THE GREEN BOOK OF MATHEMATICAL PROBLEMS
Kenneth Hardy and Kenneth S. Williams Carleton University, Ottawa
DOVER PUBLICATIONS, INC. Mineola, New York
Copyright Copy right © 1985 by Kenneth Hardy and Kenneth S. Williams. All rights reserved under Pan American and International Copy right Conventions. Published in Canada by General Publishing Company, Ltd., 30 Les mill Road, Don Mills, Toronto, Ontario. Published in the United Kingdom by Constable and Company, Ltd., 3 The Lanchesters, 162-164 Fulham Palace Road, London W6 9ER.
Bibliographical Note This Dover edition, first published in 1997, is an unabridged and slightly corrected republication of the work first published by Integer
Press, Ottawa, Ontario, Canada in 1985, under the title The Green Book: 100
Practice Problems for UndergradU(lte Mathematics Competitions. Library of Congress Cataloging-in-Publication Data
Hardy, Kenneth. [Green book] The green book of mathematical problems / Kenneth Hardy and Kenneth S. Williams. cm. p. Originally pUblished: The green book. Ottawa, Oot., Canada : Integer Press, 1985. Includes bibliographical references. ISBN 0-486-69573-5 (pbk.) 1 . Mathematics-Problems, exercises, etc. Kenneth S.
I. Williams,
II. Title.
QA43.H268 51O/.76-dc21
1997 96-47817 CIP
Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501
PREFACE
There is a famous set of fairy tale books, each volume of which is designated by the colour of its cover: The Red Book, The Blue Book, The Yellow Book, etc. We are not presenting you with The Green Book of fairy stories, but rather a book of mathematical problems. However, the conceptual idea of all fairy stories, that of mystery, search, and discovery is also found in our Green Book. It got its title simply because in its infancy it was contained and grew between two ordinary green file covers. The book contains 100 problems for undergraduate students training for mathematics competitions, particularly the William Lowell Putnam Mathematical Competition. Along with the problems come useful hints, and in the end Gust like in the fairy tales) the solutions to the problems. Although the book is written especially for students training for competitions, it will also be useful to anyone interested in the posing and solving of challenging mathematical problems at the undergraduate level. Many of the problems were suggested by ideas originating in articles and problems
in mathematical journals such as Crux Mathematicorum, Mathematics Magazine, and the American Mathematical "'1onthly, as well as problems from the Putnam competition itself. Where possible, acknowledgement to known sources is given at the end of the book. We would, of course, be interested in your reaction to The Green Book, and invite comments, alternate solutions, and even corrections. We make no claims that our solutions are the "best possible" solutions, but we trust you will find them elegant enough, and that The Green Book will be a practical tool in the training of young competitors. We wish to thank our publisher, Integer Press; our literary adviser; and our typist, David Conibear, for their invaluable assistance in this project.
Kenneth Hardy and Kenneth S. Williams Ottawa, Canada May, 1985
(iii)
Dedicated to the contestants of the William LoweD Putnam Mathematical Competition
(v)
To Carole with love KSW
(vi)
CONTI!:NTS Page Notation ...... . The Problems The Hints The Solutions. Abbreviations References
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.1 . 25 .41 169 171
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IX •
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....
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"
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..
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..
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..
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..
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....
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..
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.. .. .. .. ...... .. .. .. .. ..
.. .. .. ..
.. ........ ..
(vii)
.. ....
..
..
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..
..
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..
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..
....
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..
NOTATION
�
[xl
denotes the greatest integer x is a real number.
{x}
denotes the fractional part of the real number x, that is, {x} - x - [xl.
In
x
x, where
denotes the natural logarithm of x.
exp x
denotes the exponential function of x.
�(nl
denotes Euler's totient function defined for any natural number n.
GCO(a,bl
denotes the greatest common divisor of the integers b. a and
n k
denotes the binomial coefficient nl/k!(n-kl!, are non-negative integers where n and k (the symbol having value zero when n < k l. denotes a matrix with entry.
det A
as the (i,j)th a ij
denotes the determinant of a square matrix A.
(Ix)
THE PROBT,EMS Problems, problems, problems all day long. Will my problems work out right or wrong? The Everly Brothers
{b : n = 0,1,2, } n numbers, prove that the series
1.
If
.
.
.
is a sequence of non negative real
b
(1.0)
L
n
n=O converges for every positive real number a.
2.
Let a,b,c,d be positive real numbers, and let a(a+b)(a+2b)...(a+(n-1)b) . 11 (a,b,c,d) = c(c+d)(c+2d) (c+(n-1)d)
o
.
Evaluate the limit
3. (3.0)
•
.
L = lim Q (a,b,c,d). n n+a>
Prove the following inequality:
in
x
,
3 x -1
1
x
>
0,
x
..
1.
2
PROBLEMS (4-12)
Do there exist non-constant polynomials
4.
variable
such that
z
Ip (z)1
p (z) is monic and of degree
5.
Let
f (x)
a
For
E >
°
R> °
R ,where
be a continuous function on [O,a], where
f (x) :-:- '( '-: � -;:f'x( -') + f a
o
and
[O,a] .
a
>
0,
Evaluate the
x . d ' )
evaluate the limit x+1 2 sin (t )dt
lim
7.
=
in the complex
n?
integral
•
Izl
on
f (x) + f (a-x) does not vanish on
such that
6
Rn
<
p(z)
•
Prove that the equation Z4
(7.0)
2 2
-
2Y
z
2 2 2 2 - 2z x - 2x y
=
24
has no solution in integers x,y,z.
8. Set
Let x
°
=
a
a
and
k
and define
be pos�t�ve numbers such that •
•
x
n
2 a
>
2k .
recursively by
(8. 0)
X
n
=
x
n-1
Prove that lim n .... ex> exists and determine its value.
x
n
,til
+
k
x
n-1
,
n
=
1,2,3, .. .
•
PROBLEMS
9. b
(4-12)
3
denote a fixed non-negative number. and let
Let
be positive numbers satisfying vb
Define
x n
a
2vb .
<
recursively by ax
x
(9.0)
Prove that
<
n
+ b
n-l x + a n-l
-
n = 1.2.3....
•
exists and determine its value.
lim x n +""
n
10.
Let
be real numbers satisfying
a.b,c
a
>
O. c
2 O. b
>
>
ac
•
Evaluate 2 2 max (ax + 2bxy + cy ) . x.y e; R x2 +y2 =1
11.
Evaluate the sum [n/2]
(1 1.0)
for
n
12. ( 12.0)
..
s
I
n(n-l)
r=O
.
(n-(2r-l» 2 (r! ) •
•
n-2r 2
a posltlve lnteger. "
.
Prove that for
m = 0. 1.2 •. .
S (n)
is a polynomial in
m
n(n+l).
•
2m+l + ... + n
•
a
and
4
13.
PROBLEMS
Let
be positive integers such that
a,b,c
GCD(a,b) Show that
=
GCD(b,c) .. GCD(c,a)
l .. 2abc - (bc+ca+ab)
is insolvable in non negative integers
Determine a function
=
1 .
is the largest integer such that
bc x + ca y + ab z
14.
(13-21)
..
l
x,y,z.
such that the n
f(n)
th
term of the
sequence
(14.0)
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, . . .
is given by
15. zero .
[f(n)] .
be given real numbers, which are not all a' ' a , , a l 2 n Determine the least value of Let
•
.
•
•
where
x ' l
. . . •
x n
•
•
2 + x n
,
are real nnmbers satisfying + ... + a x .. 1 n n
16.
Evaluate the infinite series S ..
17.
•
+ ...
•
F(x) is a differentiable function such that F'(a-x)" F'(x) a for all x satisfying 0 � x � a. Evaluate f F(x)dx and give an o example of such a function F(x).
PROBLEMS
18.
5
(13-21)
(a) Let
r,s,t,u
be the roots of the quartic equation
4 2 x + Ax3 + Bx + Cx + D .
Prove that if (b) Let
rs
=
tu
then
a,b,c,d
2 A D
2
a
=
.
2 c .
be the roots of the quartic equation 2 py + qy
Use (a) to determine the cubic equation (in terms of p , q,r) whose roots are ac - bd a + c - b - d '
ab - cd a + b c - d ' -
19.
where
Let
n
p(x)
ad - bc a + d - b - c '
be a monic polynomial of degree
D
is a non negative integer and
=:
d dx
m � 1 , and set
denotes differentia-
tion with respect to x. f (x) is a pOlynomial in x of degree n UiIt-n . l.n f (x) Determine the ratio of the coefficient of x n constant term in f (x) n Prove that
( lUU
to the
•
20.
Determine the real function of
x
whose power series is
15 3 9 x x x + + ;:';:-:'" + .. . 9! 15 ! 3!
21. (21.0)
•
Determine the value of the integral I
1T
n
=
a
Sl.n nx i Sl.n x •
for all positive integral values of
n .
dx
,
- n).
6
22.
PROBLEMS (22-31)
During the year 1985, a convenience store, which was open
7 days a week, sold at least one book each day, and a total of 600 books over the entire year.
Must there have been a period of consec
utive days when exactly 129 books were sold?
23. that as
Find a polynomial m
with rational coefficients such
run through all positive integral values,
n
and
f(x,y)
f(m,n)
takes on all positive integral values once and once only.
24.
Let
m
be a positive squarefree integer. Give a condition involving
positive integers.
tees that there do not exist rational numbers
Let
R,S,m
R,S
be
which guaran
x,y,z and w
such that
(24.0)
25.
Let
k
and
h
1 � k
be integers with
<
h .
Evaluate the
limit hn IT L = lim 1 - E n2 n .... "'r=kn+l
(25.0)
26.
Let
•
be a continuous function on [O,aJ such that
f(x)
f(x)f(a-x) = 1 , where
a
O .
>
Prove that there exist infinitely
f(x) , and evaluate
many such functions
a 0
27.
The positive numbers n
L
(27.0)
r=l Is it true that
a
r
3 a = r
= r
dx 1 + f(x)
a ,a ,a , l 2 3
L
r=l
•
.
.
satisfy
2
n
for
•
a r
,
n = 1,2,3, ...
r = 1,2,3,...
?
•
28
Let
•
integer.
7
(22-31)
PROBLEMS
p
0
>
be a real number and let
Evaluate u (p) = n
(28.0)
29.
be a non-negative
n
o
-px n e sin x dx .
Evaluate
(29.0)
for integers
30.
Let
k (2
of
n
k
s <
•
•
2 .
�
n
�
2
S
n)
A selection {s =
be an integer.
elements from the set
N
W(S) = min {a If a k-selection
such that
For any k-selection
N
, what is the prob
ability that W(S) = r , where
r
is a natural number?
•
31.
Let
k a
�
n
be a fixed integer.
2 =
For
n = 1,2,3, ...
1 , if n is not a multiple of k -(k-1) , if n is a multiple of k . ,
'"
Evaluate the series
s
-a : i = 1,2,..., k-1} . i+ 1 i
is chosen at random from
S
�
= {1,2,3,... ,n}
is called a k-selection.
•
a.: i=1,2,... ,k}
L
n=1
a
n n
•
define
,
8
PROBLEMS (32-40)
32.
Prove that
r
)0 for
m 33
-
0,1,2,
.
•
m -X x e sin x dx
•
•
For a real number
•
I(u)
(33.0)
=
set
u 1T
tn(l
..
Prove that I(u) and hence evaluate
I(u)
2
2u cos x + u ) dx
-
=
for all values of k
2
�
u.
the set of natural numbers {A (k) : n - 1,2,3, n k natural numbers,
is partitioned into a sequence of sets
•
A (k) consists of the first 1 consists of the next k+1 natural Dtmbers, follows: next
natural numbers, etc.
k+2
•
1 2 I(-u) .. II(u ) ,
For each natural Dtmber
34.
m! sin (m+1)1T/4 --:-_==(m+2)/2 2
such that
as
A (k) 3 The sum of the natural numbers in
=
Let
}
•
A (k) 2 consists of the
A (k) is denoted by s (k) . Determine the least value of n n 3 2 such that s (k) > 3k - 5k , for k 2,3,... n 35.
•
n
-
n(k)
•
be a sequence of real ntmbers {p : n " 1,2,3, } n p � 1 for n " 1,2,3, Does the series n [pn ]-1 •
•
•
•
.
•
.
00
L
(35.0)
n=l
converge? 36. degrees
Let
f(x) , g(x)
n+1 , n
be polynomials with real coefficients of
respectively, where
n
�
0
,
and with positive
9
(32-40)
PROBLEMS
A , B
leading coefficients
x L
in terms of
37. where
A, B and
=
Iim x .... '"
n
g(x)
e
f(t) -f(x)
•
The lengths of two altitudes of a triangle are h" k .
and
k ,
DeterUline upper and lower bounds for the length h
and
k .
n,r
= P
n,r
(x)
(
=
n+l 1_x
n+2 n+r ) ) (1_x )(1_x r 2 (1 - x)(1-x ) (1-x )
of degree
is a polynomial in
x
negative integers.
(When P
and we have
1
Let
r = D
n,D
A, B, C, D, E
Prove that the ntmber
N
= 1
•
.
.
39.
h
Prove that P
to be
dt ,
D
of the third altitude in terUlS of
38.
Evaluate
respectively.
.
nr , where
.
•
n
and
r
are non
the empty product is understood for all
n
�
D .)
be integers such that
of pairs of integers
B" D
(x,y)
and
such that
2 Ax + Bxy + Cx + Dy + E = D ,
(39. D) satisfies
N "
where, for integers divisors of
n
1 , d(n)
Evaluate
L
k=l
,
denotes the number of positive
n . n
40.
�
2d( I F I)
k •
10
41 m
PROBLEMS
denote the sum of all possible products of
p =P (n) m m
Let
•
different integers chosen from the set
formulae for
42.
For
P (n) 2
a
>
and
.
Find
b > 0 , evaluate the integral ax bx - e e � --'::=-dx . - a x-- b x(e +1) +1) (e
o
For integers
•
{1,2,...,n}
P (n) . 3
(42.0)
43
(41-50)
n � 1 , determine the sum of
n
terUlS of the
series 2n
2n(2n-2) (2n-4) �� L�:l!::::'�� + + ... + - --'=.t.� _ "_ 2n-l 3 )(2n -S ) (2n- 1)(2n -3) 2n l )
(43.0)
44. be
2n(2n-2)
Let
k (�l)
m
be a fixed positive integer and let
45.
s=m,m+l,m+2,
•
) - (a A n ij
•
•
,m+k-1.
be the
a ij
where
46. tions
o
... +
Let
•
•
.
,z
k
complex numbers such that
(44.0) for all
z ,z , l 2
•
x > 2 .
Evaluate
Must
nXn
,
z =0 1
•
•
•
,k?
matrix where
x ,
if
i=j ,
1 ,
if
I i-j I =1
0 ,
otherwise,
D =det A n n
for i=1,2,
,
•
Deterilline a necessary and sufficient condition for the equa
PROBLEMS
(41-50)
11
x
+
y
+
z
=
A ,
2 + x
2 2 + z Y = 3 + 3 3 x y + z =
(46.0)
B , C
to have a solution with at least one of
47
•
Let
S
be a set of
n 0 -1 , where 1,2,3, ... ,1
(4 7.0)
n
<
n
x,y,z
equal to zero.
distinct integers chosen from
k
is a positive integer.
(k+1)
In
10
disjoint subsets of
S
In
it is possible to find
,
2
2
Prove that if
,
whose members
have the same sum.
48.
Let
n
be a positive integer.
Is it possible for
6n
distinct straight lines in the Euclidean plane to be situated so as to have at least
2 6n -3n
intersect and at least
points where exactly three of these lines 6n+1
points where exactly two of these
lines intersect?
49.
Let
S
be a set with
n (�1)
explicit formula for the number inality is a multiple of
SO. p (x) n
For each integer
Determine an
of subsets of
S
whose card
3 .
n � 1 , prove that there is a polynomial
with integral coefficients such that
Define the rational number
(50.0)
A(n)
elements.
a
n
by
n-1 1 ( 1) = a P (x) dx , n n 1 n 0 4
n = 1,2,...
.
PROBLEMS (51-57)
12
Prove that
a
satisfies the inequality
n
ITT
51.
- a I n
<
l --::-,=---:5n-l '
n =1,2,...
•
4
In last year's boxing contest, each of the
the blue team fought exactly one of the
23
23
boxers from
boxers from the green
team, in accordance with the contest regulation that opponents may only fight if the absolute difference of their weights is less than one kilogram. Assuming that this year the members of both teams remain the same as last year and that their weights are unchanged, show that the contest regulation is satisfied if the lightest member of the blue team fights the lightest member of the green team, the next lightest member of the blue team fights the next lightest member of the green team, and so on.
52.
Let
less than (a)
be the set of all composite positive odd integers
S
79
•
Show that
S
may be written as the union of three (not
necessarily disjoint) arithmetic progressions. (b)
Show that
cannot be written as the union of two arith
S
metic progressions.
53.
For
b > 0
prove that
,
(b .
l
o
1
x
21
by first showing that b . Sl.n x dx = x o 0
b
-ux
e
sin x dx
du
.
PROBLEMS
54.
13
(51-57)
o < a
44
be
a ,a ,···,a44 1 2
Let
< a
1
<
2
55. p
•
< a 44
� 125 .
differences
Show that for every natural number p = a
If
2 + b , where
2
n.
both greater than
a
•
b
(B)
If
a
r
-a = a j j+1 j
occurs
there exists a prime are natural numbers
b
(You may appeal to the following two theorems:
p = a s
and
n
and
4t+l
is a prime of the form
P
such that
and
d
times.
10
such that
(A)
•
43
Prove that at least one of the at least
natural numbers such that
2
2 + b .
then there exist integers
are natural numbers such that
there exist infinitely many primes of the form
GCD( r,s) = 1 ,
rk+s , where
k
is a
natural number.)
56.
Let
(1)
0
(2)
all the differences
(3)
a.
a ,a ,··· ,a n 2 1 < a < a 1 2
<
�
•
:: a
•
•
be
n
< a
,
n
a. - a �
J
•
are distinct,
� n)
•
, where
1 " a " b-l
integers such that
< �
(1 " j
•
(1 " i " n)
(mod b)
�ntegers such that
(�1)
•
and
a
b
•
Prove that n
L
a
r=1
57.
Let
A
n
- (a
-
ij
a
where
-11
< t < 11
•
)
ij
(a
r
nXn
be the
b
- 6)
n
•
matrix where
2 cos t
,
if
i = j ,
1
,
if
[i - j
0
,
-
Evaluate
D
n
otherwi se
= det
A
n
•
[ ,
- 1
,
are positive
PROBLEMS
14
58.
Let
a
and
be fixed positive integers.
b
(58-66)
Find the general
solution of the recurrence relation (58.0)
=
Xo
where
59.
=
0
Let
x
n
+ a
4ax
n
,
n
=
0,1,2,...
•
•
be a fixed real number satisfying
a
.a I
(59.0)
-
r
-a
0 < a < 1T , and set
1 - r cos u --= -="": 2 du . :"' :::" : u': :: ::s : --: c-=1 -'+ r 2r c
Prove that I , lim r + r+ 1
I
r
all exist and are all distinct.
C� w.
I::,
I
E
,
of
Let
denote the class of all isosceles triangles.
I
For
let
denote the length of each of the two equal altitudes
and
the length of the third altitude.
does not exist a function
f
of
h I::,
Prove that there
such that
•
kl::, � f(hl::,) , for all
61,
I::,
E
I
.
Find the minim.m value of the expression
(61.0)
for
x > 0
number.
- 2 (l+cos t)x +
and
o
� t � 21T ,
k(l+sin
where
t)
x
k
>
1 2
+
(3
+ 2 cos t + 2 sin t)
+
/2
is a fixed real
,
PROBLEMS
62.
Let
(58-66)
€ >
15
O .
Around every point in the xy-plane with integral
co-ordinates draw a circle of radius
€
Prove that every straight
.
line through the origin must intersect an infinity of these circles.
63.
Let
be a positive integer.
n
k = 0,1,2, ... ,2n-2
For
define k x
(63.0)
o Prove that
64.
I
Let
n-tuples
�
k
,
I
n_1
2n n + x +1 x
dx
•
k = 0,1,2,... ,2n-2 .
be the region in Euclidean n-space consisting of all
D
(x ,x ' ...,x ) 1 n 2 ,
•
satisfying
•
•
,
x � 0 n
,
Evaluate the multiple integral
(64.0)
•
•
•
D
where
65.
k , 1
•
.
.
,k
n+1
are positive integers.
Evaluate the limit n
L =
lim n +<»
66.
Let
p
and
q
.!. I: n
- 2
•
k=1
be distinct primes.
Let
S
be the sequence
consisting of the members of the set m n {p q :
m,n = 0,1,2, .
arranged in increasing order.
For any pair
.
.
} (a,b)
of non-negative
PROBlEMS
16
integers, give an explicit expression involving a b p q
the position of
67.
p
Let
in the sequence
p
elements
P with entries from
Z
N(a)
•
Let
n
•
•
P •
denote the finite ,p-1
of
p
where
A
a
-
For
.
a
2x2 matrices
an X ,
0
-
•
0
68
for
, such that 2 = A , X
(67.0)
z
0,1,2,
, determjne the number
Z
a, b, p and q
.
denote an odd prime and let
field consisting of the element of
S
(67-74)
a
be a non-negative integer and let
unique differentiable fu nction defined for all real
(f(x) ) 2n+1
(68.0)
+ f(x) - x = 0
f(x) x
be the by
.
Evaluate the integral x f(t) dt ,
o for
69
x 2: 0
,
•
f(n)
Let
denote the ntmber of zeros in the usual decimal
represen tation of the positive integer f(1009)
= 2
.
For
a
>
0
.
N
and
n , so that for example,
a positive integer, evaluate the
limit S (N) .en L = Iim ' N .en N+"" where
S(N)
=
I
k=1
f(k) a
.
PROBLEMS
70.
Let
(67-74)
k
be an integer and let
n � 2
2 � k � n .
17
be an integer with
Evaluate ,
M = max S
where
runs over all selections
S
{1,2,... ,n} such that
71.
az
Let
such that
a
2
< a < a 2 1
b
•
•
< a k
•
•
.
•
'�}
from
•
be a polynomial with complex coefficients
+ bz + c
and
S = {a ,a ' 2 l
Prove that the zeros of this
are nonzero.
polynomial lie in the region
IzI
(71.0)
72.
;
f(x)
Let
n
cl bl . f(x)
with integral coefficients
is solvable for every prime
= 0 (mod p)
is not solvable with
= 0
73.
+
Determine a monic polynomial
such that f(x)
b
x
but
an integer.
be a fixed positive integer. M =
p
Deterilli ne
max O�x k�1 ,n k-l,2, •
74.
Let
{x : i
i" 1,2,
•
•
.
•
.
,n} and {Y : i
i = 1,2,
sequences of real numbers with 2::
How must
Y '···'Y l n
•
•
•
2:: X n
•
be rearranged so that the
(74.0)
is as small as possible?
n
L
- Y ) (X i i i=1
2
S11m
•
•
•
,n}
be two
PROBLEMS
18
(75-84)
denote the finite field Let ? be an odd prime and let Z P p-1 . Let g be a given function on Z consisting of 0.1.2 P with values in Z . Determine all functions f on Z with values p P in Z which satisfy the functional equation p f(x) + f(x+1) = g(x) (75.0) 75.
• . • . •
•
for all x 1' .
