Exponential Sums and Their Applications

:1:=1

11 P - I ~ P—j_ 1~ p=1 11=1

ft

E

2·":1: "

r

e e

W.

22

= (d — = - 1l)p. )p.

p

z=1 %==1

Hence by Theorem {)5 it follows that Hence by follows that P

v;r::I .;p ~

w.,..

2 ' «:En

~e

~ (d - 1).;p. (d—

(69)

:1:=1

inequalities (69) (69) show showthat that under under fixed fixednn and and increasing increasingppthe theestimate estimate obtained obtained The inequalities in Theorem 5 has the order order which which can not be be improved. improved. Moreover, Moreover, it can can be be shown shown E19],that thatininthe theestimate estimate (6,3) (63) itit is is impossible impossiblenot not only only to to i.mprove improvethe the order, order, but but to [19], replace €)(d — replace (d (d— - 1)VP by the quantity quantity(1(1— - e)(d - l)VP under any e > 00 either. Now we shall shall cons'ider consider sums sums with a prime-power Now we prime-power denominator. LEMMA4.4. Let Let 2 ~ a ~ n, n, pp be LEMMA be aa prime> prime> n, (a1,... (al"" a1x OtZ + ... + a n 3J A • Then the thefollowing following estimate estima,te holds: holds:

+... +

pa

,an,p) and f(x) f(x) = ,an,p) = = 1,1, and =

2ff

Proof. Let y and zz run runthrough throughcomplete complete residue residue sets sets modulo modulo po-l and p, p, respecrespectively. Then the sum y11 + runsthrough throughaacomplete complete residue residue set set modulo modulo pa pO and, and, tively. + pa-l p°1 zZruns since a ~ 2 and since Rnd p> p >2,2,we we have have

f(y +

f(y) +

(mod p°).

[ch. I, §§ 4 [Ch.

Complete exponential sums Comp.fe,te sums

26

Therefore, 2 ' I(z)

pa

~

LJe

2iri 11"1

2 ' !(,+pa-t z)

,a-l P

p p ~ ~

Q

= LJ

P

LJe

2,ri 11"1

Q

P

y=1 11=1 z=1 %=1

z=1 x:l

,0-1

.

~

= LJ e

2,n

Ie,) Q

L....ti e

P

y=1

P

Cl

P . /'(11)% p ~ 211'1--

P

%=1 -

1

=P L

./(u)

27r1-

e

poe

6,,[f'(y)]. bp[f'(y)].

(70)

y=1 11=1

But then pp

,Q-t P

P

e

~P p

L 66,, pa-l Lop[f'(y)] 6,, [f'(y)] = pa-IT, p[f'(y)] [f'(y)] = p"' ,=1 ,=1

(71)

the number number of ofsolutions solutions of of the the congruence congruence where T is the

f'(y) == 0 (mod p). Since(ai, (ai,.. p) = 11 and pp isis aa prime prime> Since .... , an, > n, n, then at least least one one of of the thecoefficients coefficient,s a,,, p) is prime to p and, of the polynomial f'(y) = polynomial /'(y) = UI al + + 2a2y 2a2!1 + .,. ++ nanyn-l prime and, therefore, therefore, +... obtain the lemma assertion. assertion. T ~ n -— 1. 1. Substituting this estimate into (71), we we obtain of aa special special form. form. Let Let pp be It is is easy easy to t,o improve improve this this result result for for polynomials polynomials of b~ prime, (a,p) = = 1, and ' p0

211'i":C"

=L e ,CIl

S(a,pQ)

CIl

p

•

:c=1

Let's show that under a ~ 2 and n ~ 3 the following following equalities hold: a

(a,p ) =— S(a,pa)

{::=~S(a,pa-R)

if 2 ~ a :s.; nand (n,p) + 1. if aa ~ n +

Indeed, from (70) it follows follows that p0_I

S(a,pdl) =

e

P0 fi,,(nay"').

y=l

or

Hence under under (n,p) ==11we Hence we get get the thefirst first equality equality of (72): (72): p0_I

2wt

—

,,0_2

2iri

a(py)' PO

= 1,

(72) ( 72

,It. 4] ch. 4I, § 4J

27 21

Simplest complete sums

of p, p, which which divides dividesn.n. Then, Now let + 11 and and I'll be the greatest greatest power power of Now let aa ) n + +22 and using l)fJ + and considering considering separately sep,arately the cases + 11 ~ (p -— 1)fl using lhe the estimate estimate Qa ~ pfJ + a ~ 2{j + 2 and n + 1 ~ a ~ 2{J + 1 we obt,ain (If + + npOl-IJ-l !I"-1 z (mod pOI). (y + po-/l- 1.f)" == !1 M + Therefore, 2

S(a,j?)= y=1

z=1

pa_s_I ,.-'-1

~ = L." =

.•,n p,+1

~"".'."'" ,11" e2ft'. --., L."e

aR,

n-I

,.1

~

p5+!

,

z=I %=1

,=1

=pP+l = p,P+l

2 "

,.-'-1

2ft"1 .,.

~"",':,,"'.', L." '"e 1#

,.

£. +" I ('a"n" Up' ,', " 9,"-1)' fl·

,==1

(an,pP+') Hence" since since (an,pP+l) Hence,

= pfJ, we have =

Sea, pOt) = ,1+1

,*-,-1

E

2ft" .,"

,,8 c5,(y)

e

,=1

= pP+l

'I'"

, "-1-2,2ft',,-2wi_!!__

E

ee

,.-a = p"-IS(a,pO-R).

,,=1 y=1 Thus the assertion assertion (72) (72) is is proved proved in in full. full. TH,EOR,EM 6.6.Let THEOREM Letnnand and qq be be arbitrary arbitraiy positive int,egers integers and (a.f) (a, q) = = 1. Then Then for for the

sum Bum

— Ee 1Ft, ' 4, f,

S(a,q) = S(a,q)=>e

2 2w,,as"

.'=1 the estimate

(73) h,olds.

Proof. Since q) = = 1, then under n = 11 and Since (a, q) Wlder n estlm,ate (73) (73) follows follows from and nn == 2 the estimate Lemma 2 and Theorem Theorem 3, 3, respectively: l·espect,ively:

, 4,

,

.. =

~,"""':'", e,2.n£.-, f s=)

=0, 0,

teh'i A;' ~ Vii. e

<

s=1

Therefore it suffices suffices to consider consider the case n ~ 3.

Comp,lete expon,entia./ sums sums complete exponential

28

that for any prime p under a At first we shall show show that

[CI,. I, § 4 [Ch.

~

3, and and (a,p) = 1, n ~ 3, = 11

IS(a,pa)I where

ep(n) =

(74)

if if p
In {~

if if p>n. p >n 6

•

•

Indeed, Indeed, under a ==11 by by (60) (60) nv'i5 <

I

66 iiff p P
if p>n6. p > n6 •

=

Let 2 ~ a ~ nn and (n,p) = p. Then Then p ~ n and, using the trivial trivial estimate, estimate, we we get

IS(a,pa)1 IS(a,pa)I Let finally 22 ~

a

0:

~

~ pa ~ pa(l-~)p~ npa(l-~).

n and (n,p) (n,p) ==1.1. Then Thenby by (72) (72) IS(a,pa)1 IS(a,p a

a(1__) . a(i__) a—I = pa(l-~) = pa-l a ~ pa(l-~) p

n. Apply Thus the estimate estimate (74) (74) is is satisfied satisfied under 11 ~ a ~ n. Apply the the induction. induction. Let Let for a certain k ~ 1. 1)n ~ aa ~ kn this estimate be valid for under 1 + + (k -— l)n We shall show show that that the estimate is valid under 11+ 1)n as welE We +kn ~ aa ~ (k ++ l)n well'. Since 1)n ~ aa — + k'n kn then using equality (72), (72), by by the that11++(k(k— - l)n - n ~ 11 + using the equality it is is plain plain that we get induction hypothesis we I

=

IS(a,pa)I

for any any aa ~ I. 1. (74) is is proved proved for Thus the estimate estimate (74) of q. q. Using Let now qq = prime factorization factorization of Using the multiplication multiplication Let now = t ..•• .• p~. be the prime formula formula (34), (34), we obt,ain obtain

pr

q

a

= [I

a c

,

(75)

v=1

where the quantities bll are are coprime coprime with with PII' ps,. We Wedetermine determinea" a,,with with the the help help of of the the where equalities (zi=1,2,...,s). a"=ab,, (v=I,2, ... ,s). (a,,,p,,) = = 11 and Then, obviously, obviously, (a",p,,) and by by (75) (75)

S(a,q) = = S(al,p~l). ,.S(a8'p~').

Alfordell's Mordell's m,.thod method

Ch. Ch. I,,, § 5] 5)

29

Hence,using usingthe the estimate e8timate (74) (74) and and observing observingthat that the the number number of primes primes less less than than nO n6 Hence, 6 we get , we get the theorem assertion: does does not not exceed nn6, Qt

IS(a,q)I~Cpt(n)pl

(1-~) .. .C,.(fl)P1(1 Q, (1-~) ... C,,(n)p.

J

~n

n' 1- 1 q n.

p) = = 11 by by (72) (72) for for any any prime p under n n>> 2, Note that under 2, qq = = p" and (a, (a,p) p"

""

L..JC

2.i"~" p" p

/

11

n. . ) . =pn-l =p =pR (1- .n

:&:=1

Therefore, in in this this case Therefore,

1I

= q1- n. S(a,q) = under fixed fixed nn and increasing the order order of of the the estimate estimate (73) (73) can can not be Thus under increasing qq the be imimproved. a1x +... + anx R , (al,'" 5(q) be a complete complete rational Let f(x) 1(x) = = alX Let and Seq) rational (al,... ,an,q) = 1land general fann form exponential sum sum of the general exponential ,q 2,..;•f(x) I(~) Seq) = q. (76)

Ec

:&:=1

In Theorem 66 the estimate 1

18(q)1 IS(q)I ~ C(n)ql-i,

(77)

6

where C(n) C(n) == nn , was proved proved for for polynomials polynomials of of the special form f(x) where I(x) = = anx R • With the help of of the the significant significant complication complication of the proof proof technique, technique, Hua Rua Loo-Keng Loo-Keng showed that that under showed under certain certainC(r&) C(n) the estimate estimate (77) (77) is valid for arbitrary complete complete well. A proof of an estimate close close to (77) can be found in in [16] [16] rational sums (77) as welL and [44]. and [441. § 5. 5. Mordell's Mordell's method metho,d Let us consider a complete exponential sum with a prime denominator P,

S(p)

= Ee

211';

At

x+ ...+an xR P

•

2:==1

proposed aa method of such such sums sums estimation estimation based on the Mordell [36] [3-6] proposed the use use of of propproperties of the the system system of ofcongruences congruences

~~. ~.: :.~.~.~. ~.~~. ~.:.~ ~.~R.}) x~

+ + ...++

x: == yi +... + ++ y:

J

(mo,d p), (modp),

(78)

Complete expon,ential exponential sums Complete

30

[Ch. [C1. I, § 5

where pp is is aa prime prime greater greater than n and the variables x1,.. where variables Xl, •••. ,tin run through through complete complete , residue set,s sets modulo p independently. residue First of First of all all we we shall shall prove prove a lemma lemma about about the thenumber number of ofsolutions solutions of of aa congruence congruence system of a more general general form.

arbitrary positive int,egers, integers, q = = LCM (Ql" be arbitrary Tk (qi,..••. ,qn) , q,,) and Tk be the number numberof ofsolutions solutions of ofthe thesystem system of ofcongruences congruences

LEMMA5.5.Let Let Ql, qi,.• •• . . ;, q" LEMMA q,,

~~..~.:".:.~.~.~ ..~.~~..~.:".."~'~'.~""~:~~"~.~'~' } , x' + + ... + tlk ... + + x% == !If +···+ +... (modqi)

x~

1

1 q. l~xlI'YII~q.

(79)

(mod q,,) (mod qn) J

Then qi

qq

qn

2

L.J e

Tk=

"'"

.fajz 12:+ +4n2:") .(4

11'1

...

