0.  When x, y and z denote the roots of the cubic equation in X, X3p12+q1pq+r=0,
(1)
we find by means of Newton's formulae that (2)
x7+y7+z7=.11717'(1)4J)2q+ q2)+7p.2.
We shall see whether we can have
.r7 + t/7 + z =0
(3)
when p, q and r are rational numbers. If x, y and z satisfy equation (3), it follows from (2) that the coefficients p, q and r
must satisfy the following equation: (4)
1)771'(14 1,2q 1 q2) + 7p7.2=0.
First we suppose p = 0. If we put q = 1,2 qI and r =13 rl , it follows from (4) that
7)2  7r, (I qI  qi) + 1 =0, whence
.(1q1+ql)2;. 7'I=t(1qI+111)+ Here the square root must be a rational number. Hence, by putting 2 qI  I = t, where s and t are relatively prime (rational)
integers and t > 0, we have
DIOPHA\TINE EQUATIONS OF HIGHER DEGREE
.44 + 6)2 t2  I t4
(5)
249
= 112,
where if is a rational number. Since 7 u2 is an integer, it must also be an integer, and therefore t is divisible by 7. We distinguish two cases, according as t is odd or even. Case 1. t is odd. From (5) it follows that either s or 11 must be even. Equation (5) may be written in the form (s2
4
3 IZ)2  112 = f'_1 t4.
and by the same argument as in Sections 6366 we conclude that s2 + 3 12  11
(6)
a a}: 64,
.312
where a and b are odd natural numbers such that a t = t, (a, t) = 1
and a ° 0 (mod 7); a and fi are even natural numbers such that a#= 64. From (6) we have by addition (7) 7
This equation implies the congruence (8)
3a
.Y2
If we had a = 32 and
0 (mod 8).
2, we would have s2 =  2 (mod 8)
which is impossible. If we had a = 16 and 0 = 4, we would have s2 =  1 (mod 8) which is also impossible. For a = /3 = 8 we would have the impossible congruence s2 =  3 (mod 8). For a = 4 and /3 = 16 we would have the impossible congruence
s2 = 3 (mod 8). Thus we must have a = 2 and 0 = 32 which implies s2 = 4 (mod 8). Equation (7) may be written in the form
s2 =  3 a2 b2 + ' a4 + 16b 4 or
Hence
642(32b23a2)2= ; a4. ± 8s + (32 b23(12) = ±
± 8s(32b2$a2)= ±
C4,
d4,
250
CHAPTER VII
where c and d are odd natural numbers, and d is divisible by 7. But these equations do not hold modulo 8. Case 2.
If t is even, both s and tt are odd in equation (5).
Writing (5) in the form (S2 + 3t 2)2  td2 = 4.1 t4 = 1 (2 t)4
we obtain
52+3t2 ±tt=a4, ,.2+312+it =2b4,
where a and b are natural numbers such that a b = 0 t, (a, b) = 1 and a = 0 (mod 7). Then we get by addition (9)
s2= a a2 b2+ b4+
(t4.
Either a or b is even. If b were even, we would have (,)2 s2
=3
1 (mod 8),
whence, since s is odd, 3
2
3
(mod 8),
which is impossible. Hence a is even. Putting a = 2 al we get by (9)
3 ai + 1 (mod 8),
1 = s2
Thus al is divisible by 4. It follows from (9) that ,2(b29a2)2=1 0
and, by the usual argument,
±s+(b2va2)=+2e4,
{ s(b2'al)=±
d4.
where c and d are natural numbers such that aI = 2 e d, (c, d) =1 and d = 0 (mod 7). By subtraction we have b2=GC2d2+(C4I d4).
DIOPHANTINE EQUATIONS OF IIIGIIER DEGREE
251
Since b2 + c4 is not divisible by 7, we must take the upper sign. Thus the equation is (4+Gc2(12td4=b2.
(10)
This equation is of the same form as (5). But we have
t=alb=2hcd>d. Thus by the method of infinite descent we have shown that the Diophantine equation (5) has no solution in integers s, t and is when t 74 0. So we have proved that relation (3) does not hold for l) 34 0.
Consider finally the case p = 0. Then r + I/ + z = 0, and equation (3) takes the form x7
+ g7
(x+y)7=0.
Now we have the identity (.z + 7t)7  X.7  y7 = 7 x yt (.r + i) (x2 + x ry + y2)2.
Hence we conclude: Relation (3) holds if and only if one of the following four conditions is satisfied: x = 0; y = 0; x + y = 0; x2 + 'r y + !t2 = 0. Thus we have established the result: Theorem 124. If x, y and z are the roots of a cubic equation with rational corfpcients, the relation .i17 + y7 +27=0
holds only in the following cases: 1. one of the numbers x, y, z
is equal to zero, 2. the numbers x, y and z are proportional to the rooty of the equation ps  I = 0 taken in a convenient order.
This theorem, which was stated and proved by V. A Lebesgue,
is, of course, also true in the special case when x, y and z are rational numbers.
68. Fermat's last theorem.  The Diophantine equation (1)
x"+y"=on
252
CHAPTER VII
where n is an integer > 2, has acquired great celebrity, thanks to the famous statement made by Fermat. In the margin of his copy of Bachet's edition of Diophantos's lritlnactica he asserts without proof that equation (1) cannot be satisfied by any integers all different from zero. To quote his actual words: C'uburn autemn in duos culios, aut quadratoquadrat:vn in duos gnadratoquadrato.', polestatrm in duos et grnerali.ter nullant in infinitum ultra ejusdenr noininis fax est (1iridere, cujars rei
inirabilem
sane dctexi. Hanc inarginis exiguitas non caprrct. He thus believed that he had a really remarkable proof of the theorem ; but he never communicated his supposed proof. Up to
now Fermat's "last theorem" has only been proved for special values of the exponent n. In the case n = 4 Fermat really proved his statement; lie had only to replace zs by i in Theorem 114. Apart from the case n = 4 we can evidently suppose that the exponent n in (1) is an odd prime 1). It follows from Theorem 118 that Fermat's assertion is true for p = 3. Fermat's theorem for p = 7 is clearly a special case of Theorem 124. About 1825 Legendre and Dirichlet proved the theorem for p = 5; their proofs are based on the method of infinite descent.
Kummer was the first to succeed in proving Fermat's last theorem for p = 11, 13 and for certain other larger prime exponents. In his investigations of the equation ,rn + jj" = zr',
p being an odd prime, he applied the theory of the algebraic field K(77), where 77 is a primitive pth root of unity. In the case
p = 3 in Section 65 we have already used this mnethod. But, when p > 7, the factorization of integers into prime factors in the field K(,7) is not unique. This fact gave rise to great difficulties when applying the theory to Fermat's equation. But Kummer overcame them by creating the concept of ideals. The theory of ideals, which has turned out to be of great importance for the development of several parts of mathematics, will, however, not be treated in this volume. We shall only give the main result discovered by Kummer (1850):
253
DIOPHA7',TINE EQUATIONS OF HIGHER DEGREE
if 1% Is a J' tm nrrian prince, the Diophantine equation ai' + (ix' = yr
(2)
is not solvable in intec/ris a, (3 and y in K (77), except for aly=0.
An odd prime p is called Kini merian (or also regular), if it does not divide any of the numerators of the firsts (p  3) Bernoulli numbers:
B1= ic
B  au
B3  '
B 4 =  'u. ' Bo =
W.,
B6
B7
, etC. I
The integers in K(27) are the numbers of the form cle + al 1)  a2 ilz +
h
a, (0 < i _ p  2) are rational integers. Kumnier's proof proceeds analogously to the proof given in Section 65 in the case j, = 3. The first Dart of the proof consists in showing that one of the numbers a, (3, y in (2) is divisible by I  21. If y is divisible by 1  ht, the next step is to write the lefthand side of (2) in the form (a +#)(a + ?1#)((t+ ill #) ... (a + Yfi'1 /3).
Also in Kummer's proof the application of the method of infinite descent is an essential feature. If we write down the first (97  3) = 47 Bernoulli numbers, we see that all primes < 100 are Kummerian except 37, 59 and 67. Kuinmer showed, however, in a special investigation, that his theorem is also true for these primes and for certain other nonKummerian primes. The investigation of equation (2) has been continued along the
lines suggested by Kummer, and up to now Fermat's last theorem has been proved for all exponents at least up to ii = 600. But it has not been possible to show that there exist an infinity of Kuinniesian primes. On the other hand we know that there are nonKumnlerian primes larger than any given number. 69. Rational points on plane algebraic curves. Mordell's theo
rem.  Let K be any given field. In this section any number ' For the definition or the Bernoulli numbers see any introduction to analysis.
CHAPTER VII
254
belonging to K is said to be a rational artuaber. The point y, z) in homogeneous coordinates in the plane is said to be a rational point in K if .r., y and z are proportional to three rational num
bers. In Section 60 we considered the rational points on plane
curves of the second degree. Here we shall establish a few results about the rational points on plane curves of the third degree, called simply cubic curves.' Let the equation of an algebraic curve in homogeneous coordinates x, p, z be F(,r, y. z) = 0,
where the lefthand side is a homogeneous polynomial in
p
and z with rational coefficients. From the homogeneous equation F (x, y, 0) = 0 we obtain the rational points on the curve which
are at an infinite distance from the origin. If the curve is of the nth degree (or order), the polynomial is of the same degree.
The singular points of the curve may be obtained by the system of equations
0F _ 0,O1/=0,r7z=0. all, oil, 8
There are two different categories of cubic curves: 1. The cubic curves of genus zero or unicursal curves, i.e., the curves which have a singular point. 2. The cubic curves of genus one. i.e., the curves which have no singular point. The following result holds for unicursa] cubic curves: Theorem 125. The singular point on a turicursal ctdhie curve with rational coP/iciehts is a rational point.
Proof. Let the equation of the cubic curve in homogeneous coordinates be T'(x, y, z) = 0, where F is a ternary cubic form in x, y and z. There is no loss of generality in supposing that the singular point (xo, go, zo) is at a finite distance from the origin, thus zo = 0. We cut the cubic curve F= 0 by the conic (polar curve) 1 In this and the following section some elementary knowledge of the theory of plane algebraic curves is supposed.