'6
in
Z
P
•
Evaluate the double integral I
(760 . )
77
•
1 1 d x d Y · = 1 x 00 y
Let a and b be integers and m an integer
1 .
>
Evaluate � +
m
78.
a+b 2a+b + + m m
Let a 1.···.an be 2 S = a + ... + 1
n (>1) 2 an
•
.
•
•
(m-1)a+b + m
distinct real numbers. = min (a lSi<jsn i
M
Prove that .§. M
> -
n (n-1) (n+1) 12
Let x1,···,x be n real n n L I�I = 1 , k=l Prove that n� S (79.0) L k k=l 79
•
•
•
numbers such that n
t
� k 1 2
-
1 2n
o
-
•
•
•
Set
PROBLEMS (75-84)
80
19
Prove that the sum of two consecutive odd primes is the
,
product of at least three (possibly repeated) prime factors.
81.
Let·
be an integrable function on the closed interval
f(x)
[TI/2,TIJ and suppose that TI f(x) sin kx dx
0
-
1 ,
TI/2
i f(x) i
Prove that
82,
n
For
(82.0) where
=
s
a
n
> -
0, 1
n
,
2,
n
n+1
.
-
n
.
•
•
on a set of posl.tl.ve measure. •
, let
+
1
+
n- 2
+ .. ,
+
exists and determine its
lim s n '" n-+
Show that
•
k
•
bl 2
11
=
6n+1 =
1
1 S k S n-1 ,
,
value.
83,
Let
be a non-negative strictly increasing function on
f(x)
a
the interval [a,bJ, where the curve so that Let
y
A(a)
for all
84 .
.
be a function such that (x'
a s x
Let
Let
a
denote the area below
A(x)
<
- x)f(x) x'
and
b .
S
b
How should the number
<
F(a)
F(x') - F(x) Prove that
<
=
0
and
(x'-x)f(x')
A(x)
=
F(x)
for
a S x S b .
be two given positive numbers with r
a < b
be chosen in the interval [a,bJ in order
to minimize
(84.0)
a S x S b,
0 .
=
F(x)
(83. 0)
b .
and above the interval [a,xJ, where
f(x)
=
<
M(r)
=
max asxsb
r - x x
? •
.
PROBLEMS
20
85.
{a : n = 1,2, n
Let
lim
numbers with
a
n
n .... oo
integer such that
a
N
•
•
}
be a sequence of positive real
and satisfying the condition
= 0
-a a -a > 0 > a n+1 n+2 n n+1
•
For any
•
e: > 0 , let
N
00
s; 2e:
•
Prove that
L
L =
be a positive
k +1 (_l) � satisfies
k=1
the inequality
IL
(85.0)
86.
(85-94)
< e:
-
•
DeteImine all positive continuous functions
defined
f(x)
on the interval CO,n] for which
n f(x) cos
(86.0)
nx
n = (-1) (2n+1) ,
dx
n = 0,1,2,3,4
•
o
87.
P
Let
pI
and
circular ellipse
E
be points on opposite sides of a non
such that the tangents to
E
through
P
and
pI
respectively are parallel and such that the tangents and nOImals
to
E
.
at
P
and
pI
DeteImine the equation of
E
of
maximum area.
with respect to a rectangular coordin
ate system, with origin at the centre of
E
and whose y-axis is
R .
parallel to the longer side of
88.
R
deteImine a rectangle
If four distinct points lie in the plane such that any three
of them can be covered by, a disk of unit radius, prove that all four points may be covered by a disk of unit radius.
89
•
Evaluate the sum
S =
00
00
L
L
m=1 n=1 m nP'n
1 2
- n
2
•
(85-94)
PROBLEMS
90
If
•
21
is a positive integer which can be expressed in the
n
a,b,c
n =
form
k,
that, for each positive integer 2
form
+
A
91.
B2
Let
relations
92.
2 , where C
+
3 aba = b
satisfying
can b e expressed in the
are positive integers.
A,B,C
b5 = 1 .
and
{a : n = 1,2,3, n 0
n
2k
be the group generated by
G
Let
are positive integers, prove
a n
<
00
1
<
•
.
•
}
a
and
Prove that
b
is abelian.
G
be a sequence of real numbers 00
for all
n
L
and such that
•
Let
a,b,c
If
00
L
converges, prove that
93
a n
n- 1
If (an) 1
diverges
n=l be a function defined on [0,1]
2 a while converges. Let f(x) n n-1 such that ftl (x) exists and is bounded on [0,1].
L
subject to the
also converges.
00
L
n- 1
f(a ) n
be real numbers such that the roots of the cubic
equation 3 x
(93.0)
(2/a2-3b - a) / 3 Let
ments.
Let
the number
+
c - 0
Prove that these roots are bounded above by
are all real.
94.
+
2 ax + bx
2
•
= {0,1,2,3,4}
5 a,b,c,d N
denote the finite field with
be elements of
2
of distinct solutions in
5
with 2
5
a
�
O .
5
ele
Prove that
of the cubic equation
2 f (x) - a + bx + cx + dx3 = 0 is given b y
N
= 4 -
R
, where
R
denotes the rankof the matrix
PROBLEMS
22
A
95
•
=
a
b
c
d
b
c
d
a
(95-100)
•
c
d
a
d
a
b
b
Prove that co
(95.0)
1
L
S =
m,n=l (m,n)=l
(ron)
2
is a rational number.
96
•
Prove that there does not exist a rational function
f(x)
with real coefficients such that 2
x f x+1
(96.0) where
p(x)
97.
For
= p(x) ,
is a non-constant polynomial with real coefficients.
n
a positive integer, set n S(n) =
L
k=O
1 n k
•
Prove that n+1
n+1 S(n) = n+1 L 2 k=l
98
•
Let
u(x)
k 2 k
•
be a non-trivial solution of the differential
equation u" defined on the interval
+ pu =
0 ,
I = [l,co) , where
p = p(x)
is continuous
PROBLEMS
on
I
[a,bJ,
.
(95-100)
P rove that I
has only finitely many zeros in any interval
u
S a < b . u (x)
(A zero of
99.
23
Let
P, (j J
is a point
=
0, I ,2,
•
•
.
z ,
,n-l)
points on a circle of unit radius. S (n) where
100.
IpQI Let
L
be
, with
00
n (�2)
M
be a 2 2
3x3 =
u (z)
0).
=
equally spaced
Evaluate the sum
OSj
I P, P k J
2 ,
denotes the distance between the pOints
the finite field invertib Ie?
=
S z <
I
P and
Q
•
matrix with entries chosen at random from
{O,l}.
What is the probability that
M
is
THE HINTS The little fishes of the sea, They sent an answer back to me. The little fIShes' answer was "We cannot do it, Sir, because
--
,"
Lewis Carroll
1.
Define n <: 0,
+ b , n and prove an inequality of the type c
1 1/2 -
a n-l where
2.
c
1 1/2
a n
,
n � 1 ,
].s a constant. •
Consider five cases according as (a)
b
(b)
b
(c)
b
(d)
b
(e)
b
> -
< =
-
d , d
>
c ,
and
a
d
and
a < c ,
d
and
a .. c
d ,
•
In case (a) show that 1 " + 00 by bounding �(a,b,c ,d) from below n E. In case (b) show that 1 by a multiple of ( ) + 00 by estima d from below in terms of the harmonic series. ting �(a,b,c,d) =
•
25
HINTS (3-14)
26
1 . (a,b,c,d) �
Cases (c) and (d) are easily treated by considering The final case (e) is trivial.
A straightforward approach to this problem is to show that the
3.
function F(x)
..
-
3 ,tn x ,
x
>
0 ,
suggested by the inequality (3.0), is increasing. Apply Rouche's theorem to the polynomials
4.
Rouch€'s theorem anaLytic within and on a simpLe Ig(z)1 < If(z) I on C , where and fez) + g(z) have the same
g(z)
5.
=
p(z).
and
x
=
a - t
to
a
-
o
6.
=
states that if fez) and g(z) are cLosed contour C and satisfy fez) does not vanish, then fez) number of zeros inside c .
Apply the change of variable I
f(z)
n -z
f(x) dx f(x) + f(a- x)
•
Integrate x+1 I
..
x
by parts and obtain an upper bound for
II I .
7.
Consider (7.0) modulo 16.
8.
By squaring (8.0), obtain the lower bound
;'zk+ n
Using this bound in (8.0) obtain an upper bound for
x n
for •
x. n
HINTS
9.
(3-14)
27
Assume that the required limit exists, and use (9.0) to det
ermine its value 10.
Either set
function of 2
L.
6 ,
x
2
A(x + y ) - (Bx +Cy) 11.
y " sin 6
cos 6 ,
=
and maximize the resulting
2
2
ax +2bxy +cy
or express
2
Ixn - LI .
Again use (9.0)to estimate
in the form A,B,C .
for appropriate constants
Consider the coefficient of
n x
in both sides of the identity
-
12.
Express
•
k k (l.(l.+l» - (Cl. - Ol.) ,
(k
1, 2 , polynomial'in l. , then sum over l. k (n(n +1 » as a linear combination of =
.
•
•
=
1,2,3 , .. )
, n
.
as a
to obtain
Complete the argument using induction. 13.
Prove that the equat�on •
bc x+ca y +ab z
=
2abc - (bc +ca+ ab)+k
is solvable in non-negative integers
x,y,z
Then show that the equation with k " 0 �ntegers •
for every integer
k � 1.
is insolvable in non-negative
x,y,z,
th term of the sequence (14.0) and show 1 4 . Let un be the n (kl) k k-l , 0, 1, 2 , +1 +l., l. that u - k for n 2 n 1 and deduce that k S; 2(1 + /8n-7) < k + 1 =
=
•
•
.
"
HINtS (15-27)
28
15.
Use Cauchy's inequality to prove that n
L and choose the
16.
,
1
so that equality holds.
x.
1
Express
2 a.
-1
(n +
1) 3
in the form
An(n - l)(n - 2) + Bn(n - 1) + Cn + D for suitable constants
A,B,C,D .
17.
Integrate
1 8.
For part (b), find the quartic equation whose roots are
a -
z,
b
-
z,
c
F' (a-x) - F' (x)
-
z,
d -
tWlce. •
and use part (a) to ensure that the
z,
product of two of these roots is equal to the product of the other two.
19.
Differentiate
f (x) n
to obtain the difference-differential
equatlon •
•
20.
Consider
21.
Show that
f'(x) - p'(x) f (x) n n
sinh x + sinh wx
I - I n-l n
-
'If
o
•
2 + sinh w x , where
sin(2n i Sln x
1) x ··dx
,
1 w - -(-1 + 1-3) 2
n i: 2
•
,
and then use a similar idea to evaluate the integral on the right side.
HINTS (15-27)
29
a. (i 1, 2, , 365) denote the number of books sold � th day inclusive. during the period from the first day to the i
22. Let
=
"
'
Apply Dirichlet's box principle to
23.
Show that a polynomial of the required type is (x + y -
f(x,y) by showing that
1)(x + y - 2) + x , 2 '
f(x,y) · k , where
k
unique solution in positive integers in terms of the integers
r
1} r (r 2.:.. --= �=----=..::": ..:. _
<
x
m
and k �
is a positive integer, has a and
y
which may be expressed
defined by
r(r
;
1}
,
m
24.
Consider the complex conjugate of (24. 0) .
25.
Consider hn in II r-kn+l
1
k
_
(r --.;;.1} (;,;; r - ....;;;.;..2; "-....;. '::; _ 2
_
-�
and use the expansion
- in
(1 - x)
�
co
-
k-l
26. For the evaluation, set
k
x ..,,k
x - a - y
,
Ix l
pos1t�ve �ntegers •
•
r
.
1
•
in the integral.
27. Use mathematical induction to prove that •
<
a
r
•
r
for all
•
HINTS (28-42)
30
28.
Use integration by parts to establish the recurrence relation u
29.
n
=
n(n-1)
u
n � 2 .
•
n-2
The series may be sUilililed by using the identity tan A
30.
=
cot A - 2 cot 2A
Prove that the number of k-selections
W(S) ;;: r
r
•
=
1 2 3 .
.
• • • .
n
kq
For each + r
•
n ;;: 1
0 :> r
n n n the series in terms of =
n
lim (1 m+oo c
from
N
such that
•
and
uniquely by r n Express the partial sum s of n and q and determruline lim s by appealn n ' n-+-oo
define integers
< k
ing to the result
where
S
, is n - (k-l)(r-l) k
31.
•
•
•
•
•
+
1
m
-
- .en m)
=
c
•
denotes Euler's constant.
32.
Recognize the given integral as the imaginary part of the 00 m (i-l)x dx . Evaluate the latter integral using inteintegral x e gration by parts.
33.
For the evaluation. iterate
I(u) (n
and then let
n -+- + 00
in the case
=
=
1 2 ) I(u I
1.2, 3,
0 < u < 1
•
. • .
to obtain )
HINTS (28-42)
34.
31
Determine an exact expression for s _ (k) k l
the values of
and
s (k) k
with
s (k) and then compare n 3k3 - 5k2 •
35.
Show that (35.0) converges by comparison with
36.
Apply l'H6pital's rule.
37.
Relate
00
I
p -1 n
n=l
h,k,l
to the lengths of the sides of the triangle,
and thea use the triangle inequality.
38.
Obtain the recurrence relation
39.
Show that
40.
The first few terUls of the series are
p
=
p
n+l,r n+l,r-l and apply the principle of mathematical induction. Bx + D
is a divisor of
1
3
3 91
=
=-
1 1 1 -2 3 7 1 1 2 7
-
-
1
13
,
•
Use the identity
(1-x)(1-2x)
42.
F .
,
2 = 21
41.
•
.
•
.
(1-nx) =
For a suitable constant
show that for
t
>
0
•
C , set
•
•
eX f(x) = + C , and =eX + 1 -
-
HINTS (43-56)
32
t (I;; ax bx ( f(bx) f(ax) , ,.e e ) _ dx dx dx = ! x x ax ' bx JO z{ e +1)(e +1) o JO t
_
43.
S
Let
denote the
n the first fe-oJ values of
Stirn
n
of
terUls of (43.0).
, conjecture the value of
S
•
n and prove it by mathematical induction.
44.
Cons�der the polynomial whose roots are
use (44.0)
45.
to
z ,z , l 2 show that its constant terUl is zero.
Obtain a recurrence relation for
D
its first ro ....
n
Calculate S
•
in general,
n
•
,z
•
by expanding
46.
Expr�ss
47.
Consider the sums of the integers in subsets'of
xyz
•
k
D
n
' and
by
A, Band C .
in terms of
S
and apply
Dirichlet's box principle.
48. 49.
Count pairs of lines in the proposed configuration. Show that
A(n) =
n
I
n k
and evaluate this sum by considering
k=O k=O (mod 3) 2 n (l+w) , where
so.
w
is a complex cube root of unity.
To prove the required inequality, replace
4n _ 4n n n _ x (1 x) ( 1) 4 1 + Xz
p (x) n
by
_
1 ::; 1 l+xz
and
in (50.0), and then use the inequalities
x(l-x) <2:. -
4
I
to est�mate •
o
4n _ 4n x O x) ::"' ":' ""':': --="'- dx l + xZ
•
33
HINTS (43-56)
51,
B ,B ,... ,B 3 (resp. G ,G , ... ,G 3) be the members of 2 l 2 l 2 2 the blue (resp. green) team, ordered with respect to increasing weight. For each r (1 � r � 23) consider last year's opponents of Let
,···,B 3 or r+l 2 lighter than G r
according as
B
52,
B r
is heavier or
•
S
Each member of
for suitable integers
r
can be written in the form
�
l
and
s
� O .
(2r+l) (2r+2s+l),
Use this fact to construct
the three arithmetic progressions.
53.
For
y
>
0
prove that b -ux . e S1.n x dx
y o
o
b du
=
o
. -x Y Sl.n x (i-e ) dx x
and then show that . b -xy Sl.n x ) dx (i-e x
lim y-+oo
54,
55.
Consider
o
43
�
=
•
o
d. .
j=l
J
p
n , construct a pr1.me
For any natural number
•
n
2
p = 4k II (r +q) r'" 1 where
b . s1.n x dx x
k
is a natural number and 2 2 3 4t + a + b , 0 , so that p
q
2
- q ,
n
>
of the fotUl
is a prime of the form
Then, assuming a � n , 2 2 obtain a contradiction b y considering the factor a +q of b =
<
a
<
b .
•
b
r-r
:)0,
For
and
r
2 � r � n
obtain a lower found for
by considering the differences
a.-a. 1.
J
a -a r 1 ,
in terms of
1 < J· < i �
� r .
HIN'IS (57-68)
34
D
57. n .
n
for all
Prove your conjecture by using the recurrence relation which may •
be obtained by expanding
58.
D
n
by its first row.
Prove that x .. x + a n n+l
,fb2
+ 4ax
n+l
and use this to obtain the recurrence relation - 2x + x 1 n+ n
59.
For
r
x
n- 1
=
2a .
0
>
1
2 I .. a + - (l-r) 2 r
a -a
du -:--= -:;..;::.- -�2 ' 1-2rcos u + r
and evaluate the integral using the transformation
60.
t
=
tan u/2
•
Construct a class of isosceles triangles whose members have
two equal altitudes of fixed length
h ,
while their third altitudes
are arbitr arily long.
61.
Recognize the expression in
(61.0)
ance between a point on a certain circle and a point on another plane curve.
62.
When the line
L
through the origin has irr ational slope,
use Hurwitz's theorem to obtain an infinity of l attice points whose distances from
L
are suitably small.
In 1881 Hurwitz proved the following basic result:
If
b
'/..s •
an irrational number then there exist infinitely many pairs of integers
(m,n) with n
'"
0
and
GCD(m�n)
..
1
such that
36
69
HIN'IS
•
70. 71
•
m S(lO -1)
Evaluate
SeN) .
n-l
M"
Show that
exactly and use it to estimate
(69-81)
k-l
•
az
Express the roots of
2
+ bz + c
in terms of
a, b and
c
and estimate the moduli of these roots.
72..
Choose
a, b and
•
lntegers
2
+ a :: 0 (mod p) 2 p'" 2 • x + b :: 0 (mod p)
such that
c
lS solvable for prlmes
x
p :: 1 (mod 4) and 2 + c :: 0 (mod p) lS solvable for p :: 3 (mod 8) ; x •
•
•
p :: 7 (mod
8)
;
is solvable for
and set 2 2 2 f(x)'" (x +a)(x +b) (x +c)
73.
•
•
Assume without loss of generality that •
and show that
S
•
�
llOi<jlOn
I x.l
- x·
J
l
n -
L
x (2k-n-l) k k-l k �
Consider those terms in the sum for which that
74.
2 M'" (n /4] .
i(n+l)
•
and deduce
Show that the smallest sum (74.0) is obtained when the
y.
l
are arranged in decreasing order.
75.
Replace
x
by
x+k
(k'" 0,1,2,
•
•
•
the alternating sum p-l
L
k-O
k (_l) g(x+k) .
,p-l)
in (75.0) and form
HINTS ,
76.
(69-81)
37
Express the improper double integral
as a limit of proper
I
double integrals over appropriate subregions of the unit square and use standard methods to show that
77.
Use the identity k-l
x
� k
+ e
x=O where
78.
k
- [ekJ
Reorder the
b
i
- a
j
j
,
is any real number.
in ascending order and define
a's .
+ lM(i - j)
(i
- 1,2,
•
S
Deduce the required inequality from
79.
e
is any positive integer and
for a fixed subscript Set
I -
2 n /6 .
•
�
•
min a i lSisn
n
L b�
1
.
Establish and use the inequality
1 2 -1 n k
:0
1
-
-1 n
'
1
:0 k :0 n
•
th
80.
, and show that if n prime by Denote the Pn k R. 2 p , for some odd prime p , then k + R. � 3 + Pn Pn+l a
81.
=
and prove that
,n)
i",l
2
.
Estimate the integral
n n/2
I f(x) I
n
�
k-l
sin kx
from above under the assumption that except for a set of measure O. and derive a contradiction.
Use
dx
1 nR.n2
I f(x) 1 < (81.0) to obtain
on
"n L2,nJ
"
a lower bound
38
(82-96)
HINTS
s
82.
Show that
83.
Assume that
is non -decreasing and bounded above.
n
A(x) and F(x)
differ at some point
and obtain a contradiction by partitioning [a,c]
c
in (a,b]
and using (83.0) on
each subinterval.
84.
85.
n S
Let +
86. cos nx
n
=
L
k=l
k+l (_l) a k
-
1,1
-�)
and
r .
and show that
and
I Sn_l-L I .
Express
2
+ 1)
(cos x + cos x
(n = 0,1,2,3,4)
o
2
as a linear combination of
and consider
1T
87.
�
max(
with an appropriate choice of
M(r)
then minimizes
as
M(r)
A direct approach recognizes
2
f(x) (cos x + cos x + 1)
Begin by determining
2
dx .
when the ellipse is in standard
R
position and then rotate the axes through an appropriate angle.
88.
Given n(�4) aonvex regions in the
Recall Hetty's theorem:
ptane suah that any th ree have non-empty interseation, then aU regions have non-empty interseation.
89.
Use partial fractions and the result
N lim
L
1 k
-
N+oo k=l
-inN
::
C
,
n
HINTS (82-96)
where
c
39
is Euler's constant, to evaluate lim
N+oo
90.
N
L
n=1 n"'m
1
2 m -n 2
•
Use the identity 2 2 2 2 2 2 2 2 2 2 ) + (2yz ) . (x +y +z) = (x +y -z) + (2xz
91
Prove that a and 1 2 -1 in the form b- ab = b a •
b
commute by using the relation 4 4 to deduce ab = b a .
92.
Start by applying the extended mean value theorem to
93.
Let
f
on
[O,a J. n
inant of
94.
p
be the largest root of (93.0). 3 2 (x +ax +bx+c)/(x-p) .
Let
B
be the Vanderlllonde matrix given by
and consider the rank of
95. in r
B =
BA .
Collect together terms having the same value for 00
J
96.
Consider the discrim1 1 1 1
1 2 3 4
12 22 32 42
13 23 33 43
,
GCD(r,s)
1
2 • (rs) =1 Suppose that such a rational function
f(x)
exists and use
the decomposition of its numerator and denominator into linear factors to obtain a contradiction.
HINTS (97-100)
40
97.
Sum the identity (n+l) !
for
k
98.
=
•
•
•
, n- l
-
n
1
=
n-l
k ! (n-k) ! - (k+l) ! (n-k- l ) ! ,
k
•
A s sume that the set of zeros of
[a , b ] with
u (x)
S
on [a , b ] , 1
Deduce the existence of an accumulation point u(c)
Take
P
=
u ' ( c) = 0 , and then show that
(j
j on the unit circle 2 I P k I in terms of
/
lOO.
-
�
(n+l)
0, 1 , 2 ,
is in,finite .
99.
2
Let
0, l , 2 ,
=
I I z
= 1
•
•
•
u(x)
to be the point
,n- l )
=
0
a c
<
b , in
on [a , b ] .
exp (2nji/n)
in the complex plane , and express
exp ( 21T (k- j) i/n)
•
( 1 S i , j S 3) and with the usual notation ij let det M a 1 A1 1 + a A + A a . Begin by counting the number 1 1 1 l l 3 2 2 3 of triples ( a , a , a ) for which det M 0 , dis tinguishing two I 3 ll I2 cases according as (A , A , A ) (0,0,0) or not. l l I2 I3 M
=
(a
)
=
=
=
TO E SOLUTIONS It seemed that the next minute they would discover a solution. Yet it was clear to both of them that the end was still far, far off, and that the hardest and most complicated part was only just beginning. Anton Chekhov
1.