91

2k 2k

II

'lfn

x1

ai=1

:1:=1

Since the product Proof. Since

equals unity, unity, if numbers numbers Xl, x1,.... . . ,,Yk satisfy the congruence system (79), (79), and vanishes equals congruence system Uk satisfy otherwise, then, then, obviously, obviously, 9

Tk= T, =

~

6'1(Xl

+... - Yk) ••• 6,.. (xi +... - Yk)'

Xl,.··",=1

Hence, using Lemma 2, we get the assertion Hence, assertion of of Lemma Lemma 5: .fa,(x1+...—yk)

q

Tk =

1

qi . . . q,, al....,an Z1,...,ykrl

~

1

e"

L.Je

qj

21ri(.,t z + + ,,:a:") A

o ••

1

2k

9"

r=1 x=1

In particular, under kk = and qlq1==... = n and ... = qn = pP ·itit follows follows from from Lemma' 5, 5, that that 1

i'

2w,

(80)

where Tn is the the number nwnberof ofsolutions solutions of of the the system system(78). (78). LEMMA LEMMA 6. Under any n n ~ 1 and a prime prime pp>>n,n,the thenumber numberof ofsolutions solutions of the the system syst,em (78) satisfies satisfies inequality in,equality T" ~ n! ph. p".

31

MordeJi's method MordeJl's method

5) Ch. I, §§ 5]

. . ,An) , A,, be , A,,) be be the be fixed fixed integers, 0 ~ A" ~ p -— 1, and let proof. Let A1,.. AI, .... ,An let T(A1, T(A I , .••• number of solutions of the the system system of ofcongruences congruences

~~,,~.:".:.~.~.~. . ~. ~.1})

}

xf +··.+ x: == An

(modp) (mod p)

1

~ ~ p. x"

(81)

We shall show show that that We shall (82) T(AI, ... ,An) ~ n!. (82) Indeed, we intro,duce introduce the notation for for the elementary symmetric the following following not,ation symmetric functions functions and the sums sums of powers powersof ofquantities quantitiesXl, xi,.•••.. ,X , and n:

0'1

= Xl +... + 5,,,... 51+ ... +X ii..•• .IX5,,, n, n , ••• ,0'" = Xl

81

+ ... + ... ,sn=x~+I + xn• =Xl+,".+Xn, .. +x:" = .

Let Xl, arbitrary solution solution of of the the system system (81). (81). Then, Then, obviously, obviously, Let ii,. •.•• ,Xxnn be an arbitrary (modp), 81 == At, "'" 8 n == An (mod p),

and using the the Newton Newton recurrence recurrence formula formula

=

110.,, vO'" = 810'''-1 -— 820'''-2

+..."' T 8,,-10'1 ± s", +. J

under = 1,2, 1,2,... under vji = ... ,n ,nwe we have have A,, (mod p). +... (83) + ... =f A,,-lO'I ± A" Since Since pp isis aa prime prime greater greater than n, n, then then (v, (lI,p) p) ==11 and andthe thecongruence congruence (83) (83) is is soluble soluble 110'"

==

A10'1I-l -— A20'1I-2

for 0'". oP,,.FraIn From (83) (83) we we get successively for B'uceessively ,0.,, O"l=""l"",O'n=J1.n

(0 p,, p — 1), (mod p) . (O~JLII~p-l), (modp)

where the the values valuesPI,. ni,. .. ,ltn uniquely by bysetting setting quantities quantities AI, A1,..., where are determined uniquely ... ,A,,. Ani , with one one of of the the permut,ations permutations of But then every every solution of the the system system (81) (81) coincides coincides with of the roots of the the congruence congruence f

•

x

n

fl

I-'lX -

-

1

+... ± Itn == 0

(modp) (mod p)

with fixed coefficients cOlefHcients and, therefore,

T(Al" · . ,An) ~ n!. Now, since P

T,, TR

= =

L

+ V:), +... + + ... + + tin,· y,,,... yf +

T(y1 T('II1 +...

f ' ,,

y1 ,... =1 111 "",'n=l

we get get the lemma assertion: we Tn ~

'P

L

'1 ..... ,,,=1

n!=n!p". n! = n! p•.

f.

I

(ch. I, § 5 (Ch.

Complete exponential sums Complete

32

immediately, that that under any Not,e. the equality equality (80) (80) it it follows follows immediately, any Note. From this lemma and the n ~ 1 and a prime prime p> p >nnthe thefollowing following estimate estimat,e holds: holds:: P P

P P

2:

2: e

Gl1 ••• ,I1.=1

TUEOREM atX

7. Let n

~

2n

alz+...+anr" 2 ' tltx+ ...+an z " 2n p

lI'l

= pllTII ~ n! p211 .

:1:=1

greater than n, (ai, . . ,an,p) = 1 and and f(x) f(x) = = 2, p be a prime great,er (at, ....

+... + anx n • Then

p I'

2 '•f(x) /(,;)

2: e "'1-,-

1 ~ npl-n.

,;=1

Proof. At first we shall consider consider the the case case (an, (as, p) p) == 1. 1. Let Let integers integers A ,\ and J.t vary in the bounds bounds 11 ~ ,\A ~ p - 1, 11 ~ Jl ~ p. Arrange Arrange the the polynomial polynomial f(Ax + + It) in the ascending order of of powers powers of x

+... +

= bo(A,/1) + b1(A,

f(Ax +

p)z"

(84)

observe that that and observe bn -

and and

= aMA"

1 ('\,

= (nanJ.t + J.t) = an-l)'\ n-l. + an....i

D,enote number of of solutions solutions of the system system Denote the number bi(A,1i)

p)

(85)

b,, , n ). is plain, plain, that that H(b1, H(51,. . . ,b It is . . ,,bn ) does by H(b1, ... ... does not not exceed exceed the the number nmnber of solutions of the system made up up of of the the last last two two congruences congruences of the system system (85): (85):

+

1

aDA"

j

b,

(mod p),

since (nan,p) (nan,p) = and, therefore, therefore, since =11 and and (A,p) ('\,p) = =1,1, (86)

By (25) for complete sums the equality . f(x) /(z) 2n 2n p '"" 211'1LJe P :1:=1

p

=

. f(Ax+p) I(~:I:+") 2n 2n

E21r* LJe

'""

:1:'=1

2'1f1

,

3,3 33

AAordell's Morddll's m,ethod method

Ch. I,4 §§ 5]

Hence by by (84) (84) after after the summation with respect respect to A we have holds. A and JI"p, we holds.. Hence p

f(x) 2t*

2ir; —

P—i

=

>2

(p

2n Pp . I(>':z;+~) f(Ax+p) 2ft 2,rs '"" 271" L..Je P

P

>2

p1 r=i :&:=1

A=1

P' p

2n p P _.. 61(>',I')Z+ ... +6.. ('\.~)a:R 2ft bi(A,p)x+...+b,,(A,p)x" '"" 2,...

L..Je

>2>2

P

x1

A=lp=1

:1:=1

ii), ... , n (A>Il) Grouping summandswith withfixed fixedvalues valuesb1b1(A, (A,I-&), ,b Grouping the summands /1) and and using using the the esti. mate (86), we get .

.

2n

p

2n

>2

p

z=i

x1 2n

p . >2e27rt

P

(

Hence by the note note of of Lemma Lemma 66 we we obtain obt,ain the the theorem theorem assertion assertion for for the the case case

(an,p) = = 1:

f(x) 2n ./(z) 2n

p

Ee

211"1-

~ ~

p

:.:==1

Ee 11'1-,p

2 ' I(z)

n n'p2n < n 2 n p 2 n-2 (p -1) , p(p—i) ·

P

1

< npl-n.

x=1

Now we show show that that the general Now we general case (a1,. (al'.'.', an,p) == i1can can be be reduced reduced to to the the case case when leading coefficient p. coefficient of the polynomial is prime to p. Indeed, let (a.,p) = 1 and a s +l == ... == an == 0 (mod p), 1 ~ s ~ n. Thenwe Then we obtain obt,ain . .

pP

~

L..JeS

2 ' f(z) f(z)

p ~ 271'i 2,rs

-71'1-

= = L..J e

P

z==l

atx+...+aax 41:&:+ ••• +a,x' P

1

1——

1

1——

x=l

The theorem is proved in i.n full. full.

Not·e. AAsubstantial subst,antialimprovement improvementofofMordell's Mordell'sestimate estimatewas wasobtained obt,ainedby byA. A.Weil Weil[48], [48], Note. who showed,that thatunder underprime primepp> (an,.... . ,, an, p) = who showed, > nn and (an, =11 the the estimate estimate .

p

is is valid.

Complete exponential sums Comp,/ete

34

(ch. I,I, §§ 6 (Ch.

§ 6. Systems of of congruences congruences

is the use of the estimation for One of the main main points points of ofMordell's Mordell's method method (§ 5) is the number number of of solutions solutioll8 of of the the congruence congruence system ) ~1.~ ~'.~ ~~.'~ ~~ ~ ~.~.n.} n_ni ft :

.... :

X11...VXn.Y1r...TYn xi +···+ x: == yf +···+ y:

(modp),

fl

L

I

where pp is is a prime Hereafter, congruences congruencesofofthe the same same form form but where prime greater greater than than ,z. n. Hereafter, with respect to to distinct distinctmoduli moduli being bei.ng equal equal to togrowing growing powers powers of a prime p will will be be great importance. import,anee. For For the thefirst first time timesuch suchsystems systemsof ofcongruences congruences were applied by of great Yu. V. V. Linnik [34] for the the estimation of Weyl's by Vinogradov's Vinogradov's method. Yu. [34] for Weyl's sums by n(n4+i) 1 ~ 1, 1, kIe ?~ n("4+ ), p be a prime prime great,er be greater than than n, n, and let let Tk(pn) Tk(p') be tbe of solutions of the system syst,em of congruences the number of congruences

LEMMA LEMMA7.7. Let Let n

(modp)

Then

)

n

(87)

n(n+1) 2

Tk(p'2)

Proof. Under Undernn==1 1the thelemma lemmaassertion assertionisisevident, evident,so soititsuffices suffices to consider the case n ~ 2. Take Take qlqi=p,...,qn=p" = p, ... ,qn = pnin inLemma Lemma5. o. Then we obt,ain obtain

"

n(n+1) 2

ajl

P'

e

I

x=1

of the the summation summationover overat, al,... We split up the domain domain of ... ,a'n , into two parts: part,s: p't

p

aj'tl where the summation in where

a1,...,a't

2:1 is extended over over n-tuples n-tuples a1 at .... an sa~i,sfy an which salisfy . .

p\a2,p

\a3,...,p

and for for 2:2 n-tuples are taken t,aken into account, for for which which at least for for one one of of E2 only those n-tuple8 ,i in the interval IJ interval 22 ~ v1J ~ n p"-l is not a divisor divisor of all.

__________ ___________ Ch. ch. 1,I,

S6]

Systems o(cosgruenccs

35

case,determini,ng determiningb1 b1,. In the first flrs't case, t • • •. t,b,, 6ft with the help of the t,he equalities equalities

..., a,,

a1 = b1, a2 = b2p,

we get

2irs (!-1!+ ~.".' '., '. .211'i---+ LJe .. "

( . 811:

II

4

=

. ''.' " 2wi "1~ =~ LJe

Sa )..

" P"

pU +...!L;, ."

z=1

.I.

+

.I.

"

II+.,,,a

•••

.=1

,P

~.' . 211'" =Pn-l LJe

.=1

61.+···'+...." . ,P ' ·

Therefore

f:

_~z +...+ -;:") P p"

2iri ( e21l'i

2k

2k

=

p2R1t-2ft

~-1

t

p

e 21I'i

'l.+..~H.. z..

2k

211

.=1

bix+...+&,,x" 2n

P

and using using the note of obtain and of Lemma Lemma 6, 6. we obt,ain

.n). 2'

2 " . .!!.=.+... +~ . . ~.".""" ..... Le .. , ,n

PD p.........

(....

. ...

2k

.

x=1 al,···,4" 11:=1 P

P

(88)

P

b1,...,b,,=1 x=1

In the second v in the second case, case, there exists exists an integer integer" the interval interval 22

v ~ ~ II

p.-l ~ a"pM-II. Therefore" Therefore, ,,-1. . . . ( alP ,G2P. ..-2 •...

where 2

~

") ,a,,,p") ,G,.,P .= = PR-a ,

n. But a ~ R.. But then

Q

a-I (lIP

= ClIP"-a , • • • ,tit. = VaP.a-a , 1..

(bi,...,b,,,p)=1, (b'l' .... ,b., p) = 1,

~

and using Lemma Lem,ma 4, 4, we we get ~ LJe '..••..."

..