DIOPHANTINE EQUATIONS OF RICHER DEGREE
2555
I'+cI'=0, J where the parameters a, h and c are rational numbers. Two of the six points of intersection coincide with the singular point. It is obviously possible to choose the numbers a, b and c so that the other four points of intersection are distinct from each other and from the singular point. The coordinates 'r of the six points z
of intersection are the roots of the equation of sixth degree with rational coefficients
which has the double root
'°.
z°
Then the greatest common divisor
of the two polynomials with rational coefficients f (u) and J '(u) is v  z°, which must also have rational coefficients. Consequently, 0
the number r° is rational, and similarly we can prove that '0 z0 to is also rational. Thus Theorem 125 is proved. Every rational straight line through the singular point cuts the curve in a third point which is necessarily rational. In this way we obviously obtain all the rational points on the curve. There are infinitely many of them. Theorem 126. If a cubic of genus one with rational coefficients has a rational point, by nremis of a transtbrnration with rational coejWeients it can be trau.formed into a cubic eurre of which the equation is I'z = X s 'I X  B. where A and B are rational numbers.

Proof: Let f (x, y) = 0 be the equation of the curve in nonhomogeneous coordinates, and let PO be a rational point on the curve. There is no loss of generality in supposing that Po is at
a finite distance from the origin. The tangent to the curve at P0 cuts the curve in a third point Pl which is obviously also
CHAPTER VII
256
110 is a point of inflection, P1 and PO coincide.) By taking P1 as origin and the tangent at Po as the t/axis, we define a linear transformation with rational coefficients which transforms the equation of the curve into the form
rational. (If
(1)
T3 (.c, !I)
I
q'2 (', J) + 971 (r.
y) = 0.
where T; (.r, y) denotes a homogeneous polynomial in .r and y of
degree i with rational coefficients. Then, cutting the curve (1) by the straight line p = lx.. the number .x°, apart from the value :r = 0. is determined by the equation of second degree .c2 q`3 ( 1 .
() + ' 92(1, t) + q`1(1, t) = 0.
From this we obtain the following parametric representation of the curve (2)
92(1, t) ±11R(t(t) 9q'3(1, t)
tX.
where 11(t) denotes the polynomial q.(1, t)"49=111, t)
(1, 1).
This polynomial has rational coefficients and is of the third degree. For its zeros are obviously the slopes of the tangents to
the curve which pass through the origin, except the tangent .r = 0 at the point Pa. Now, if we put a t = 1  .',. L in the expression
R(t)=a13+1) 12+et+ (1, where a. b. c and d are rational numbers, we iret
x211(1)= T3_11111 with rational A and B. By means of relations (2) we then arrive
at the following relations between the coordinates X, Y of a point on the curve (3)
12=:3.1 \ B
and the coordinates c, i, of a point on the curve (1) :
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
.1._ (4)
257
a4'2(a,Xsb)a2I' 2T3(a.XS b)
y=al(ib)
and ,L2 (5)
ay
The system (5) represents a socalled biratiunal traw'formation, which transforms the curve (1) into the curve (3): the system (4) transforms the curve (3) into the curve (1) and represents the birational transformation which is inverse to (5). These transformations define a oneone correspondence between the points of the curves. Since the transformations have rational coefficients, each rational point on one of the curves corresponds to a rational point on the other curve; the rational points at infinity are also, of course, taken into consideration. Thus Theorem 126 is proved. The theorems proved in Sections 62 and 63 may be interpreted as results concerning the distribution of rational points in K(1) on certain algebraic curves of the type (6)
a
,4 + b 13 + (. X2 + d 1 + P. = k 1'2.
If [.r, y, z] is a solution in integers of the Diopliantine equation ax4 + bv3t/+ cx2j/2 + 11x?/3+ er/4=kz2, (7)
the point X = ' , I" = is a rational point on the curve (6). Conversely, if (1, I) is a rational point on (6), and if S is a natural number such that X X and X Y are integers, then x = N X, y = AT, z = 'V2 I' is an integral solution of (7). Theorems 114, 116 and 117 may be formulated in the following way : On the curves
s`141=k12, where k is either I or 2 or a prime  3 (mod 8), there are, apart from the points (1, 0) and ( 1, 0), no rational points at a finite distance. 17  616670 Trpgre 11 agell
CHAPTER VII
258
It is evident from the result in Section 63 that we obtain all the rational points in K (1) on the curve
2X41=I.2 by means of the recursive formula
±1=
(a)
Ii11 + 1)2 + (X 1 ±
)'1)2
(2 X + 1)2  . , ] (11 7 1'1)2
when we start with X1 = ]"1 = 1.
In a similar way the theorems in Sections 65 and 66 may be interpreted as results concerning the rational points in K (J 3) on certain cubic curves of the type
,3
3,3 .
= (V.
The curve
l'2=,3_11B
(9)
is of genus one if and only if 1)=4 _1327B2,F 0.
The cubic curve (9) has the following parametric representation by means of elliptic functions:' X
Y=v
(a; 1.1. 4B),
where 4A and 4B are the invariants. Every point on the curve (9) corresponds to a certain argument it which is uniquely determined apart from addition of periods; for the sake of brevity we then speak of "the point tt." The point It= 0 is the point of inflection at infinity; it has the homogeneous coordinates 0. 1, 0 and is therefore a rational point. If tt1, 112, . ., it., are rational points on the curve, it is easily seen by means of the addition theorems for the functions V(i) .
and Sd' (it) that the point (10)
7t11 tt1
+ ttr2 t12 +
1 See for example Goursat, Coors d'.1,
+ ))I" its
tome II.
DIOPHANTINE EQIATIONS OF HIGHER DEGREE
259
is also rational when nil, i112i .... 1)1 are rational integers. In 1922 MIordell proved the following theorem on the rational points
in K (1) on the curve (9) when A and B are ordinary rational numbers and L 0: There are on the eurre (9) a finite number of rational points nl, .. if, such that the formula
1(2i . .
))111(1 + 1)?2 1(2 + ... 1 Mr n,.
(11)
lli2,
.
.
the whole set of rational point, on the c,()re when x)11. ., 11r 0a01 through all rational integer;.
Well (1930) has shown that this result is also valid in any field K (t,) where $ is an algebraic number.
The least possible value of the number r is called the rank
of the curve in the
field. If
r has this value, the set
u1,
YLr form a baz is of the rational points on the curve: the points 711, 112, .., 11,. are the gencvato),S. The point it on the curve (9) is said to be e.rcclitiollal when the number if is commensurable with some period of Y' (u). Otherwise the point if is ordinary. If at least one generator point in the basis of the curve is ordinary, there are an infinity of rational points on the curve. If all the r generator points are exceptional, there are only a finite number of rational points on the curve. The following theorem, which was proved by the author in 1(2 i
.
. . ,
1935. gives a method for determining all the exceptional rational points on the curve when the field is the ordinary rational field K(1):
Let X and 1' he the courdinales of an l.rrcptional raliwlal point in K(1) on the curve (9) )there _1 and B are integer,. Then X and I are i ntel/er.., ai )'2 is either a d i ri.,or of L = 4 A3  27 B2
o, =0.
F. C'hatelet (1940) obtained a similar result for algebraic fields. Billing (1937) and other investigators have determined the basis
for a great number of curves with an infinity of rational points in K(1). For instance, all the rational points on the curve
260
CHAPTER VII
are given by the arguments /r it , where it, is the argument of the point (3, 5) and where k = 0, ± 1, + 2, ± 3, etc.; the point ul is ordinary. This curve has the rank 1. It was shown by Euler that there are exactly six rational points on the curve I'2= c3 + 1, namely, besides the point of inflection at infinity, the following five points: (2, 3), (0, 1). ( 1, 0), (0,  1), (2,  3); the corre
sponding arguments are k b, for k = 0, 1, 2, 3, 4, 5, where w is
the least positive real period of the function 6' (u; 0,  4). In this case all the rational points are exceptional, and the rank is 1.
A great number of similar results have been established by other investigators. However, no general method for determining a basis of a given cubic curve has been found. If a rational point on the curve (6) is known, it is easily shown that there exists a transformation with rational coefficients which transforms the curve into a cubic curve of the type (9). Remark. The definition of an exceptional point can be extended
to the case of the general cubic curve C of genus one as follows. If P0 is a point on C, the tangent to C at P0 cuts C in a point Pl, the socalled tangential point of F0. Denoting by P2 the tangential point of P1i by P3 the tangential point of P2, etc., we ;;et an infinite sequence of points
A, P1, P2,P3,.
,
where Pis the tangential point of P,R_1. If, in this sequence, there are only a finite number of distinct points, P0 is said to be an exceptional point. Otherwise P0 is ordinary. 70. Lattice points on plane algebraic curves. Theorems of Thue
and Siegel.  In Sections 5659 we developed a theory of the lattice points on conics. There does not exist so complete a theory of the lattice points on plane algebraic curves of higher degree. Investigations of the integral solutions of Diophantine equations of the type
DIOPHANTINE EQUATIONS OF HIGHER DEGRE]
261
f(x, y)=0,
(1)
where f(.r, y) is an integral polynomial in x and y of degree 3, lead to problems concerning algebraic numbers which can
not be solved by means of elementary number theory. In this exposition we shall only mention, without proofs, something of the main problem in the theory of Diophantine equations of the type (1), namely: Which equations have an infinity of integral solutions, and which have only a finite number of solutions, including those which have no solution at all? Thanks to the important work of Thue, Siegel and Maillet this problem is now completely solved. It is easy to indicate algebraic curves of an arbitrary degree
which
pass through an infinity of lattice points. For instance,
the curves given by the parametric representation
X =f (t), y = g (t)
where f (t) and y (t) are integral polynomials in t, have this property.