If
{b : n = 0, l , 2 , .. . } n numbers , prove that the series
is a sequence of non-negative real b
00
L
( 1 . 0)
n
n=O
converges for every positive real number Solution:
For 'a > 0
a .
we set •
•
+ b
•
n
,
nil: 0 ,
so that a - a = b ' n n-l n As
b
n
;;: 0
we have , for a
O
nil: l .
n il: 1 ,
;;: a > 0
and
41
a ;;: a 1 > 0 . nn
SOLUTIONS (1-2)
42
n � 1 , we may deduce that
Now, for
b
n
( a +c. O+b 1+ ' . . +bn
=
3 /2 )
'"
a - a n n-1
/2 3 a
1
n
a n-1
-
1 /2 a
/2 3 a
n
=
n
a 1 n-
1
1 /2 a
1
-
n
=
a 1 n-
1
1/2 a
-
1/2 a n-1
m �
1
n
1 /2 a
n
Hence , for
1
1 /2 a
1 2 / 1 a
-
m
1/2 a
-
1/2 a n-1
1 1 /2 a n
1
-
1 /2 a n
•
n
we have
m s
1
1
n-1
n
n- 1
,
. /2 1 a
1/2 a n- 1
1 +
1/2 a 2
+
1 /2 a n- 1
n
1
m
=
+ n=O
2 L n=l
b =
O
3 72 a
+
<
O
/2 3 a O
1 2 1 /2 a O
O
b
1 1 /2 a
+
-
n _1 -
1 1 /2 a n
1 1 / 2 a m
2 1/2 a O
2 . 1 /2 a As the partial sums
(1 .0)
s
m converges for every
of
(1 .0) a > O.
are bounded , the infinite series
43
SOLUTIONS (1 -2)
2.
Let a , b , c , d be positive real numbers , and let (a+(n-l)b) a(a+b ) (a+2b ) Qn (a " b c , d) - c(c+d)(c+2d) . . . (c+(n- l)d) . .
_
.
•
Eval uate the limit L = lim � (a ,b ,c ,d) . n + OO Solution : Considering the five cases specified in THE HINTS , we show that +00 , in cases (a) , (b ) , L = o , in cases (c ) , (d) , 1 , in case (e) . c We set k = d + 1 , so that c < kd •
(a) When b
>
d , as c + j d
(j + k)d , we have for n n-1 a(n-l ) ! b Q (a,b ,c , d ) � k(k+I) . . . (k+(n-l) drl n n-1 ! b a(k-l) �!::. !. .:. .. ..:....:. �-= n (k+n-l) (k+n-2) n d <
�
k
-
.
a (k-l) ! ;: k (k+n-l ) d
•
.
•
n l b , d
-
which tends to infinity as n tends to infinity , showing that L = + (b ) When b = d and a
>
c we have
n- l a + Q,b Qn ( a ,b , c , d). = II c + Q,b Q,=O n-l a c = II 1 + c + )',b Q, = O n-l >
II )',=0
>
n-l a - c L b x'=0
Q.
1 +k
,
co.
SOLUTIONS (3-4)
44
which tends to infinity as n tends to infinity since the harmonic series diverges, showing that L = + co
•
(c) (d) By considering the reciprocal of Qn (a ,b ,c ,d) , we obtain from the above results L 0 , if b < d , or if b d and a < c =
-
•
(e) Clearly � (a,b,c ,d) '" 1 in this case , so that
L =
1.
3. Prove the following inequality :
.in x
(3 . 0)
3 x -1
Solution : For x
>
<
(x+!) 3 (x 3+x)
.!.
,
x
>
0, x
;I!
1.
0 we define
3 (x -1) (x+l) F(x) '" 3.in x , 3 (x +x) so that - 3 .in x
F(x) '"
•
Differentiating F (x) with respect to x , we obtain 2 3 2 3 3 4 ( -x-l) (3x +3x -l) (x +x) (x +x 4x +1. ) 3 F' (x) '" ' 2 3 x (x +x) -
x that is
6 5 4 3 2 x +3x -2x +3x -3x+1 -3x (3 . 1 ) 3 2 (x +x) The polynomial p (x) in the numerator on the right in (3 . 1 ) has the 1 6 -3 property that p (x) '" x p (-x ) , and so x p (x) can be written as a cubic polynomial in x + l/x . •
SOLUTIONS (3-4)
We have
As
45
F' (x )
�
3 3 x !.) (x+ 2 x +x 3 (x )
_
3 2 X -3 X+4 " (x..-bl) ( X-2 ) we obtain
2 2 t1 ) 1 +x+ (x x ) 2 ( F' (x) 2 2 x + (x l ) 2 2 x ( t �) + i) (x-I ) 2 2 x 2(x +1 ) ..
•
so that F'(x ) > 0 for x > 0 , while F'(l) .. O . Thus F(x) 18 a strictly increasing funct ion of x for all x > 0 . Hence 1n part1cular we have F(x) > F ( l ) for x > 1 , and so .lnx > for x 1 , (3 . 2 ) 3 x -1 1 Rep lacing x by x in (3. 2 ), we obtain .in x for o < x < 1 , (3 . 3 ) 3 x -1 •
•
•
,
•
•
Inequalities (3 . 2 ) and ( 3 . 3 ) give the required inequality . 4. Do there exist non-constant polynomials p ( z ) in the complex n variable z such that I p (z) 1 < R on I zl - R , where R > 0 and p(z) is monic and of degree n? Solution : We show that no such polynomial p (z) exists , for suppose there exists a non-constant polynomial + an- l such that
I p (z ) 1
<
n R on
n-l z
Izl " R .
SOLUTIONS (5-7)
46
Then we have
and so, by Rouche's theorem, z
n
+ (a _ z n l
n-l
and
+
-z
n
have the same number of zeros counted with respect to multiplicity n-l inside I z I = R , that LS, a _ z has n zeros, + ... + alz + a a n l which is clearly a contradiction. Hence no such polynomial p(z) .
eXLsts. •
5,
Let
such that
be a continuous function on
f(x)
f(x) + f(a-x)
does not vanish on
[a,al , where [a,al .
a
>
a ,
Evaluate the
integral a a Solution:
f(x) � '-'="' dx . �f�(x�) + f�(a---x�)
Set I=
a a
f(x) dx , J= f ex) + f(a-x)
a
f(a-x) dx . f(x) + f(a-x)
Clearly we have I + J=
a a
1
dx = a .
On the other hand, changing the variable from obtain I =J . Hence we have
1a.
I =J=Z
x
to
a-x
Ln •
I , we
SOLUTIONS (5-7)
Solution:
47
Integrating by parts we obtain -
so that for x
2 1 .:c=.;0:. S ::.;Co-: �� d t 2 2 t
-
> 0
2 2 x+1 2 cos -cosCx+1 ) x sinCt ) dt = 2Cx+1 ) + 2x
_
2 x+1 co s ( t ) dt 1 2 2 t
_
glvlng •
•
x+1 2 sinCt ) dt
1 = x
-
so that x+1 £ 12 sinCt ) dt x 1 Since �+ £ x
0
,
as x + +
00
lim x + oo 7. ( 7 . 0)
1 :O--=-C
£ x
•
we deduce that
,
x+1 1-£ 2 sinCt ) dt = x
0 .
Prove that the equation 2 2 2 2 4 4 2 2 4 2Y z - 2 z x - 2x y x +Y + z
has no solutions in integers x , y , z .
__
24
SOLUTIONS (7-8)
48
Solut ion: Suppose that. ( 7 . 0) is solvable in integers x , y and z . 4 4 4 Cl ear 1 y X +y +z must be even . However x,y,z cannot all be even, as 24 is not divisible by t6 . Hence exactly one of x,y , z is even� and , without loss of generality, we may suppose that x Thus we have
� 0
4 x 2 2 -2y z
and so
( 7 . 0)
(mod
16)
-2 (mod
16)
� 0 �
(mod 2) , y
z
�
4 , y
� 1
4 z
�
(mod 2) . (mod 1 6 ) , 2 2 2 - 2x y -2x (mod � 1
=
16) ,
gives 2 - 4x
that is
2 x
� 8 �
(mod
16)
,
2 (mod 4) ,
which is impossible. Second so lut ion : We begin by expressing the left side of ( 7 . 0) as the product of four linear factors. It is easy to check that 2 2 2 2 A + B + C - 2BC - 2CA - 2AB (A+B-C) - 4AB ( A+B-C) - 2/AB) (A+B-C) + 2/AB) . 2 2 2 Replacing A , B , C by x ,y , z respectively , we obtain ,
=
=
4 2 2 2 2 2 2 + z - 2y z - 2z x - 2x y 2 2 2 2 2 2 = (x + y _ z _ 2xy) (x + y - z + 2xy) 2 2 2 2 «x_y) _ z ) «x+y) _ z ) = (x - y - z) (x - y + z) (x + y - z) (x + Y + z) , =
so that
(7 . 0)
becomes
(x - y - z) (x - y + z) (x + y - z) (x + y + z) = 2 4 .
SOLUTIONS (7-8)
49
In view of the form of the left side of ( 7 . 0) , we may assume without loss of generality that any so lut ion (x , y , z ) satisfies x � y � z � 1 , so that x - y - z
S
x -y
+
S
z
x
+
y -
z
S
x
+
y
+
z .
Moreover x - y - z and x - y + z cannot both be 1 As 24 we have (x-y-z , x-y+z , x+y-z , x+y+z) (1 ,2 ,2 ,6 ) ,( 1,2 ,3 , 4) or (2,2 ,2 ,3) . •
=
However none of the resulting linear systems is so lvable in positive �ntegers x , y , z . •
2 8 . Let a and k be positive numbers such that a > 2k. Set Xo - a and define xn recursively by k (8.0) xn - xn-l + ·x , n 1 ,2 ,3 ,. .. n-l Prove that xn lim n + oo Iff exists and determine its value . -
=
-
•
Solut ion : We wi ll show that x
l im ; " 12k . n + oo Clearly xn > 0 for all n � o. Since xn xn-l 1 , 2 . . , we have =
n
=
(8 . 1 )
,
.
2 Xn
..
2 xn-l
+
2k
+
+
k for xn-l
SOLUTIONS (8-9)
50
and so >
2 xn_2+4k
>
>
•
•
•
2 2 x O+2kn = a +2kn ,
>
that is xn � /2kn + a2 , n = O, 1 , 2 , . . .
(8 . 2)
•
On the other hand , we have, us ing (8 . 1 ) and (8 . 2) , 2 2 2 k xn -� xn-1 + 2k + --�-""2" , n = 1 , 2 , . . . 2k(n-l)+a
,
and thus
<
-
n-1 2 1 + 2kn + k L 2ki + a2 i=O n-l 2 2 dx a + 2kn + �k . + a 2 2kx -1
2 + 2 a 2k (n-l ) . - 2kn + a + 2 tn a2 -2k -
glvlng •
•
( 8 . 3)
•
Hence , from (8 . 2) and (8 . 3) , we obtain 1 and thus
$
xn hkn+a2 ::; lim n+oo
Since
lim n-+oo
k 1 +2
2kn (a2 -2k) + ( tn ) a2 -2k 2kn + a2 = 1
hkn + a2 = hk , we obtain In
•
xn l im m= 12k . -+
"n
00
51
SOLUTIONS (8-9)
a
-'
b
,
deno te a f ixed non-negat ive number , and let
Let
a
and
be posit ive numbers satisfy ing
/h Define
recursively by
x n
x = n
( 9 . 0) Prove that
l im x n n + oo
Solut ion :
As
so that
L
L
2
,
<
2 /h
+ b 1 -;- a + x n-l ax
n ..::. : .-.:..
n
'
= b
,
•
L -
If
l im x n n + oo
aL + b , L = L + a
+/h . I xn - /hI
=
-
=
Ixn - Ibl
:>
-
-
Ib
- /h) (a - /b) (x n- l x n- 1
I Xo
+a
Ib) IXn_ l x a + n- 1
(a -
- Ib) I x a n-l
I xn-1 so that
+b +a
ax n-l x n- l
(a
n
•
b > 0 , the recurrence re lat ion shows ,
,
1,2,3, .. .
•
exi s t s and determine its val ue .
Xo � 0 , a > 0
Next we have
Letting
•
x > 0 for n = 1 2 , . • • n then from ( 9 . 0) we obtain
that equal to
a
<
-
2
2
n
Ibl
/bl
tend to inf inity , we obtain l im x = n n+oo
;; . "0
•
,
-
-
.
Ib l
"
Ib l
eX l st s , say •
52
10.
SOLUTIONS (1 0)
Let
be real numbers satisfying
a,b , c
a > O Evaluate
Solution:
max (ax x,y E R xz+yz"l A 1 1 pairs by
(x ,y)
>
c
,
2
0, b
2
>
ac . 2
+ 2bxy + cy )
E RxR
satisfying
x
•
2
+ y
X " cos 6 , y .. s in 6 , O S 6 S 2n .
2
.. 1
are glven
Hence we have
2 2 max (ax + 2bxy + cy ) .. max F ( 6) , (x,y) Eaz OS6S2n Xl +yz-1 where
2 F (6) .. a cos 6 + 2b cos 6 s in 6 + c s in 6 2
..
%( 1
+ cos 26 ) + b s in 26 +
t(a+c)
1 = Z (a+c ) where
C learly
+
11 ���ffiT
z" Ca-c)2 14b2
a - c tana .. 2b
max F (6) OS6S2n required maximum is
s in ( 26 + a) ,
is attained when sin ( 26 + a)
We seek real numbers
A,B,C
Equating coeffic ients we obtain ( 10.2)
A - B
( 10 . 3)
-2BC
00 . 4)
A - C
2
.. a
,
.. 2b
,
.. c
•
1 , and the
..
•
such that (Bx+Cy)
( 10 . 1 )
2
- cos 26 )
•
+ 4b z S e�ond solution:
1-0
t(a-c) cos 26
+ b sin 26 +
•
2 •
53
SOLUTIONS (10)
Subtract ing ( 1 0 . 2) from ( 1 0 . 4) we obtain ( 1 0.5)
•
C
-
a
•
Then,from ( 1 0 . 3 ) and 00 . 5 ), we have 2 2 2 (B + C )
•
2 (B
_
2 2 2 C ) + (2BC)
.. (c - a)
2
+ 4b
2
,
so that
=
00.6)
•
Adding and subtracting ( 1 0.5) and ( 10.6 ) , and taking square roo t s , we get ( 10 . 7)
B ..
l( a-c)2 + 4b2 - (a-c) 2
,
C ..
l( a-c)2 + 4b2 + (a-c) 2
Then, from ( 1 0 . 2) and ( 10 . 7) , we have
2 2 ax +2bxy+cy
Final ly, from ( 10.1 ) , we see that the largest value of 2 2 on the c ir c l e x +y ';'1 occurs when Bx + Cy .. 0 , that is. at the po�nts •
(x ,y) ..
±C
+B
,
and we have 2 2 max (ax +2bxy+cy ) .. A X2+y2 "' I
•
i (/(a-c)2+4b2 +
(a+c) )
•
•
54
SOLUTIONS 01-12)
11. Evaluate the sum [n / 2 ]
( 1 1 . 0) for
n
\ £
n(n- l ) . . . (n- (2r-l » 2 r=O (r !)
S =
n-2r 2
a pos � t � ve � nteger.
So lution:
•
•
•
We have S
=
=
=
[n/2]
n! n-2r 2 ( r!) 2 (n-2r ) !
L r=O
[ n /2]
n n-r
L r=O
n
s
n-2r 2
s 2s-n
2s-n 2
s t
n s
L L s=O t=O
n-r n-2r
2s-t=n n x
which is the coeff ic ient of F ( x)
=
F (x )
=
n
�n •
s
L L s=O t=O
s t
n s
•
Now
= = -
n
� L s=O
s
2 n-s ( 1 + 2x) (x )
2 n ( ( 1 + 2x) + x ) ( 1 + x)
2n
.
n 2n is As the coefficient of x in ( 1+x)
) (2n! , we have S = (n! ) 2
•
55
SOqJTIONS ( 1 1 -12)
12.
m " 0 , 1 ,2, . . .
Prove that for
2m+1 2m+1 2m+1 .. 1 + .. . + n + 2 S (n ) m
(12.0)
is a polynomial in
n(n+1 ) .
We prove that
Solut ion:
1 2 S o (n) " 0
,
n(n+l)
2 2 S (n) .. (n(n+l) ) , 2 1 l 3 3 3 2 S ( n ) ,. (n (n+l ) ) S (n) + 2 0 l 2 2 2
4
4
S ( n ) .. S (n) + 2 1 2 :3 3
,
4 (n (n+l) )
,
• • •
and genera l l y for
k .. 1 , 2 , 3 , . . . k
k n) S 2 1 d r ( r+k'_ raO 2 r+k odd
(12. 1)
.. (n (n+l)
An easy induct ion argument then shows that a po lynomial in
n (n+1) .
We noW prove ( 1 2 . 1 ) . k
k S (n ) 2 1 r r+k- l r"O 2 r+k odd
We have ..
..
k
1
2
;
r-O r+k odd 2
k
n
t
t.. l
L reO
r+k odd
S (n) m
k •
(m ,. 0 , 1 , 2 ,
•
.
•
)
�s •
SOLUTIONS (13) .
56
=
n
k
L L l=l r=O
=
= =
n
L (
l= l
k k .e (l+1» - (
(n(n+1) )
k
as required . 13.
a,b,c
Let
be positive integers such that GCD(a,b) = GCD (b,c) = GCD ( c ,a) = 1 .
Show that
l
=
2abc - (bc+ca+ab)
is the largest integer such that
bc x + ca y + ab z = l is insolvab l e in non-ne gative integers So lut ion :
x,y,z
We begin by proving the fol lowing simp le fact which wi l l be needed be low:
Let
A,B,C,
•
be reaZ numbers suah that
A + B + C Then there exist integers
t,u,v
<
-2 .
satisfying
t - u > A , u - v > B , v - t > C .
57
SOLUTIONS (13)
To see this , choose t
=
[AJ + 1 ,
u
..
0 ,
v - - [BJ so that t , u , v
-
1 ,
are integers with t - u = [AJ + 1 u - v
>
A ,
[BJ + 1 > B ,
=
v - t = - [AJ - [BJ
-
2 � -A - B - 2 > C
•
The required result will follow from the two result s b elow : (a)
If
k
is an integer � 1 be x
+
ea y
+
then
�
abz
is always solvable in non-negative integers (b)
The equation
be x
+
ea y
+
ab z
=
As
GCD (ab ,bc , ca)
=
such that
x,y,z .
2abe - (be
Take
Xo
-
Yo
-
A = - a -
+ab)
is
'
+ ca YO + ab Z = o o
•
aa
1 , there exist integers x Y , z o o o
bc X
B
+
x,y,z.
insolvable in non-negative integers
Proof o f ( a ) :
+ ab) + k
=2abe - (be -I'- ea
b
1
,
1
,
k .
SOLUTIONS (13-14)
58
so that
xo YO A + B + C = - --'- + b a =
-
<
- 2
k - 2 abc
-2 .
Hence , by our init ial s imp l e fac t , there are integers
t,u,v
that
Yo
u - v > b
-
1
,
•
Thus we have a + x + at - au > 0 , o b + YO + bu - bv > 0 ,
Zo + cv - ct
> 0
•
Set x = a-I + x + at - au , o y '" b-l + Y O + bu - bv
,
Z '"
so that
x,y,z
+ Z
-1
o
+ cv - ct ,
are non-negatlve lntegers . •
•
Moreover bc x + ca y + ab z =
2abc - (ab + bc + ca) + k
as required . Proof of ( b ) :
Suppose the equation is solvab l e , then 2abc = bc (x+l ) + ca(y+l) + ab ( z+l) ,
where
x+l , y+l , z+l
are poslt lve lntegers . •
•
•
such
SOLUTIONS (13-14)
Clear ly , as divide s
59
GCD ( a , b )
x+l , b
GCD (b , c )
=
divides
are posit ive integers x+l
=
=
GCD ( c , a)
y+l , and
r,s,t
c
, we have that
divides
z+l .
a
Thus there
such that
y+1 = bs ,
ar ,
1
=
z+l· ct .
Hence we have 2abc = abc ( r + s + t ) , that is 2
=
r + s + t >. 3 ,
which is impos s ib le. This comp l e tes the solution .
14.
Determine a function
f (n)
such that the n
th
term of the
sequence (14.0)
1 , 2 , 2 , 3 , 3 , 3 , 4 , 4, 4 , 4. 5, . . .
is given by Solut ion :
[fen) 1 .
Let
u
n lnteger •
be the n
th
term of the sequence ( 1 4 . 0) .
f ir s t occurs in the sequence when
k
n = 1 + 2 + 3 + . . . + (k-1) + 1 Hence ( 14 . 1 )
u = k n
=
(k-l ) k + 1 2
for
n =
;
(k- ) k
+ 1 + t.
t = 0 , 1 .2
• • . . •
From (14 . 1 ) we ob tain (kl ) k o :0 n 2
_
1 :0 k 1 -
and s o ( 1 4 . 2)
2 k - k + 2 :0 n 2
•
k-l .
•
The
60
SOLUTIONS (15-16)
Mul t i p l y ing (14 . 2) by S and comp let ing the square , we have (2k-1) (2k-1)
2
2
� Sn - 7
�
�
�
(2k+1 )
( 2k+1)
2
- 1 , <
- S
/Sn-7
<
2k+1 ,
1 + /Sn-7 2
<
k + 1 ;
2k-1 k
�
+ 7 � Sn
2
(2k+1)
2
,
that is u = k = [ ( 1 + /Sn-7) /2] . n
15.
zero .
where
a ,a , ,a be given real number s , which are not al l 1 2 n Det ermine the least val ue of Let
x , 1
• • •
• • •
,x n
are real numbers sat i s fying + a x = 1 n n
So lut ion :
•
We have , us ing Cauchy's inequal ity , n
2 a. L . � � =1
1/2
n
2 . L X.� � .. 1
so that n
2 � L X.� i= l
1
---=:....-
n
•
2 a. L � • = 1 �
If we choose x. �
..
a. � --=-n 2 a. L . � 1 �=
( i .. 1, 2 ,
•
•
.
,n) ,
1/2
,
SOLUTIONS (15-16)
61
we have n
= 1
a.x. L . 1 1
1= 1 and n
2 x. = L . 1 1= 1
so the minimum value of
16 .
n
2 x. L . 1 1=1
1 �n � a2 . l.. . 1 1=1
-
,
n
a.x. L . 1 1
subj ect to
1= 1
- 1 1S •
n
1 •
2 a. L . 1 1=1
Evaluate the infinite series 3
2 S = 1 1I S o l ution :
•
We have (n+1)
3
=
n (n-1) (n-2) + 6n (n- 1 ) + 7n + 1
so that 00
s =
L
n= O
n
(-1) (n+1) n!
3
n
00
(-1) (n(n1 ) (n-2) + + 7n + 6n(n1 ) 1 ) = L ! n n= O 00
=
L n=3 00
=
-1:
m=O
n
( _1) + 6 (n-3)! (-1) m!
00
= - L m-O
m
(-1) m!
00
L
n-2
n
( _1) + 7 (n-2) !