6)

•

. ..........•... 61ft

.:==1

= =

(·.·. . •

a.• +

'

f: z=1 .:==1

.•••

+., .".").

2wi ehi

,It bix+...+b,,z .t.+.;:......

= p,.-a

' Ee ,"

z=1 .=1

211'"

'11:+ ... +6,,11"

,. P"

that nn such such tha.t

[Ch. (rh. I,I, §§ 6

Complete exponential sums Complete

3,6 36

n(n+1) it follows that Hence, since k ~ n(~+l), follows that pn "

>22 >2e Le L z1

anx") ,n

2iri. (al~ -+...+ P

211'1

2

41 ••••• 4n

2k 2k

~=1

n(n+1) —

observing that that under n Now, observing

—

(n

2

(89)

~ 2

l)2k_1 > 2k -— (n — 1)21c ~ 2k(n -— 1)2k-l nfl2k - l)n-l ~ n! (n _1)21: n(n — > n(n

from (88) and (89) (89) we we obtain the the lemma lemma assertion: assertion: n(n+1) 2 Tk(pn) =p---2-

Ll + L2 >22)

n(n+l) (

)

(

Al J•• 'J4 n

~n!p

41 J' •• ' . "

n(n+1) 2nk2nk— n(n+l)

n(n-4-1) 2nk2nk— n(n+l)

+(n_l)2k +(n—1) pp

2

2

n(n+l) n(n+1)

2nk— 2k 2 nk--
~

Note. Let Not,e. Let Tk(P) Tk(P)denote denotethe thenumber numberof ofsolutions solutions of of the systcm system (87), when the domain variables variation has the form of variables form 1~

1 ~ Xj ~ P,

Yj ~

(j=1,2,...,k). (j = 1,2, ... ,k).

P

If m m is a positive integer, then then under under PP = mpn mp" IT positive integer, 2nk- n(n+1) n(n+l)

(mn)2kp2' Tk(mpn) ~ (mn)2k p

(90)

22 .

Indeed, using the the complete complete sums sums property property (26), (26), we we get get m,n 2 .• """

L...J e

11' I

(~+ ••• +4,.X

n

,n

P

p"

)

"""

x=1

0(8 .2:+ 1

2

=m m L...J e =

11' I

p

•••

+a.nx

n )

pn,

;1:=1

and, therefore, n(n+1) n(n+l)

,Hp"

p

T,(mpn) = p - -22- """ L...J ..• """ L...J ap=I 41=1

=m

0

11' I

(!!=.+ \p P

+ «-ax") •••

"n

2k

1

pH 2,ri 2 (fi!+ (!!!+ + A"x" 1 i" ,n " ... ,," >2 ... >2 >2 e

,"

L'" L Le

&1==1

=

""" L...J e

a,,=1 4 n =1 x=1 x=1

n(n+1) n(n+l), 2k p - - 2 2 -

mp" m pn .2

4,,:z1

x=1 x=1

n

0

)

2k 2k

____________________ Sys:tems of ofcongruences congfuen,ces Systems

Ch. i,I, § 6] 6] ch.

37 37

Hence, since by Lemma 77 Hence, 2 k—

Tk(p")

2

we obtain the estimate estimate (90). (90). we . ,, X n run Let L.J"'1,. ~!Q •• , . denote the sum, sum, in in which which the the summation summation variables variables x1,. Xl, ••• through complete complete residue sets set,s modulo modulo pa pO' and and belong belong to to different different classes classes modulo p. .

= a1x 1(x) I(x) = ala: + + anx n , + ... +

LEMMA8. 8.Let Letppbe beaaprime primegrea,ter greaterthan than n, n, aa ~ 2, and LEMMA

with the help of the equality Let Sa( aI, . .., . ,,an) be defined defined with e,quality Let Sa(ai,. ,ol 2m /(%1)+ •.• +/(x,.) 2ir, 5'a(al,'" ,an)

L

=

,0

e

Then a ) -S eN (a 1,···,n

{

P(a-i)ns1 (b1 , • • ., bn )

if a"=pa-1b,, otherwise. otherwise,

o0

(v=1,2,...,n), (v=1,2, ... ,n),

Proof. Let Let us us change change the the variables variables X"

(v=1,2,...,n). (v= 1,2, ... ,n).

= YII + P0'-1 Z.,

different classes, classes, then then the Since Sinceby bythe theassumption assumptionthe the quantities quantities Xl, xj,.••• ,X , x,,n belong to different belong to to different classesmodulo moduloppas as well. well. Therefore, Therefore, using using quantities Yl,' different classes quantities Yi,.... ,Yn , yn belong that (mod pa), + f(yv we obtain .

.

Sa(ai,. .. p

e

=

>2

n =p

e

%1 ••••• %n=1

111,···.lIn

,.-1 P

2"'; /(1/1)+ ...

L

2

e

+/('n) 0, [f'(Yl)] ... 0p[/'(lIn)].

pO

(91)

lit •• ... 11,.

and (at, (ai,...... ,, an,p) Sincefl(y) f'(y) = at a1+2a2Y+ +2a2y+.. Since ... +nany"-l, then under prime p > n and p) = = the congruence congruence f'('II) f'(y) == 00 (mod p) can values of 11y 11 the can be be satisfied satisfied by at at most most nn— - 11 values from different different classes classesmodulo modulop.p. In In the the sum (91) (91) the quantities Yl,.,., Yn from belong to y, belong different different classes classes and, and, therefore, . .

1

.8

=

{

0

((al,. )= , an,p) at, . ,an,p = p, (as,... ,an,p) if (at, ,an,p) = =1.1. 'f lif

.

[Ch. I, § 6 [Ch.

Complete exponential expon,ential sums

,a,,) ,an,p) But then then by by (91) (91) the sum S'UDl Sa(ai,. Sa-Cal"~ ... ,a under (a1,... (al," .,an,p) n ) vanishes under (ai,... (al , ... ,, a,,,p) an, p) = = p, then .

pa_I pClr-t

= 1.1.

If

2wi

E

Sa(ai,...,an)=p Bo(al"" ,an) = pR >2

e

't,··"Y" Thus

.. ,anp-l) f ... -— {pnSa_l(alP-l, S o (at,·· · , an ) 0 —

if (a1,... (at, .. . ,an,p) =P 'f (a1,... (aI," .,a,.,p) = 1if = 11..

0

1

92 (92)

Applyingthe theequality equality(92) (92) ,anp') we get Applying to to 5 0 Sa_i(aip',. - 1(alp-I, ... . ,anP-I) .

S (a a)Sa(ai,...,an) 1, • • • ,n Q

{

- 2 , ... ,anP- 2 ) ' f (al,···,a n ,p2) =P, 2 P2n S 0-2 (alP 1if (aI,...,an,p2)=p2, 00

otherwise. otherwise.

Continue st,ep we we obtain obtain the thelemma lemma assertion: assertion: Continue this this process. process. Then after a-I — 1 step . Sa-(al"'" an) = = fp(a_1)nSi(aip_(a_1),...,anp_(a_1)) l)n S1 (atp-(a-l), ... , anP-(a-l») .

{p(a1

=pO_l, if (al"'" an,pa-l) = pQ-l, otherwise.

O 0

LEMMA9.9.(Linnik's (Linnik's lemma). lemma). Let A1,. LEMMA At, ..... ,,A,, An be fixed fixed int,egers, integers, pp aa prim,e prime greater greater than T*(A1,. . . n and let T'*(Al, .. " , As,) be the number An) be numbcl' of of solutions of the tl1e system system of of congruences congruences

~1.~ '.~~~.~ ~~ '«~n'lOo.d .Pp'>R) } , (modp)

:'.

•

xi +···+ x: == An

(modp")

vaJ.·ia,bles run through complete complet,e residue sets sets modulo pn and belong belong to t,o variables different class,cg modulo p. Then different classes mod ulo p.

where the

n(n—1) n(n-l)

··T*(,xt, . .. ,A n ) ~ n!p-2n!p 2 n - 2 x 2 + ... + anx Proof. Let 1(x) = alpn-l x + a2p a,,x"n and according to the notation Let lex) according to a2p"2x2 of Lemma 8 •

2ir,

>2

e

~l, .... ;l:n

Using Lemma 2 we we obt,ain obtain Using pfl p"

n

~ll ••• ,Xn

v=1

L II 6pv(x~ + ... + x~ >2 pfl p"

n(n+l) n(n+1) 2 T*(A1,.... . . p-2-T'*(Al' p

,,=1

,An) = =

,

>

at,... ,4"

L,:I:"

>.

Xlt •••

2' (

2 11'1•

e

41

'

Xl

Av ),

+...+2:" -..\1 ++ ••• :1:;.'+•.. +:1:: -..\. ) ..• +4" + P p" 1,,

_____________ ________ 39 39

Systems o(congn,ences Sys,tems of ~

Cit. I, § 6] '] ch.

.. ,I aM is wbere the summation summation with with respect respect to to a11. 61 I ••• i,8 extended extended over over the the domain domain where 1 ~ (11 E:; P, . · · ,1 ~ a.. ~ pM.. Henceobservingthat Hence observing that

E .".•. .-.

.' .,(. . .1+.··+·. .. .:+,.+.:). , +...+..

e21F'·1

Sl ••• ·,·" -

2

= =

pfl

e

Sit···.·..

we have a T*('\l' , ... An) . . I

.L

A ?' -2 __ A" ).. . —2 •(.. . 41 l + +.... ~ S (.. ..-1 \. 1ft k , 00. ,. P" ••. = P L . . , ' ... L..,,,alP , • ·..· I a'Rr'

n(R+1) ,.(R.+l)

.pp .

- -2 - ~ 2

'.........

ajl

at=1

.'

a.=l

Determi.ne t,he ... ,b. with the help help of of the the equalities equalities Determine thequantit,ies quantities~,...... (11 = bl , ai=bi,

"2

,a" =

= pb21.'"

p,,-lb._

Acoordi,ng to Lemma 8 According

S.(alp·-l, ... ,a.) .

,b,,) = {. pfl(tI-. l)Sl (b'I' _. - ,b,.)

(v == 1,2, 1,2,... =p"1b,, (II =p.,-lb" ... ,n), otherwise, otherwi.se, if

o U

I,

a,, 0"

, E

where

P

-

Therefore, n(rt—3)

p

2

T(A1,.. .,A,,) p

=

Si(bi,. ..

, L

p

p

=

2W,

P

-

Now, using Lemma 2, we obtain n(n—1) nCn-l)

T*(Ai,...,A,,)=p 2 T:*(;\l, · · · , ~ ..) p-2-

..ii

,p

L II 6"(%1 +·... + a:: - A.,)

=

.,=1 tj,...,Xn.. v=1

.1,.0.,~

,p

3:1

nii

L E I16,(:,;r +... + %: - A,,),

"'0,... =1

,.,=:;:1

_(ft-l) n(n—1) 2 = p-2-T(Al' . _. .. ,Aft),

•

[Ch. I, §§ 6

Complete exponential sums Comp,/e't••xpan,ential sums

40

where T(Al"'" nwnber of of solutions of the system of of congruences . . , An) is the number

1 ~ x" ~ p. 72i

I

n! by (82), the . . ,, An) ~ n! Hence, because T(Al" .. the lemma lemma assertion assertionfollows: follows: n(n—i) n(n-l) 2 T*(Ai,. T*(Al"'" An) ~ n!pn(n-l)/2. . .. An) ~ p-2-T(Al"'" .

COROLLARY. Let of solutions of the system of of congruences COROLLARY. Let T:(mpn) be the number of

x1 +

—

0

(modp)

1~

i

(modp)

Xj,Yj ~

mpn,

i- j => Xi '¢ Xj,

Then

Yi ¢.

Yj Y, (modp).