On the other hand, a closed curve which has no infinite branch
can pass through only a finite number of lattice points. Thus the curves .<.2n + y2n = 1,
where n is a natural number, have only the four lattice points (1, 0), ( 1, 0), (0, 1) and (0, 1). For certain classes of Diophantine equations it is possible by considerations with respect to convenient nioduli to show that the equations have no integral solutions. For instance, the curves 2x3+ 1 =7yn,
where n is a natural number, have no lattice points, since the number 2 is not a cubic residue of 7. Fermat and Euler gave several examples of a complete solution of nontrivial Diophantine equations of higher degree. Thus Fermat stated that the equation (2)
11:3 = ) + i/2
CHAPTER V11
262
[/ 5. To prove it we may apply the theory of the field K I V 2). The two fac
has no other integral solutions than x = 3,
tors g + l  2 and i  V 2 on the righthand side of (2) are relatively prime; and therefore, by Theorem 118. they must both But from be cubes in
y+ where it and r are rational integers, it follows that
[/=3(i[t[ 1 = )[[2r2 ['3 The latter equation is only satisfied by n = ± 1. r = 1. Thus, we ;lave established the result of Fermat. There are a ;,Treat number of equations of the same type as (2) which may be solved coinlpletely by similar methods. Now let (a
(3)
0)
be a homogeneous, irreducible integral polynomial in aiid y of degree n ? 3.. my algebraic number (see Section 12) which is a root of an irreducible equation of nth degree with rational coef
ficients is said to be of degree n. Thus the algebraic numbers ;1. SE, ... , n are of i,th degree. Further consider the Diophantine equation IF (x, h)  (:
(a)
!l),
where G (.r,, af) is an integral polynomial of degree m less than )i
in v and /. If we suppose I?/ I ? I .r 1 there is a positive con,
stant c, such that we have for sufficiently large (5)
I
( (r,
I /;
1)1 < e1 11/ I,,,
It follows from equation (4) that we must have, for at least olle of the factors  [ y, u
m
DIOPIIANTINE EQt ATIONS OF HIGHER DEGREE
263
where c2 is a positive constant. If sr is distinct from 5, it fol
lows that m
IX $iyI=1(5a)J+.rSLY I>r31YI
'21?/I/1>(.41YI
where r3 and r4 are positive constants. Hence from (4) and (5): I
and finally r (6)
17, l am
l;
where c' is a positive constant which depends on the coefficients of (4). By supposing 1 XI > I !/ I we should obtain an analogous
inequality in which 5 was replaced by
x by y and y by
Thus we conic to the following conclusion: If for every the inequality (6) and the analogous inequality have only a finite number of integral solutions ;r, ij, the number of integral solutions of the Diophantine equation (4) is also finite. We are led to the problem of finding especially. good approximaP5
tions to a real algebraic number by rational numbers. An im
portant contribution to the discussion of this question was given
in 1908 by Thue, who established the following result: If ; is an algebraic number of degree n 3 and if c and 6 are positive numbers, the inequality (7)
Ir

.'I/
c
I<
1 J 77+I+E
holds for only a finite number of integers ,r and Y. From this it follows that equation (4) has a finite number of integral solutions if ni < u  1. In particular, if in = 0, the equation Y) = C'
where C is an integer, has a finite number of solutions. In a paper published iii 19:30, Siegel succeeded in proving that Thue's result holds good even when the exponent I it + I + e
CHAPTER VII
264
is replaced by 2'V7? in inequality (7). Thus equation (4) has a finite number of solutions if n2 < n  2 Yn . By means of Thue's result for equation (8) Maillet (1919) determined all the unicursal curves which pass through an infinity of lattice points. A unicursal curve is characterized by the property that its coordinates x, :I can be expressed as rational functions of a parameter f. llaillet proved the theorem: A unicursal curve passes through an infinity of lattice points if and only if there exists a parametric representation of the form X __ 1'(t)
(h
(t))"'
't
__
g (t)
(h (t))"'
where n is a. natural number, and where f (t), g (t) and h (t) are integral polynomials in f satisfying one of the following conditions: 1. Either h (t) = a t + h with (a, b) = I or I, (t) = 1; f (f) and g (t)
are both of degree n; 2. h (t) = a t2 + b t + c is irreducible,
and a > 0, h2  4 a e > 0; f (t) and g (t) are both of degree 2 n; the form a rr2 + b rr c + r. r2 can represent for integral values of
u and v a certain integer k
0 such that /:" divides all the
coefficients of both f (t) and g (t).
In a paper published in 1930 Siegel proved that Maillet's curves are the only curves which pass through an infinity of lattice points. The proof of this theorem is very complicated; it is based on a generalization of the theorem on inequality (7), and on a generalization of 11:lordell's theorem (Section 69) due to Weil. This result of Siegel signifies that it is always possible to decide whether or not a given algebraic curve passes through an infinity of lattice points. When there are only a finite nuIIIber of lattice points, it is possible, by means of the methods of Thue and Siegel, to determine an upper limit for this number as a function of the coefficients of the curve. However, their methods give no algorithm for determining all the lattice points on the curve. Such algorithms have been found for special classes of curves by means of quite different methods. Thus, in 1925 the author of this volume showed that the prob
lem of determining all the integral solutions x and y of the Diophantine equation
DIOPIIANTINE EQUATIONS OF HIGHER DEGREE
265
A.z$+By2=C, A and B being natural numbers and C = 1 or 3, can be reduced to the problem of finding the fundamental units in certain cubic A_ fields K (Vm), being a natural number. There is, apart from a special case, at most one solution with x y 94 0. In 1938 Ljunggren showed how the complete solution in integers r and y of the Diophantine equation 77?
Ax4B/,4= C, where A and B are natural numbers and C= 1, 2, 4, 8 or 16, can be obtained by determining fundamental units in certain quadratic and biquadratic fields. By the work of Mordell and other investigators the Diophantine equation
y2=x2+D has been solved completely in integers x and y for a great number of integral values of D. For instance, the complete solution of the equation
y2=x!+ 17 is
given by: x=2, y=3; x=1, y=4; x=2, y=5;
..r=4, y=9; x=8. y=23, x=43, y=282; x=52, y=375;
x = 5234, y = 378661 (shown by the author in 1930). Other types of Diophantine equations may be solved by means of a general method developed by Skolem (1936). Exercises
123. Show that every prime p = I (mod 6) can be expressed in the form p = (a2 + 27 b2), 4
where a and b are natural numbers. It can also be written in the form
1)=,.2+;T + /2, where x and y are integers.
266
CHAPTER VII
124. Prove the theorem: Every prime p = I or = 7 (mod 24) can be expressed in the form 1) = ,.2 + 6 t/2,
where .i and y are rational numbers; no other primes have this property. 125. Prove the theorem: Every prime 1,  7, 13, 23 or 37 (mod 40) can be expressed in the form 1) = 2 ,c2 + 5 t/2,
where .i and i/ are natural numbers; no other primes have this property. 12(3. Show that every priliie p = 1, 3. 4, 5 or 9 (plod 11) can be written in the forin p = .r2 i .4,y + 3 +/2,
where r and q are integers: no other primes
11 have
this property. 127. Determine the number of solutions in integers i and r/ of the Diophantine equation L j/2
where if is the product of r primes which are = 1 (mod 4)
or twice such it product. 128. Determine the number of solutions in integers the Diophantine equation
.
and y of
where n is the product of r priules which are = 1 (mod 6) or thrice such a product. 129. For what primes 1) is the Diophantine equation
rigpt2 solvable in integers if and The same question for the Diophantine equation #r21i12=  &).
DIOPH . NTINE EQUATIONS OF HIGHER DEGREE
267
Show that the Diophantine equation is solvable (mod u).
in integers r and y when p is a prime  ii
Let J, and q b( two primes
3 (mod 4). Show that the
Diophantine equation J)a29?12=
1
is solvable in integers .r and ?; either for the upper or for the lower sign. Prove the following theorems: 1. Every prime 1, = 1 or 19 (mod 24) can be expressed in the form 1,= 3.r.2
where ,r and y are natural numbers. No other primes = 3 have this property. We cart choose .r < Vp and g/ < YT. 5 or 23 (mod 24) can be expressed in 2. Every prime the form 2
and ?1 are natural numbers. No other primes T 2 where have this property. ate can choose .r < 4 :, J, and < 1 p. consecutive positive integers have the property that the sung of their squares is itself a square. For what values of < 25 is this possible? Prove the following theorem of Sti rmer: Let r and y be natural numbers which satisfy the equation it
r2  I)?;2 = ± 1, where I) is a natural number which is not a perfect square. If all prime factors of y divide T), the number .r + y l !J is the fundamental solution of this equation. Suppose that rri and r, are the fast positive integers which satisfy the equation
CHAPTER VII
268
U2Dr2= ±4, where D is a natural number and not a perfect square. Show that all the positive solutions u and r of this equation may be obtained by the formula if + ,_1_D
 (UI + r, VT
Y'
where n = 1, 2. 3, etc. What connection has the number I (llj + fundamental solutions of the equations r2 Dy2 = 1
_
1J) with the
1Z2 DIy2=  1:.
and
136. Show that there are an infinity of integers of the form 2 x2 + 1.
(.
integer)
which are divisible only by primes  1 (mod 8). Similarly for the integers of the form 2 .2x2  1,
(x integer).
137. Show that there are an infinity of integers of the form 3.r..2 + 1,
(.r integer)
which are divisible only by primes
1 (mod 12).
138. Prove the theorem: If D is an integer, the congruence
.21)/2+1=11 (mod 31) is solvable for all moduli 31 if and only if IJ can be expressed in the form 1) = zj2 + 1.2,
where it and r are relatively prime integers. 139. Show that the Diophantine equation
J'=2Y+32 has no other integral solutions than the following:
.=y=,.=1; .,=1. r/=2. i=0;.r=2,y=4.z=2.
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
269
140. Find all relatively prime integers x and y and all natural numbers n such that the number x"  y" is divisible by no prime > 5. 141. Show that the Diophantine equation xy  Js = 1 has only the following solutions in positive integers: x = 3,
y=?;
?/=1.
142. Prove the theorem: If x, y and ii are natural numbers. the number 4
ltie,.
9
is never a perfect square. 143. Show that the Diophantine equation .r4  1/4 = 2 p z2,
where p is a prime _=5 (mod 8), has no integral solution with z76 0. 144. Show that the Diophantine equation .,..s + y3 = 4 ,s
is not solvable in integers in the field K (Y 3) if
r Y6 0.