00
1:
(-1) + 6 m! e m O
m
=
-e
m
-1 •
00
L
n=1
00
- 7 L m= O
(-1) n!
?
(- 1 m.
m
n
00
+
n=O
00
+
1:
L
m=0
(
(-1) n!
t,) m.
-
m
n
SOLUTIONS (17-18)
62
17.
F (x) x
for a l l
is a dif ferentiab le func t ion such that F ' (a-x) = F ' (x) a satisfying 0 � x � a . Evaluate F (x ) dx and give an
examp l e of such a funct ion So l ut ion :
�
F (x) .
As F ' (a-x) = F ' (x) ,
0 � x � a ,
we have by integrat ing -F (a-x) e
where
F (x) + e ,
=
x = 0
Taking
� s a constant . •
we obt ain
e
-F (O) - F ( a) ,
=
so that F (x) + F ( a-x)
=
F ( O) + F ( a) .
Integrat ing again we get a F (x) dx +
a
o
F ( a-x) dx = a(F (O) + F ( a»
.
As a
o
a F (x) dx ,
F(a-x) dx =
the de s ired integral has the value
I(F (O)
+ F (a»
.
Two examp l e s of such func t ions are 'TTX
k cos a where
18.
k
is an arbitrary constant . (a) Let
r,s,t,u x
Prove that if
rs = tu
4
be the roo t s of the quartic equat ion
+ Ax then
3
+ Bx
2
+ ex + D = 0 .
2 2 A D = e .
63
SOLUTIONS (17-18)
(b) Let
be the roots of the quartic equation
a,b,c,d
Y
4
+ py
2
+ qy + r = 0 .
Use (a) to determine the cubic equation ( in terms of p , q , r ) whos e . roots are ad - bc a + d - b - c •
ac - bd a + c - b - d '
ab - cd a + b - c - d '
(a) As r,s , t , u are the roo t s o f the quartic equation 3 2 4 O , we h ave x + Ax + Bx + Cx + D
Solut ion:
=
r + s + t + u = -A ,
r s t + rsu + rtu + stu = -C , rstu = D
Since
rs
=
tu
•
we have 2 2 2 2D A ( r + s + t + u) r s 2 2 2 - ( r s + r s + r s t + rsu) 2 (rtu + stu + rst + rsu) 2 - C -
=
-
•
(b) As the root s o f the equat ion 4 2 y + py + qy + r
•
( 1 8 . 1)
a + b + c + d
0
( 1 8 . 2)
ab + ac + ad + be + bd + cd
are
a , b , c , d , we have
(18 . 3 )
z
•
abc + abd + acd + bed
be a real or comp lex number .
equation who se roots are
, =
abcd = r
(18 . 4 ) Let
0
=
-q
p
, , •
We begin by finding the quartic
a-z , b-z , c-z,d - z
.
SOLUTIONS (18-19)
64
From (18 . 1) , we obtain (a-z) + (b-z) + ( c-z) + (d-z) = -4z .
(18 . 5 )
S imilarly, from ( 1 8.1) and ( 18 . 2), we obtain (a- z ) (b-z) + (a-z) (c-z) + ( a-z) (d-z) + (b-z) (c-z) + (b-z) (d-z) + ( c-z}(d- z) 2 = (ab + ac + ad + bc + bd + cd) - 3 (a + b + c + d ) z + 6z , that is ( a-z) (b-z) + (a-z) (c-z) +
(18 . 6 )
+ (c-z ) (d-z )
. . •
=
p + 6z
2 •
Next , from ( 18 . 1) , (18 . 2) and ( 18 . 3) , we have (a-z) (b-z) (c-z)+(a-z) (b-z) (d-z)+(a-z) (c- z ) ( d-z)+(b-z) ( c-z) (d-z) 3 2 = (abc+abd+acd+bcd) - 2 (ab+ac+ad+bc+bd+cd) z + 3 ( a+b+c+d) z - 4 z , so that (a-z ) (b-z) (c-z) +
(18 . 7 )
.
•
+ (b-z ) (c-z) (d-z)
•
= -q - 2pz - 4z
3
•
A l so , from (18 . 1 ) , (18 . 2) , (18.3 ) , ( 18 . 4) , we have (a-z) (b- z ) (c-z) (d-z) =
abcd - (abc+
. . •
+bcd) z + (ab+
•
.
.
+cd) z
2
- (a+b+c+d) z
so that ( a-z) (b-z) (c-z) (d-z) = r + qz + pz
(18 . 8)
2
+ z
3
+ z
4
4 •
Hence the desired quartic equat ion , whose root s are a-z , b-z , c-z , d-z , �s •
4zy
3
2
+ (p+6z ) y
2
3
2 4 + (q+2pz+4z ) y + ( r+qz+pz +z ) = 0
To f inish the prob l em we take
ab - cd z = a + b - c l
= (c - z 1) (d -
_
s o tha t ' d
•
,
SOLUTIONS (18-19)
65
and thus by (a) we have 2 16z ( r + q Z 1 so that z
=
3 2 4z ) , 1
( q + 2p Z
i s a root o f
l
8q z
(18 . 9)
3
2
2 2 + 4 ( 4r - p ) z - 4pqz - q
0
•
•
ad - bc ac - bd and z = --�7--7�-Similarly z a + d - b - c d a + c 3 2 b roo ts of (18 . 9 ) , wh ich i s the required cubic equation . =
19 . Let
where
n
_
_
be a monic po lynomial of degree
p (x)
i s a non-negative integer and
t ion with respect to x .
:
D = d •
•
cons tant term 1.n •
Solution:
f (x) n
m � 1 , and set
deno tes differentia
f (x) 1.S a po lynomial 1.n x n mn-n Determine the rat1.O o f the coeff icient o f x Pro ve that
•
are also
of degree 1.n •
f (x) n
(mn - n) to the
•
Differentiating
f (x) n
by the product rul e , we obtain
so that
and so (19 . 1)
f
(x) :; f� (x) - p ' (x) f (x) . n+1 n
f (x) :; 1 i s a polynomial of degree 0 , f (x) :; -p' (x) 1.S 1 O 2 a po lynomial o f degree m-1 , and f (x) :; -p"(x) + p ' (x) i s a po1y2 nomial o f degree 2m-2 . With the induct ive hypothes i s that f (x) n i s a po lynomial of degree mn-n , we eas i ly deduce from (19 . 1 ) that Clearly
•
•
SOLUTIONS (1 9-20)
Ii
m(n+1) - (n+1) , and hence the
i s a po lynomial of degree
princip le of mathemat ical induc tion implies that nomial of degree
(mn-n)
Sett ing
for al l
m p (x ) - x +
P
n .
m-1
m-l
+
X"
•
•
.
f (x) n
i s a p o 1 y-
+ P O
and f (x) n
•
mn-n x a mn-n
+ . . . + aO '
we obtain , from ( 1 9 . 1) , mn-l1n-n-1 a x + mn-l1n-n-1 =
•
.
.
+
a
O
mn-n-2 mn-n- 1 X x + .,. + a + (mn-n-1) a (mn-n ) a mn-n 1 mn-n- I mu-n mn-n-1 x + a ( amu-nx mn-n- l
Equat ing coef ficients of
mn+\D-n-1 • we Qbt ain x a mn-l1n-n-1
ma mn wu-n
•
Solving this recurrence relation, we obtain a mn-n
•
n
('Om) a ' 0
that i s - ( 'Om)
20.
Determine the real function of
n •
x
15 x x x + + + 1 5! 3T 91 3
whose power series is
9
• • •
•
SOLUTIONS (1 9-20)
Solut ion:
67
We make use of the comp lex cube of unity -1 + R w " 2
,
so that 3 w
(20,1) Now, for al l real
2 w + w + 1 - 0 ,
1 ,
_
x , we have 3
x sinh x - x + + 3! s inh wx
5
7
x + 5!
9
x x + + 7! 9!
, ,
,
,
'
,
,
•
•
•
3
=
wx
7 9 5 2 w x x x wx + + + + + 3! 5! 7! 9! 3
9 2 7 X x 2 2 w x x w w w , + s �nh X " x + + + + 3! 9! 51 7!
,
5
•
Adding the se equations and using (20,1 ) , we obtain ' h x + s �n ' h wx s �n
3 9 x x 2 ' + s�nh w x .. 3 + + ... 3! 91
,
Now s inh
wx
=
ix, 3 -x s inh + 2 2
x l3 x , .. -s�nh - cos 2 2 and s im i l arly
-x -x ix/3 cosh �<-=- + cosh .. s inh. s inh 2 2 2 + i cosh
I
s in .::;,.=-
xv3 x , - � co sh - s�n 2 2 ,
and s o s inh wx
,
,
2 , + s �nh w x - -2s inh � cos .::.:,..::: 2
,
g�v�ng ,
,
9 3 x x + + 9! 31
.
.
,
=-
x
cosh 2"
x. 3 - cos 2
,
SOLU'I'IONS (21-22)
68
21
•
D��ermine the va lue of the inte gra l
(21 .0)
1T Sln . nx,2 ! dx . sin x J
.. n
I
for a ll positive integra l va lues of n . 3 n I .. n1T We wi ll show that 1 .2. Soluti on : . n F ro m (21 .0) . we have fo r n � 2 •
D
I - I n n-1 n
• • . •
•
1T(Sln 2nx •
..
o
1T (Sln nx - Sln (n- l) x )(sin nx + Sln (n-1 ) x ) .. d x sin2x o •
•
•
1T 2 s1.n x ) 2 Sln . x cos (nx -2" x x . nx -2" cos 2" ( ) 2" .. dx 's in2x o
1T s in x s n (2 1 .::.:� :...:::-_- ' ..::;. :.:�:: ...;� sin2x o
•
dx
Sln x
•
•
dx
•
that is .. J n ' n 2n- l
(21 .1 )
D
whe re
1Tslnmx Jm .. Sln . x dx •
Now. for m
�
2
•
we have
•
�
2
•
m - 0 ,1 , 2,
1T( s lnmx - sin (m- 2)x) Jm - J m- 2 .. dx . Sln x •
• . •
•
69
SOLU'I'IONS (21-22)
'"
112 sin x cos (m 1 )x dx s �n x o •
11 - 2 c os (m l)x dx 11 .. 2 sin(m 1 )x m-l o -
=
so that
0
•
Jm - Jm 2 '" Jm-4
..
• • •
-
•
•
..
•
-
Hence . f rom (21 . 1 ). we obtain Dn as I 1 1T n� 1 •
..
Jo '" 0 if m even . Jl 11 if m o dd. 11 n � 2 so that In '" n 11 •
•
•
•
During the yea r 1985. a convenience sto re . which was o pen 7 da ys a wee k. so ld at least one bo ok each da y. an d a tota l of 6 00 boo ks ove r the entire yea r. Must the re have been a pe rio d of c onsec utive days when exact ly 1 29 b ooks we re s old? 22.
365. denote the nu mbe r of b ooks Soluti on: Let a � i .. 1 .2.3 so ld b y the sto re du rin g the pe ri od f ro m the f i rst day to th the i da y inc lusive . s o that 1 � al < a2 < < a 365 .. 6 00 and thus .
• • • ••
•
• • •
a 365+l 29 Hence a l a 365.a l+1 29 between 1 an d 7 29 inc lusive . Thus . b y two of these nu mbe rs must be the s ame . a 365+l 29 a re distinct an d a l+1 29 • • . ••
• • • • •
• . • ••
•
a re 7 3 0 positive intege rs Di rich let 's box princi ple . a 365 a re a ll As a l a ll distinct . one of the a �. • . . .•
SOLUTIONS (23-25)
70
must be the same as one of the a.+1 29 , sa y, ak a l + 1 29 , 1 � l < k �
=
� 365 .
th Hence ak - afh- 1 29 an d so 1 29 books we re so ld between the (l+l) da y an d the k day inclus ive . Fin d a po lyno mia l f (x ,y) with rational coefficients such that as an d n run th rough all positive integra l va lues, f (m,n) takes on a ll positive integ ral va lues once an d once on ly . 23.
m
Sol ution : Fo r any positive inte g e r k we can define a uni que pai r of inte g e rs (r, m) by (r-l )(r- 2) < k .. r(r- l) 2 2 ' (r-2) l) (rk = 2 �
m
_
•
Clea rl y we have o
<
m
.::.r� =--....: �) - (r- 1 )(r-2) 2 2
�
that is
1
=
r- l ,
m< r,
�
so that r an d m are pos�t�ve �ntege rs . Mo reove r, we can def ine a positive inte ge r n uni que ly by r = m + n , which gives •
k-
•
•
(ml n-1 )(ml n- 2)
2
+ m ,
an d a po lyno mi al of the re qui re d t y pe is the refore f (x , y )
=
(x+ y- l)(xly-2) + x . 2
71
SOLUTIONS (23-25)
24.
Let be a positive squarefree integer . Let R , S be positive integers . Give a condition involving R, S , m which guaran tees that there do not exist rational numbers x , y , z and w such that m
(24 .0)
•
Solution : If there exist rational numbers x , y , z and w such that (24 .0) holds then 2 2 R - 2SIri = (x-ylri) + (z-wlri) ;:: 0 , •
•
and so a condition that wil l guarantee the non-solvabil ity of (24 . 0) is R - 2S Iri < 0 , that is
25.
Let k and h be integers with 1
l imit
�
k < h . Evaluate the
hn
r IT 1 - n2 L .. l im n-+-"" r-kn+1 So lution: For I xl < 1 we have in.
( 1 - x)
..
_
"" x s '\ =t. s ' s .. l
and so x+
in.
( 1 - x)
•
"" s x :.:. - L ... s , s-2
g�v�ng •
•
I x + in ( 1 - x) I ..
"" s =-x l: s s= 2
�� s s=2
•
SOLUTIONS (25-26)
72
Taking x
..
rn2
(kn < r
�
hn) we obtain
r- + in ( 1 - -r ) n2 . n2 �
2 r
h 2n ( l - n) 4
•
-
Thus we obtain I I.
hn L !r=kn+1 .
-r2 + n
hn 2 Lr r=kn+1
r in ( 1 n2 )
,
hn 2 Lr r=l
+ 0 as n + 00 , showing that (25 . 1 )
hn l im L �2 + in n +00 r=kn+l
1
-
-r2 n
.. 0
Next we have hn
L �2
r..kn+ l
-
1- hn(hn+l) - kn(kn+l) n2 2 2
•
SOLUTIONS (25-26)
73
so that l im
(25 . 2 )
•
Thus from (25 . 1 ) and (25.2) we obtain l im n .... ""
hn
(1 r=kn+1 IT
In.
and so hn
r IT ( 1 - -2 ) = l im n n .... "" r=kn+1
•
Let f (x) be a continuous function on [O , a] such that f (x) f (a-x) = 1 , where a O . Prove that there exist infinitely many such functions f (x) , and evaluate 26.
>
a o
dx 1 + f (x)
•
a x - -2 Solution : The function f (x) = e is continuous for all x and n satisfies f(x) f (a-x) = 1 , so that f (x) (n 0 , 1 , 2 , gives an infinite family of functions of the required type . Setting x = a - y we obtain z
I
a
dx = o 1 + f (x) = =
a a o
..
dy 1 + f(a-y) 1
o
-dy + f 1 ( y a ) a a f(y)dy f (y) + 1
1 - 1 + f (y) dy
. • •
SOLUTIONS (27-28)
74
a
d y = a - o 1 + f ( y) = a
a so that 1= -2 .
- I ,
27.
satisfy The positive numbers a 1 ,a 2,a 3, n n 3 2 (27 . 0) L a r .. L a r , n = 1,2,3, r=1 r= 1 •
•
•
•
Is it true that a r .. r for r = 1,2,3, . . .
•
•
•
?
Solution : The answer is yes . We proceed by mathematical induction, making use of the identity 3 3 2 3 1 + 2 + . . . + n .. (1 + 2 + . . + n) 3 2 Taking n " 1 in (27 . 0) gives a l .. a 1 ' which means that because a 1 0 Next, assume that � k for k .. 1,2, . . . ,n-l . The equation (27 . 0) gives .
>
•
•
..
3 3 + + 2 1
•
•
•
3 3 + (n- l ) + an .. ( 1 + 2 +
•
•
•
2 + (n-1) + an )
so that 2 3 2 {1+2+ . . . +(n- l)) + an .. ( 1+2+ +(n-l)) + 2( 1+2+ that is, 3 2 (27 . 1 ) an .. (n- 1 ) n an + an . •
•
•
•
•
•
2 +(n-l)) an + an ,
As an 0 , we see that (27 . 1 ) gives an .. n , thus completing the inductive step. >
75
SOLUTIONS (27-28)
Let p > 0 be a real number and let n be a non-negative integer. Evaluate 28.
un (p ) = o
(28 . 0 )
Solution : For n � 2 and p parts we obtain
>
-px n e sin x
dx
•
0 , 1ntegrat1ng un .. un (p) by •
•
co
un
I n n-l . -px n S1n x + cos x sin x - -e p o o
=
=
•
px e dx p
n sinn- I x cos x e- p x dx . p 0
-
Integrating by parts again, we get un
=
n
-
p
-px nI - -p e sin x cos x 1
+
=
=
o
co
o
-px e n-2 2 n (n-I) sin x cos x - sin x p dx (
n -px n-2 2 (n- l ) sin x ( I-sin x) - sin x e dx
n 2 p
2 -px n n-2 e sin x dx - 2
n(n- l) p
P
that is un
n -p x e sin x dx ,
-
n(n- I ) u n-2 p
2 n
n
,
�
Thus we have un
=
n(n- I ) u n-2 2 2 n +p
,
n
�
2 ,
2
•
SOLUTIONS (29)
76
giving Un
,.
n(n- l) 2 2 n +p
(n-2) (n-3) 2 2 (n-2) +p
•
n(n-1) fi2+p 2
•
•
(n-2 ) (n-3 ) 2 2 (n_2) +p
One easily sees that Uo
..
•
•
•
•
2·1 2 2 Uo 2 +p
,
if n even,
•
3·2 u 2 2 1 3 +P
,
if n odd
1 and u 1 P
-
..
1 2 1 +P
so that
n! if even, n , n/2 2 2 p II « 2i) + p ) i-I
un ..
nl
(n- l ) /2 2 2 IT « 2i+l) + p ) i-O 29.
,
•
, if n odd.
Evaluate 11
(29 . 0 ) for integers n
' n-r 2 �
2
•
Solution : We use the identity tan A - cot A - 2 cot
2A ,
which is easily verified as 2 2 A A cos 2A cos 2 sin 2 cot ZA ,. sin 2A " sin A cos A .. cot A - tan A Then we have 1T
n-r 2
=
n-2 r L 2 cot r O =
11
- 2 cot n-r- 1 2
•
SOLUTIONS (29)
77
n-2 1T 1T - L 2 r+1 cot - - '::' n r n-r r=O 2 2 1T n-r 2
1T r n2
1T n-1 1T = cot n - 2 co t -2 2 .. cot
•
Second solution (due to L . Smith) : For 0 < B < 21T we consider the integral l(B ) .. -
ciA
n-2 r B A tan n-r I.. 2 2 r-� O n-2 r L 2 r=O
•
r n2
bl. cos
n-2 B n L bl. cos - 2 n-r r 0 2 n-2 B n 2 bl. IT cos n-r r 0 2 B s i n Z n .. 2 bl. --:---=:. ,... ..,..., · B =...: l sin i'l n 2 •
ciA
A
n-r 2
B o
SOLUTIONS (30-3 1)
78
.. 2
n
. B tyl s:tn "2 -
Differentiating we ob tain I'(B) Taking
B
rr
-
=
B B n-1 cot 2 ' cot "2 n 2
we get rr
n-r 2 Let
30. of
s
k (2
<
k •
•
n
�
S
n)
be an integer.
2
I f a k-se lection
a : i'" 1 , 2 , , kJ i N - {1,2,3, , n } such that =
•
.
is called a k-selection .
S
{s
A selection
elements from the set
<�
•
•
•
•
•
For any k-selection
is chosen at random from
where
r
f (n , k) , 2 S k r k-selections S from
where
Let
•
•
•
•
m k
.. 0
s
n ,
N
denote the number of
such that
�
W(S)
r ,
We will show that
f (n , k) .. r
(30 . 1)
r ,
is a natural number?
Solution: r '" 1 , 2 , 3 ,
=
n-(k- 1 ) (r- l ) k
for any integer
enumerates all k-selec tions
m
<
r - 1,2,3,
, k
•
When
N , so that
and hence (30 . 1 ) holds in this case .
Now suppose
S .. { a , a "" '� } l 2
be a k-selection from
N
•
.
•
,
r .. 1 ,
from
S
S ,
N , what is the prob
ability that W(S)
•
with
f (n ,k) l n f (n ,k) '" , k 1
r
�
2 .
Let
W(S) � r ,
SOLUTIONS (30-3 1 )
79
so that a. + r � def ined by
for
1.
(30 . 2 ) F (S)
a
•
1.
=
1 , 2 , . . . , k-l . The mapping F
{a I ' a 2 - (r-l ) , a3 - 2(r-l) ,
assoc iates with a k-selection S from selection from the set M
=
N
• . • ,
ak - (k-l) (r-l) }
having W(S)
{ l , 2 , . . . , n - (k-l ) (r-l) }
�
r , a k
•
Clearly F is one-to-one and onto , so that f r (n , k) is j ust the number of k-selections from M , which is given by the right side of (30 . 1 ) . Thus the required probability is f r (n,k) - f r+l (n,k) f l (n,k) 31 .
n- (k- 1 ) (r-l) - n-(k- l ) r k k
=
•
Let k � 2 be a fixed integer . For n = 1 , 2 , 3 , . . . define 1
an =
, if n is not a multiple of k ,
-(k-l) , if n is a multiple of k . an n n=l 00
Evaluate the series
•
solution : Let s n be the sum of the first n terms of the given series . For each n � 1 we have uniquely n = kqn + rn ,
o � rn
<
and s ince -(k-l ) /tk for t
=
1 , 2 , . . . , q , we have n
a
l/tk - lit
k ,
SOLUTIONS (31-32)
80
•
Now, n
q (k + r /q ) so n n n
•
in n - in q + in (k + r / q ) n n n
,
and hence
1
(l + - + 2 :l u - v + w n n n
•
.
•
1 +...!...
- .in q ) + .in (k + r /q ) n q n n n
•
{u : n .. 1,2,3, converges to Euler ' s constant c; } n n .. 1,2,3, the sequence {v } also converges to c as n + oo n imp l ies q + 00·' and the sequence {w n = 1,2,3, converges to } n n 00 an is bounded . Thus , we have ink as r r .. in k n n=1 n The sequence
• •
.
.
•
•
.
.
• •
. • •
•
32.
Prove that m -x x e s1n x dx •
o
for
m · 0,1,2,
Solution:
•
•
•
•
ml
....,... ., :.:... ..= ....':' sin (m+1 ) rr/ 4
(ml. 2 )/2 2
•
We set for
m " 0,1,2, S
c
m
m
..
..
E .. m
•
•
•
00 m -x . x e S1n x dx ,
m -x x e cos x dx ,
m (i- l ) x x e dx .
SOLU'I'lONS (3 1-32)
81
As loX •
e
cos
a
X
+
•
1.
S 1.n X , •
we have
Em Integrating
Em
- C + is m m
•
by parts , we obtain
Em
=
=
(i- l ) x m .: e.. :: --:...,.. _ X i-I -m
.