,z(n+i)

2 —

2

(93)

. . ,, Yn runs runs through a complete Proof. Since Sinceeach eachvariable variable among among x1,. Xl, ••• complete residue system modulo p" m m times times (under (under the the additional additional conditions conditions i f; jj => Xi ¢. Xj, Yi tem modulo pR y1 ~ y, (modp)), then using the lemma we obtain Yj (modp», then using the lemma we T:(mpn) = m 2nT:(pn) ph

=m

L

2n

+···+Yn, .. o,yj+ ... +y:)

T:*(YI

>

1I1,.. ·,Yn

n(n—1)

p

2

2 2_ n(n+1) 2 = n! m2tlp

Sums with exponential function §§ 7. Sums Let a be an Let an integer, integer, in m form

2 ~ 2

and q

~ 2

be coprime positive int,egers. integers. Sums of the coprime positive P

aqr:

S(P) = ""'" LJe 211'im :.:=1

rational exponential sums cont,aining containing an an exponential exponential function. function. In the invesare called ra,tional invessome properties properties of of the the order order of qq for for modulus~ modulus m. tigation of such sums we shall need some

Sums Sums with exponential ~ponent;al function function

7] Ch. ch. I, §§ 7)

Let p be a prime, prime, m

41

==pm1, pm!, rr and andr1rl be bethe theorders orders of ofqqfor for moduli moduli m m and and m1, ml,

respectively. We shall shall show, show,that that ifif rr respectively. We

~ r1 rl and

p\ml, then the equality equality

rr=pr1 =pTl

(94)

holds. (mod m) we 1 Indeed, since since ml mi\m, 1 (mod Indeed, \m,then thenfrom fromthe thecongruence congruenceqT qT == 1 we get get qT qT == 1 (mod ml) and, \r. On (mod and, therefore, therefore, r1 71 \r. Onthe theother otherhand, hand,frQm fromthe thecongruence congruenceqTi qT l == 11 + uimi, is a multiple of of (mod ml) mi) we obtain = 11 + (mod obt,ain qTl qTt = Ulml, where where u1 UI is an integer and m1 ml is assumption. But then p by the assumption. But p qPfl qP T l

= (1 + Ulml)P == 1

(modm) (mod m)

:F rr and and pp is is aa prime, prime, the equal~ty (94) follows: follows: 1 rrj'\p, TT1 \p, T1"l-1 = p, r=pr1. r = PTlrrr'=p, the prime factorizationof ofm, m,Tr and r1 now ~ rn be be o,dd, odd, m m == pfl ... P~' be the Let now prime factorization 71 be

and r\]JTlr\ pr1. Since \r, Ti Since r1 71 \T, Tl

. . .

orders of of qq for for moduli moduli m m and PI pi .... .-P"~ respectively_ respectively. the orders We determine the quantities flu,.. We det,ermine PI, .... ,,P. with the help of the conditions conditions _ qT' qTl —

11 --=

U '0 p{Jl . . . pP,', 1 .•• 8

(u p1··· 1· (u0, 0, P1 -- 1. .. . p , ) =

(95)

For definiteness definitenesswe wesuppose supposethat that in in the prime factoriz,ation factorization of of m in those primes, which which For satisfy the the inequality inequality a" a,, > so a" a,, > /9,, satisfy > f3,,, {3", are put at the the first first rr places places (0 (0 ~ rr ~ .s), s), so P" ur.der IJz' > r, r. Further let under v1/ ~ rr and a,, a" ~ /9,, p" uI~der _ (JI m1 -=p1 mI PI

(Jr

.. 'Pr

. .

Qr+l Pr+l •.

a, a

·P. ·

From the the definition of Tlr1 and and the equality (95) it follows that the order From definition of equality (95) follows that order of of qq for for modulus ml m1 isis equal equal to to 71 r1 and ~odulu8 qri + u1m1, qTt = 1 + Ulml,

(uj, Pu (Ul' PI ... .pr) . Pr) ==1. 1. .

Let us show the validity validity of the equalities equalities

m = + (u, PI .. . Pr) = 1 and T = - T l . (u,pi...pr)=1 ml Indeed, let m2 = pm1, where p is any number among the primes Pu,... Indeed, m2 = pml, where p is any number among the primes PI , ... ,Pr.

qT=1+um, qT 1 urn,

(96) Let 'T2 r2

denote the order denote order of of qq for for modulus modulusm2. m2' Obviously ObviouslyT2 T2 ~ r1 7"1 (for (for otherwise otherwise we we would would 1 have m2\qT' pm1\Ulml, \uimi, which m2\qr — -11 and pml which contradicts the condition (uj, (UI' p1 PI .. ..p,.) Pr) == 1). 1). . Since, besidesthat, that, p\ml, p\rn1, then by (94) But then Since, besides (94) r2 72 = pTI. PTI. But .

qP7i1 ~ = (1 (1 + + Ulml)' q,,'r qT2

qT2

where u2 'U2

+ u1m2 (modp1 ... .. Prm2), == 1 + Ulm2 (mo,dpl .

= 11 ++u2m2, = U2m2,

u1 (mod pi == Ul PI .. . p,.), Pr), and, a.nd, therefore, (u2, (U2' Pj PI .. . Pr) Pr) = = 1.1. Thus Thus . . .

T

qq

2

= 1 fU2m2, +u2m2, = 1

.

m2 ml

2 (u2, Pl Pu.. = —Ti. (U2' .. ..Pr) Pr) = 11 and T2 T2 = m 7I •

Repeating this being equal equal to to each each P" p', (IJ (ii = 1, 2,. , this process processa,, all— - /9,, P" times with pp being 2, .... ,r), obtain the we obt,ain the equality equality (96). (96). .

(Ch. I,I, [Ch.

Complete exponential sums

42

117

2, pp be a prime, 2, prime" m = pmj, pml, rT and andr1 Tl be be the theorders orders of of qq for for and p2\m, then under moduli mt, respectively. respe,ctive1y. If Tr 1 Tl &l1d undm- any a not not divisible divisible moduli m m and andm1, by p byp

T'HEOR,EM THEOREM8.8. Let Let m ~

7r

. at/' —

2w, '""'. . . . . . . . . . . 2wi - m

L..."e

= O'. ,

•

(97)

1:=1

Proof. Let LetTTdenote denotethe thenumber numberoforsolutions solutionsof ofthe t,hecongruence congruence

== fl"

qUsi,n~ Using

(modm), (mod m),

Lemma 2, we ~ obt,ain obtain

I

7

— qV) =

T=

In

7

2w;

I

In

m a=I z=1

T. But then On the other hand, hand, obviously, obvi,ously, TT = T. In

L '"

2

~ 271 2tri -(' L..." e '"

mT = = mr. = mT mT".

a=I 4,:=1 x=1 ~=1

(98)

Therefore, by (94) 2

(e,p)=I In

=

E s1

= mT mr =

r

in 1ft

Ee27tmn

—

L a1 (CI,,)=, 4=1

USj "'1

,Ta

E Le

. Cll'. 22 21ft -;;-

gjI z=I .:=1

r .a.L..." e ,. '""' . . . . •.:. . . . . 2wi -.-

2

1M

.,=1 x=I

= mT = mT -

—

=

p'Jml TI = 0. O.

4t=1

Hence for for any any CIa not divisible by pp we we obtain obtain the theorem assertion Hence divisible by r

T

.,.

" ,. . . . ~hr' --L..."e ..

E

=0. = o.

1:=1

Let us show that ifif at at least 1e'astone one of of the the theorem theorem conditions conditions and

p2\m

Sums expon,ential function function Sums with exponential

Ch. ch. I, §§ 7]

43 43

satisfied, then the sum is sum (97) (97) might be be not not equal equal to to zero. zero. is not satisfied, Indeed, > 22 be be an an arbitrary arbitrary prime, prime, m and qq be be aa primitive primitive (a, m)=l, m)=1, and Indeed, let let pp> m = 2p, (a, root of 2p. we have 2p. Then we rr

49#1:

2,ri— 'Jr. -;;;;m

" ' " e2 L...i

r". =

x=l

211'1' _4(_2_x-_l_>

P

L

e

"i

2,

= -e =—e

a 211'1 2

2=1. = 1.

x=1 z~p+l 2

r1 is obviously satisfied, but In this example but p2 p2 ~2p, and the the second second example the condition 7r ~ 71 condition is violated. Let now primitive root of p2, = pP — - 11 and, p2, and and q = gP. Then rT = Let now m m = p2, p2, 9g be a primitive using using the the equality (98), we obt,ain obtain

p—i 2ir1—

p2

p(p-l) (a,p)=1

p1 2,ri — L L e ,2 1) pp22

=

1

p-l 211'1 ag"· 2

= pep ( - 1) a=i 4=1 x=1 :1:=1

(a,p)=1

Le

%=1

,

"'"

L...i e ( — 1) aj=i x=1 :1:=1

—

r

max

2 ajgW 2 .. al'z

1'-1 1

e

2tri

211"1--

P

— 1, = pP -1,

aq21

m ~.;p:::l.

In this case the condition condition p2\m p2\m was was fulfilled, fulfilled, but 7 = 1"1. conditions, under under which which the the complete completesums sums S( S(r) Another form of conditions, T) vanish, is shown in the following following theorem.

pr

.. . P~' be THEoREM9.9. Let Let m m == 1 ••• be prime prime factoriza.tion factorizationof ofodd oddm, m,7 r be be the the order of THEOREM for modulus modulus m m and and the quantities /3k,... determined by by the the equality equality (95). (95). If If q for PI, ... ,,{jIJ be determined there exists exists vi/ such such that that Q II > PI! and aa¢.O0 (mod p~l1-fJP), then there

rr

a9~

"'" L...i e211'1m :a:

Proof. Chose Proof. Chose that value value

II,

alJ

=0. = O.

=1

which satisfies satisfies the conditions

> {ill'

a

=

and write a in the form a = = pap -fJp -"'( at, where '1 ~ 11 and Let a1, where and (a', (a', p,,) p,,) = 1.1. Let m == P~" -p" -'Y m', rn' r' and for moduli moduli m' m' and m", m m' = puinU, pI/mil, 7' and i-" T" be be the the orders orders of q for respectively. Since p~., \m, then respectively. then pe l1 +"'t\m'. But then pe., \m" and and by by (96) (96)

T,, =

m" T1, mi

m'

m'

mt

mil

,, = —Ti m' m'" T = -". —r ,, =p,,7". PvT 7 = -71 = mj rn"

Complete exponential sums Comp,/ett

44

[Ch. I, § 17 [Ch.

divisibilityofofm'm'bybype,,+'Y p' it it follows follows also, also, that p~ \m'. Thus From the divisibility

r' t= r",

in' =p"m", = p,,rd', m'

and

TI

(a',pp) =

1.

Therefore, by Theorem 8 T'

Le

.

•

a't/'

21r'-·

m'

=0.

~=1

Since qT in') and in) and m'\m, then Since qr == 11 (mod (modm) then qT qT == 1 (mod (modm') andT1 r' is a divisor divisor of of T. T. Now, using the property property (26), (26), we we obtain obt,ain the the theorem theorem assertion: assertion: r 2,ri. '" L...Je m = L...Je T

('

T

•

' " 211';.!!........

a' ,:II

211'1 - ,

m

.,.1.

= T"

~=1

x=1

a' ,"

r ' " 211"1-,

L...Je

m

= O.

x=1

that the in Note that theTheorem Theorem99requirements requirement,s can can be be relaxed, relaxed, namely, namely, the condition of m being odd may suffices may be h,e omitted. omitted.InInorder ordertotoprove provethat thatit it suffices(see (see[321) [32]) in determining the quantities the. PI, .... . ,,P. equality fl8 to use the equality 1 -- u 0p(Jt 1

q(IL+l)11 q(l&+l)Tl _

pfJ,

" •• ,

where IJ = 1, if m 1 (mod (mod2), in == 0 (mod 2), r1 ;1 == 1 2), q == 33 (mod4), and It = = 00 otherwise, instead instea.d of the the equality equality (95). (95).