145. Show that the Diophantine equation x$ + y3 =))z3,
where p is a prime = 5 or 11 (mod 18), is not solvable in the field K(1`3) if r 0. 146. Show that the system x + 1 = 2y2,
J,2+1=212 has only the following solutions in natural numbers:
and x=7. t/=2, r=5. 147. Prove the following theorem: Apart from the point of in
flection at infinity and the point x =  1, y = 0, there is no rational point on the cubic curve
CHAPTER I'll
1=1)y'. where I) or  D is the product of ditlereut primes which are = 5 (mod 12). The same result holds for I) 1. (Compare Section 69.)
148. Show that the Diophantine equation 4
is solvable in integers only for xy = 0. 149. Show that the cubic equation
r3ff
1 =0.
where a is rational, can never have three rational roots. 150. Show that the cubic equation
_0. where a is rational, has three rational roots onl% when a = 3.
151. Show that the cubic equation .13  it :% + iJ = 0
has three rational roots for an infinity of rational values of a.
152. Show that the polynomial in
.
where a is rational, is reducible only for a = 0. Similarly for the polynomial
"+
,
a".
153. Prove the theorem: There are an infinity of rational numbers a such that the polynomial in
'3:).efl" is reducible.
DIOPIIa\TINE EQUATIONS OF IIIGHER DEGREE
271
154. Show that the polynomial in where a is rational. is reducible only for a = 0 and a = ± 155. Show that the polynomial in
where a is a rational number 0, is always irreducible. 156. Show that the polynomial in .C .,' + .r  rt
where a is a rational number  0, is always irreducible. 157. Show that the polynomial in a:.2.,  u2,
where a is a rational number
0 and
6, is always ir
reducible.
158. Determine all the solutions of the Diopllantine equation J .4 4 _ .2
in integers x, t/ and z. 159. Determine all the integral solutions a and y of the Diophantine equation .r i4=J3.
160. Show that there are no integers
p and z which satisfy
the conditions x j 71 0, (x, y) = 1 and the equation 4 + r/4 = 3
161. Show that there are no integers ?J and r which satisfy 0, (r, /) = 1 and the equation the conditions ry .r3  3  A 162. In what quadratic fields do there exist two numbers .,' and z/ which satisfy the equation
272
CHAPTER VII
163. If it is an odd integer 3, determine all the integral solutions x, y, z with (.e, y) = 1 of the equation ,
expressed as functions of two variables it and r. Show that x = + I only for y = ± 11, 3, u = 5. 164. Show that the Diophantine equation
x2+a:+ l =3y", where r is an integer > and y fory34±1.
has no solutions in integers x
165. A number in the field K(V
)
a+b1 +V:: 2 i3 said to be an integer when a and b are rational integers. Show that the factorization of integers is unique in K Use this to prove that the Diophantine equation
x2+x+2=y3 has no solution in rational integers x and y if y / 2. Show that the number x2 + 7,
x being a positive integer, is a power of 2 only when x has one of the following values: 1, 3, 5, 11 or 181.
166. Let F(x) be a primitive integral polynomial of the second degree which has distinct zeros. Denote by q (x) the number of squarefree numbers in the sequence .F(1), F(2), 1''(3). ..., F(x),
x being a positive integer. Show that there exists a positive constant k such that q (x) > k x .
DIOPHANTINE EQUATIONS OF HIGHER DEGREE
273
167. Prove the following theorem: The Diophantine equation ;r4  f4 = rl ?2
where A is a positive integer, has either no solution or infinitely many solutions in relatively prime positive integers c, y and z. Suggestion. If [;, ij, 5] is a solution of the equation, show that another solution is given by the formulae ' = 54  9j4 ± 2
2
?/=L4)j42 2p1 168. Let p be a prime which is not = + 1 (lnod 1G). Show that p cannot be expressed as the difference of two rational biquadrates, apart from the case p = 5 = (3)4  (21)4.
169. Show that every natural number it can be represented in the form ll = .t 2 ! 2 y2 ! 3 ,2
a2,
p, z and it are integers. Sugg estion. Apply Bachet's theorem and the identity where 25
2 t/2 + 3
+ G uc2 = (i + z
11)2
a)2 +
+ (z  2 u)2.
170. A partitiun of a natural number it is a representation of n as the sum of any positive integral parts. Denote by U(u) the number of partitions of a into unequal parts and by Ul (n) the number of partitions of it into odd parts which are equal or unequal. Prove Euler's formula (valid for I x I < 1) 11 (1 + :ck) a=1
I 00
H 0 x2h1)
and show by means of it that U (n) = UI (ll). 18  516670 Trygre lVagell
,
274
CHAPTER VII
171. if a, b, c are integers in the field K(V1) such that a b c is squarefree, find the necessary and sufficient condition for the solvability of the Diophantine equation a.)'2 + b?i2 + e.:2 = 0
in integers x, y, z (not all = 0) of the field K ()/1). Same question for the fields K (1' :.?) and K (V 3).
CHAPTER VIII
THE PRIME NUMBER THEOREM
71. Lemmata on the order of magnitude of some finite sums. 
Let x be an integral or continuous variable which tends to infinity. If g (x) is a positive function of we denote by o (g (x))
any function of x which has the property that the quotient u r,(q(x) (XI
tends to zero when x > oo. Thus x = u (.c"), sin .r = o (1 x), lob x = r, (x). Theorem :
tion 17) may be enunciated:
.,r (x) = o (x).
(Sec
The prime number
theorem (Section 16) may be formulated as follows: r
()_ 1'
r
log
+u
x log x
Furthermore, we denote by 0 (g (x))
any function of x which has the property that the quotient (I (.g (x))
g(x)
is less than a positive constant when x  oo. Thus 2x + Y:v = 0 (x). sin x = 0 (1), log x= 0 (Vx). Formula (11)
in Section 17 may be written log 1)
(1)
=x
1)
=1obx + 0(1).
CHAPTER VIII
276
Lemma 1. There e.wsl.v a positive absolute con: tant y such that
we hart
' 1=log>>+y+0\1/ t! it
(2)
the suns extending ore), all positive integers n < y.
Remark. The number y is known as Euler's constant. Proof: Let z be the least integer > y. If we put (511
=1logll+l), n ` It
it is plain that loge  tit, lU 1 + ( u1
(3)
11_
i
)t
n
1
i

lSa .
=1
It is known from the elementary theory of logarithms that
I
0<&<;
(4)
12, 
It follows from this that the infinite series
bn 00
is convergent and has a positive value y. Further we have Do
1 `, 1. <
IIJ rgn<2n_'n
(n1
;r)
2 (?'1)
Thus we conclude from (3) that
`.
Z1 iwI ;?
= low C + y +
0 7
where 0 is a function of z such that 10 is less than a positive constant. This formula clearly leads to (2). Lemma 2. There exist.oz an absolute constant c such that we hare lo(* n
(J) nSy
%
_ (log y)24" + e + 0 (LOK J
where the sum extends over all positive integers n < y.
THE PRIME NUMBER T1IEOREM
277
Proof. Let z be the least integer > y. Clearly (log z)2 =
[(log ;21 + 1 )2  (log n)"], n=1
and since log (u + 1) = log JJ + log n
v (log z)2 = n_I
n

1
 bn. we have
Jt
n1
_ 2 JJ2 _l 62 1
1
16nlog72+
bn
By means of (4) we see that the latter sum on the righthand side tends to a finite limit c when z > oo. Further we have
loa I
u00=z `
J2 00
loa z
log x
log z
.r..
t
I
log z
z
z
Thus we conclude that 1077 JJ

(log :)"r r + (1 (logz z
where c is an absolute constant. This formula at once gives (5).
Lemma 3. If v(n) denotes the number of positire divisors of n, we have
I
'r (n)
= A (log y)2 + 2
logy + y2  2 c ' I) log t/
nsv n where the sum extends over all positire integers n :55 y, and where y and c are the same absolute constants as in Lem2ata 1 and 2.
Proof. Since r(i) is equal to the number of pairs of natural numbers a and b such that a b = n, we have LEI)
ngy It
'Y1,
16.j
11
278
CHAPTER VIII
where the sum on the righthand side extends over all natural numbers a and b such that ab y. Now let S, denote the part of this sum in which a < 1 y, let SQ denote the part in which b I y, and let S3 denote the part in which both a:5; and l y. Then the value of the required sum is clearly S, +
b
Putting ,a = 1 'y and t =
109
we have by Lemmata 1 and 2
1 0+7+ yT0
a9z
L ,l V
Y.
lo ,a
a_. a
a9z
(°)as. (11
as= a
a
_[logr,+y]1ogz+y+001 =(loaf,)"
?1
ass
(log z)"r (lobzj+o(1)
;ylogy+y'c
tJ
og I\
It is plain that S. = S1. Further we have by Lemmata I and 2 ((log!  y a:iz
0 (l)/Z (logy)2
i
y log y
y2  U (log
Hence
N=S1+S3S3
ab
=:
(lobyy)2+3ylogy
:.?y= 2r;(logy)2ylogy ' , 0 logy (10_
'+?ylogy+ yy '2 c + 0 lor 1,
y)2
Vy
which proves Lemma 3. 72. Lemmata on the Mobius function and some related functions.
The Mobius function u (na) was defined in Section 9. For any integer h > 0, we define an arithmetical function by the equation
279
THE PRIME NUMBER THEOREM
u (d) (log d)h,
97h (n) = d
where the sum extends over all positive divisors d of the natural number n. (log d)° means the number 1.
Lemma 4. If the natural number n is divisible by more than h different primes, we hare 97r,(n)=0.
Proof. This is true for h = 0 according to Theorem 14, and so we can suppose h ? 1. We use mathematical induction and suppose that Lemma 4 is true for all the functions gle(n) when e :S: h  1. If we suppose n = p" )n, where a > 1, and where the integer nz is not divisible by the prime p, we obtain
id: u (dl d2) (log d, + log
a (d) (log d)h =
Ti, (n) = d
Cl,
where the outer sum on the righthand side extends over all positive divisors (1, of nz and the inner sum over all positive divisors d$ of p". Then h
921, (n) _ 8=0
1i
(s)
Ft (di) (log d,)"
It (d3) (log dy
(1,
h
(h)qs(ni)
Th8 (l)").
has more than h different prime factors, nz has more than h  I different prime factors. Therefore by hypothesis Since
it
¢,e (m) = 0 for s = 0, 1,
. .