.1;-
1
c
m-
o
1
IX)
00
e m- ':::"'''' --1 - dx mX i
1
m ;;;
Hence we have
Em Clearl y
EO
m � 0
=
I = so that for ' i-I
•
m;;; 0
Em
•
Fina l l y , we obtain for
m;;; 0 S m
1 = . (E 2 m 1.
- E ) m l l _ m+ m+ ( l_ i) (l+i)
,
and so , by Demoivre's theorem, we have m+l
S = m
2 2
*
(COS(m+l) + i s in(m+l )
=
*)
7T m! s in(m+ 1 )- ' 4 (m+1)/2
2
m+l
-
2 2
(CO S(m+l)
m ;;; 0
•
* - i sin(m+l )i)
SOLUTIONS (33)
82
33.
u
For a rea l number I(u)
(33 . 0) Prove that
rr 2 {n(l - 2u cos x + u ) dx 0
I(u)
and hence eva Solution:
.
•
I(u)
e
•
I(-u)
•
2 1 -I(u ) 2
for all values of
•
, u
•
We wi l l show that ,
if
2 rrtn l u l '
if
o
I(u) ..
(33. 1 )
First , we prove that
I(u)
(33 . 2) Setting
set
•
I(-u)
lui � lui
1 ,
> 1
•
•
in (33 . 0) , we obtain
x - rr - y I(u)
•
rr 2 {nO + 2u cos y + u ) dy
o
.. I(-u)
•
Next we show that (33 .3) We have 2
2
(1 - 2u cos x + u ) (1 + 2u cos x + u )
•
•
•
2 2 2 (1 + u ) - (2u cos x) 1 + u
2 2 + 2u (l - 2cos x)
4 2
1 - 2u cos2x + u
4
,
so that 2
2
2
{nO - 2u cos x + u ) + .(,.I',(l + 2u cos x + u ) " tn(1 - 2u cos 2x + u4 ) D..
and thus I(u) + I(-u) ..
rr 4 2 tn(l - 2u cos 2x + u ) dx ,
,
SOLUTIONS (33)
and setting
83
y " 2x , we ob tain l(u) + 1(-u)
..
1 2
271 2 4 {nO - 2u cos y + u ) dy
2 71 2 4 1 2 .. - I(u ) + .!. (l - 2u cos y + u ) dy {n 2 71 2 Setting
y = 271 -
in the last integral , we obtain
z
271 2 2 4 bt(l - 2u cos y + u ) dy .. l(u ) , proving (33 . 3) a s required . From ( 33 . 2) and (33 . 3) , we deduce (33 . 4) For
lui
= 1
,
that is
u = tl , we have
1(1)
=
1
1(- 1 ) = "2 10 ) ,
so For
lui <
1(1 ) = 1(- 1 ) 1
=
0 .
we have 1
2
l(u) = "2 l(u ) for a l l positive integers as
lui <
1 , and
l(u)
=r
n
•
Letting
• • •
::II
n + + 00 , we have
being continuous gives lim n + oo
1(u
2
n
[ In fact it fol lows from (33 . 4) that we obtain lim n + oo
) .. 1(0)
•
1(0) = O . J
Hence , as
•
SOLUTIONS (34-35)
84
g�v�ng •
•
for
I(u) = 0 , Next , for for u
>
Iu I
>
0 , we have
=
I
•
I u = - , so that v
I , we set
I(u) ..
I uI <
o
< Iv I <
I
Then ,
•
1T I 2 In(l - v co s x + "":;;' ) dx 2 v 1T o
2
{In( l - 2v cos x + v ) - 21n v} dx
.. I(v) - 21n v
1T dx
= 0 - 21T in v = 21T in u F ina l l y , if
u
<0
•
, we have I(u) .. I(-u) .. 2 1T ln (-u) ,
so that for all
u
with
lui
>
1
we have
I(u) .. 21T in Iu I
34.
For each natural number
k � 2
i s partit ioned into a sequence of sets
.
the set of natural numbers {A (k) : n " 1 , 2 , 3 , n k natural number s ,
• • •
next k+2 natural numbers , etc . A (k) n
is denoted by
such that
s (k) n
>
3k
3
s (k) n _
Sk
•
2
as
A (k) 2 cons i s t s of the
A (k) con s i s t s of the first l cons i s t s of the next k+l natural number s , fol lows :
}
A (k) 3 The sum of the natural numbers in
Determine the l east value of , for
k .. 2 , 3 ,
• • •
•
n .. n(k)
SOLUTIONS (34-35)
Solut ion:
85
A (k) n
The last el ement in
k + (k+1 ) + (k+2) +
.
.
is the number
+ (k+n-1 ) " nk + ( 1+2+ . . . +(n-1 ) )
•
•
Since there are A (k) is n
k+n- 1
(n-1 ) n nk + 2
(n-1 ) n nk + 2
•
A (k) , the f irst e lement in n
numbers in
1 2 (k + n - 1 ) + 1 " (n - 1 ) k + 2" (n -3n + 4 ) .
_
Hence we have (k + n - 1 ) s (k) = n 2
(34.1 ) Taking
k = 2,3,4,5
required value of and
s (k) . k
(
2
2n-1 ) k + (n - 2n + 2) )
in (34 . 1 ) suggests that n .
We have
n = k
•
may be the
To prove this conj ecture we calculate s (k) k-1 2
3 2 s _ (k) = ( k - 1 ) (3k -7k + 5) " 3k - 1 0k + 1 2k -5 k 1 and 2
( 2k - 1 ) (3k -3k + 2) 9 2 3 7 = 3k - 2" k + 2" k - 1 = 2 One easily checks that for
so that 35.
n (k) " k Let
converge?
. . •
is the required minimal value .
{p : n " 1 , 2 , 3 , n for n = 1 , 2 , 3 , .
co
(35 . 0)
k " 2 ,3,
•
L
n=l
.
•
. . •
be a sequence of real numbers such
} .
Does the series
[ p ] -1 n
SOLUTIONS (35-36)
86
So lut ion :
The answer a n
Let Then we have for
yes .
•
loS
Pn - 1 , P l ' "Pn
-
n = 1 , 2 , •• .
•
m � 1 , S
.m m
�
-
n=l
a
m n
1 P 1 . • • Pn P l " ·Pn-1 1
L
-
n=l -
1 1
-
1 P l' • • Pm 1 --=P l" ' Pm+1
__
__
so that {S • m = 1 , 2 , 3 , • • . } loS an 1.ncreas1.ng sequence which loS m eX1.st s , showing that a Hence l im S bounded above by 1 n� l n m m-+oo converge s . Fina l l y , a s •
•
•
•
•
00
•
•
n = 1,2,3,•. .
(35 . 0)
the series given by
36.
f (x) , g (x)
Let
degrees
n+1 , n
converges by the comparison test . be polynomia l s with real coef f icient s of
respectively , where
leading coef ficient s
A, B
qo 1ut ion :
n �
respe c t ive l y .
L = l im g (x ) x-+oo in terms of
x
e
A>
0
and
B>
0
0
, and with posit ive
Evaluate
f ( t) - f (x)
A , B and n . As
,
, we have
dx
,
SOLUTIONS (35-36)
87
li m x + oo and li m x + oo
More ove r
x f ( t) e d t .. + 00 f (x) e
lim
•
) x f( d e dx g(x)
x + co
..
- + 00
2 (x) li m f ' (x) g�x) - g ' (x) x + oo
2 2n B x + = li m n x + oo AB( n+l)i +
. • .
..
so,
by
B ( n+OA
• • •
'
L 'H8pi ta 1's ru le , we have L ..
=
..
lim
li m x + oo
x d e f ( t)d t dx Uo d e f x !g (x) dx
B (n+l)A
•
SOLUTIONS (37-38)
88
The lengths of two al titudes of a trian gle a re h and k , whe re h � k . De te rmine uppe r and lowe r b ounds f or the length of the third altitude in ter ms of h and k . jJ,
Soluti on: We show tha t hk < l < hk h+k J
(37 . 1 )
•
Le t the poin ts P, Q, R be chosen on the sides BC, CA, AB ( possibly extended ) re spec tively of the trian gle ABC s o tha t AP,BQ, CR a re the alti tudes of the trian gle . Se t a IBCI, b = I CAI , c = I ABI, h = IAPI , k IBQ I l ICRI . Clea rly ' ah bk = c l , s o tha t b l a =l c h' c k · =
=
=
=
-
-
:;:
-
Wi thout loss of gene ral i ty we may s uppose tha t h < k . F rom the ineq uality a
� <� + �<�+ l , h k c c
tha t is Als o f rom the inequali ty we ob tain
hk l < k-h
•
c
1 < �+� , c c
1
<
1
'
89
SOLU'l'IONS (37-38)
tha t is .f..
>
hk . h+k
This comple tes the pr oof of (37 . 1 ) . 38.
Pr ove tha t Pn , r
n-r n+2 n+1 . . . ( 1_x ) ( 1_x ( 1_x ) ) Pn , r (x) = r 2 ( 1-x) ( 1-x ) ( 1-x )
=
• • .
•
is a polynomial in x of de gree nr , where n and r are non ne gative in te gers . (When r 0 the e m pty prod uc t is unders tood to be 1 and we have Pn, 0 1 f or all n � 0 . ) =
=
Soluti on : F or n
and r � 1 we have
� 0
n+1 (1_x )
r Pn+1 , r _ x pn,r
( I-x)
=
s o tha t
n+2 n+r ( 1_x ) ( 1 _x ) r-1 (I-x) . . . ( I-x ) • . •
r Pn+1 , r _ x pn , r
(38 . 1 )
•
. . • • . .
n+r ( 1_x ) r ( I-x )
,
Pn+1 , r-1
•
We now make the ind uc tive hypothesis tha t pn , r is a polynomial of degree nr f or all pairs (n , r) of non-nega t�ve �n te gers sa t�sf yin g n + r $ k , where k is a non-ne gative in te ger. This is clearly true when k 0 as , in this case , we mus t have n = r = 0 , and Po , o = 1 . Now le t (n, r) ' be a pair of non-ne gative in te gers s uch tha t n + r k + 1 . For n � 1 and r � 1 , b y (38 . 1) , we have •
=
=
,
•
•
SOLUTIONS (39 40)
90
P
(38 . 2)
r = x P
n,r
n-l , r
+ P
n , r- 1 .
k
, by the induc t ive hypo the s i s , P n- 1 , r i s a polynomial of i s a polynomial of degree (n- l ) r and P n , r-l is a polynomial of degree degree n ( r-l) . Henc e , by ( 3 8 . 2 ) , P n,r
As
(n- l ) + r = n + ( r- l ) =
max ( r + (n- l ) r , n ( r- l ) ) = nr . P
n,r n = 0
i s cl early a polynomia l of degree and
ln the remalnlng cases
0
•
•
•
The resu l t now fol lows by the princi p l e of
r = 0 .
mathematical induct ion .
39 .
Let
A, B , C , D , E
N
Prove that the number (39 . 0)
Ax
2
be integers such that
of pairs o f integers
B
(x,y)
�
and
0
such that
+ Bxy + Cx + Dy + E = 0 ,
satisf ies
N ;;; 2 d ( I F I ) , n � 1 , d (n)
where , for integers divisors of
Solution :
n .
(x , y)
Let
be a solut ion in integ ers of ( 39 . 0) , so that
- (Bx + D ) y = (Ax
(39 . 1 ) We def ine an integer
z
(39 . 2 ) C l early 2
have
denotes the number of p o s i t ive
PJ) 2 B
z � -
z
2
+ Cx + E )
•
by
= - (Bx + D) ,
0 , for otherwise
so that
x =
-
(z
+ D)
•
and from ( 3 9 . 1 ) we wou l d
x = -D/B
CD + E = 0 , contradict ing B
1 B
F
�
0
•
From ( 3 9 . 1 ) and
91
SOLUTIONS (39 40)
( 39 . 2 ) we have tha t is
2 2 2 B zy - A(z+D) - BC(z+D) + B E , 2 z( B y - Az - ( 2AD- BC»
•
F ,
s o tha t z is a divis or of F Thus the total number of possibili ties f or z �s ::i 2d ( IF I ) . F or each such z there is at mos t one possibility for x , 1 na me ly , x = - -B (z + D) if this is an inte ge r. As (39 . 1 ) i mplies tha t each x de ter mines a t most one y , the t otal nu mbe r of pairs (x, y ) is ::i 2d ( IF I ) •
•
•
n Evaluate L k= l Solut i on : We have f or k
k
�
•
1
and
whe re s o that
= f (k-l ) - f (k ) , l f (x ) = -"-2--,�_ , x +x+ 1 - f (k» =
i
( f (D ) - f en» �
SOLUTIONS (41-42)
92
1 = 2 1 = 2
-
L!1 . m
1 _ 1 _ -,;----:'-2 n + n + 1 2
n + n 2 n + n + 1
•
P = P (n) deno te the Sl� of a l l pos i t ive produc t s of m m different integers chosen from the set { 1 , 2 , . . . , n } . Find Let
Solut ion : P
2
We wil l show that =
1
24
n(n+l) (n- l ) (3n+2) ,
P
1
2
2
= 48 n (n+l) (n- l ) (n-2) .
3
We begin by considering ( 1 -x) ( 1-2x) . . . ( 1-nx) = 1 so tha t , with
Po = 1
and n
lYl
� L.
r=O
.
x
•
+
.
suff iciently smal l ,
r
r
(-1) P x . r
=
n
L
k=l
= -
lYl ( l - kx)
n
L
k=l
•
Hence we obtain n
I
r
( _ l) p x r r=O
r
= exp
,
SOLUTIONS (41-42)
93
that is 00
L
(41 . 1 )
h.. O
and s o , for
pr
1 hI
,
r " 0,1,2, . . . ,n , (_l )
'"
r
Thus we have P
2
x
coefficient o f
2 . x �n
= coeff . of
� {.
h=O
r
on the r ight of ( 4 1 . 1 )
(_ 1 ) h!
h
.!. n ( n+ 1 ) .!. n (n+1 ) ( 2n+1 ) _ + = 6 2 2 2
2
1 n ( n+ 1 ) ( n- 1 ) (3n+2) , = 24 and P
3
= coeff . of
1 = +3
n \' {.
k=l
k
3
1 _ 2!
1 + 31
n
L
k= l
k
3
1 3 3 1 2 2 1 2 2 n (n+1) ( 2n+1 ) + = n (n+ 1 ) n (n+ 1 ) 24 48 12 2 2 1 .. n (n+ 1) ( n- 1 ) (n-2 ) 48
42 .
For
•
a > b > 0 , evaluate the integral co
(42 . 0)
ax bx e ::" -=e:"' --': - ax-- b'-x- dx o x ( e +1 ) (e + 1 )
•
•
SOLUTIONS (42-43)
94
Solut ion :
C , the func tion
For any constant
f (x )
=
x e --'--- + C , eX + 1
X ::: 0 ,
i s such that f ( ax) - f (bx) t > 0
For
we have t
J ( t)
=
ax bx e - e -------7--- , ax bx ( e +0 ( e +l )
=
o t -
o
t
ax
bx e - e ---=---.::,-- dx bx ax x ( e +0 (e +1)
t
f ( ax) dx x
f ( ax) - f (bx) dx x
= o
f (bx) dx , x
o
provided b o th the l a t ter integral s eXl. s t , f ey) ' eXl. s t s , which holds if and only This l.S guaranteed if I l.m y -+ o y ,
,
•
if
C
=
-
1
-
2
With the cho ice
,
J(t)
at
=
o
f ey) dy y
-
t l im f (x) x -+ OO
real number
If
=
Xo
1:. 2
=
so that g iven any x ( e: ) O
such that
x > x o
.
t > x /b , then O
>
=
1
we have 2 '
-
bt
f e y) dy .::.-'.-"c y
at
Now
C
-
o
f ey) dy y
�.:...
,
e: > 0
there exi s t s a po s i t ive
1 1 - e: < f (x) < ' + e: , 2 2
-
at > bt > X
o
' and so we have
95
SOLUTIONS (42-43)
1 ( 2
-
- e:)
a t (!. 2 ...c::..
a bt - b
_
e:)
y
_ _
t
dy
< J(t) at (.!. + e:) 2 --=--::dy y t
<
e:
S ince
-
is arbitrary we obtain l im t + oo
=
.!.. e n � b 2
'
that i s
=
43 .
n
For integers
�
a 1 - In b 2
.
1 , determine the sum of
n
terms of the
ser�es •
(43 . 0)
2 n ( 2n - 2 ) 4 )� 2n 2::.:n ( 2 n- 2.!.. ) (2 n-..:..; o;.",,:r.::-= �;::;� -r::' + + + ( ••• ( 2n-l ) ( 2n-3 ) (2n-5) 2n-l) ( 2n-3 ) 2n-l
Solut ion :
Let
denote the
S
n (43 . 0) . S
of
n
terms of the given series
We have 2
= T= 2 , I =
S
SlW
3
=
4 8 + 3 3
-
-
=:
12 3
=
4 '
6·4 6 6 . 4 ·7 2 18+24+48 � + -;:-�� = - + 5·3·1 5 5·3 15 �
•
=
90 -:-:- = 6 • 15
SOLUTIONS (43-44)
96
These values suggest Firs t , as ers n Assume that S n = 2n Then we have •
the conj ecture Sn = 2n for all po sitive integ S I = 2 the conj ecture holds for n = I . holds for n m •
==
•
2m+2 + 2m+2 2m + 2m(2m-2) + 2m+1 2m+1 2m-1 (2m-I) (2m-3) .
. . •
(m terms)
2m+2 + 2m+2 S 2m+1 2m+1 m 2m+2 2m+2 = 2m+1 + 2m+1
•
=
2m+2 ( 1 + 2m) 2m+1
=
2m+2
2m
•
showing that Sn = 2n is true for n = m + I Renee , by the prin ciple of mathematical induction , Sn = 2n is true for all positive l.ntegers n . •
•
44 .
Let m be a fixed positive integer and let z l ' z 2 ' zk be k (� l) complex numbers such that s (44 . 0) . . . + Zk o for all s = m, m+I , m+2 ,m+k-l . Mus t z . = 0 for i = 1 . 2 , k? • . . •
=
• . . .
•
1.
Solution : The answer is yes . To see this , let z l ' z 2 ' roots of the equation k k-l z + ak_ I z +
. . . •
. • • •
zk be the
. . . + a lz
We wil l show that aO = O . Suppo se a O � O . Clearly z l ' z 2 are also roots of m+k-2 m+k-l m + a z + . . . + a + = 0 Z k-l l Z
•
• . . . •
zk
97
SOLUTIONS (4344)
Hence for i .. 1 , 2 ,
.
•
•
, k we have
m+k-1 z �.
(44 . 1 )
Adding the equations in (44 . 1) and appeal ing to (44 . 0) we obtain k m-1 aO � z �. �. 1
-
0
•
·
As aO
0 we have
�
k. m-1 L z �. .. 0 �.. 1
•
·
are roots of
C learly
m+k-3 m+k-2 + z + ak_ 1 z
(44 . 2 )
z . , i = 1 ,2, Taking � we obtain as before Z
..
•
•
•
, k , in (44 . 2 ) and adding the equations k. m-2 L z �. .. 0 �.. 1
•
·
Repeating the argument we eventual ly obtain �
L z. • 0 , � �.. 1 ·
and one more appl ication then gives aOk . 0 , which is impossib le . Hence we must have aO • 0 , that is
and so at least one of the can then be applied to 1 these �s 0 say zk-1 .. 0 z l .. Z
•
,
'
The argument z �. �s 0 , say k .. 0 , zk-1 to prove that at least one of Cont inuing in this way we obtain z2 0 .. k •
.
.
Z
•
•
•
..
•
•
•
..
Z
•
SOLU'IlONS (45-46)
98
where x
>
2 . Evaluate D n
=
det An
•
Solution : Expanding Dn by the first row, we obtain Dn x Dn-l - Dn- 2 ' so that =
(45 . 1 )
Dn - x Dn- 1
+
Dn- 2 " 0
•
The auxiliary equation for this difference equation is 2 t - xt
+
1 = 0 ,
which has the distinct real solutions ,
as x > 2 . Thus the solution of the difference equation (45 . 1 ) is given by for some constants A and B . We now set for convenience a so that , as
=
t(x + Ix2 -4)
t(x + Ix2 -4) t(x •
1 a
-
=
Ix2 -4)
=
, 1
1-(x - IX2 _4 ) , 2
,
SOLUTIONS (45-46)
which give s Now
99
n -n Dn .. Aa + Ba , n D1
and
so that (45 . 2 )
•
x
•
•
1,2,3, . . . .
1 B a + -a .. Aa + -a
2 2 1 1 - 1 = a + -a - 1 = a + 2 + 1 a
a
,
4 4 2 a A + B .. a + a + 1
•
Solv in g (45 . 2 ) for A and B y ield s 2 a 1 :... . . -= A .. ....,2=, B " ...". 2 ' a - 1 a - 1' _ _
so that
tha t is
46 .
t�on s
2 n a Dn .. 2 a a -1 2n+2 a 1 Dn .. n 2 n " 1, 2 , 3 , a (a -1 ) '
• • •
•
Determ ine a nece ssary and suff icient cond it ion for the equa-
•
(46 . 0 )
x + y + 2 x +l+ 3 3 x + y +
z A , 2 z .. B , 3 C , z •
..
to have a solut ion with at least one of x, y,z equal to zero .
SOLUTIONS (47-49)
100 •
So lut ion : Let x , y , z be a solution of (46 . 0) . Then, from the ident ity 2 2 + 2 (xy+yz+zx) , (x+y+z) = x and (46 . 0) , we deduce that xy + yz + zx
(46 . 1 )
=
.!. (A2 - B) 2
•
Next , from the identity
we obtain us ing (46 . 1 ) 3- AB 2
C - 3xyz so that (46 . 2)
Hence a solut ion (x, y , z) of (46 . 0) has at least one of x , y , z zero if and only if xyz = 0 , that is, by (46 . 2) , if and only if the 3 condition A - 3AB + 2C = 0 holds . 47 .
Let S be a set of k distinct integers chosen from n 1 , 2 , 3 , , 10 -1 , where n �s a pos�t�ve �nteger. Prove that if •
.
(47 . 0)
•
•
•
•
.
n < .en
(k+O 2
.en
10 ,
it is possible to find 2 disjoint subsets of S whose members have the same sum. n Solution : The integers in S are a l l :> 10 -l . Hence the sum of the integers in any subset of S �s •
SOLUTIONS
(47-49)
101
The number of non-empty subsets of
S
is
k
2 _l .
From
(47 . 0)
we have
n 1 k 2 - 1 > k · l 0 - Z k ( k+1) , and so , by Dir ichlet ' s box princip le , there mus t ex ist at least two which have the same sum. ' 2 l are d i s j o int the problem is so lved .
diff erent sub s e t s of If
S
l
and
S
S , say
S
and
S
2 If not , remova l of the common elements from S' I
two new sub s e t s 48 .
Let
n
and
S
and
S
2 l with the required property .
S2
I s it po s s ib l e for
be a positive integer .
yie lds
6n
distinct straight l ines in the Eucl idean plane to b e situated so as 2 to have at l east 6n -3n point s where exactly three of these l ines intersect and at least
6n+l
points where exactly two of thes e
l ines inter sect? So lut ion :
Any two distinct l ines in the po �nt . •
l ine s .