THEOREM 10. Let Let m ~ 2 be in) = TUEOREM 10. be an arbitrary arbitrary integer, int,eger, (a, (a,m) 1, (q,m) = 1, (q, m) == i, 1, , and Tr be modulus m. m. Then the order of of q for modulus Then the the estimate estimat,e r

T

Le

(s,. mT

.(aq' +bZ)

2 1r' •

()

~.;m

z=l

holds under any any integer integer b. b.

parts Proof. Since Since the fractional part,s

{a::} m

and and

{b:} T

have the same by (28) same period period T, r, then by (28) the sum (99) (99) is aa complete complete exponential exponential sum. sum. But then under under any any integer integer z

r Le r

2'

11'.

r 2 ' (492+.1: + 6l1:+6Z) bz\ bx+bz\ (49.:6 + 6X) m T = Le 1r. ---m= T

-T-

:1:=1

te x=1

:1:=1 21ri

(ll:; +":)

=

=

te

z=1

. 21ri

(Il':" +":)

,

Distribution of ofdigits di6its in in complete comp./ete period period of 0.(periodic p.riodic fractions fra·ctions

8J Ch. ch. I,I, §§ 8)

Therefore,

~ 211'i ". L...Je

(ata: +6:t) r

m

,211'i

=E

r

~~1

r Ee

r

2

(4f~, r + 6~ )

45

2

r

m

%=1 :.:=1

rr

~

L-Je

z1

2 ' #.

faqzz bz\ 2 (4',##%+":1:) 2 -' m

T

:1:=1

because the congruence Hence the theorem assertion assertion follows, follows, because (modm),

q:t=q"

1 ~ x,y

~ T,

=

satisfied for for xx = y only: is satisfied 2 1~ Ee (8""iiI +6:1:) r ~ - L...J r

9':

2 2iri' 11'1

~ 211'1 L-Je

(a,Zm

%

+6:t) 2 T

z=1 x=1 x1

x=1

T %=1

1

r

T

b(z—y) m "b,(x-,> m

,

. (q:D -q~)z

211'1 -, - ~ 21r1 =— = - ~ L-J e T L...Je

=m =:;:

m m

%=1

T x.,=1

b(x—y) ~ 211'; 2wi "(x-,) r On1 ( qX L...J ee T T

)

-—

q11 q11)

= m, in, =

x%.11·=1 ,y= 1

+hX) Ee2.(4'" m r rr

11'1

2

2

~

y'iii..

x=1

§ 8. Distribution of of digits digit,s in in complete complete period p,erio,d of periodic fractions 8. Distribution Let prime to to m. m. In Let;; be an irreducible irreducible fraction and qq ~ 2 an arbitrary arbitrary integer int.eger prime In writing the q-adic following infinite infinite pure pure recurring q-adic expansion of the number number ii, the following "decimal" to the base "decimal" base qq arises: arises:

~ == [~] ••, ['J ++ 0."Y1"Y2 · · ·."Yx7x· ...,

7x+r "YX+T = =7x "Yx

. .

(x (x

1), ~ 1),

(100) (100)

with a period rT being beingequal equalto tothe theorder, order,totowhich whichqqbelongs belongs for for modulus modulus in. m. Let N!:)(6 1 .••• 6n ) denote the number number of the times times that that the thefollowing following equation equation is is .

.

satisfied: "')'~+1

• • • 7x+n

= 61 • • • 6n

(x=0,1,...,P—1), (x = 0, 1, ... ,P - 1),

where p ~ T and 61 ••• 6,, where P 6n is an arbitrary arbitrary fixed fixed n-digited number in the the scale s·cale of q. In In other words, N!!)(61 .••• on) 6,,) is the number number of of occurrences occurrences of the given given block block 61 ••• On .. .6,, of digits digits of of length length nn among among the first P blocks of blocks . .

.

. .

"')'1 •• ·"')'n, 72...7n-f-I "')'2.·.')',.+1 , ,..., •.• , 71...7n,

"'YP.··"')'P+n-l, 7P".7P-4-n—1,

Complete expon,ent;al exponential "ums sums Complete

46

[Ch. 8 [rh. I,I, §§ 8

by successive successive digit,s digits of the expansion (100). formed by fonned (100). The question question about about the the nature of the distribution of digits digits in in the the period period of of the the The distribution of fraction is is closely closely connected connected with with properties properties of of rational rational exponential sums containfraction;; function. This connection is based based on on the the possibility possibility to to represent represent the ing exponential function. connection is .. . 6n ) in terms of the number quantity N!:)(6 1 ••• number of of solutions of the congruence congruence aqZ

== y + b (modm),

o ~ x < P,

1 ~ y ~ h,

(101)

where bb and and h depend . where depend on on aa choice choice of of the the block block of of digits digits D1 ." . D We denot,e denote the n • We number of solutions of the congruence congruence (101) (101) by ~P)(b,h). (b, h). .

LEMMA10. 10.Let Letquantities quantitiest,t,b, LEMMA b, and h be defined defined by the equalities t

-s—, 0.61 ••• Dn = -, qfl qR

b = [:,::]., b=

•

(2 + 1)m

h

—

[tin

qfl

Then . .

=

.

h).

Proof. Let xx be any any solution solution of the equation 7x+1

=

. . .

•

.

S (0 ~ x

.


(102)

Then we we obtain obt,ain from from (100) (100)

Iaqx) 8x aq ,;} = O."')'x+l • •• 'Ya:+n · • · = 0.1':1:+1 · • ·1X+71 + -qn { -m

= 0.61 ••• 6n + OJ: = t + OJ: qn

I

q"

where 0 < < 8:s; <1. < 1.Hence Henceititisisplain plainthat thatthe theequality equality(102) (102)isissatisfied s.atisfied for for those those and and only those x, for only for which which

t. { -a,qZ } <--, t +1 -< qfl qR m qR

1mj

qfl

O~x
(103) (103)

and h itit follows Since from the definition definition of bband follows that t < b b+1 !—b ~ — .!.+ 1and and b+h b + h ~ 2+1 t + 1 < bb+h+1 +h +1• — qfl qfl 111 qn in m m m then the the inequalities inequalities (103) (103) are equivalent equivalent to the inequalities inequalities

!m

<

{a'Ininqx )} ~ b +\ m (

O~x
(104) (104)

Ch, cii. I, § 8]

Distribution of 0" digits digits inin complete compo/ete period period of ofperiodic periodic fractions (racaons

47

aqz to to modulus m. Then We use use y Y to to denote denote the theleast leastnon-negative non-negative residue residue of of aqz modulus m. Then We

for those those and and only only those x, which !!l- == inequalities (104) (104) are satisfied s,atisfied for \vhich = ~ and the inequalities '" m congruence satisfy the congruence

o ~ x < P,

(modm),

b < y ~ b + h,

or, that is just just the the same, same, the thecongruence congruence aq~

o ~ x < P,

=y+b (modm),

1~ y

~

(105)

h.

But then the coincideswith with the the number the number number of ofsolutions solutions of of the the congruence congruence (105) (105) coincides solutions of of the the equation (102). of solutions (102). The T'he lemma lemma is is proved. proved, of m, rn, Tl r1 the order . . . P:' Let m prime factorization factorization of order of of qq for for Let m be be odd, odd, m rn = p~1 ••• p80' the prime be determined determined as as in § 77 with with the help of . . . ,P8 modulus PI quantities/3k, PI' ... be pi .. ·. ·. Ps and quantities , the conditions _ qrl = pP1 .. P, (uo, PI •. . Pa) = 1. qrt —1 - 1 -uoI···P., P1..

s). Choose We assume assume that that a,, v > r (0 We alJ >>13,, (J" under ii II ~ r and and a,, all ~ {JII (0 ~ rr ~ 8). Choose 13,, under II th _— P1

m1 ml

Pr

a, aa

O'r+l

.Ps · PI .. . . 'P r P Pr+i r+l .. ·P.

Then the order of q for modulus Then

ml

should be be equal equal to should

'Tl

and

r=

T

:1

Tl

by (96).

h1. Then LEMMA 11. Let b=b1 (modmi),h=hl(modml),andh~hl' hmh1 (modmi), Letb::bI(modml), LEMMA1!.

= hh_hiri ']'<':)(b, h) = - hlTI

+ T!,.r:)(bt, h.}.

ml

Proof. Using Using Lemma Lemma 2 we get r-l

T~)(b, h) = =

h

L L6

- b) m ( aq a: -— y —

x=O 1J:=1 y=i z=O

1 m =m z=1

/

(I

h

A

L L

/ Lr 22,v— (,+b)Z)If(r 4%'11:) ,m- · e"m .

.(y+b)z\ —2ai•

1e - 2 1 tm m

,=1

s=l / \x1

By Theorem Theorem 99 the the inner sum of the right-hand side of of this this equality equality may may not vanish By only for for values values z, z, which satisfy the congruences only congruences z

0

(mod

IJ

= 1,2, .. . ,r,

i.e., for i.e.,

z=

. .

=

in1

Complete exponential sums sums

48

= Therefore, using using Tr =

::1

Tl

(Ch. I, §§ 8 [Ch.

obtain (modmi), we obtain and bb == b1 bl (modmt),

mj

aziqr

r

(y-1-b)zi

h

mi m1

=

/

(y+b)zi \ / Ti

h

>2

>2

y1

zt=1 Ti

h

E x1

y=1

x=1

ri

h

—y—bj).

X1 yl

and, therefore, Since difference h -— h1 h t is is a multiple of of m1 ml and, Since the difference A-hi h - hI h—h1 6m1 (aq~ - y_bi)=>2ömi(y)= y - bt) = 6m1 (Y) = - - ,

11

I: >2

L

y=kj+l1 ,=11+

fit

,=1

we the Lenima Lemma assertion: assertion: we obtain obtain the

r,

Tl

rj Tl

hi

>25m1(aq5 y — bi) T!:>(h)=L: — +>2 =>2 Le5ml(aQZ-y-bl)+L

y1

h.h

L Dmt(aq:F:-y-bl) —y >2 Smj(aq2' —

b1)

x=1 y=hi+1 2:=1 ,=hl +1

x=1 11=1 z=1

h - hi ( ) == - T l +T,:t (bl,h t ). fit 1

us consider the question digits in the question concerning concerning the distribution distribution of of blocks blocks cif df digits Since there exist q" distinct blocks complete period of the fraction ii. qn distinct blocks 61 ••• bn , then the mean block of of nn digit,s digits equals :,. T. mean value value of of the the number number of ofoccurrences occurrences of a given block Let Let

. . .

.. . bn ) from Let R n be the deviation of the quantity N!:)(DI ... from its mean 11. Let THEOREM 11. value: N!:)(61 ••• 6n )

Then Then under under odd odd m, m, any n estiIna,t,e is valid: valid: estimate is

~

= q: T + R n •

1, and any any choice choice of the digits digits Si Dl .,. following 5,, .. . D n the following

IRnl ~ IR,,I

1\ ( (1:n )Tt,

(106)

—

where Tl is the order order of of q for modulus modulus being being equal equal to to the the product product of primes primes ent,ering entering the prime prime factorization fact,orization of of m. int,o into the

Proof. By By Lemma 10

. .6,,) = T!:)(b, T,(,,')(b, h), h), N!:)(Ol ... on) =

where T~)(b, h) is the the number number of ofsolutions solutions of of the the congruence congruence aqX

== y + b (modrn), (modm),

o ~ x < r,

1~ y

~

h,

Distribution of 0.(digits digitsinincomplete comp,/eteperiod periodof 0.(periodic p,eliodicfractions fractions Distribution

Ch. I,I, §§ 8] 8]

and the quantities bb and hh arc are determined determined with the the help help of the equalities equalities and 0.61 ••• 6n

t = qfl -, qn

h = [((t+1)ml t + 1)m] h= qfl qn

qfl

J

_[tm]. qfl qt&

—

hi denote the least least non-negative non-negative residue residue of h to modulus mj ml: Let h1

o ~ h < mt.

(modmi), Observing Observing that that h;;;~l is an integer and

rnI(t+1)ml+Itm {(t+l)m} + {tm}, qfl

hh= =m _ q"

we we obt,ain obtain

qn

h—h1

m

h—h1 m nil — mlq?*

qn

J

ml

fm

m —

—

[hi + 1 (f(t+1)mj

mikj

Lmi

ftm

J But then itit follows follows from Lemma 11 by virtue of of the the equality equality i-r = .l!t. Tt , that ml

= =

—

ml Since 0 ~ hI ~

fit -

mi

1qfl

qfl

/

1, then

h1 1 (f(i+1)mjftm [~ll + ~l ({ (t ~;)m }-{ :r:})] qfl m1 J

_{-l —1

_

=

{ Q0

if h1 hi =0 = 0 and otherwise

and and therefore

h1=o and 0

otherwise.

<{tm} {@ qri {(t+l)m} < {tm} qr , }

49 49

exponential sums Comple,'te exponential sums

50

[Ch. I,/, §§ 8

Since Since

o ~ T~,~l)(b, hI) ~ Tl and T$:t1 >(b,0) = 0, we obtain the the theorem theorem assertion: assertion: we

It is easy to ascertain as,cert,ain that that the theestimate estimate (106) (106) can can not not be h,e substantially subst,antially improved. improved. and m m == 22?r -— 1.1. The 2-adic Indeed, let T >1, 1, aa = = 1, and 2-adic expansion of ;; has Indeed, let qq = 2, r> period Tr period 1 a 1 =O.(O...O1)O..01.... -m = -r _ 2 1 =0.(0, .. 01)0 ... 01 ....