.,
h  1. The remaining term Th (an) 9'0 (p")
is also equal to zero, since its last factor is zero. Thus the lemma is proved. Lemma S. ii'hen x is a 1)ositice number, ire put
A(d)=,It(cl) logd) and A(d),
.f (n) d
CHAPTER VIII
280
where the sum extends over all positive divisors d of the positive integer n. Then we have f (1) = (log x)°;
(log p)2 + 2 (log x) (logy)
when p is a prime and a an integer
1;
.f (1)" qP) _ 2 (log Ir) (log q)
when p and q are dilrerent primes, and a and fi are integers
1;
f(n)=0 when n is divisible by three or more different primes. The proof follows immediately from the definitions and Lemma 4
(for h = 0, 1 and 2). Lemma 6. For every natural number x tee have g(d)I<
rr=1
Proof. It follows from Theorem 14 that
1=
Y, g (d), 1,1 'r
where the inner sum extends over all positive divisors d of the positive integer n. Hence, since the number of multiples of
d ? 1 and < x is equal to I'd (d)[rl]=1.
d=1
Consequently
I
d=1
Thus
x1
=I Ng(d)(d[d]) d1
rr=1
.r " ,[ (cd) I < 1 + .c  1= :x. r1=1
which proves the lemma.
d
Till, PRIME i\lTMBER THEOREM
281
Lemma i. I or mere/ positive x we have (1)
2:P((l)logx=0(1), "d d
,rfix
over all positive integerx d < x.
where the stun
Proof Applying Lemma 1 we find that the lefthand side of the assertion is equal to 1
_'U(d)
dx
0i
d
where t =
l
and where 10, 1 is less than a positive constant r,.
For d n = to this becomes !
mgz
Y
J
(d)
rl d_s µr!)
/;Ex
X
where 6 runs through all positive divisors of m. By Theorem 14
the first sum has the value 1, and by. Lemma 6, just proved, the second sum has an absolute value < y. The absolute value of the third sum is at most 14J1=r,.
das
This gives the desired result. Lemma 8. For ever l natural number tt we have (2)
/A
d
(d)r( )=1,
the sum extending over all positive divisors d of n.
Proof. Since v (2) = v 1, the sum extending over all positive J
divisors 6 of 1, the lefthand side of (2) becomes d
J
J
J,
CHAPTER T HI
282
where the inner sum on the righthand side extends over all positive divisors al of
By Theorem 14, this inner sum is equal
.
to zero when d : n and equal to 1 when a = n. Thus the righthand side is equal to 1. Q. E. D. Lcnrnw .rl. l'or
1rositire x ire hare
1 /Z R) °lo.r.2 (l J
(3)
rrSS
log X + 0 (1), b
``
Cl
where the sum extends over all 1%ositire integer.v d < x.
Proof. Applying Lemma 3 for r/ _ , the lefthand side of formula (3) may be written C (n)
2
rl
(,gas:
t C
Jory

)2
rl
y
JJ
Ill(d) f0 (
'
dsz d
x log d) ,
where 101 is less than a positive constant c.. For all sufficiently large x the absolute valve of the last sum is smaller than l(x)l (x)1.._,,
4cs(
ln(>'cd
.SZ
=xiU(f z
47
cir)=0(1).
d6X
Further, by putting 1. = n d, we have T(7r)_.r
((l)
.3=2 (him
ksm
71
rr(d)T
1
xs.e L'
d
where the inner sum on the righthand side extends over all positive divisors d of 1. Hence, by means of Lemniata 8 and 1,
S =2logx+ 0(1). Finally, applying Lemmata 7 and 6 we see that the lefthand side of formula (3) is equal to 2 log .x + 0 (1).
Thus Lemma 9 is proved.
283
TILE PRIME NUMBER THEOREM
73. Further lemmata. Proof of Selberg's formula.  By means of Theorem 30 we prove Lemma 10. The scan being extended over all princes p < x, we hare (log 1,)
(1)
(log) = o (x lo,, x).
n==
Proof.
If J = log x, the suns on the lefthand side is equal to
(log p) (log
r
i'1
)
(log 1)) (log x)1
i
< (log
log p
log p + (log log x) pzx _ (log X) 99
(
a
lox
T) + (log log.') 8 (x). 25
Applying Theorem 30 we see that this function has the order of magnitude U (x log log .r'),
which is somewhat better than (1). Lemma 11. The sum being cxtrnded over all prime powers p° < x, where a is a natural ncanber, we have log 17 = 0 W.
(_))
Proof. The sum on the lefthand side is obviously equal to O (.r) + fl (Vx) + v9 ('Vx) + 
+
(lax},
where k is the greatest integer such that 2'' S x; therefore k :!g Log x
log 2
This sum is at most equal to 79(x)+ l,'0(lax).
From Theorem 30 it follows that it has the order of magnitude o (x) + log x 0 (l x) = 0 (x).
log2
CHAPTER VIII
284
Lemma 12. If f (n) is the function drfined in Lemma :;, we have
j, f (n) = (log x) 0 (x) + 2 I 0 (xl log p + o (x log x), na.r
_+l
where the sum on the lefthand side extends over all positive
integers n < x, and where the stun on the righthand side extends over all primes p <
Proof. It follows from Lemma 5 that (3)
, '(n) _ (log x) + } (2 ;log x) (log p.  log p ') + 2
(log p) (log q).
Here the first sum on the righthand side extends over all prime powers p" < ,r, a being a. natural number; the second sum on the righthand side extends over all prime powers pa and 0 such that p" q13 < x and p < q, where a and P are natural numbers. In the first sum on the righthand side we first consider the terms with a z If we denote by g (.x) the number of prime powers p":_5 x with a 2, we see that the contribution of these terms to the sum is at most equal to 2 (log.,)* g (x)
2 (log x)' (V__ + l/x + ... + 1''.x.) .1
A
when k is the greatest integer such that 2k:5 x. Thus the contribution does not exceed 2 (log x)=  k lG  `Z (log X)2.
log
x YX_
l0 g 2
= o (x log x).
Consider next the terms with a= 1 in the first sum on the righthand side. The contribution of these terms is equal to (2 (log x) , logp'
logp E)
P3;
_ (log x) 2 log p +.1 (log p) (loge') = (log x) V (x) + u (.r log r), p6z
Paz
according to Lemma 10. Thus the first sum on the righthand side in formula (3) is equal to (4)
(log x) 0 (.2) + 0 (x log x).
THE PRIME NUMBER THEOREM
285
Finally consider the second sum on the righthand side. Applying
Lemma 11 for X3 instead of for x, we see that the contribution T
of the terms with f > 2 and a % I has the order of magnitude q=0 log q  0 X) = 0 (x) 2. 'Oa q,
for the infinite series
log q
2. Thus the second sum on the
clearly convergent, since f righthand side is equal to 2
(5)
extending over all primes q is
(logy) (log q) + 0 (.
).
the sum extending over all primes p and q, such that p q < x and p < q. If we put y = Y.ce, the latter sum is equal to I (log p) (log q)  }. (log p)pat/
p qix
= v (log p) (log q) + I (log p) (log q) pall
qay
pgax
pqz:

'**1
(10,,.p)
pav
(log q) 
(log; 1))2.
psy
q9Y
According to Theorem 30 the last two sums have at most the order of magnitude ( .Vx))E = 0 (r') and
(log V X) 0 (YX) = 0 (Yx log x)
respectively. Hence we conclude that expression (5) is equal to (log q) 6. I x)
(log p) 0 GO + pay
qSy
q
`
+ 0 (.r) = 22 (log p) 0 ('x) + 0 (x). p5y
Introducing expressions (4) and (5) into formula (3), we finally obtain Lemma 12.
CHAPTER VIII
286
We finish with the proof of Selberg's basic formula. Theorem 12i'. Putting i/ = J /X' we hare
z9(x)log x + 2 }`.d ) p
Proof According to Lemma 12 the lefthand side is equal to
f(n) 2xlog.e + u(x logs). nar
According to the definition of the function f(n) we have S n=t
14:9:e
d
where the inner sum extends over all positive divisors d of n. Hence d
(d)(1'
d.:x
where 0 < e j < 1. By putting z = (log x)x we have (log
j n (d) I
;
i
x d
a
_
J
lour
°d
og, \
a
° dJ
Z (log x)° + 4 x (log log x)2 = 0 (.,.) + () (x log log x. °) = o (.r. log x).
Hence
S=
2 (cI)
dSx
x cl
+ 0 (x log x) _
c
.t;
cclix
(d) I lob I d \ d/ x
+ 0 (x log
and, by Lemma 9,
S=2xlog x+o(xlog.r). This proves Theorem 127. 74. An elementary proof of the prime number theorem.  Every
thing is now prepared for finishing the proof of the prime number theorem. We have seen in Section 17 that this theorem is equivalent to the proposition (1)
lien 0 ('')
t. Q x
= 1.
THE PRIME 'NUMBER THEOREM
287
It follows from Theorem 30 (Section 17) that for increasing X the quotient 0 (J') :r
has a positive lower limit a and an upper limit A: then 0 < a < _1.
Thus, to prove (1) we have to show that
ti=1=1.
(2)
The following proof is mainly based on Selberg's formula which was proved in the preceding section: (x)
(3)
x
+
x
. ,9
1og . 'ZL!,
log 1)  _' = (1).
the sum extending over all primes P < y = V:r. We also need the following formula which was proved in Section 17: log p (4)
p5x P
= loh .c
0 (1),
the sum extending over all primes 1, S .r. We first prove Len:ana 13. If lira sup V (x) x r.Co
=.:I and i f lieu ii,f (x) = n, tee /race x X.. :c
a+_1='I.