There are altogether
mo st one '" 3n ( 6n- 1 )
A triple intersection account s for
pairs of
of the s e pairs of
l ine s , and a s imp le intersection accounts for one pair .
As
2
( 6n -3n) 3 + (6n+I ) 1 2 .. 1 8n - 3n + 1 > 3n (6n - 1 ) the conf iguration is impo s s ib l e . 49 .
Let
S
be a set with
n ( �1 )
exp l icit formu la for the number cardina lity is a mul t i p l e of
3 .
A (n)
e l ement s .
Determine an
of subsets of
S
whose
SOLUTIONS (49-50)
102
Solution :
� 3
, l = 0 , 1 , 2 , . . . , [ n/3 ]
Let
[ n /3 ]
L
A (n) =
(49 . 1 ) .!. w = 2 (-1
+
i/"3)
contalnlng
S
The number of subsets of
•
3�
l=O
Thu s , we have n
n k
L
=
•
k=O k==O (mod 3 )
so that 3
'TN
and , for
3l e l ements i s
•
•
= 1 ,
r = 0 , 1 , 2 , define
Then , by the binomial theorem, we have .
(49 . 2 )
=
( 1 + w)
(49 . 3 )
n
=
n
L
k=O n
L
k=O n
L
(4 9 . 4 )
k=O
n k
= S
+ o
n k w = S o k n 2k W = So k
Adding (49 . 2 ) , (49 . 3 ) , (49 . 4) , we obtain , as
1 +
1.1
2
+ 1.1
= 0 ,
so that n 2 n 1 n A (n) = - (2 + ( -1.1 ) + (-1.1) ) 3
•
Hence we have A(n) =
i (2
n
n
+ 2(-l) ) ,
.!. ( 2n 3
(_ l )
n
) ,
if
n == 0
(mod 3) ,
if
n 1. 0
(mod 3) .
SOLUTIONS (49-50)
103
For each integer n � 1 , prove that there is a polynomial pn (x) with integral coefficients such that 2 ( 1+x ) pn (x) Define the rational number an by 50 .
•
n-l 1 _ ( l) an = n- 1 Pn (x) dx , n = l , 2 , . . . 0 4
(50 . 0)
.
Prove that an satisfies the inequal ity , n = 1,2, . . .
•
Solut ion (due to L . Smith) : Let Z denote the domain of rat ional integers and Z [ i] = {a+b i : a , b £ Z} the domain of gaussian integers . For n = 1 , 2 , 3 , set .
.
•
(50 . 1 ) so qn (x) £ Z [x] . As qn (±i) = 0 , qn (x) is divisible by x+i and 2 x-i in Z [ i ] [x] , and so pn (x) = qn (x) /Cx +1) £ Z[ i] [x] . However pn (x) £ R[x] , and as R[x] n Z [ i ] [x] Z[x] , we have pn (x) £ Z [ x ] . This proves the first part of the question. For the second part , we note that m
, so that 1 dx
Now, as
1 dx 1T 4 = o 1+X 2
'
=
n n pn (x) dx + (-1) 4
we have us ing (50 . 0)
1
SOLUTIONS (51-52)
104
1
I'IT - an I = n-l 4
dx . o
Now 1
' 4n 4 as x ( l-x)
< -
2:. , and thus we have
4
I'IT
-
an I
<
1 :......,- ' -:-= 5n-l
4
comp leting the second part of the question. In last year ' s boxing contest, each of the 23 boxers from the b lue team fought exactly one of the 23 boxers from the green team, in accordance with the contest regulation that opponents may only fight if the absolute difference of their weights is less than one ki logram . Assuming that th is year the members of both teams remain the same as last year and that their weights are unchanged , show that the contest regulation is satisfied if the lightest wember of the b lue team fights the l ightest member of the green team, the next l ightest member of the b lue team fights the next l ightes t member of the green team, and so on . 51 .
So lution : More general ly we consider teams , each containing n mem bers , such that the absolute difference of weights of oppo nents last year was less than d kilograms . Let B l ,B 2 , , Bn deno te the members of the blue team with weights •
•
-.
SOLlITIONS (51-52)
and let G 1 , G2 , weights
. • •
105
, Gn denote the members of the green team with
For each r with 1 � r � n , we consider this year ' s opponents Br We show that gr l � d and We treat only the case as the case b r is similar . If there exists s with and t with r < s such that B s fought last year , then b r If not , then every boxer B s with r < s � n was paired with an opponent G t with r < t � n last year , and thus Br must have been paired with some Gu with 1 � u � r last year . Thus we have •
b r - gr
�
b r - gu
�
d
•
This completes the proof . 52 . Let S be the set of all composite positive odd integers less than 79 . (a) Show that S may be written as the union of three (not necessarily disj oint) arithmetic progress ions . (b) Show that S cannot be written as the union of two arithmetlc progresslons . •
Solution :
•
(a) Each member of S can be written in the fotm (2r + 1 ) (2r + 2s + 1 ) ,
for suitable integers r � 1 and s � 0 and so belongs to the 2 arithmetic progres sion with first tetm (2r+l) and COllilllo n diff erence 2(2r+l) Taking r 1 , 2 , 3 we def ine arithmetic progres sions A 1 , A2 , A3 as fol lows : ,
•
=
SOLUTIONS (52-53)
106
A l = { 9 , ls , 2 1 , 2 7 , 33 , 39 , 4s , s l , s7 , 63 ,69 , 7s } , A 2 = { 2s , 3s , 4s , ss , 6s , 7s } , A 3 .. { 49 , 63 , 77} . It is easily checked that Suppose that
(b)
S = Au B , where A .. { a , a+b ,
.
•
, a+(m- l ) b }
.
and B = {c , c+d ,
, c+(n-l)d}
• . •
•
Without loss of general ity we may take a = 9 . Then we have either In the former case A = A 1 and so c .. 25 a + b = 15 or c = 1 5 This is impossib l e as 49 loS neither c + d 35 glovlong B = A2 In the latter case either a + b = 21 or c + d = 2 1 . lon A nor B If a + b .. 21 we have A = { 9 , 21 , 33 , 4s , s 7 , 69 } and so c + d - 27 Th1S is impossible as 49 belongs glov1ng B .. { l s , 2 7 , 39 , s l , 63 , 7s } If c + d = 21 we have B = A l - {9} so neither to A nor B a + b .. 25 glv1ng A = { 9 , 2s , 4 1 , } which is impossib l e as 4 1 1S prlme . •
-
,
•
•
•
•
•
•
•
•
•
•
•
•
•
. . •
•
•
53. For b
>
0 , prove that b S o
•
J.n
X
dx x
by f irst showing that b S ln x d x .. x •
_
2!:.
2
b -ux . e Sln x dx du
o
•
SOLUl'IONS (52-53)
107
Solution : We begin by showing that for b > 0 ( 53 . 1)
b s nx � dx - l im x. y + <»
s �n x dx x •
•
We have for y > 0 b s �n x dx x
s �n x dx x •
•
�
b �xy s � x !l dx e x •
s x �n max x O:oxlOb �xy b e M(b) �y 0 �by ) � e ( 1 - M(b) y :0 M(b) y :0
•
,
b � e XYdx
•
•
Let ting y + <» we obtain (53 . 1 ) . Next we have b
s�n x dx x •
(l
b y �xu s �n x xe du dx x •
b �ux . e s�n x dx du . Letting y + <» we obtain b s �n x dx x •
=
b -ux . e s�n x dx du
•
SOLUTIONS (54-55)
108
00 =
o .
-bu ( 1 - e (u sin b + cos b) du 2 1 +u
'IT
2
-
o
-bu e (u sin b + cos b) du
•
Hence we have . x b Hn _! dx x 2 o 00
=
-bu e (u s in b + cos b) du 1 +}
00
-bu l1+u2 e du 1+u2 o o
-bu e du
1 = -b , as required . 54 .
Let
be 44 natural numbers such that
Prove that at least one of the 43 differences occurs at least 1 0 t�mes . •
Solution : We have j =l
d J.
SOLUTIONS (54-55)
109
If each difference d . occurs at most 9 times then J
43 L dJ. j=l
�
9 ( 1+2+3+4) + 7 ( 5)
z
125 .
This is clearly a contradict ion so at least one difference must occur at least 10 tlomes . •
Show that for every natural number n there exists a prime 2 2 p such that p = a + b . where a and b are natural numbers both greater than n . (You may appeal to the following two theorems : (A) If p loS a prime of the form 4t+l then there exist integers 2 2 a and b such that p a + b (B) If r and s are natural numbers such that GCD ( r , s) 1 , there exist inf initely many prlomes of the fotUi rk+s , where k loS a natural number . ) 55 .
•
=
•
-
•
•
Solution : Let n be a natural number . By (B) there exists a prime q > n of the fotUi 4t+3 . Set Clearly we have GCD(m,q) .. 1
•
Hence, by (B) , there exists a natural number k such that the number 2 p = mk- q is a prime . Clearly p is of the form 4u+l . Hence , by exist natural numbers a and b such that
CA) ,
there
Without loss of generality we may assume that a < b . Suppose now that a � n . Then we have
SOLUTIONS (56-57)
110
. 2 c
= p - a
Z
Z Z = m k - q - a n
Z Z Z ( a +q) = 4k IT (r +q) r=l Z
Z
n
Z
= (a +q ) 4 ( a +q) IT (r +q) r=l r>'a
Z
- 1
,
where the factors on the right hand s ide of the equal ity are copr ime . Consequently they must be squares , but thi s is impos s ibl e as the second factor i s of the form
56 .
(i)
(ii) (iii)
4v- l .
Thus we mus t have
b > a > n .
be n ( � l ) integers such that a ,a " " ,a n l Z < an , o < a l < aZ < a . ( 1 � j < i :ii n) are di stinct , a l l the differences a . 1. J a . - a (mod b ) ( 1 � i � n) , where a and b are po s l. t 1.ve 1. 1.ntegers such that 1 � a � b-l
Let
•
•
•
-
•
-
•
-
•
•
Prove that n
L ar
r=l Soluti'on :
Let
r
be an integer such that Z � r � n For r I I � j < i � r there are = - r (r- l ) d i s t inct diffZ Z erences a . - a . , and thes e are a l l divisible by b . Thus the l ar1. J gest difference among these , namely a - a ' must be at l east r l r (r- l ) , that is •
!
and so a
r
�
Z � r � n .
SOLUTIONS (56-57)
111
As a 1 a (mod b) there 1S an integer t such that a 1 " a + bt 1 then a 1 " a + bt :i a - b < -1 , which loS impossible as If t ;;; Hence we have t Ii: 0 and so a l Ii: a glov1ng al > 0 -
•
•
•
-
-
•
,
ar
r
:0
n .
The inequality (56 . 1 ) clearly holds for r ·
1
.
(56 . I )
n
I ar r=l
�
2
an
:0
•
•
Thus we have
n b2" dr-I) .. an I r"l
+
,
so that n
z: a r r=1 57 .
Let An
-
-
-'!r
<
t
<
'!r
+
(a 1 J ) be the n X n matrlox gloven by •
.
•
•
.
a lo. . J where
Ii:
b-n 3 6
..
2 cos t 1 0
Evaluate Dn
, , ,
•
if lo J if Ii - j l otherwise •
•
-
, ..
1
,
,
det An
•
Solut ion : Expanding Dn " det An by the first row, we obtain the recurrence relation (5 7 . 1 )
Dn = 2 cos t D n- 1 - Dn-2 ' n
Ii:
2 .
We now consider two cases according as t � 0 or t " O . For t � 0 the values of D l and D 2 may be obtained by direct cal cula tion as follows : sin 2t D l '" 2 cos t .. ,
and
SOLUTIONS (57-58)
1 12
2 D 2 - 4 cos t - 1 2 4 sin t cos t - Sln t Sln t 2 2 1 ) t + 2 ' sin t Sln t cos t (2 cos Sln t -
•
=
•
•
-
sin 2t cos t + sin t cos 2t Sln t Sln (2t + t) Sln t sin 3t Sln t •
•
•
•
•
These values suggest that Dn
(57 . 2)
..
sin (n+1) t , n Sln t .
=
1 ,2,3,
• • •
•
In order to prove (5 7 . 2) , aSS1!me that (57 . 2 ) holds for n Then we have, by the recurrence relation (57 . 1 ) , Dk
=
1 , 2 , . . . ,k-l,
2 cos t Dk-l - Dk-2 s in sin (k1 ) t kt 2 cos t sin t ) n s 2 ( i s l o i n c t t s k t :: :.. :: · k.::. :.: . :... :. ..: _ :,; .:: � :..: :. .: ::. :. :. .: .. ::.; ..,:: :..: .. -:-= :;::Sln t + s n i k .::. =.; � ::. .. ::...:.: :...:. l�).::.t Sln t =
=
•
Thus (57 . 2 ) holds for a l l n by the principle of mathematical inductlon . For t 0 we have •
=
which suggests that Dn n + 1 , n � 1 . This can al so be proved by mathematical induction . We note that =
SOLU:rIONS (57-58)
1 13
sin (n+l ) t " n + l . lim . t s�n t .... O Remark : The auxil iary equation for the difference equation (57 . 1 ) is which has the roots X "
i - 2 (cos t)x+ l - 0 , exp ( it) , exp (-it) , t ;l! 0 , , t '" 0 , 1 (repeated)
g�v�ng •
•
Dn
=
A l exp (nit) + B l exp (-nit ) , t , t A2 + B 2 n
;I! =
0 , 0 ,
for complex constant s A l and B l and real constant s A 2 and B 2 , which can be detettuined from the in itial values D l 2 cos t , 2 D 2 4cos t - 1 , using DeMoivre ' s theorem in the case t ;l! O . One f inds A l = i (1 - i cot t) , B l i ( 1 + i cot t) , =
=
=
A2
=
B2
=
1
•
58 . Let
a and b be fixed positive integers . Find the general solut ion of the recurrence relat ion 2 ,. xn + a + Jb + 4axn , n ,. 0 , 1 , 2 , . . .
(58 . 0) where xo
=
0 .
Solution : From (58.0) we have
,
SOLUTIONS (58-59)
1 14
so that (58 . 1 )
•
Hence , by (58 . 0) and (58 . 1 ) , we have (58 . 2) Replac ing n by n-1 in (58 . 2) , we get xn-l
(58 . 3)
=
2 xn + a - � +4axn
•
Adding (58 . 0) and (58 . 3) we obtain = 2xn + 2a , and hence 2xn + xn- 1 = 2a
(58 . 4) Setting
= x x we obtain in (58 . 4) , n-1 n n-1 Y - Y = 2a n n- 1
Y
(58 . 5)
•
•
Adding (58 . 5) for n = 1 , 2 , 3 , . . . yields (as yn = 2an + (a + b) , 2 and so xn = an + bn 59 .
•
Y
o
= xl -
X
o
= a + b)
•
Let a be a fixed real number satisfying 0 < a < 1T , and set
(59 . 0)
a
r cos u 1 Ir = :. 2-r"-c-'0...;s""u....;+:-;. r""?2 du . .1....:---;;' -a
Prove that l im I r , r + 1+
Ir
SOLUTIONS (58-59)
115
all exist and are distinct . Solution : We begin by calculating
I
We have
I '
a 1 1 - cos u du du 2 2 cos u 2 a -a
a
(59 . 1 )
=
a ,
where we have taken the value of the integrand to be !.2 when u Now, for r > 0 and r � 1 , we have 2 1 - 2r cos u + r
>
2r - 2r cos u
=
2r ( l - cos u)
;:
0
=
O.
•
so that the integrand of the integral in (59 . 0) is continuous on [-a , a ] . We have a !. l_--= r _ 2:... ) . --, ( du � + Ir ,.,..; u r2 ) 2 ( 1 2r cos 2 + -a -:-"""7''<"
-
=
a
du ' u + r2 2r cos 1 -a
a+
which gives ( 1 I .. a + r
(5 9 . 2 ) where (59 . 3) Let t
J
=
a
..
r
U
du cos u r2 1 2r + -a
tan -2 with -a cos u
=
�
•
u � a so that
2 1 - t 2 1 + t
•
du
=
2 --"--""2 d t . 1 + t
Using the above transformation in (59 . 3) , we obtain
SOLUTIONS (59 60)
1 16
2
(59 . 4)
dt ' 2 l-r l+r
where t l .. tan a/2
•
Evaluating the standard integral in (59 . 4 ) , we deduce that (5 9 . 5)
tan
-1
l+r tan a/2 l-r
•
From (59 . 2) and (59 . 5 ) we get
Hence for r
2 2 ( l -r ) -1 l+r I r .. a + tan 2 l-r I l-r I we have
> 1
Ir and for 0
<
r
<
•
r+l tan a/2 r-l
,
-l a + 2 tan
l+r tan a/2 l-r
•
a - 2 tan
we have
1 I
..
-1
r
,.
Taking l imits we obtain =
1T . 2 a 2
-
l im I r .. a + 2 . 1I2 r 1-
=
a - 1T ,
=
a + 1T
•
-+
Thus the quant ities distinct .
, l im I r a l l exist and are a l l r 1 -+
1 17
SOLUTIONS (59-60)
60 . Let I I:; e: I let and of does not exis t
denote the denote the the length a function
•
class of all isosceles triangles . For length of each of the two equal altitudes of the third altitude. Prove that there f of hI:; such that
for all I:; e: I . Solut ion : Let h be a fixed positive real number . For t
>
1 let
(60 . 1 ) As b < Za there A , B , C such that the cho ice (60 . 1 )
exists an isosceles triangle I:;(t) with vertices IAc l a , I Bc l b . It wi l l follow that I AB I is such that =
=
=
1 t + t 1 t - -t
•
Let P , Q , R be the feet of the perpendiculars from A to BC , B to CA , C to AB respectively . Then we have 2
hMt)
-
BQ'
=
CR' .. a' sin' A
=
a' ( 1 - cos2 A)
Applying the cos ine law to Mt) we ob tain 2a' b2 cos A Za2 Hence it follows that '"
(60 . 2 ) Next , from (60 . 1) , we see that 2a-b h
=
2a+b , h
=
t C t + i) I (t - -t )
•
•
SOLUTIONS (60-62)
1 18
so that
and thus from
(60 . 2 )
we have
App lying Pythagoras ' theorem in triangle ASP , we have
so that ..
h
2
-
.!. t + t t
1
•
-
t
F ina l ly , suppose there exis t s a function
for a l l
�
e:
I .
f
=
f (h ) �
such that
Then , in particular , one sees that ,
t
> 1 ,
which imp l ie s .!. t + h t
2
that
t
-
,
t
> 1 ,
f (h) , h
t
> 1
:> f (h)
1
-
-
t
1. S •
.!. t + t
(60 . 3)
t
As the left s ide of
-
<
1 -
2
-
t
(60 . 3)
•
tends to infinity as
t + + oo while the
right s ide i s f ixed , we have obtained a contrad ict ion , and therefore no such function
f
can eX1.st . •
SOLUTIONS
61
•
( 6 1 . 0)
Find the minimum value of the exp ress ion 2
x +
k x
for
x
1 19
(60-62)
>
2
k ( l+sin t ) _ 2 ( l+cos t ) x + x
2
and
0
o � t � 2n , where
>
k
� 2
+ ( 3 + 2 co s t + 2 s in t )
12
+
,
is a f ixed real
number .
Solut ion :
The exp r e s s ion g iven in ( 6 1 . 0) can be wr i t t en in the form ( x - ( 1+cos t ) )
2
+
(� x
(l+s in t ) )
2
,
k (x > 0) which is the square of the d i s tance between the point (x , -) x on the rectangular hyperb o l a xy = k in the f irst quadran t , and the poin t (l+cos t , l+sin t ) with radius
1 .
( 0 � t � 2n)
The condit ion
curves are non-intersect ing .
k
>
on the circle centre ( 1 , 1 )
�
12
+
ensures that the two
C learly the minimum distance between
these two curves occurs for the point ( Ik , Ik) on the hyp erbola and 1 1 1 + 12) on the cir c l e . Hence the required minithe po int ( 1 + 72 '
mum 1. S •
2 ( 1k -
62 .
E > O
Let
.
•
Around every point in the xy-plane with integral E
co-ordina t e s draw a circle of radius
.
Prove that every straight
l ine through the origin mu s t intersect an inf inity of these circles .
Solution :
Let b
s a t i s fy ing
If be a l ine through the or1.g1.n with slope b k are 1.ntegers , where k and 1.S rat iona l , say b L
•
•
•
•
•
.t
>
-
1
and
- 1 (k , .t) -
= I ,
then
passes through the
L
centres of the inf init y of circles a l l o f radius l a t t ice p o in t s
(.tt , kt)
, wh ere
t
E
1.S an 1.nteger . •
•
,
centred a t
SOLUTIONS (62)
120
If b is irrational , then by Hurwitz ' s theorem there are infini tely many pairs of integers (m, n) with n � 0 and GCD (m,n) = 1 for which m < 1 -n - b 2 n •
Choosing only those pairs
(m,n) for which ,
we see that there are infinitely many pairs (m ,n) for which m -n - b
,
n
that is , for which <
(6 2 . 1 )
e:
•
Since the left side of ( 62 . 1 ) is the distance be tween the line L and the point (m,n) , L intersects the infinity of circles all with radius e: centred at these lattice points . ,
Second so lut ion (due to L . Smith) : Let L be a line through the origin with s lope b . The case when b is rational is treated as in the first solut ion . When b is irrat ional , we construct an infinite sequence of lattice points whose distances from L are less than any given pos�t�ve e: Pick any lattice po�nt (x l' Y1 ) with d 1 I y 1 - bx 1 1 < 1 Clearly d 1 �s pos�t�ve as b �s irrat ional . Set •
•
•
•
•
•
•
=
•
> _
so that
1 ,
•
SOLUTIONS (62)
121
(62 . 2 )
•
Let (x 2 ' Y 2 ) be the lattice (a 1 x 1 (x2 ' Y 2 ) (a 1 x 1 -
point g�ven by •
' a 1 Y 1 - 1 ) , if Y l if Y l < ' a 1 Y 1 + 1)
>
,
and set d 2 = IY 2 - bx 2 1 . Clearly , as d 2 (62 . 3)
�
0
,
bX 1 , bX 1
,
we may set
•
It is easy to see that d 2 = 1
, so that by (62 . 2 ) we have
(62 . 4 )
•
Thus , from (62 . 3) and (62 . 4) , we obtain
Cont inuing this process we obtain an infinite sequence of lattice whose vertical distances dk } points { (xk ' Yk ) : k 1 , 2 , from L satisfy 1 --'-7 -= "7 dk < k ;: Z , ' ak-1 + 1 =
where ak =
• . .
,
, k <: 1 . Furthermore {ak : k
=
1 ,2 , . . . } �s a
s trictly increasing sequence of positive integers . 1 Finally choose a positive integer N such that -N < for al l n ;: N + 1 , we have < .!. N <
dn < and the latt ice points (xn , yn ) (n
E:
•
E:
,
-
>
N) are as required .
•
Then,
122
SOLU'I'IONS (63-64)
63 .
n
Let
•
•
•
k - 0 , 1 , 2 , . . . , 2n-2
For
be a pos �t �ve �nteger .
-
def ine I k
(63 . 0)
I � I k n-I '
Prove that
For
Solut ion :
k
=
k
0 , 1 , 2,
=
°
•
•
0 , 1 , 2 , . . . , 2n-2 .
=
•
k x dx 2n n x +x +l
(63 . 0) can be written
, 2n-2 ,
•
a I
k
=
l im
•
a ... b ... O+ co
Applying the transformat ion
x
=
1 , we obtain y
-
l ib I k
so that
I k
=
l im
a b ... O+ ...