2'—l

Choose

01 •••

Dn

= 0 ... O.

Then we get N$:>(61 '" 6n ) =

T -

n and

1 it follows from Theorem Theorem 11, 11, that that Let also, that under under Ti Tl = 1 follows from Let us us note also,

1

Q and mm = 33a• under qq = 4 and where )8n I < 1. It is is so, so, for for example, example, under • In general, for for m m == PIal ••.. .• p.a, under fixed fixed primes primes Pv PII and and growing growing all, the magnitude magnitude Ti T} 18 bounded and the the following following asymptotic formula formula is is valid valid by by (106): (106):

. .

.

5,,)

=

+ 0(1).

Now, let let us establish the occurrence of aa given given block block of of Now, est,ablish the correlation correlation between between the occurrence of digits in period of the fractions fractions + O.'Yl'Y2 •••.'Yz 'Ix... : = [~] + •.• . .

=

and

,, a = [— a ] -—= -fit rni mi +0'1112"'lx'" ml a

a

I

I

I,

(,z+r ('Ix+r

= 'Ix) IZ)

(7x+r,=7x),

where the the quantity ml where ml is is determined detennined as 88 in in Lemma Lemma 11. 11.

Distribution of 0'(digits di,its in in complete complete period ptJriod o( p.riodic (ra,ctions of periodic fractions

Ch. 8] ch. I,I, § 8j

m1 under a THEOREM 12. If Ifqflo\m qRG\m— - ml

u.

certain cert,mn n0 no

~

51

1, then

T_nTl

T - Tl Nm(T)(~UI··· ~) = ----q;-+ N(Tt)(£ '"1 (Il

•••

u.

~ )

6,,. n0 and .. . 6,.. under any n ~ no and any any choice choice of ofaa block block of ofdigits digits fi1 61 •••

b, h, b1, proof. Determine integers t, b, bI , and h1 hI with the help help of the equalities Proof. 0.61 ••• 6n

[tm],

t = -, qR

[imi 1)m] _[tm], qn Lqj

bb=1—-1, = qR h = [(t+1)ml + h=1 qR + 1)nzi 1 hhi=[ = [(1 + 1)m 1 ] _ — 1 qR qn J

[(t

iqi

[tmil b1 ], b1=1——, qR

[tm J.

[(t

=[tm

Then obviously

q"(b—bi) qn(b - hI)

= t(m -

mt} - qn({:::} —

{t;l }),

m1 h1) = qR(h -— hI) =mm—- ml

n(f(t+1)mjf(i+1)mil+ftmilJtm _qR( {(t:~)m} _{(t +q~)ml} + {t;l} _{:':}). J

—

qfl

congruence m Using the congruence

m1 (mod q"), which is satisfied satisfied under under n == ml (modqR), which is

= {tm { tm} qR qn' 1

Therefore,

J

}

n0, no, we get

f(t+1)ml_f(t+1)mi (t + l)m} = {(t + l)ml }. {

qn q"

I

q"(b— qR (b - bbi) =t(m t (m— - mi), ml), 1) =

and, since sinceml mi\m and, \m and and (q,rn1) (q,ml)

~

1.

—

qfl qR

h1) = m — m1,

(107)

= 1,

bb == bb11 (modmi), (mod ml),

hh=:hh11

(modmi). (modmt).

But then, according according to to Lemma Lemma 11, 11,

= Multiply the second equality of (107) observingthat thatTr == Multiply (107) by ~. Then, observing mtt obt,ain obtaIn h—h1 rn—rn1 r—r1 h - hI m - ml T - Tl Tl= - - TT1= I = 1"1 = - - , rn1 m1q" n1l mtqR q" and, therefore,

..!!L1"1' mt

we

Comp,/e,te Complete expon,ent;al exponential sums

52

[Ch. I, §§ 8 [Ch.

applying Lemma 10, 10, we we get get the the theorem theorem assertion assertion Hence, applying ~) = T - Tl ~ ) fi,,) . . Nm,(.,.)(~(,1t··· + N("'l)(~ V n = -n-.- + ml vI··· on · .

.

q

.

m= = pO, p19,where where p is a prime j/', and m1 = 2, m = pP, Let us notice particularly the case q = ml = prime greater than 2. for instance, instance, under under pp = 3, greater 2. Suppose Suppose further /3 {j = 11 (it is so, for 5, 7) and 3,5,7) under n ~ n,o no compare of any any n-digited n-digited block block in in the under compare the numbers numbers of occurrences occurrences of period of 2-adic expansion of the fractions fractions ,Ia and ~. The former former exceeds exceedsthe the latter latter by one one and the same by same quantity (being (being equal to "'-:1). 2 So, for example, under rn = 27 we we get get ml m1 = 3, Tr = 18, Tl r1 = 2 and So, for example, under m = 3. and no n0 = digits of of length length 1, 1, 2, or 3 in the Therefore, the number number of of occurrences occurrences of any block block of digit,s the period of the fraction period 1

27 = O.(0'O'OOlOOlOllllOllOl)O'O... 0.(000010010111101101)00. · ·

exceeds by 8, 4, and 2, of the same exceeds by 2, respectively, respectively, the number of occurrences occurrences of s,arne block of digits digi ts in the period of the fraction 1

3" = = 0.(01)01 ... · Analogously under m m= = 25 we we obt,ain obtain no n0 = 2, a.nd and the of oFcurrences o~currences of Analogously under the number of any block of digits digits of length length 11 or or 2 in the period of the fraction 1 = 0.(00001010001111010111)00 0.(0'O'O'OlOlO'O'OllllOlOlll)O'O... 25 = ...

is 8 or or 4, 4, respectively, respectively, more more than the the number number of of occurrences occurrences of the same s,arne block block of of digits digit,s in the period p,eriod of of the fraction fraction 1

5 == 0.(0'Oll)O'O... 0.(0011)00 ... · These relations relations can be observed under pp = = 3,5, These observed under 7, nn = 1,2,3, 1,2,3, and pO ~ 125 12,5 in the 3,5,7, t,able given below. \ table given below. *

5,3 53

Expon,ential Exponential sums sums with with recurrent recurrent function function

Ch. I,,, § 9]

Table of values of

81 ... 8 -~ 0 P

-

--;-

l1

J...1

1

1 1 25

_I_ 1 125 125

27 27

2 22

10 10

2

9 9 4

11

5

11

11

00 00

2

55 4 2 22

14 13 13 14 77 77

50 50 25 25 25 2,5 25

3

6

11 11

2 2

7 77

11

0

0 00

11

3 2

6 7

1

2

7

1

11

3"3

'99

0

11

11 00 01 10 11 000

11

3 3

0

11

n

001 010 011

0 11

1010 100 101 110

0

0

111

11 11 00

27

81

1

5

5 5 5 5 5 3

11 11 11 11

00

2 2

11 49

2 11 1 1 1 1

11 10

1 1 1 1

3 3 3

00 00

3

2 2

3

12

11

2

13

00

3

12

0

3

6 6 5 5

11 00 0

13 12 13 13

11 0 11 1 1

0 11 0

12

I1

"77

2 22 2 33

Exponential sums with recurrent function § 9. Exp,onential function Let us us consider consider functions functions .,p(x) satisfying s,atisfying the linear linear difference difference equation with conconsta.nt coefficients stant coeflicients

= aItP(x -— 1) + n) tjJ(x) = ... ++ antP(x -— n) +...

(x > n). (x

(108)

that any function "p(x) determined by the recurIt is j,B known (see, for example, [11]) [11]) that recurrence equality equality (108) (108) can can be represented in in the form rence

=

+... +

where rr ~ n, AI, ... ,, Ar are distinct roots where roots of of the characteristic characteristic equation

An

= al,xn-l + ... + an,

(109)

and Pi . . ,, P,.(X) whose degrees degrees are are unity unity less less than than the :PI (x), ... :Pr( x) are polynomials whose the multiplicmultiplicity of the the corresponding corresponding roots roots of of the the equation equation (109). (109). In In particular, particular, if the characteristic equation has no multiple roots, then .

(110) (110)

where where C1,. C 1 , •••. ,, C,, On are are constants const,ants depending depending upon upon the choice choice of initial values values of the the function t/J(x). If coefficients of the the equation (108) and initial . coefficient,s of initial values values 1/J(l), ... ,1/J(n) .

.

. ,

[Ch. [Ch. I,I, §§ 9

Complete exponential sums

54

are integers, then, then, obviously, obviously, under any any positive p,ositive integer x the the function function sb(s) ""(x) takes takes on on integer values. = 1, and at . . , 1/J(n) be not > 1, 1, (an, m) = at least least one one of of the the initial initial values values 1/J(1), .... Let m > a multiple of in. m. In the the equation equation (108) (108) we we replace replace xx by by 3; + + n and and transit tr'Rnsit to to the thecongruence congruence to the modulus in: m:

(modm). ,p(x+n)=a11P(x+n-l)+ + n) + n—i) + ... +a,,,1/J(x) +

(111)

can be expressed in terms of t/J( x + Since (a1,, (an, m) m) — = 1, so in this this congruence congruence t/J( x) can expressed in + , —1, —2,..., we may extend the function 1/J(x) for + n) and, setting x = 0, -1, -2, ... , we may extend 1),.... . . , t/J(x +n) 1), integers x a; ~ 0. O. determined for for integers integers xx by by the congruence (111) and and initial A function function ,p(x) determined congruence (111) . . . ,, "p(n) (see (see [21]) [21])isiscalled calledaarecurrent recurrentfunction functionof ofthe then-th n-th order order t,o to the values .,p(1), ... modulus m, m, and the sum modulus 8um 2 ' t/J(~)

= Ee tramP

S(P)

= s=l

an exponential seenthat that under under nn = 11 exponential sum Bum with with aa recurrent recurrent function. function. It is easily easily seen sums with an exponential function. these sums coincide coincide with with considered considered in in §§ 7 sums Let us show that that aa sequence sequence of of least least non-negative non-negative residues residues of the function t/J(x) to rn'1 —1. modulus in m is periodic perio,dic and that thatits itsleast leastperiod perio'ddoes doesnot notexceed exceedthe thequantity quantity'm" -1. In fact, let let us us denote denote the theleast leastnon-negative non-negative residue residue of 1/J(x) to modulus m by 1:1:: tjJ(X)='IX (modrn), (modm),

o ~ 'rz ~ m -1.

Then by virtue virtue of of (111) (111) 'YZ+R

(modrn). +... + an'rs (mod == al "Y:I:+R-1 + ···+ m).

(112)

Consider blocks blocks of of nn digits digits with with respect respect to the base base m 1,;+1 ... 1z+n

(x =0, 1,...,rn'1).

(113)

Since the number of distinct distinct blocks blocks of n digits is equal to m'1, m R , then among among the the blocks blocks (113) there exist (113) exist two two identical identical blocks blocks

(x2 > x1).

We determine determineTr by by the the equality equality Tr = X2 We

— - x1 Xl

will show show that that under any xx ~ and will

(114) Xl

(115)

Expon,ent;al Exponential sums sums with with recurrent recurrent function function

Ch. ch. ,,I, § 9]

55 56

fact, under x ==x1 In fact, Xl this thisequality equalityisis fulfilled fulfilled by virtue of (114). (114). Apply the induction. x1. In the Let us us suppose suppose that the Let the equality equality (115) (115) holds holds for a certain xx ~ Xl. the congruence congruence (112) we replace replace x by Z + + rT ++1.1. Then (112) we Thenusing using the the induction induction hypothesis hypothesis we we obtain obt,ain 7x+r+n+1 /$+r+n+l

.. . + anl'x+r+l == all'x+r+n + ··· (modm), =all'x+n+...+an7x+l Yx+n+1 (mod = all'~+n +··· + llnl'x+l == 1'2:+n+1 m),

therefore, 12:+r+n+l 7z+r+n+1 = = 1:c+n+l. 7x+n+1. But But then and, therefore, 7x+2

7x+n+1 = 7r+r+2

. .