(5)
It is possible to let .r tend to infinity in such a way that (x) tends to A. If F is a given positive number, we have Proof.
x
61 I >(as) for every .r sufficiently large and for every prince p  y = Vx, and therefore 2 (a  e) ` log' p gp x log; x 1):911 p
CHAPTER V111
288
It follows from (4) that the righthand side of this inequality oo. Thus, applying formula (3) we get tends to a  s when 2  A a  e. Since this holds for every positive s, we conclude that A+a `?.
(6)
On the other hand, it is possible to let .v tend to infinity in such a way that
t9 (x)
tends to a. If s is a given positive num
ber, we have 0 (x) < (A + e) 1)
P
for every ., sufficiently lapse and for every prime p S y = Y.r, and therefore 2
x log x
pJl0n"p
log p
2 (Al + s
logx
P_V
p
It follows from (4) that the righthand side of this inequality tends to A + s when x  0. Thus applying formula (3) we get 2  it < A + e. Since this holds for every positive e, we conclude that
A+a?2.
This inequality together with (6) leads to (5).
In the following we alrrays let the variable x tend to infinity in such a manner that
10(x)
x
tends to A.
Lemma M. If A is a given number > it, and if the sum
S(x)loa2 P
extends over all primes p < v and such that 0 the quotient o(xx tends to zero for x
oo.
2 x, then ? (PX) p
THE PRIME NUMBER THEOREM
289
Progf: We have, by putting y (X)
v
log;, = v log p , log q
P
re,=r
1,9.r
x) 1 P`9Y
1
loge+q(iogr)2 q;;Y
P
PRY
p
Pay
Since, by Theorem 30, the last term has the order O (x), we see that Selberg's formula (3) may /also be written 'Y '0
z9 (.c) 1
(7)
x
1
,
.r,logxl`=r
(x) log p  2 = o (1). p
Let a be a positive number. For every
x
exceeding a certain
1111
number it which depends on a we have 0 (X) > (a  ) x. There exists a positive number b depending on it and so on a such that

(8)
.r
>(ae)1x
P
b
for all the primes p such that p < u. Thus the latter inequality holds for every p < x.
If the sums (x)
extend over all primes p < x such that
x, p we have
1
2 0Ipx logp?Ax rlog1) p > (Aa)x S' rlogp r
(9)
p
+
`
if/ the sums vrr extend over a9 lrl < fix, we have by (8) 19  516670 Trggce Nagell
all
(ae)x2rlogp p
primes p < x such that
290 (10)
CHAPTER VIII
Y"'#(e)x ',"l
p
La9 ( x.)
From (9) and (10) we deduce that
80 log p =
19 (')iogp + N', g (')iogp
> (ae)x2Log 1)
J
Y,'logPL19(x). p
P
Substituting from this result into formula (7), we obtain for x
x)
tending to infinity in such a manner that x tends to A. ' ' log )) A + a8 + (A  a) lim sup .r.:
) <=.)
log x
Hence, recalling that a + A = 2,
N' log P lim sup W
p
log .,r
Since A  a > 0, and since e can be chosen arbitrarily small, this gives the desired lemma. Lemma ir3. If to is a giren positive )mmber < A, and if the suns
R (x) =I' (!2 U ,/
11o2 ?!
extends over all the princes p and q which satisfy the folloFei,)g conditions: x7 x < S µx) x, $ p
`q VP
PP
then the quotient
R (4 (log.z )2
teiids to zero for x
00.
pgi
pq
THE PRI3IE NUMBER THEOREM
291
Proof. Replacing x by p in Selberg's formula we obtain, by
putting v = V
p't
x+0 1p/
`l;)l log x
`p
`pql log q
p
00 into the same formula we
Introducing this expression for get, by putting .11=1/x,
O(x)=2x+o(x)111logp(2
4V
logxPisy p
logx
where
v= 20(x)(logp)(logq) A9
logx
P9
1)
the sum extending over all primes p and q such that p < Vx, q< Since, by formula (4),
11og'' = I logx + 0(1),
PAY p
it follows that 4V
log x
+ 0 (x).
In every term of the sum V we have p < Vx and q <
x, and
V p
therefore pq = p) (pq2)* < x3. Thus, if b is any positive number, we have 0<(A+8).r
Julq
pq for x sufficiently large.
Let us write N'
Y
CHAPTER VIII
292
where the first sum extends over all primes 1) and q such that p S V x, q
X and 0 (
q) < )q , and the second sum over all
primes 1) and q such that p
VX1, q < I/P and 0 (j)
> 1'q
Then we have log p log q
1
p
to'
log p lob q
1
7
p
log
p
q
1)
where the sums are taken as above. We further get
V<(A+6).).TV(A +b  z)x
1
loge logq q
1)
loo,x
where 1
TV=
n,
log X
log p log q q p
1_ log p /,z`y log1
p
log q y__t
q
,
the sums extending over all primes p, < It = Vx and over all primes q < z =
W=
x.
VP
I
Applying formula (4) we obtain
log p
Psy P
Hence
O(;r)S(A+&)x
4
log x
+ o (1)) _ log x + o (log x).
l (A+5g)x N' I.Ci log x
lob)) log 9 +?IX, p
q
where 77 tends to zero for r oo. From this we deduce that 1
(log, x)s
(+b)
,'loge logy!
< .4 + 8  0(x) .r
where the sum is the same as in Lemma 15. Hence, for x tending to infinity in such a manner that 6 (x) tends to A, x
THE PRIME NUMBER THEOREM
l i'i sup X 00
'loge logq =
1
(log
x)=
q
1,
293
b
4 (A  u)
Since A  u > 0, and since 6 is an arbitrary positive number, this proves the lemma.
By means of Lemmata 13, 14 and 13 it is now easy to prove the relation
it= A=1.
(11)
a. Let a be a positive number > 1 such that as < _1, and let 6 be a positive number so small that Suppose that .4
Ait a?Sa+26.
(12)
Further. denote by N a natural number. We consider the suns ,g
where

log p logy 1) q 2
3 1lo; r
extends over all the primes 1, and q such that
pq? 1. $
p
(_1(eq)
1)
and where Z3 extends over all the primes r such that 1)9 r 5 apy a
If there are no primes r, the sum 23 is = 0. For every term in the sum 51,3 we have r
apq == ap'(1)g2)1 S ax}xi= ax, =<,q
when x is sufficiently large. For the same terms we shall prove the inequality (13)
V `xl > (a +
)x
when x is sufficiently large. This inequality is true for all r < p q, since
CHAPTER VIII
294
,(x ?0 (xO91 ( A6)p>(A6)ar=(a+)r in virtue of (12). x
Consider now the terms with r > p q. If we put = it and = r, we get u < v ;;S am If in Selberg's formula
PQ
(log x) 0 (x) + 2 2 0
log p = 2 x log x + o (x log x), (;X))
P9!'
where y = Vx, we first replace x by v and then x by u, we obtain on subtraction (log r) 0 (r)  (log it) 19 (u) < 2 r log v  2 it log u + o (it log u), or
0 (to >
lon it
(r,  u)  2 v log
t9
log u to
gl
+ 0 (to.
"
Thus
t9(a)>(AS)v2(vu)+o(rt)=2ut(2A +S)v+0(rt) and, since a+11=2 and Aas>ba+2S, 0 (tt) > (a + 2 S) it + o (u).
if x is sufficiently large, we therefore have t9
(X) )=6(tt)>(a+6)u=(a+S)x
This proves inequality (13) for all r. Consequently
r S < yj log r. r
log 1) log q p
q
where the first sum extends over all the primes r < x and such
that 0 (x) r
(a + 6) X, and where the sum
r primes p and q such that p < 1'x, q < V , 1)
extends over all +
tr
q < ar .
THE PRIME NUMBER THEOREM
Thus we have, by putting y =1"x and t =
c locrp p
}
295
a r. 1)
ogq
q
glst
p)6 a))
logp
P2 ,J
where cl and c, are positive constants. Consequently
S<e.logx L r
logy )
where the sum extends over all the primes r < x and such that
' \x!r ? (a + 6) x) . By Lemma 14 we then obtain S = i (log x)2,
(14)
where 27, tends to zero for x  oo. Now consider the sum 10L1) log9
p
q
extending over all primes p and q such that
pq<
1/7x., pq> V.
Putting y =1 x, z = Y and n =Y7V, we then have T
12^P')1)
r" n
`q
logg1.
Hence by formula (4) T > e, (log x)=,
1.5)
e, being a positive constant. Let us put
T=
log 1) log q
p
q
+
log? p log* q q
.9' 1)"
CHAPTER VIII
296
q'
where the latter sum extends over all the primes 1) and fying the conditions X.
q
satis
<(16)f pq
c9px
)
1)
q
This latter sum, in virtue of Lemma 15, is equal to ?1, (log where 91, tends to zero for r  oo. Hence we have
V q
1'
and, in virtue of inequality (15), for p to<' q > cy (lo" x)s q p
(16)
for x sufficiently large. In the sum `'. we now consider for a fixed value of x the
primes N and q which have the property that the sum
log
takes its minimum value µ; u depends on x only. Then we have, by (16), /Z
lo;; p log q !J' p q
Y V:; (log x)
If we compare this result with inequality (14), we obtain for .c' > 00 that
Consequently, to every positive number a and to every natural number N there corresponds a number t =.p q ? N such that F;50 log r
,u
r
the sum extending over all primes r which are > vt acrd S at. Hence
rR nt
log r < eat, rn t
THE PRIME NUMBER TIIEOREM
297
and /
(17)
t)
<SOr t.
If X, and therefore also t, are sufficiently large, we have 0(al) > (a  )at and (.1
? t)a.
Hence. it follows from (17) that .TE <sa. (a  e)a a
This inequality holds for every positive number E. Hence we obtain a a2  A < 0. On the other hand, we have a a < A and a > 0. Thus, every number a < "I has the property that a= < I If .
U
a tends to
a
we obtain that
z
(a
<
a
or a
and a + _1 = 2, we clearly have a = A = 1.
1.
Since a< A Q. E. D.
The basic new thingt, in the above proof is the asymptotic forIuula (3) (Theorem 127). From this formula there are several ways to deduce the prime number theorem. The first proof of Selberg dates from 1'J18. The proof developed in this section is related to it; it follows an exposition given by van der Corput
and based on notes of some lectures held by Erd6s in 1948 (Demonstration glc%mentaire du tlreorenme sur la distribution des nonzb,v..