2n-k-2
y . ..
dy
,
co
=
Now, us ing the arithmetic mean-geometric mean inequality , we have for
x �
° , k 2n-k-2 X + x n-l � x , 2
(6 3 . 1 ) As
x
2n
n +x +l
> °
k
=
0,1,2,
, we may divide ( 63 . 1 ) by
the resul t ing inequal ity to obtain
from which the des ired inequality fol lows .
x
•
.
.
, 2n-2 .
n +x +l , and integrate
2n
SOLUTIONS (63-64)
D
Let
64 .
be the region in Eucl idean n- space con s i s t ing of a l l
(x , x , 1 2
n-tuples
123
. • .
,x ) n ,
,
satisfying
�
x n
,
• • •
0
+ x � 1 n
,
Eva luate the mul t iple integral (64 . 0)
• • •
D are pos�t�ve �ntegers . •
Solution (due to L . Smith) :
We begin by considering
I(r,s) where
r
and
s
Integrating
•
•
a =
o
r
s
x ( a-x) dx ,
are pos i t ive integers and I(r,s)
>
a
by part s , we obtain
I ( r , s)
=
s I(r,s) r+l
•
!
r l
so that
s-l r+2
I (r+l , s-1) ,
• • •
1 I ( r+s , 0) , r+s
that i s r!s! I ( r+s , 0) I (r , s ) - ( ) r+s ! As
I (k , O )
(64 . 1 )
=
k a , for k+l
0 .
k
� 1
,
•
we obtain
, r+s+l . r! s. a I (r , s ) ( r+s+O !
•
Now, applying ( 64 . 1 ) succes s ive ly, we obtain
•
124
SOLUTIONS (64-65)
•
•
D
•
I-x 1
1 =
xn =0
xn-1 =0 1 •
x 1 =0 x 2 =0
=
! k� ! ! k k I n � � +1 n � n� _ � � (kn- l +kn+kn+1 +2) ! __
•
•
l- x 1 - . . . -xn-2 kn+kn+1+1 kn-1 xn- 1 ( 1-x 1 - · · ·-xn- 1 )
xn-1 =0
1
__
•
•
•
x 1 =0 x 2=0 •
•
1-x 1 - · · · -xn- 3 kn_ 2 xn-2 ( 1-x 1 xn-2 =0
.
.
•
kn- l+kn+ k +2 -xn_ 2) n+1 dxn_ 2 · . . dx 1
•
k 1 ! k 2 ! . . . kn ! kn+1 ! = (k +k + · · ·+k +k 'ln) ! . 1 2 n n+1 65 .
Evaluate the l imit n 2 ft 1 L = lim -n L n .... k-1 Ii
•
'"
i Solut i.on : We show that L = 3 [ 2r ] - 2 [ r ] = Hence we obtain
o
,
3
•
For any real number r we have
if [ 2r]
1 , if [ 2r]
lS •
even ,
is odd .
125
SOLUTIONS (�5)
- 2
A(n)
Tn Ik
n - L 1 k-1 odd [ f en) n 1 = L L 5=1 k=l E. [ 2 k] - 2s+1 [ 2 Tn] - 1 2
where f en) =
,
2
Next we see that
•
and only if 4� 2 (25+2)
<
:i
k
n ".. 4:.:.._..:. 2 (25+1)
�
•
and thu5 4n 2 (25+1)
4n 2 (25+2)
-
n [ 'k] = 2 5 + 1
where
I E1 1
<
1
•
1n
4n 4n :.:..". . _ _..:. E " + ' 2 2 l (25+2) (25+1 ) Hence we have f (n) 1 1 ...:... A(n) = 4 i 2 2 (2s+2) s-l (2s+1) -
.."..
where :i
Letting n -+ oo gl.ves •
f (n) n
•
I E1 I
<
f (n) n
:i
rn " 1 . . n ,Iii
,, 25 + 1 if
SOLUTIONS
126
00
L
-
-
1 4 L 2 s=l (2s+1)
-
that �s , •
L
-
2 7T 4 8 -1
-
1 , 2 (2s+2) J
2 7T 1 24 4 -
-
2 7T - 3 - 3 -
,
2 7T + . . . = 24
and
as
(66-67)
•
66 .
Let p and q be dis tinct primes . Let S be the sequence cons isting of the members of the set { pmqn : m,n = 0 , 1 , 2 ,
. . •
}
arranged in increasing order. For any pair (a,b) of non-negat ive integers , give an explicit expres sion involving a, b , p and q for b a the position of p q in the sequence S . Solution : Without loss of general ity we may suppose that p < q . a b . th C learly P q �s the n term of the sequence , where n is the n mber of pairs of integers (r , s ) such that ••
r s p q Set
:>
k
a b p q =
a+
,
r ;: 0 , s ;: 0 . b bt q In. p
,
k a b so that p is the large st power of p les s than or equal to p q Then we have k n = ! 1 r , s=O r a b s p q <=p q
•
1 27
SOLUTIONS (66 67)
k
L
..
1
r,s 0 r bt p + s bt q :;; a bt p + b bt q k
..
�r=O b +
so that the po s i t ion of (k+l )
67.
Let
a bt p + b .tn q - r .tn p
L
"
,
+
�r=O
(a-r) bt p hi q
elements
0,1,2, N ( a)
•
z
denote the finite
p . . . ,p-1 . of
an
matr�ces •
..
A ,
where
A "
a
0
x ,
o
a
•
-
1 , 0 ,
":1 ,
2 b 0 ,
if a if a otherwise ..
..
for some non-zero e lement
•
We wil l show that
(6 7 . 1)
a
For
It is convenient to introduce the notation
Solu.t ion :
k ( a)
,
Z , such that p
2 x
(67 .0)
1
�s
determine the ntwber
p with entries from
+
•
(b+l)
p
1
+
deno te an odd prime and let
p Z
(a-r) .tn p ln q
a b P q
field cons is t ing of the element of
..
N ( a) ..
2+p+2 , P 2 p , 2 p - p ,
if
k ( a)
if
k ( a)
if
k (a)
1 , = 0 " -1 ..
,
.
b
of
Z , P
SOLUTIONS
128
We note that if k(a) 1 , 2 of Zp such tha t a. = b , 1.S x = -b x y , where Let X = z matrix such that -
(67-68)
so that there 1.S a non-zero element b 2 then the only other solut ion of a = x •
•
•
x , y , z and w are element s of
Z
a , be p
Then we have
2 2 x + yz = yz + w = a , (x + w)y = (x + w) z = ° .
(67 . 2) (67 . 3 )
We treat two cases according as (i) x + w = ° or (ii) x + w In case (i) the equat ions (67 . 2 ) and (67 . 3 ) become 2 x + yz
(67 . 4) If k(a) = 1 , say a = are given by
b �
°
=
�
° .
a .
, then a l l the solut ions of (67 . 4)
2 2 -1 (x , y , z ) = (±b , O , O) , (±b , t , O) , (±b , O , t) , (u , t , t (b -u »
,
where t denotes a non-zero element of ZP and u denotes an element of ZP not equal to ±b . Thus there are 2 + 2 (p- l ) + 2 (p-l) 2 + (p-2) (p-l ) = p + P solutions (x, y , z , w) in this cas e . If k(a) = ° , so a = ° , then all the solut ions of (67 . 4) are g1.ven by 1 2 (x,y , z) = (0 , 0 , 0) , (O , t , O) , (O , O , t) , ( t , u , -t u ) , •
where t and u denote non-zero element s of Zp Thus there are 2 2 1 + (p-1) + (p-1) + (p-1) = p solutions (x , y , z ,w) in this case . If k(a) -1 , so that a 1.S not a square 1.n Zp , then all solut ions of (67 . 4) are given by 2 -1 -1 (x , y , z) = (O, t , at ) , ( t , u , (a-t ) u ) , •
=
•
•
where t and u are non-zero elements of Z p Thus there are 2 2 (p- l ) + (p- l) = p - p solut ions (x, y , z ,w) in this case . •
129
SOLUTIONS (67-68)
In case (ii) the equations (67 . 2) and (67 . 3) become x
=
w
�
2 0 , x
=
a , y
=
z
=
0 ,
which clearly has two solut ions if k(a) - 1 , and no solutions if k(a) 0 or -1 Hence the total number of solut ions is given by (6 7 . 1 ) . •
=
68. Let n be a non-negative integer and let f (x) be the unique differentiab le function defined for all real x by 2n+1 ( f (x) ) + f (x) - x
(68 . 0)
=
0
•
Evaluate the integral x for x
'=
o
0 .
f (t) dt
•
Solution : The function y f (x) defined by (68 .0) pas ses through the origin and has a positive derivative for all x . -1 Hence , there exists an inverse function f (x) , defined for all x , -1 -1 2n+1 and such that f (0) O . Clearly we have f (x) " x +x . When x > 0 , the graph of f (x) lies in the first quadrant , as f (O) 0 and f 1.S 1.ncreas 1.ng . Thus for al l x '= 0 we have =
=
•
=
x
•
•
f (x) -1 f (t) dt ,. x f (x) f (t) dt +
•
Now f (x) 2n+1 (t + t) dt so that
=
2n+2 (f (x)) + 2n+2
,
SOLUTIONS (69)
130
2 2n+2 f (x) f (x) ) ( ( f (t) dt x f (x) 2n+2 2 0 2 · 2n+1 - 2n+2 x f (x) - 2:+2 ( f (x) ) x
-
-
-
•
69 .
Let f en) denote the number of zeros in the usual decimal representation of the positive integer n , so that for examp l e , f ( 1009) = 2 . For a > 0 and N a positive integer , evaluate the limit iY'. S eN) L = l im in N N + oo where .
,
N
f (k) S eN) = 1: a k=l
•
be a non-negative integer . The integers between Solution : Let t i+1 lO and 10 _1 have i+1 digits of which the first is necessarily non-zero . The number of these integers with i (0 � i � t) i-i+1 of their digits equal to zero is � 9 •
�
Choose m to be the unique non-negative integer such that m 10 - 1 so that m = Then we have
:ii
m+ 1 N < 1O - 1 in (N+1) in 10
,
•
m m 10 . -1 f (k) m-l 10 . -1 f (k) = 1: a 1: a = 1 k=l i=O k=l f (k)=i
131
SOLUTIONS (69)
m m-l . l O - l � = i a i 1 k=l i=O f (k ) = i m-l
- i
i-a
� .
a
m
�
t+l 1 lO _l
t
taO
1 .
t ka l O
f (k ) - i Appeal ing to the f irst remark we obtain
i-O m-: l =
l:
t=O
t= O
9
t+ l
t=O
g-
l:0 �� .
.a
i
9
�= •
m-l � J +l
.. i
.e
t
l + .!!
9
m- l t .. 9 i ( a + 9 )
•
t.. O
that is
where
9 c = a+8
.
As ,
we obtain c ( a+9)
m
� S (N ) +c < c ( a+9)
,
Taking l ogari thms and dividing by
m , we get
m+l
•
SOLUTIONS (70-71)
132
.e.n c m Letting N
_
. 'r,
(m+l) .e.n(a+9) m
.e.n (a+9) ::; .!.m .e.n(S (N)+c)
•
so that m .... "' , we deduce that
,
l im .!. .e.n(S (N)+C) m N ....
=
.e.n(a+9)
•
""
Hence we have .!. l im m .e.n S (N) N ""' CO ,.
lim N ....
""
,. .e.n(a+9)
,
and so l im N ....
""
.e.n S eN) = .e.n(a+9) bi N bi 10
•
70 .
Let n � Z be an integer and let k be an integer with Z ::; k ::; n . Evaluate M where {l,Z,
S .
.
.
•
max
S
(
l
min ( a 1.· +1 -a 1.. ) 1:i\i:i\k-l
runs over all selections S ,. { a l ,a Z , , n } such that
, . • .
, ak } from
133
SOLUTIONS (70-71)
Solut ion (due to
J.F.
Semple and L . Smith) :
We show that M ·
•
We consider the selection S*
given by which has - a �.
:II
n-l k-l
,
1 :iI. i :iI k-l .
Thus for S* we have min (a �+ . 1 l:ili:ilk-l
-
n-l a �. ) = k-l
,
n-l In order to prove equality , we suppose that so that M <: k-l there is a selection S with n-l min ( a�+ . 1 - a . ) <: k-l + 1 . � l:ili:ilk-l Then we have k-l n-l +1 > n-l -a ) ( a · (k-l) <: � . .�= 1 �+l � k-l •
,
which is impossib l e . Hence any selection has min (a �. +l -a �. ) l:ili:ilk-l which proves the required result . 2
<
n-l +1 k-l
Let az + bz + c be a polynomial with complex coeff icients such that a and b are non-zero . Prove that the zeros of this polynomial l ie in the reg�on c b (71 . 0) Izl < a + b 71 .
•
-
-
-
•
,
SOLUTIONS (71-73)
134
Solut ion : Note that l ib' -4ac I - I b I -
<
IbI
-
< IbI
-
Ibl
=
4ac 1 - b2 4ac 1 + b2 2ac 1 + b2 2ac + b ,
so that 1b2 -4ac I � b + b b 2a 2a 2a 2a I 2 and hence the solutions of az + bz + c -
+
=
0 satisfy ( 7 1 . 0) .
Second solution (due to L . Smith) : Let w (;10-1) be a comp lex number . The inequal ity w l W+l l + w+1
( 7 1 . 1)
is easily established , for if l W+l l whi le if 0 < l W+l l < 1 then w I w+l j + -r'W+l
>
> _
�
1 ,
1 then ( 7 1 . 1 ) clearly hold s ,
1W+1 1 + I wl
1
>
•
Let z l , z 2 be the roots of the given quadratic chosen so that As a and b are non-zero , z 2 ;1O 0 and z l+z 2 ;1O 0 , I Zl l � I Z 2 1 and setting w z l /z 2 (;10-1) in ( 7 1 . 1 ) we obtain I zl z2 1 + I z +z 1 1 2 •
=
•
The inequality ( 7 1 . 0) follows as
-
b and a
-
=
c a
-
•
135
SOLUTIONS (71-73)
Determine a monic polynomial f (x) with integral coefficients such that f (x) = 0 (mod p) is solvab le for every prime p but f(x) = 0 is not solvable with x an �nteger . 72 .
•
2 Solution : If p 2 or p = 1 (mod 4) the congruence x + 1 = 0 (mod p) � s solvable . If p 3 (mod 8) the congruence 2 x + 2 0 (mod p) � s solvable . If p 7 (mod 8) the congruence 2 x - 2 0 (mod p) � s solvable . Set �
•
-
-
•
-
-
-
-
-
-
•
-
Clearly f (x) is a monic polynomial with integral coefficient s such that f (x) = 0 is not solvable with x an �nteger . •
73 .
Let n be a f ixed positive integer . DeterlDine
M=
max L I x �. -xJ. I . 1 < ' < J' <-n O:O�!Ol k=1 , 2 , . . . , n _ 1.
Solut ion : Without loss of general ity we may assume that xn
:0
1 ,
so that S =
L I x �. -xJ· I " l:oi<j :on
The sum S has n (n- 1 ) /2 terms . For each k , 1 :0 k :0 n , xk appears in k-1 terlDS in the left position and n-k times in the right position . Hence we have n n S = L xk « k- l) - (n-k» ) = L xk (2k-n-1 ) k=l k= l As 2k-n-1 <
0 ,
n+1 for k < 2
,
we have
•
SOLUTIONS (73-74)
136
L (2k-n-l ) . n+ l
n � L ( 2k-n-l) = l+3+5+ n k= -+l 2
• • .
2 +(n-l) = n /4 ,
and for n odd s :;;
n 2 L (2k-n- l ) = 2+4+6+ . . . +(n-l) = (n -l) /4 n+l k= 2
•
Thus we have For n even, the choice .. x n2
g�ves
..
0 ,
=
x
�+l 2
= xn
• • •
1
=
•
S
..
2 n /4
and for n odd , the choice :II
g�ves •
• • •
=:I
xn-l = 0 , xn+l 2 2
This shows that
2 M = [n /4]
-
.
.
.
s
�
:
• • .
1
•
Let { x . i = l , 2 , ,n} and {y . : i - l ,2 , sequences of real numbers with 74.
xn =
�
• • •
, n} be two
SOLUTIONS (73-74)
137
How m�st Y l " " ' Yn be rearranged so that the sum n 2 (x . y . ) ( 74 . 0) 1. 1. L . -1 1. is as small as pos s ible? . th . th term y 1.. is smaller than the J Solution : Suppose the term Interchanging y . and y . produces y J. , l :;; i < j :;; n J a new sequence 1.
•
1.
with z
k
=
Yk , if k ;z! y . , if k J y . , if k ,. =
1.
1.
or ,
•
•
1.
J
,
J •
•
•
Moreover we have 2 2 2 2 (x.1._y . ) + (x . -y 1.. ) - (x 1.. -y . ) - (x .-y . ) ,
=
J
J
J
1.
J
where 2 2 2 2 « x . -y . ) +(x . -y . ) ) - « x 1.. -y 1.. ) +(xJ.-y . ) ) l.
J
J
l.
J
,. (y .-y . ) (2x .-y . -y . ) + (y .-y . ) (2x . -y . -y . ) l.
•
J
1.
1.
2 (y . -y . ) (x1.. -x . )
:$ 0 ,
1.
J
J
J
J
so that •
l.
J
l.
J
SOLUTIONS (75-76)
138
Hence every such transposit ion decreases the size of the sum (74 . 0) , and the smal lest sum is obtained when the y �. are arranged in decreas ing order . 75 . Let p be an odd prime and l et Z denote the finite field P Let g be a given function on zp cons isting of 0, 1 , 2 , , p-l with values in zp . Determine all functions f on zp with values in Z , which sat isfy the functional equation P f (x) + f (x+l ) = g (x) (75 . 0) •
• . •
for all x �n •
z
p
•
Solut ion : Replacing x by x+k (k = 0 , 1 , 2 , . . . , p- l ) obtain (75 . 1 )
f (x+k) + f (x+k+l ) = g (x+k)
in (75 . 0) , we
•
Hence, us ing (75 . 1 ) , we have p-l p- l k k L (-l) g (X+k) = L (-l) (f(x+k) + f (x+k+l» k=O k=O p-l k+l k = L « -l) f (x+k) - (-l) f(x+k+l » k=O = f (x) - (-l) P f (x+p) = 2f (x) , so that there is only one such function, namely, l p-1 k f (x) = 2 L (-l ) g(x+k) k=O
•
1 39
SOLUTIONS (75-76)
76 .
Evaluate the double integral
(76 .0)
I
l I dxdy x 1 y o
•
•
Solution : Set S - { (x , y) I 0 � x � 1 , 0 � y � 1 } For 0 < e < 1 set R 1 (e) = { (x ,y) I 0 � x � l-e , 0 � y � 1
•
}
,
R2 (E) = { (x , y) I l-e � x � 1 , 0 :;; y :;; l-e
}
R(e) = R l (e)
U
,
R2 (e) ,
so that R(e) = S - { (x,y) I 1-e :;; x � 1 , l-e :;; y � 1 } . Then, for
j
a
1 , 2 , we set 1.
J
(e)
dxdy l - xy ' -
=
R. (d J
1 The function 1 - xy is continuous on the square S except for a discontinuity at the corner (x,y) .. (1 , 1 ) , so that (76 . 0) becomes I ..
e
l im . ....
o+
dxdy .. l im 1 - xy e .... o+ R(e)
For n a positive integer and for J . ( e , n)
J
J
..
2 2 O + xy + x y +
= R . ( e) J
K . ( e , n)
j
..
R . (d J
.Xn n Y dx dy . I - xy
1 , 2 , we set • • •
n-l nl + x y ) dx dy ,
140
SOLUTIONS (76)
2 2
n-l n-l
As l + xy + x y + . . . + x
y
1
..
-
1
-
+
n n xy
+ xy
1 -== -:-1 - xy we have , for j = ( 76 . 1 )
; 0 <
1,2
<
E
+
J . (E,n)
J
1
n n xy
xy
-
1
n n xy
xy
-
,
1 ; n � 1 ,
K.
J
( E , n)
=
1.
Next , as the largest value of xy on both l- E , we have 1
n n xy
xy
-
on
,
J
(E)
.
and
R (E) 1
and
so that
Hence for 0 <
E
< 1 we have l im
n � oo
Next , for j =
1,2
K.
J
( E ,n)
= 0
,
•
, we have J . ( E , n) ..
J
n-. l
m m I x y dx dy m=O R . (d J n-l
.. L m=O
m m x y dx dy , R.
J
(d
R (E) 2
�s •
SOLUTIONS (76)
141
so that n-l l-e: m 1 m x dx y dy .. L m-O n-1 O _e: ) m+l 1=-=- L "": m+l 1 m+ m-O •
n-l
.. L
,
maO
g�Vl.n g •
•
l m+ _ e) O lim J ( e , n ) " L 1 2 OO -+ n m=O (m+1) 00
,
and s imilarly 2m+2 e: _ ( O-: ) 2 m=O (m+l )
l im
•
Letting n -+ co in (76 . 1) , we obtain m+1 _e: (l ) .. I I (e:) L 2 111'"0 (m+1) m+l 2m+2 _e: _e: ) O ) O I 2 (e) - i i 2 2 m"O (m+l) m-O (m+1 ) 00
,
00
so that
co
I 1 Ce:) + I 2 (e) .. 2 L m-O
00
00
- i
,
m-O
Hence , by Abel ' s l imit theorem , we have 1 1 dxdy 1 o 0
vv . OJ
l im e " 0+ co 1 .. 2 i --,:.2 ...". m=O (m+l)
_
00
-
L maO
1
,
SOLUTIONS (77-78)
142
00
1
= L m=O
,
Let a and b be integers and m an integer > 1 . Evaluate (m1 ) a + b � + a+b + 2a+b + + m m m m 77 .
.
•
•
.
•
Solution : Our starting point is the identity k-1
L
(77 . 1 )
x=O
�+
e
�
[ ek] ,
where k is any positive integer and e is any real number . As {y} = Y [y] , for any real y , (77 . 1 ) becomes -
(77 . 2 )
k-1
1 L .!k + e = -2 (k-1) + {ek} . x=O
For fixed k and e , .!k + e is periodic in x with period k . If c is chosen to be an integer such that GCD ( c , k) = 1 , the map ping x .... cx is a bij ection on a complete res idue system modulo k . Applying this biject ion to (77 . 2) , we obtain (77 . 3)
k- 1
�
x=O
cx + e = "21 (k-1) + { ek } k
•
We now choose c = a/GCD (a,m) and k = m/GCD (a,m) , so that GCD ( c , k) = 1 , and e = b /m . Then (77 . 3) becomes (keeping k �n place of m/GCD (a,m) where convenient) •
143
SOLUTIONS (77-78)
k-:l ax+b b 1 m 1 = + ::-TL a , m """" ) ( -= G-= CD GCD (a,m) 2 m x=O and so
,
t
GCD (at)-l k m-l ax+b + b a (y+kz ) L m m z�O y-O x=O GCD (af m) -l k-l ay+b = L m z=O y=O 1 b - 2' (m - GCD(a ,m) + GCD(a ,m) GCD(a ,m) Finally, we have .