. 7z+r+n+1,

hence the the equality (115) hence (115) is proved proved for any xx ~ x1. Xl. By By means of of such considerations considerations weget get this this equality equality for for xx < <x1 we Xl as as well well (but in this case, 1X should be expressed from congruence (112) (112) in terms of 1'%+1, ... beforehand, and and that could the congruence could be done .. ,,1'x+n 7x+n beforehand, = 1). because of (an,m) = Hence itit follows follows that that the least residues of of the the function function "p( x) have Hence least non-negative non-negative residues have a

period T, r, where period where 11 ~ Tr ~ m n. Let Let us assume 88sume that least period perio,d is is equal equal to to m's. mn• that the least Then any any block of nn digit,s digits should should occur occur among among the the blocks blocks (113), (113),and, and, in in particular, particular, Then block of . .00 being formed by zeros zeros only onlyisispresent presentamong amongthem. them. But then the block 00.... then by by (112) (112) all terms of the sequence will equal equalzero zeroand and its its least least period equal 1, all sequence of the residues residues will that contradicts contradict,s to the the assumption. assumption. Therefore, Therefore, the least least period period of the function function t/J( x ) does does not exceed m n -— 1. sequence of the least least nonnonHenceforward Henceforwardwe welet letTr denote denote the least period of the sequence negative residues of the function negative residues modulusm. m. It It is is easily easilyseen seenthat that Tr is is the function ¢( x) to modulus period of fractional of the the function .function tP~): period fractional part,s parts of

= 7z+r =

I

7x

=

rn

Therefore, the sum

S(r) =

rT

Le

2. "'(x) 'll"tm-

x=1

is a complete exponential sum" sum. Since is Si.nce under integer a

fin'

fa(x+rfl r

1

—

faxl

then by (28) under any integer a the sum ax

D

e

is aa complete sum as is as well. well.

complete exponential Comple,te expon'ent;a' sums sums

56

[Ch. I, § 9 [Ch.

recurrent functions functions satisfying satisfying the the equation equation (108) (108) and de(si),. ,tPn(x) , Let tPl(X)"" be recurrent termined by initial initial values values . .

x=j, = i,

if

3:

if if

1 ~ a: ~ n,

x

~

(j

j

= 1, 2, ... , n).

It is i.s easy to show show that

+ 1)""1 (x) +···+ + ... + t/J( z + + z) = t/J( z + + n )tPn(x). t/J( x +

(116)

In fact by by virtue virtue of of the the linearity linearity of of the the equation equation (108), (108), any any linear linear combination combination of its solutions is aa solution too. In it,s solutions i,B solution too. In particular, part.icular, the the sum sum in in the the right-hand right-hand side side of of the the equality (116) is a solution of the equation (108). (108). From the definition of the functions , n the initial values that under xx = 1, 2,. .. ,n of this this sum are equal to tPj(:J:) it is seen tha.t 2, ... values of + 2),. . . ,, t/J(z + + ii), + z) has tjJ(z + 1), 1/J(z + 2), ... n), respectively. respectively. The The solution solution 1/J(x + has the the same same + 1), initial values. But But solutions, solutions, which which have have the same same initial values, values, coincide. coincide. Hence the equality (116) is is proved. proved.

=

THEOREM 13. Let 1/J(x) be be aa recurrent recurrent function function afthe of the n-th n-th order to the modulus in, THEOREM 13. m, r. Then be its its least least period, and P ~ T. Then we we have the estimates estima,t,es r be

T

r 2 . "'(~) ~

L-J e

>

11"

-;;;-

~ 2 · ,,(:t)

!!.

n n

L..Je 71"-;;;- ~m2(1+nlogm). m2 (1 + n log m).

~ m2, m2,

z=l

:.r:=1

Proof. Since Since under under an integer integer a the the sum sum rr

Le

Sa(r) =

211'i

ax (""(3:) +.!!) r

m

3:=1

is a complete sum, then under i.s under any any integer integer zz

r

.fv,b(x+z)

(,p(x+z)

r

m

Squaring and summing suuuning over over zz yields yields

r L Le (",e +

r—1 r-l

rlSa(rW =

r

x z) 2 '. (e,&(x+z) 11'1

z=O ~=o

x=1 %=1

m

-;;;-

ax\ 2 +82:)

2

T

.

(117)

57 67

Expon,ential Exponential sums sums with with r«urrent recurrent fu-n,ction function

Ch. ch. I, § 9]

least non-negative residue of of the the function function We let 1': 'Yz denote denote the least non-negative residue We

t/J(z) to modulus modulus

m. in. Then by (116) + 1)""1 (x) + + z) = = tjJ(z + +... +n)v5,,(x) ,p(x ... ++ tjJ(z + n)t/J,,(x) v5(x + (modm), == 1~+11/Jl (x) + ... + 1z+n 1P,,(x) (mod m), and, therefore,

rr y,'(s) n ~ 271'1L..Je m = ISa(T)I IS.(r)1 ~ m 2 • = I

(118)

.2:=1

1,.... . . ,, T,- — modulusm, m, then then under under zz == 0, 1, Since r'r is the the least least period period of of 1% to modulus - 1 1 all n digits are distinct. Therefore, blocks 'Yz+l '1z+1 ... Therefore, extending the summation to to . ;.r+n of n . of nn digits, we obtain . . . z,2 all possible blocks %1 ••• all Zn of .

I

r

p—I

ax\

2

m

TISa(T)I2 x=1

rr

=

~

L..J e

m-l

a(x—y) .4(X-II) 2,ri 211'1--

L

r

x ,y=I z.,=1

"("'1(.1:)-.1/11(')

e

2 11"1

m m

%1

+ + t/J,,(Z)-.pR(J/) '""

m m

ZR

)

%1 ' •• "'Z" =0

r

~ rn n

L

=m9', bm [1/Jl(X) -1Pl(Y)] · I. b,n [tPn(X) - tPn(Y)] = mnT,

(119) (119)

Z I z,,=l

where

is the number T number of of solutions solutions of of the the system system of of congruences congruences T is

~.(~~~.~(~~} 1

(modm),

1 ~ x,y

~

r.

(120)

tPn(x) == tPn (y ) Let us us assume assume that this this system system has has aa solution solution with y Y ~ x. Wiihout Withoutloss los'S of ofgenerality, generality, we may may assume assume that that yY >> x. we x. Using Using the theequality equality(116), (116), we we get get

+ n)"pn(x), = t/J(z -— x + + 1).,pl(X) + ... + + tP(z -— x + tjJ(z) = t/J(z + y - x) = t/J(z - x + l)1/Jl(Y) + + tjJ(z - x + n},pn(Y). Hence by by (120) (120) it follows that under Hence follows that under.any .any integer integer zz

1/J(z + + y -— x) == t/J(z) (modm). is the least period, But then theny'II— - x is a period perio,d of 1%, and since since rr is period, then theny'II— - x ~ r, which leads to to a contradiction. leads contradiction. Thus, Thus, the thecongruence congruence system system (120) (120) has no no other other solutions solutions except for for solutions solutions with with yy = x and, therefore, T = T. r. Now from from (119), (119), we get except therefore, T

rJSa(r)12

m"T = in"r,

n

ISa(r)1 ~ m 2 , ISa(r)I

(121)

Comp,/e,ta sums Complete exponential sums

58

[Ch. I, §§ 99 (Ch.

= 00the thefirst firstassertion assertionofofthe thetheorem theoremfollows: follows: Hence under aa = rr n . t/J(:t) 11 LJe2"& -m = = ISO(T)I lSo(r)I ~ m 2 •

~

z=1

followsimmediately immediatelyfrom fromTheorem Theorem22 and and the The second assertion of the theorem follows estimate (121): r

az\

(1 +logr)

max

x1

x=1

n

= max IS.(T)I(l = 1~a~r

+ logr) < m 2 (1 + nlogm).

Note that that in in the thegeneral generalcase case the the order order of of the the estimation estimation r . f/J{z) 211'1L..-,e m

~

n -

~m2

z=:l

from the theory of finite be improved improved further. Indeed, Indeed, using using considerations considerations from can not be fields (see, (see, for forinst,ance, instance, [33J), [33]),ititcan canbe beshown shownthat thatunder underany anyprime primepp> ~fields > 2 and

positive integer integer n

=

(122) (12'2)

By By virtue of properties of symmetric symmetric functions functions there there exist,s exists an an equation equation with with integral coefficients pR = == b1pn-l + +... ...++ b,., . , A~, respectively, whose roots equal A¥, .. ., and the the free free term is relatively root,s PI, respectively, and , pj,.... . . ,IJR prime to p. Consider from (12'2) (122) that Consider the the functions functions tjJ(2x) and t/J{2x + + 1). ItIt follows follows from

+ .... .. + CRP:, 1/J(2x) = := C1IJf + 1/J{2x + 1) := CtAtP: +···+ CnAnp:.

+

of the the n-th n-tb order to the modulus Thus, t/J(2x) and t/>(2x + 1) are recurrent functions functions of modulus satisfying the equation p satisfying t/J*(x) = = bl 1/J*(x -— 1) +... + batP*(x -— n). +.. . +

p. Under r1 = iT Denote by 'Y~ the least non-negative non-negative residue of "p(x) to modulus modulus p. Under Tl we get* get· 72(x+ri) '1'2(:I:+rl) = 'Y2~+T = p,eriod being equal to Ti. TI. 72x and, therefore, 'Y21: has a period = 72x+r = 'Y2:r; ·8' · 0odd, dd is Since IDee p IS ' J then t h,en 11

=

p"-I" 2 - is IS

an integer. Integer.

59 59

Exponential sums sums with with recu"ent recurrent function function Exponential

ChI 9] ch. I,I, § 9)

Let us assume assume that r1 T1 is not the least period. Then Then we we can can find a positive positive integer such that under <<Ti, T1, such under any any integer integer xx

1'2 1•2

(modp).

(123)

Applying the equality (116) we obt,ain obtain Applying "p(2x + +2r2) 2T2) = = t/J(2,X)t!J1(2T2

+ nn— + 1), + 1)1)+... - 1)tPn(2T2 + + tjJ(2x + +... + + n - 1),pn(1). t!J(2x) = = ""(2$)"'1(1) +... + .,. ++ t/J(2x + —

But then by (123) (123) the congruence congruence

+ ... +

[#i(2T2 + 1)—

+ 1)—

0

(modp) (124)

under any any integer integer x. x. From should be fulfilled fulfilled under From properties properties of solutions solutions of the system (120), itit follows followsthat that a,t at least one is not congruent to (120), one of of square brackets in (124) (124) is to zero zero to modulus modulus pp and, therefore, therefore, the the number number of of solutions solutions of the congruence congruence (124) (124) does does exceed pn-l — - 1. On the other other hand, hand, according according to the the definition definition of of the the function function not exceed under xx = = 1, 1,2,... yield distinct solutions t/J( x) under 2, ... ,r1 ,"'1 n-tuples n-tuples1'2z, 12%, 12:.:+1, .•• ,12s+n-I yield solutions . .,72s+n—i of this congruence. Since Since obviously obviously Tl

= p"— n-11 pfl_i = P_- > > pn-1 2

-— 1, 1,

then arrive at aa contradiction tpen we arrive contradiction and, and, therefore, therefore, r1 Tl i8 is the last last period p,eriod of of 72x. 12x' Analois gously we get that the least period of '12z+i equal to r1 as well. Now, gously we get the least period of 1'2:1:+1 Tl well. Now, in the same s,ame fashion as in the deduction of (118), we arrive at the equalities fashion as of we arrive equalities

Ee Tl

Tl

,""(2:1:) 22 •

2iri 2111

e

- ,' -

ri—i Tt-l

x=1 2:=1 Tl

Tl Ti

Ee

2

,';(2z+1) 22

•

"'.,

x1 z==l

Ti Tl

=E = =

Ee

. ' "2.+1 "'1 (2%)+"'+"2.1+" "n(2:':)

2

e

8".

L Ee

2iri 2

z=O ,=0

z=i ~=1

Tl-1

Tl

z=O %=0

x=1 s=1

.

p

2

2 ,

112.1+27/11 (2z)+ ••• +11,.. +I+" 1/1" (2:1:) 22

,"

,

•

Hence by by virtue virtue of the choice Hence choice of the function ,,(x), it follows follows that that

/

Ti Tl

~

ri( LJe >e

Tl

•. ,,(21:) 2irs 2 71'1--

" P

(

22

Tt

+ +

z=1

= =

~

LJe

• 2irt 271'&

1/1(2a:+l)

,"

2) 2

z=l

It-I,

Tl Ti

%1 .... "n=O

:1:=1

>:I Ee E z=I

•.