Scriptum 1, Centre Mathematique, Amsterdam 1949).
In a paper published in 1949 (A n elementary proof' of the prime number theorem, Annals of Mathematics, Vol. 50) Selberg uses
another method for deducing the prime number theorem from his basic formula (3).
It is easily seen that the proofs given in this chapter can be modified so as to avoid completely the application of concepts and results from analysis.
A new epoch has been inaugurated in the theory of prime numbers by the ideas of Selberg.
CHAPTER VIII
298
Exercises
172. Prove the approximate formula
I
92 n)
_
x + 0 (log x),
nsr
the sum extending over all positive integers n < x. 1 73. Show that there exists a positive constant k such that, for all natural numbers n, u
(n) > log log n
174. Prove the approximate formula
1 =loglogx+ C +0(1),
per P
the sum extending over all primes p:5 x. C is an absolute constant. 175. Prove the approximate formula
11 ( PAX
 p) _ log x r ° 1
1
e
1
log x
the product extending over all primes p < x. y is Euler's constant. 176. Denote by rv (n) the number of different prime factors of n.
Prove the approximate formula CO (n) = x log log x + Cx + 0 (2'), aaz
where C is an absolute constant, the same as in Exercise 1 74.
The sum extends over all positive integers n < x. 177. Denote by p the nth prime. By applying the prime number theorem prove that lini
A,
oc Mlogn
n
_ 1.
Conversely, deduce the prime number theorem from this relation.
THE PRIME NUMBER THEOREM
299
178. Show that the infinite series 1
2 p (Log log p)h ' °0
extending over all primes p, is convergent or divergent according as li is > 1 or 1. 179. By applying the prime number theorem prove that Y, A()n)=o(x), nSx
the sum extending over all positive integers u < x. 180. By applying the prime number theorem prove that 0. n=1
300
Table
indicating the least positive primitive root g of the first 150 odd primes p. An asterisk attached to the prime means that it has the primitive root 10. 1'
I
N
F
I+
I
P
1l
I
4
1'
II
1'
13
Ui3
I
U
3
2
131`
2
2
137
3
293 307
2
5 7
3
139
2
311
17
479 487 491
11
2
149*
2
313
10
499'
131
G
317
2
ilea'
i
509;`
2
521
3
7u9'* 719
523
2
72_7
541*
2! 733
U
547
2
731.1
3
II
13
I
5
17,
3
157
5
331
19
2
163
2
337 Y
23" 29*
5
167
5
2
173
2
31
3
179"
2
37
2
181"`
2
41
6
191
19
43
3
193
47* 53
3
it
:3
I'
10
3
3 2
2
683 (J77
7
691
3
701
2 2 11
347 349 353
2'
7
5557
2
743'
5
6
56:3
2
751
3
5
359 367 373
2
569
:3
i i57
2
197
2
379`
2
511"
:3
,01
6
2
199
3
383'
5
57i"
;i
769
11
59'
2
211
2
773
2
61
2
223"
3
787
2
67
2
227
2
7
797
2
71
7
229'
7
S09
3
73
5
607
:3
811*
3
79
h3
I
1
2 3
i
I
1
389k
2
587
2
397 401
:i
51)3
3
3
6
409
21
509 Gul
233*
3
419'
2
3
239
7
421
2
613
2
821
2
2
241
7
431
7
617
:3
823*
3
89
3
251
6
433*
5
619"
2
827
8
97"
5
3
439 443
13
631
3
2
041
829 839
11
440
3
643
853
2
457
13
647
5
461
2
653
2
G5 (;GI
I
"
1
I
101
2
257" 263`
103
5
269'
:3
107
2
271
G
109" 113"
U
5
3
277 281
3
463
3
3
283
3
467
2
127
1
5
I
I1
1
i
II
1
.I
11
2
857"
3
859
2
2
8133*
5
2
S77
2
i
301
The fundamental solutions of the equations x2  D5'= ± 1 for D 103. The table indicates the fundamental solution of equation .c2  D<12
 I or of z`  Dy2 =  1 according as an asterisk is attached to the number D or not.
I)
I
I
<
r
71
I
,
I
y
11
l
,
I
y 413
2
1
1
11
38
37
6
IN
341,11
:3
2
1
II
3:)
4
72
17
2
2
1
40
25 19
3
73'Y
10iis
(i
5
2
41*
32
R
74°
7
8
3
42
13
2
75
43 26
125 5 3
43
3482
531
76
57 799
6 (53(1
199
40
.53
6
24 335
3 588
80
9
48
7
77 78 79 80
331
161
30 24
2
44 45 46
9
1
82
9
1
83 84
82 55
0
378 10 405
41
8
3
1
10*
3
1
11
10
3
12
7
13*
18
5
47
14
15
4
48
7
1
15
4
1
5i(*
7
1
17*
4
1
31
50
7
18
17
52
649
170
53'R
183
99 25
85°
19
4 39
CI
86
0
1 122
I
9
2
54
485
60
87
28
3
55 197
12
89
12
88
15
89*
23
24
151
2 20
197 500
5
55 56 57
91,
10
21 53 2
24
5
1
58*
99
13
91
1 574
165
1151 12151
1260
"10
21
22
42
26 127
24
29 718
3905
2 543 295
221 064
29*
70
13
62
63
8
95
39
4
30
11
2
63
8
1
06
49
5
31
1 520
273
97*
17
3
66
5 604 99
569
:32
33
23
4
:34
35
6
67 68
27 28
5
59 60
1
5
35
6
1
37*
6
1
i
61*
05*
69 70
530
69
31
4
120
92 93 94
26*
8
1
65 48 842
8
5 967
33
4
7 7 75 2S1
936 30
98 99
10
10 1
10
1
102
101
10
103
237 52,14
22 419
101*
NAME INDEX
BACHET DE MFZIRIAC, C. G., 191, 227, 252
BAUER, At., 168 BERTELSEN, N. P., 54
I I oNECKER, L., 144 Kr'mtER, E. E., 262, 253
BERTRAND, J. L. F., 67, 125 BILLING, G., 232, 269 BOHR, H., 57 BRUN, V., 68, 67
LAGRANGE, J. L., 5, 99, 192, 197, 235 LAVBERT, J. H., 40 LANDAU, E. G. H., 57 LEBESGUE, V. A., 194, 251 LEGENDRE, A. M., 5, 64, 133, 219, 246,
('ATALA. , E. CH., 124 CAUCHY, A. L., 144
I.EIr\rER, D. N., 51 LIND, C. E., 232
CHILTELET, F., 259 CORPUT, VAN DER, J. G., 297
LINDEMANN, C. L. F., 40 LIOUTILr.E, J., 36 LITTLE1VOOD, J. E., 57 I.JUNGGRE\, W., 266 LL CAS, E., 62, 128
262
DIOPHANTOS, 6, 32, 227, 252 DIRICHLET, P. G. L., 38, 66, 197, 262 EISENSTEXN, F. G. M., 141, 241 ERATOSTHENES, 51 ERDUS, P., 297 EUCLrn, 6, 14, 16, 22, 34, 44, 228 EULEII, T.., 6, 23, 38, 44, 71, 188, 144, 199, 246, 261
FAREY. J., 48 FERMAT, P., 5, 44, 71, 197, 227, 229, 248, 262, 261
MAILLET, E. TH., 261, 264 MANGOLDT, H. C. F., 57 MErSSEL, D. F. E., 54 MERSENNE, M., 44
MERTENS, F. C. J., 127 MILLS, W. H., 85 MINKOwsKi, H., 34 \IORDELL, L. J., 235, 253, 259, 285 MiiBIUS, A. F., 27 NIVEN, 1., 40
GAUSS, C. F., 5, 54, 68, 113, 139, 144,
177, 241 GOLDBACH, CH., 66
HADAWARD, J. S., 54 HARDY, G. H., 57 HERAIITE, CH., 40, 126
ORE, 0., 90
PEI.I., J., 197 I'IPI'ING, N. J., 66 PYTHAGORAS, 6, 34
RIEMANN, G. F. B., 56, 66
INGHAM, A. E., 67 SCIIOLE, A., 122 JACOBI, K. G. J., 113, 144, 146, 241 .TACOBSTHAL, E., 182
SEI.nERG, A., 67, 88, 288, 297 SIEGEL, C. L., 260, 263, 284
NAME INDEX
SKOLE)r, Tit., 265 STURMER, F. C. M., 267
Sux TsE, 78
303
VALLEE POLSSIN, DE LA, CH. J., 56 `INOGRADOV, I. 1MI., 66
SYLVESTER, J. J., 50
WEIL, A., 269, 264 WILSON, J., 99
TCHEBYCHEr, P. L., 55, 60 THUE, A., 122, 260, 261, 262 TITCHarAasH, E. C., 57
WOLSTE\HOLME, J., 128 ZASSENIIAUS, H., 66
SUBJECT INDEX
A Algebraic congruences 73, 79, S3, 8S, 90
Algebraic curves, 254, 260 numbers, 35 Algorithm, Euclidean, 21, 22 Ambiguous classes, 205 Arithmetical functions, 26
Combination of at things. 49 Complete residue system, 70 Composite numbers, 1:i Congruences, 68 algebraic, 73 binomial, 115 exponential, 114 functional, 74 in a quadratic field, 24(
progressions, 66
linear, 713
Associated numbers, 237
quadratic, 132 simultaneous, 77
solutions, 204
Y,
Congruent numbers, 68
B Bachet's theorem, 191 Basis of a modul, 10 > of rational points, 259 Bauer's theorem, 168 Bernoulli numbers, 253 Bertrand's conjecture, 67 Binomial coefficients, 49 congruences, 115
Binary quadratic forms, 215 Biquadratic congruences, 73 residues, 115, 130
Birational transformation, 257 Box principle of Diricltlet, 38, 123 Brun's sieve method, 66 theorem on prime twins, 67 C Canon arithmeticus, 113
Classes of residues, 69 in a quadratic field, 240
Classes of residues, prime to a, 71 Classes of solutions, 204 , ambiguous, 205
>
»
, conjugate, 205
polynomials, 74 67 Conjecture, , Goldbach's, 66 Conjugate classes, 205 numbers, 235
Cubic
73
Cubic curves, 254 of genus one, 254 of genus zero, 254 unicursal, 254 I)iopliantine equations, 241, 246, 254, 255 Cubic residues, 115 Curves, algebraic, 254 Maillet's, 264 nnicursal, 254 Cyclotomic polynomial, index of. 158 irreducibil
ity of, 160 Cyclotomic polynomial, 158 » prime divi> sors of, 164
D Decomposition into prime factors, 15 Degree of an algebraic congruence, 73 of an algebraic number, 262
SUBJECT INDEX
Diophantine equations, 32 of the first
305
Exponent of an integer modulo n, 102
Exponential congruences, 114
degree, 29
Diophantine equations of the second (egree, 188
Diophantine equations of the third