21r1

1 (125)

.ll1J1t(2s)+... +.a'n tPn(2:1:) 22 l'

,

________

Complete exponential sums Complete sum's

60

[Ch. [Ch. I, 5§ 9

,

indicates the deletion where the sign sign ', in the the sum sum "".. deletion of n-tuple z1,. Zl, ••• .. ,Zn, LJ.,l.·· ....." zeros entirely, entirely, from from the the range of of summation. summation. Let T1 formed by zeros T 1 denote the number number of of solutions of the system system of of congruences congruences 40'

~.(~~).:.~l.?~.)}

(modp)

1 ~ X,y

= tPn(2y)

tPn(2x)

In the same a,arne way way as as in inthe thesystem system(120), (120),we wehave haveT1 T1

= T1

rj

p—i

~ T.

and, therefore,

2

=p'2T1

>

x=i

But then (125) (125) can be rewritten in the the form form

/

Tt

Le

rj

•

t/J(2~) 22

2'J1"1--

P

2

2

.

Le Tl

+ +

• 271'1

"'(22:+1) 22

"

P

>

x=l

pn+l == = Pp"n -— Tl = - .. 2

x=l

(126)

We We determine 18*(Tl)' equality (r1 ) with the help of the equality

IS*(Tdl = max

/

— " ~e >e rj

Tt

"'"

(

.

t/i'(2z) "'{2:t)

2'J1"1 - -

p

Ti Tl ,I

•

2ira

~e 2'J1"1

"'"

t/t(2~+1) ) p

•

Then from (126) (12:6) we we get get • 12 IS (Tl) ~

pR + 1

-4-'

IS*(rl)I >

Hence by > 22and > 11 there there exists a by (117) (117) itit follows followsthat thatunder underany anyprime primep p> andany anynn> function of of the the n-th order to the modulus such that that for recurrent function modulus pp such for the exponential S*(r) sum S*( Tl)the thefollowing following estimates estimates

1

!!.

"2 p2 < IS*(Tdl hold.

!!.

~ p2

________

61 61

Sums of of Legendre's S.ums Legendre's symbols

Ch. I, § 10] ch. 4

io. Sums § 10. Sums of of Legendre's L·egendre's symbols symb,ols Let Pp>> 22 be = an a0 + alx polynomialwith withintegral integral Let be aa prime, prime, 1(x) f(x) = a1x ++ ....., + + anx R be aa p,olynomial coefficients,n n<
denote the number number of of solutions solutions of of the the congruence congruence and Tn denot,e y2 =. f(x)

The quantities Tn and

Un

(modp).

(128)

by a simple relationship are connected by

=P+7n.n . TR=~[l+(f~))] =P+CT

(129) (12'9)

This relationship reduces reduces the question on the number of solutions of the congruence (128) (128) to studies of of sums 8ums of of the the Legendre Legendre symbols. symbols. The sums (127) for polynomials polynomialsof ofthe the first first and and the second (127) are easily easily evaluated evaluated for second degree. function with with aa period perio·d p degree. Indeed, Indeed, since since Legendre's Legendre's symbol symbol (;-) is a periodic periodic function and under under (a1 (at,, p) = 1 the linear function a0 ao + a1 al xx runs runs through through a complete residue system modulo p, when x runs through a complete residue system modulo p, then system modulo p, when x a complete then

tE(ao+alx) t (=-) = o (a +alx) = P

x=1

O.

P

2:=1

In order to evaluate the sum Bum CT2

at:

(aoo+ a1; + a2;2) = ~ (a + + a2 x ) ,

=

2

we consider consider the congruence we y2

== x 2 + a (modp) (mo,dp)

and denote the number number of of its its solutions solutions by by T(a). T(a). Obviously, Obviously, p

T(a)= T(a) =

L

2 tS,(x2—y2+a) c5p (x - y2 + a)

x, y=1 2:.11==1

= 11 p-l

2 . as GZ 2,r, —

=-LeD"p p z=0 z=o

1

2 2 ' Z(x _,2) elM II

pP

E

2,rt

:1:,,=1 —1 p-l

=p+-Ee P z=l

211'1!.! II

P

Le 2;:=:1

22

271'1 .!L II

22

•

[Ch. I. , 10 {Ch.

complete exponential sums

62

we obt,ain obtaln Using the fact that that the themodulus modulusof ofthe theGaussian Gaussian sum sum equals equals ~, we p—I ,-1

.••

= p+ Le '" p = p + ,,6,.(a) -

T(a)

2

1, I,

~l

and, therefore, by (129) and. (12,9) (.. . a: 2

,

+

a) = p6,.(a)-1.

L -P. . - •

(130)

:1=1"

Let a2

~

0 (modp). (mod p). Then Then observing observing that that o

(~)(4;2)=11 \pJ p 1 =1, we get

= =

0'2 a2

a,) t (.(4aOaa • 4aoa ++4a1a2x+4t4r2) 4ala2% + 4a~%' )..

(.(aa)

2

p . • =1

P

.'

((2a2x + ai)2 + 4a0a2 + al)2 + 44oa, -— a~ )..

=

= (a,) L.il'. (.. (242 % p" .,=1 .'

= 2 = (a ) p

t (.

p P

+ 4aoa, - a~)....

%2

P P

*'·1 .

Hence by by (130) (130) itit follows followsthat that under under (02, (a2, p) p) = 11 Hence 0'2 -_

"' LJ'. ( ao

2 c (.'.4". aoG2. - at - .1] · -_ ( a. )·.·.· .· [.'po, '. .' p

+ al% + a2%' )

.=1 ..

P

2.' )'..

—

(131)

Note that) that, in particular, Note

~ ('.'% - 0)·.·· (·····S -

L J -.:=1'

-

P.

P

6) = ,,6,,(a - 6)-1.

- •••

(132)

Indeed, this thi,8 equality equality follows follows at once once from from (131): (131):

t

p

• =1

(% -a).. (:I: - 6) = t

p

P

. P

.'=1 z=l

((ab—(a+b)x4-x2 • + +

2

6):1:

a 6- (a

:1:

)•.

P

:::: p6,[46," - (0 + 6)2] - 1 = ptS,,(a - b) -1.

Under n ~ 3 the investigation Under i.nvesti~ation of the sums sums a,, (lR is i,amuch much more more complicated, complicated, except except cases. for some special cases.

of Legendre's Sums 0.( Legendre's symbols symbols

ChI 10) Ch. I, § 101

63 63

cases. Let n ~ 3 be be odd, odd,pp> (a1,p) > n be be a prime, prime, (aI, p) = 1, Consider one of such special cases. and

(xi' +aix x=1

show that Let us show

~ ( zn :

alZ )

I~

(133)

(n - I)JP.

In fact, complet,c fact, since under z ~ 0 (mod (modp)p) the the linear linear function function zx zx runs runs through aa complete residue system system modulo modulo pp when when xx runs runs through a complete residue residue system system modulo modulo p, p, residue then P /flIZi-rcZiZZ ç—fZX

x=1

Therefore,

Iun(alz n- l )/= I

=

~(znzn:alznz) ==

t,(zn: alz ) ==lun(at)l. Iun(ai)I.

and summing summing over z, we we obtain Squaring this equality and

1)Ian(ai)12 (p (p-l)lu n(al)1 2 =

,-1 P—i

p-l p—i

~=1

~=1

Elun(alz n- 1 )1 2 == Et('x)lu n ('x)1 2 ,

where teA) is the number number of of solutions solutions of of the the congruence congruence n 1 alz -

== A (modp).

that Since t(A) teA) ~ n -— 1, then from from (134) (134) it follows follows that 1) (p — -1)

lun (al)1)122

1) ~ (n — -1)

p—i ,-1

E IU (A)1

2

n

~=1

=(n-l)?; ztl (zn;,Xz)

(yn;,Xy)

x,y—i

= -1) 1) = (n —

E (:l: t Y)

~=1

p

x,y=1 z,y=1

+ (.\ + (A + x"-l) (A + yn-I). p p P

P

Hence, using the equality equality (132), (132), we we get the estimate estimate (133): (133): Imn@i)12 lun(at)l 2

— yfl_I) ~ n=~ ,-1 E ( zy ) [p8p (zn-l_ -1]1] y n-l) —

I

p

£ (?)

:1:,11=1

(n—l)p = (n = __ ?p P Icrn(ai)I lu,.(al)1

~ (n

P

.E (X )6 (x z.,=1

-l)JP.

Y

p

p

n-

1_— yn-I)

~ (n _1)2 p, —

(134)

[Ch. [Ch. I,I, § 10 io

Complete expon·ential exponential sums CompJe·te

64

Tinder odd n Under odd

~

for the general case also: p) = 11 the same estimate holds for 3 and (an, (an,p) also:

t

(ao + alX + ... + anx

~=l

~ (n -1)y'P.

n )

(135)

P

For n = 33 this under an arbitrary nn ititfollows thi.sestimate estimatewas wasobtained obt,ainedby by Hasse Hasse [13], [13], under follows from more general results of A. Weil [48]. One can acquaint with elementary methods result,s Weil [48]. One can acquaint with element.ary for obtaining thesums sums(135) (135) by bypapers papers[35], [35], [42], [42], and and [31]. [31]. for obt.aining estimates estimates of ofthe for polynomials polynomials of of the the second As it was was shown shown above, above, sums sums of of Legendre's Legendre's symbols symbols for degree Gaussian sums. sums. Let Let us us show show that that Gaussian degree can be evaluated with the help of Gaussian used in in estimating estimating the thesimplest simplest incomplete incomplete sums sums of of Legendre's symbols: symbols: sums can be used p

o(P) u{P) = = ~ (;)

(P < p).
Indeed, in the same Indeed, same way way as as in in the the estimation estimation of of incomplete incomplete exponential sums (see (see Theorem 2), we obtain p—I p-1 (

u{P)=?; x=1

p—i(/ ). ,-1

P P

)

~ ~6,{x-y)=~?; y1

;

x=I

P P

p—I p-1

y=i

z=O

z(x—y) . %(X-II)

~~e2WI-p-.

Hence, alter interchanging and singling out the summand Hence, after interchanging the order of summation summation and singling out that with z = 0, 0, by by (41) (41) and and (42) (42) itit follows follows that u{P)

=! ~

(t

.t=1

11=1

=p

P P

1 p—I ,-1

Ia(P)I ~ p~ ~e lu{P)1

e-

Z:) (E (~)e27fi z:), P

2:=1

In,

P1 ,-1 ( ) ~ ~

2 ' zfI

zIyl

~

21Fi

1

p-l

mm y'P ~min

(

P,

2 . zx Z'X

— e WI,

xi

1)

211~11 ~

y'P log log p. p.

Thus we we obt,ain obtain the estimate Thus P

?; (~)

~ y'P log p.

log pp this this estimate estimate isis better better than the trivial Plainly, > y'P log trivial one. Plainly,under underPP> The availability availability of a nontrivial estimate P E . (:.)

z=1

p


(136) (13-6)

65

Sums 0" of Legendre's Sums L'cendre's symbols

Ch. 10) ch. !,I, § 10]

signalizesthat that on on the the interval interval [1, [1, P] P] there is is at at least one quadratic nonresidue modulo signalizes

Po denote the least that p. least quadratic quadratic nonresidue. nonres'idue. From From (136) (13-6) it follows follows that p. Let Po 1

O(p22 log p). p] = = O(p + [v'P log p] Po ~ 11 + p).

under any e > 00 the According to the conjecture conjecture enunciated by I. M. M. Vinogradov, Vinogradov, under estimate Q(pe) p == O(p~) Po

is valid, valid, where wherethe the constant constant implied implied by by the the symbol symbol "0" "0" depends is depends on e only. only. In this direction, the strongest result has the form In Po =

0(p7),

where any numb,er number greater greater than ~. A of this this result result [4] based upon upon the the where 'Y'y isis any A proof of [4] isis based use of the estimate (135). the Hasse—Weil Hasse-Wei! estimate Let Let N1 N 1 and N2 N 2 be, be, respectively, res'pectively, the number number of quadratic residues residues and quadratic quadratic nonresidues nonresidues to to modulus modulus pp among the first PP positive positive integers. integers. It it is is easy easy to obtain obt,ain asymptotic formulasfor for Nt N1 and and N N2 with help of the estimate (136). Indeed, asymptotic formulas with the estimate (13,6). 2 observing that that the the number number of of solutions solutions of of the the congruence congruence observing y2

== x (modp),

1~ y

~ p,

1~z

~

P,

under P