F Farey series, 46 Format's last theorem, 261 primes, 44 theorem, 71
degree, 241, 240, 264, 266
Diophantine equations of the fourth degree, 227, 232, 235, 257
Field, 21 Gaussian, 241 imaginary quadratic, 21, 236 quadratic, 21 rational, 21 real, 21 real quadratic, 21
Diophantine equations of higher degree, 227
Diophantine problems, 32 Diophantos's Arithmetica, 5. 32, 226, 252
Dirichlet's box principle, 38, 123 Dirichlet's theorem, 66 Disquisitiones arithmeticae, 6, 68 Divisibility by g fix) modulo n, 76 of numbers, 12 » of polynomials, 93, 160 Divisor, 11, 12
Fixed divisors of polynomials, 122 Formula, Hermite's, 126 » , Selberg's, 283, 288 Function $(x), 60 C (a), 66
, greatest common of two or
»
more numbers, 17 Divisor, greatest common of two polynomials modulo p, 98
»
»
Divisors, fixed, of polynomials, 122 of f(x) modulo n, 76 proper, 12 trivial, 12 trivial of f(a) modulo p, 94
E Equations, Diophantine, 32 " , indeterminate, 32 Eratosthenes's sieve method, 51. 52 Euclid's algorithm, 21, 22 » Elements, 5, 14, 16, 34, 44, 226
Euler's constant, 120, 276, 298 criterion, 133 ffunction, 23 identity, 55, 192 » theorem, 72 Even numbers, 12 »
»
Exceptional point, 259 20  5 18670 7'rrigve Nagell
Li (x), 55 µ (n), 27 ?6 (x), 54
T (n), 28
"
p(n), 23
Functional congruences, 74 Functions, arithmetical, 26 Fundamental solution of a class, 206 Fundamental solution of an equation, 1:47, 201
Fundamental theorem of number theory, 16
G Gaussian field, 241 "
sums, 177
Gauss's lemma, 139 " polynomial identity, 174
Generating system, 19 Generator, 269 Genus of an algebraic curve, 264 Goldbach's conjectures, 68 Greatest common divisor, 17, 240 of poly
k
nomials modulo p, 96 Greatest integer
c, 13
306
SUBJECT INDEX
H Height of a solution, 244 Hermite's formula, 126 Homogeneous coordinates, 217
L Lattice, 32 Lattice points in the plane, 32 »
I Ideals, 252
Identical congruence, 74 Identity, Euler's, 65 A
, Lebesgue's, 194
Imaginary quadratic fields, 21, 235 Incongruent numbers modulo n, 68 roots of a congruence, 73
solutions of a congruence, 73
0, 218
Induction, mathematical, 11 » , multiplicative, 16 Inequalities, Tchebychef's, 80, 62 Infinite descent, 229
Initial solution, 235 Integers, 11, 236 Integral logarithm, 65 Integral polynomials, 73 Inversion formula, 31iibius, 28 1. r. (integer representing) polynomials, 120
Least nonnegative remainder, 12 Least positive common multiple, 16 Lebesgue's identity, 194 Legendre's symbol, 133 Lemma of Gauss, 139 Linear congruences, 78, 76
Irreducibility of cyclotomic polynomials, 160
Irreducible polynomial modulo p, 95
J Jacobi's symbol, 146 Jacobsthal's theorem, 182
K Kummerian prunes, 253 Kummer's theorem, 268
Diophantine equations, 29
M Maillet's curves, 264 A theorem, 264 Mathematical induction, 11 Mersenne's primes, 44 Module, 19 Modulus of a congruence, 68 Mordell's theorem, 263 Multiple, least positive common of two or more numbers, 16 Multiple of an integer, 12 » of f (x) modulo n, 76 Multiple root of a congruence, 86 Multiplicative induction, 16 Miibius function, 27 A
Irrationality of e and n, 38 Irrational numbers, 34
on plane algebraic
Least absolute remainder of a modulo b, 13
A
of a natural number, 112 of the equation a x'+b y'f c z=
»
curves, 260
Indeterminate equations, 32 Index calculus, 111 Index of a cyclotomic polynomial, 168
on conics, 212
A
inversion formula, 28
N Natural numbers, 11 NonKummerian primes, 263 Nonresidues, quadratic, 115 Norm of a number, 236 Normal polynomial modulo p, 94
nth roots of unity, 166 Number axis, 32 Numbers, algebraic, 36 , associated, 237 , Bernoulli, 268 , composite, 13
SUBJECT INDEX
Numbers, congruent, 68 conjugate, 236 even, 12
»
, incongruent, 68 , irrational, 34 , odd, 12 , perfect, 44 » »
, prime to n, 23 , relatively prime, 23
»
, relatively prime in pairs, 23 , squarefree, 27
»
, transcendental, 36
V
307
Prime divisors of a cyclotomic polynomial, 164
Prime divisors of polynomials, 81 » of quadratic polynomials, 149
Prime factors of an integer, 15 Prime function tnodulo p, 95 Prime number theorem, 55, 275, 286 Primes, 13 »
, Fermat, 44 in arithmetical progressions,
66, 153, 168, 173
Primes, Kummerian, 253
O Odd numbers, 12 Order of an integer modulo n. 102 Orders of magnitude o (x) and o (x), 275
P Partitions of numbers, 273 Poll's equation, 197 Perfect numbers, 44 Point lattice, 32 Points, exceptional, 259 rational, 216, 254 singular, 254 tangential, 280 Polynomial coefficients, 49 identity of Gauss, 174 Polynomials, cyclotomic, 158 divisibility of, 160 integral, 73
Mersenne, 44 nonKummerian, 263
of the form ny  1, Prime twins, 52
Primitive polynomials, 73 roots of unity, 157 Primitive root of a congruence, 104 of a modulus, 107
Principal remainder, 12 Products, trigonometrical, 173 Proper divisors, 12 solutions, 219
Q Quadratic
73, 132 fields, 21, 235 nonresidues, 132 residues. 115, 132
Quadratic reciprocity law, 141
R
i. r. linteger represent ingi, 120
Polynomials, irreducible, 160 irreducible modulo p, 9:5
normal inodulo p, 94 primary modulo p, 94 primitive, 73 reducible, 160 reducible modulo p, 95 Power residues, 1 J 5 Primary polynomials modulo p, 94
Prime divisors of an integer, 13
1 73
Rank of a cubic curve, 259 of a modal, 19 Rational field K(1), 21 Rational number in a field, 251 Rational points in a field, 216 in the plane. 21f; on conies, 216 on plain algehraie
curves. 254
Real field, 21 Reciprocity law, quadratic. 141
SUBJECT INDEX
308
Reduced residue system, 71 Relatively prime numbers, 23, 240 »
polynomials mod
ulo p, 96 Remainder, least absolute of a modulo b, 13
Remainder, least nonnegative of a modulo b, 12
Remainder, principal of a modulo b, 12
Residue classes, 69, 94 in a quadratic field, 240
Residue classes, prime to n, 71 Residues, biquadratic, 116 P cubic, lib modulo n, 69 quadratic, 116 Residue system, complete, 70 w » , reduced. 71
Riemann's zeta function, 56 Rings, 20 Root, of an algebraic congruence, 73 primitive, 104, 107 Roots, incongruent, 73 Rules for congruencec, 68, 69, 76 » for index calculus, 112
S Scholz's theorem, 123 Selberg's formula, 283, 286 Selberg's proof of the prime number theorem, 288
Siegel's theorem on approximation of algebraic numbers, 263 Siegel's theorem on lattice points on algebraic curves, 264
Sieve method, Eratusthenes's, 61, 62 » » , Brun's, 66 Simple root of a congruence, 85
Singular point, 264 Solution, height of a, 244 initial, 236 of an algebraic congru
ence, 73
Solution, proper, 219 Solutions, associated, 204 Solvability of congruences. 73 Squarefree number, 27
StSrmer's theorem, 297 Sums, Gaussian, 177 » of integral squares, 188
Sylvester's theorem, 50 Symbol [C], 13 Symbol, Jacobi's, 346 » , Legendre's, 133 Symbols o (x) and O (x), 276 System of residues, complete, 70 »
»
»
, reduced, 71
T Tangential point, 260 Tchebychef's inequalities, 60. 62 Theorem, B,achet's, 191 Bauer's, 168 Brun's, 67 Dirichlet's, 66 Eisenstein's, 141 Euler's, 72 Ferwat's, 71 Jacobstbal's, 182 Rummer's, 253 5laillet's, 264 Mertens's, 127 Mordell's, 259 Scholz a, 123 Siegel's, 263, 264
St6rmer's, 267 Sylvester's, 60 Tchebyehef's, 60, 62 Thue's, 263 Wilson's, 99 Wolstenholme's, 128
Theory of ideals, 252 Thus's remainder theorem, 122 Thus's theorem on approximation of algebraic numbers, 263 Thus's theorem on Diophantine equations, 263
Totient of n, 23
SUBJECT INDEX
Transcendental numbers, 36 Transformation, birational, 267 Trigonometrical products, 173 Trivial divisors of f (x) modulo p, 94 » » of an integer, 12
U Unicursal curves, 264
Unique factorization, 16, 239, 262 Units in quadratic Holds, 286
309
Unity, nth roots of, 156 , primitive roots of, 166 ++
w Wilson's theorem, 99 r
, generalized, 100
z Zeta function, liiemann's, 56