•
ax+b = �(m- 1 ) + b - 1 (m-GCD(a,m) - GCD(a ,m) b � ";:' � , 2' :-:: :'::: '::' m GCD (a , m'"')' 2 that is m-l L x=O
m
b = j(am - a - m + GCD (a ,m) + GCD (a,m) GCD(a ,m)
•
Remarks : The identity (77 . 1) is given as a problem (with hint s) on page 40 in Number Theory by J . Hunter , Oliver and Boyd , 1964 . 78.
Let a l ,
• • .
, an be n (>1) distinct real numbers . Set +
Prove that
.
•
.
2 2 + an , M min (a . - a . ) J l:iOi <j :iOn '"
•
1.
�
M
�
n(n- 1 ) (n+l ) 12
•
Solution : Relabel ing the a ' s , so that preserves the values of S and M .
<
• • •
< an ,
144
SOLUTIONS (78-79)
2 l. Let m n a 1.. l;;; i :>n •
-
2 a. J
where J loS a fixed subscript . Then , we have > J , a . > 0 , for 1. a 1.. < 0 , for 1. < J •
,
•
•
1.
•
•
•
•
min ( a . +1 - a 1.. ) = I min ( a . +1 - a . ) 2 = I mi� (a. - a . ) 2 1. 1. 1. 1. J lSi�n-1 lSiSn- 1 l�i<j �n
Next , =
r. "1'1
•
Def ine b . = a . + 1M ( i - j ) , i = 1 , 2 , 1. J b i = b + IM ( i - 1) . 1 Then , for i > j , we have a . = ( a1.. - a . - 1 ) + (a . - 1 - a . - 2 ) + 1. 1. 1. 1. � 1M ( i - j ) + a . J = b.
.
•
.
•
.
, n , so that
+ (aj+1 - a . ) + a . J J
•
1.
a . > - a 1.. , J that is a 1.. � b . ;;: - al.. , S imilarly we have �
-
1.
1. •
> J •
•
1. •
Thus , we obtain n 2 S = � a . ;;: .1. = 1 1.
(i = 1 , 2 ,
.
.
•
< J
•
•
, n) , and so
n = . �(b 1 1.= 1 ) 2 1 (n-1)n (n+1) (n ; ) + = n (b 1 + 1M M 12 (n1 ) n(n+1) >_ M . 12
SOLUTIONS (78-79)
145
79. Let x l " " ,xn be n real numbers such that ,
Prove that (79 .0)
;$
12
-
1 2n
•
2n Solution : For 1 :;; k :;; n+l we have 1 1 - n :;; 1 - n ' 2n and for n+l :;; k :;; n we have o :;;
-
1n '
so that 2- - 1 - -1 :;; 1 - -1 , n n k
( 79 . 1) Thus , as
n L � = 0 , we have <
k=l
.
-1 2 n 2- - 1 - -1 1 l t k I X -2 k-l n k 1- l 2(
-
n 1 by ) 1 . 79 ( . x. I , t -) I n k= l K
n The inequal ity ( 79 . 0) now follows as L I Xk l = 1 k= l
.
146
SOLUTIONS (80-81)
on QU .
Prove that the sum of two consecutive odd primes is the product of at least three (possibly repeat ed) prime factors . th prime rn1mber , so that P I = 2 Solution : Let pn denote the n For n � 2 , we consider P 2 = 3 , P3 = 5 , Clearly qn l.S even . If has exactly one k prime factor then qn = 2 for some posl.tl.ve l.nteger k , and as 3 qn � 3 + 5 2 we have k � 3 proving the result in this case . has exactly two distinct prime factors , then we have If • • •
,
•
•
•
•
•
=
and an odd prime p . for positive integers k, or l � 2 the result holds . If k = l = 1 then , as we have P n pn <
If
,
and
which is impossible as completes the proof .
are consecutl.ve prl.mes . This •
•
Let f (x) be an integrable function on the closed interval [ rr/2 ,rr ] and suppose that 81 .
(81 .0) Prove that
rr /2
f (x) sin kx dx =
1 :0 k 1 , k =n
a
•
n 1 -
,
•
1 I f (x) I � rr ln 2 on a set of positive measure .
Solut ion : From (81 . 0) we have rr
(81 . 1 )
:fO
n f (x) sin kx dx = 1 . L k=l /2
SOLUTIONS (80-81)
147
Interchanging the order of summation and integration , and using the identity x 1 cos "2 - COB-en + "2) x n , 21T , o < x < L sin kx x k=l 2 S ln l (81 . 1 ) becomes 1T f (x) -
.
12
1
1T Suppose I f (x) I < ] , except on , for a set of measure [ o . 1T l 2 1T-= ln� Then we have 1T I cos I - cos (n + i) x dx , 1 � I f (x) 1 x 2 sl.n "2 12 .
gl.Vl.ng •
•
(81 . 2)
1T
1
1 <-:i 1T .e.n'--.:2 :-
dx . x Sl.n 2 12
•
Now Jordan ' s inequa lity implies that �
1T
� sin
and so , on [1" 1T ] , we have (81 . 3 )
(0 � x � 1T) 2
�
---:1:. ..� .!!: . x x S l.n -2
•
Using (81 . 3) in (81 . 2) we obtain
1T dx 1 - 1 , 1 < ln 2 x 12
•
SOLUTIONS (82-84)
148
which is impossible. Hence I f (x) I � measure . 82 .
(82 .0)
For n = 0 1 2 ,
,
, • • •
sn =
6n+l where an = n+l value .
•
1
1T
on a set of positive .in 2
, let
+
+
Show that
an_2 + . . . +
l im sn exists and determine its n + oo
Solut ion : First we show by mathemat ical induct ion that the sequence } { non-decreasing . 0 1 Note that n is 2 : .. s 3 ="9 n < s = Then we have s Assume that s = 1 < n- 1 n l 2 So ,
,
, • • .
•
-
•
s n .. 3 fan + sn- 1 < 3 /an+l + sn -
Next we show, also by induction, that the sequence Clearly S o - 1 < 2 { sn n = 0 , 1 , 2 , } �s bounded above by 2 Assume that sn-l < 2 Then we have . . •
• •
•
-
•
•
•
sn = 3 /an + s n- l < 3 /6 + 2 = 2 . 3 Thus L ,. l im sn ex�st s . Letting n + oo �n s .. a + s n n n- 1 n + oo 3 we obtain L = 6 + L , so that L " 2 . •
•
83 .
Let f (x) be a non-negative strictly increas ing function on the interval [ a, b ] , where a < b . Let A(x) denote the area below the curve y " f (x) and above the interval [ a , x ] , where a <. x < b so that A(a) a . Let F (x) be a funct ion such that F (a) a and -
..
=
(83 . 0)
(x' - x) f (x) < F(x ' ) - F (x) < (x ' - x) f(x ' )
,
149
SOLUTIONS (82-84)
f or all a � x < x ' � b . Prove that A(x) - F (x) f or a � x � b
•
Solut i on: Clea rl y A(x) satisfies the inequality ( 8 3 . 0) . Assu me that A(x) and F(x) are not identical on [ a,b] . Then the re exists c with a < c � b such that A(c) � F(c) We partiti on the interval [a,c] b y + k(c-a) a � • n ' •
whe re n is a positive intege r and k -
S'P'M"ing f rom k -
to k - n
1
( 83 . 1 )
•
we
<
A(c)
<
<
F(c)
<
0,1 ,2,
.
•
•
, n . Then we have
obtain n c-a f (� ) L n k-l k
•
Simila rl y we have (83 . 2)
F rom
( 83 . 1 )
and
(83 . 2 )
we obtain
IA(c) - F(c) 1 s o that (as A(c)
�
<
« c-a) /n ) ( f (c) - f (a»
F(c) ) n
<
(c-a) f (c) - f (a» Ac -Fc
This is a c ont radict on f or sufficientl y la rg e positive intege rs n . 84. Let a and b be two given positive nu mbe rs with a < b How sh ould the Ullmb e r r be ch osen in the interval [a,b] in orde r t o mlnl.ml.Ze r x M(r) max ? (84 . 0 ) x a::iix�b
•
•
•
•
150
SOLUTIONS (84-85)
Solution : For a � r � b we have
and so M(r)
max
=
(i -
1 , 1
-
�)
•
Thus , for any c in the interval [ a , b ) , we have min M(r) a�r:ilb
max ( E.a - 1 ' 1
=
=
. max ( E.a _ 1 ' 1 _ E.b) ' max ( E.a - 1 , 1 - E.b )
•
m�n
Choos ing c to be the point in [ a , b ) such that c- - 1
=
a
c 1 - b ' -
that is c
=
a+b '
we have m�n max ( E.a - 1 ' 1 a�r:ilc •
-
E.b )
=
m �n a�r�c •
=
and max ( E.a - 1 ' 1 - E.b )
=
m �n c:ilr�b •
so that min M(r) 1 � ( � - 1) a:ilr:ilb and the required r �s (2ab) / (a+b) . =
•
-
=
=
b-a a+b '
c 1 - -b
•
SOLUTIONS (84-85)
151
} be a sequence of positive real Let {an : n = 1 , 2 , and satisfying the condit ion numbers with l im n .... an - an+1 > an+1 - an+2 > O . For any E > 0 , let N be a positive
85 .
•
.
.
'"
'"
integer such that � the inequality
S
2E
•
k+l _ sat isfies Prove that L '" L ( l) � k=1
N k+l I L - L (_l) ak I < E k=1
(85 . 0)
•
n k+l Solution : For n a pos itive integer, we define Sn = L (_I) . � k=1 We have L = Sn + and 00
n-l \" L = Sn-l + (-1) L (an+r+l - a n+r ) . r=1 As a n+r-l - an+r
>
an+r - an+r+l , we have
(85 . 1 ) Since have
and
L
lies between Sn-l and Sn , we
(85 . 2) Taking n = N , where aN < 2E , we obtain I SN - L I < E as required .
SOLUTIONS (86-87)
152
86. Determine all positive continuous functions f (x) defined on the interval [ O , lT ] for which IT
(81j . 0)
o
f (x) cos
nx
dx
=
n (-l) (2n+1) , n
=
0,1,2,3,4
.
So lut ion : We begin with the identities cos 2x
=
cos 3x
=
cos 4x
=
2 2 cos x - 1 , 3 4 cos x - 3 cos x , 2 4 8 cos x - 8 cos x + 1
•
Hence cos 4x + 4 cos 3x + 16 cos 2x + 28 cos x + 23 2 3 4 8 cos x + 16 cos x + 24 cos x + 1 6 cos x + 8 2 2 8(cos x + cos x + 1) , =
=
and so IT
2 2 8 f (x) (cos x + cos x + 1) dx o 9 + 4 (-7) + 16(5) + 28 (-3) + 23 ( 1 ) =
=
0 ,
which is impossible as f (x) is positive on [ O , lT] . Hence there are no pos itive functions f (x) satisfying (86 .0) . 87 . Let P and p I be points on opposite sides of a non circular ellipse E such that the tangents to E through p and respective lY are parallel and such that the tangents and normals p' to E at P and p I determine a rectangle R of maximum area . Determine the equation of E with respect to a rectangular coordin ate system, with origin at the centre of E and whose y-axis is parallel to the longer s ide of R .
SOLUTIONS (86-87)
153
Solution : We choose initially a coordinate system such that the X 2 y2 The po ints equation of E 1.S a2 + b2 - 1 a > b > 0 Q (a cos t b sin t) (0 $ t $ 2iT) and Q ' - (-a cos t , -b sin t) l ie on E and the tangents to E through Q and Q ' are parallel . We treat the case 0 $ t $ iT/2 as the other cases iT/2 $ t $ iT , iT $ t < 3 iT/2 , 3iT/2 $ t $ 2 iT can be handled by appropriate ref lect1.ons . Let the normal s through Q and Q' meet the tangents through Q ' and Q at T and T ' respectively . Our first aim is to choose t so that the area of the rectangle QTQ ' T ' 1.S maX 1.mllm. The s lope -b cos t . of the tangent to E at Q ' 1.S a S1.n . t , and so the equations of the l ines Q' T and QT are respectively b cos t x + a sin t y + ab = 0 2 2 and a sin t x - b cos t y - (a _b ) sin t cos t = O . Thus the lengths I QT I and I Q ' T I are given by •
-
=
,
•
,
-
•
•
•
2ab � ,g, !ib: , I QT I = �';;= == =; := ":' + 2 =: 2 := '=' 2 2 = = = a s in t b c 0 s t The area of the rectangle QTQ ' T '
•
is clearly
b is attained when tan t = a
whose maximllm value case
-
•
In this
2 2 QT = a +b > 1 , Q 'T a 2_b 2 so that R 1.S not a square . Thus and the s lope of the tangent at P through iT/4 clockwise by means of (x,y) (X , y ) , where X 1.. (x-y) ' v'2 tion of the required ellipse is •
..
=
2 2 a b P is the po int -/a2 +b2 ' la2 +b2 1.S -1 . Rotating the axes the orthogonal transformation y =� (x+y) , we find the equa•
SOLUTIONS (88-89)
154
1
.
88. If four distinct po ints l ie in the plane such that any three of them can be covered by a disk of unit radius , prove that all four points may be covered by a disk of unit radius . Solut ion : We first prove the following special case of He l ly ' s 4 2 theorem: If D . (i 1 , , 3 , ) are four disks in the plane such that any three have non-empty intersect ion then all four have non-empty intersection . Choose points W,X, Y , Z in D l'l D{ D3 ' respectively . We consider two cases D l'1 D2r! D4 ' D l' � D3�D4 ' D f D 3' � D4 according as one of the po ints W,X,Y,Z is in or on the (possibly degenerate) triangle formed by the other three points , or not . In the first case suppose that Z is in or on triangle WXY . Then the l ine segments WX ,WY ,XY belong to respectively, so that tr iangle WXY belongs to D l ' and thus belongs to D l . Hence Z is a point of D l D2 n D3 " D4 . In the second case WXYZ is a quadrilateral whose diagonal s intersect at a point C ins ide WXYZ . Without loss of general ity we may suppose that C is the intersect ion of WY and XZ . Now the l ine segments WY and XZ belong to D l D3 and D2 " D4 respectively. Thus C is both in D l n D3 and in D2 n D4 and so in D I ll D2 I) D3 " D4 . To solve the problem let A , B , C ,D be the four given distinct points . Let G be the centre of the unit disk to which A , B , C belong . Clearly the distances AG , BG , CG are all less than or equal to 1 , and so G belongs in the three unit disks UA ,UB ,UC centred at A , B , C respectively . Thus any three of the four disks UA ,UB ,UC , UD have a non-empty intersection , and so by the first result there is a 1.
=
f1
"
'
SOLUTIONS (88-89)
point A, B ,
155
The unit disk centred at
P C
89 .
and
P
contalns
D .
Evaluate the sum 1 2 2 · m - n
Solut ion : For positive integers m and N with N > m , we have N
N 1 1 1 1 A(m, N) = I -;2�2 = - 2m I n-m - n+m n=1 n=1 m -n n;z!m n;z!m =
3 - S (N-m» ) - 2 ' 4m
� ( S ( N + m ) 2
where for r = 1 , 2 , 3 , . . . we have set 1 -k = ln r + c + E(r) , S (r) = k=1 c denoting Euler ' s constant and the error term E (r)
sat isfying
for some absolute cons tant A . Then lim ( S (N+m) - S (N-m» ) N + oo = lim N + oo 0
N+m .en N-m + E(N+m) - E (N-m)
=
and so lim A(m,N) = N + oo
,
•
SOLUTIONS (�92)
156
and thus S = lim
M + oo
..
M
L lim A(m,N) m=l N + oo
3 lim 4
-
-
•
If n is a positive integer which can be expressed in the 2 c , where a , b , c are pos �t �ve �ntegers , prove form n .. 2k can be expressed in the that , for each positive integer k , n 2 2 2 form A + B + C , where A , B , C are positive integers . 90.
•
•
•
Solut ion: We begin by showing that if m = x , y , z are positive integer s , then where X , Y , Z are pos�t�ve �ntegers . Without loss of generality we may choose x � y � z . Then the required X,Y , Z are given by •
•
•
2 - z , Y .. 2xz , Z
..
2yz
•
r Letting 2k .. 2 (2s+1 ) , where r � 1 , s � 0 , we have and applying the above arg1!ment successively we obtain
where X, Y , Z are pos �t �ve �nteger s . •
91 .
•
•
be the group generated by a and b subj ect to the S 3 relations aba " b and b " 1 . Prove that G is abel ian . Let
G
SOLlITIONS (90-92)
Solut ion :
157
It suffices to show that aba .. b
3
b
g�ve s •
-1
-2
b
a
b
and
The relation
COIlDDute .
2 -1 , and so ab .. b a
ab
2
•
•
•
-1
b b
-1
(b
-1
ab ) b
2 -1 (b a ) b
2
-1 -1 b (b a b ) 2
.. b (b
-1
ab )
-1
2 -1 -1 = b (b a ) 2
2
.. b ab
-2 ,
81v�ng •
•
ab Hence , as
92 ,
b
S
4
4
.. b a
•
4
S
ab " b ab .. b (b a) b .. ba
= 1 , we obtain
•
be a sequence of real numbers "" diverge s � a sati s fying 0 < a < 1 for a l l n and such that n n n=l 2 � an converge s . Let f (x) be a funct ion def ined on [ 0 , 1 1 while n=l � f (an) If such that f " (x) ex� s t s and is bounded on [ 0 , 1 1 ne1 also converges . converge s , p rove that {a : n = 1 , 2 , 3 , n
Let
.
.
.
}
""
""
•
•
""
Solution :
Applying the extended mean value theorem to interval [ O , a 1 , there exi s t s n
and
w n
such that
f
on the 0
< wn < an
""
If
�
n=l
f (a ) n
converge s , then we must have
l im n ....
""
f ( a ) .. 0 , and so n
SOLUTIONS (93-94)
158
Next , as I f" (x) 1 � M , 0 x 1 , we have b y cont1.nu1.ty f eO) = 0 2 2 2 co co co a f " (w ) ) a f" (w ) w C a " f . 2 n n n n n n !:! and a that both , so L L I I 2 2 n=l n 2 2 n=l n=l n=1 co a 2 f " (w ) co co n n converges , and converge . Hence f ' (0) L an L f (an ) - L 2 n=l n=l n=l <
•
•
-
•
<
-
00
<
co
Thus an diverges , we must have f (0) 0 I n=l co co a 2 f " Cw ) n n converges . L I f (an ) 1 L 2 n=1 n=l
so as
I
-
•
..
93.
Let a , b , c be real numbers such that the roots of the cubic equat1.on 3 2 x + ax + bx + c = 0 (93 . 0) •
are a l l rea l . Prove that these roots are bounded above by 2 (2 � -3b - a) /3 . So lut ion : Let p , q , r be the three real roots of (93 .0) chosen so 3 2 that p t: q t: r . Then , as p + ap + bp + c = 0 , we have 2 2 c :: (x-p) ( x + (p+a)x + (p +ap+b» . 2 2 The quadratic polynomial x + Cp+a)x + (p +ap+b) has q and r as its two real roots , and hence its discriminant is non-negat ive , that 1.S •
(93 . 1 ) Solving (93 . 1) for p we obtain which comp letes the proof .
159
SOLUTIONS (93-94)
94 .
Let
Z
ment s .
Let
a,b ,c,d
the n1!mber
5
=
{O , 1 , 2, 3 , 4 }
•
f (x) is given by
N - 4 -
R A
Solut ion :
Z
be e l ements of
of distinct solut ions l.n
N
Define
B
=
a + bx + cx
with
s
2
Z
a '"
0
e le-
Prove that
•
of the cubic equatl.on •
s
+ dx
3
0
=
, where
R
deno tes the rank of the matrix
a
b
c
d
b
c
d
a
c
d
a
b
a
b
c
=
•
to be the Vandermonde matrix 1
2
1
1
1
B
5
denote the f inite f ield with
2
2
2
-
1
3
1
4
3
1 2
3 3 ,
2
so tha t f(l)
BA
f (2)
-
a '"
0
1 2
-
f (3) f (4 )
As
-1
, the matrix
3 4
-1 -1
-1
BA
f(1)
-2
1
-2
f (2)
2
f ( 3)
3
f (4 )
4
has
N
-2
-2
1
f(l)
2
f(2)
3
f ( 3)
4
f (4 )
-3 -3 -3 -3
f(l) f ( 2) •
f (3) f (4)
zero rows , so that
rank
BA �
4-N .
Let [f(r. ) 1.
b e the
4-N
-1 r . f (r . ) 1.
1.
-2 r . f (r . ) 1.
non-zero rows of
1.
BA ,
-3 r . f ( r 1. ) 1 1.
where
(i
.
1
�
r
1 , . . . , 4-N)
=
l
<
• • .
< r
4-N
�
4
•
SOLUTIONS (95-96)
160
Clearly
•
•
•
•
•
•
•
•
•
•
•
•
-1 r 1
1 f (r ) 1
-
•
•
•
f (r _ ) 4 N
•
•
-1 r _ 4 N
1
•
•
•
•
•
•
•
•
- ( 3- N ) r 1 •
- ( 3 -N) r _ 4 N
•
0 ,
;<
BA
so that the rank of
i s exactly
4-N .
Finally , s ince
B
inver t ib l e , we have
R that
1S,
N = 4 -
•
95 .
R
= rank
A
= rank
BA
= 4-N ,
•
Prove that co
( 95 . 0)
S
1
I
=
m , n=l ( mn ) (m , n ) = l
2
i s a rat ional numbe r . Solution :
We no tice that 00
1 I 2 r=l r Sett ing
r = dm ,
5
00
2 -
�
1
r , s= l (rs) = dn , so that
2
,n.
-
00
00
L
�
1
d= l r , s= l (rs) GCD ( r , s ) =d
2
•
GCD (m , n ) = 1 , we obtain
1S •
SOLUTIONS (95-96)
rr
161
2 2
6
ex>
ex>
1 = L 4 d-l d
1
L
(mn) m , n=1 GCD (m, n) =1
2 '
that �s •
4
rr 36
96.
-
4
rr S 90
-
S
,
5 2
-
•
Prove that there does not exist a rational function
f (x)
with real coef f i c ien ts such that f
(96 . 0) where
2
;"1
= p (x) ,
is a non-con s tant polynomial with real coeff ic ients .
p (x)
Suppo se there exi s t s a rational func t ion
Solut ion :
non-constant polynomial such that ( 96 . 0) ho lds .
As
there exi s t comp lex numbers with
a. �
;t
b.
J
p (x)
a
(both with real coeffic ient s )
p (x)
is the quot ient of two polynomials ,
(;to)
, a
such that
l
'
• . •
a (x-a ) · · · (x-a ) r l f (x) = ( x-b ) · · · (x-b ) l s
( 96 . 1 ) S ince
f (x)
is a non-constant polynomia l , f (x)
constant nor a po lynomial , and so
and a
f (x)
�
s
1
, a
b ' 1 r
'
•
.
•
, b
s
•
can neither be
•
From ( 96 . 0) and ( 96 . 1 ) , we obtain 2 a ) · · · (x -a x-a ) x-a (x r r s -r 1 l = p (x) . a(x+ O 2 2 (x -b x-b ) · · · (x -b x-b ) 1 s s 1 2
(96 . 2 )
••
s-r < 0 , x+l
2
x -a . x-a . for some � � � 2 so (_0 - a . (-1) - a . = 0 , that �s , 1 + a . - a . � � � � c learly impo s s ible , and thus s � r
If
divides
•
•
•
, =
1 < � :; r , and •
-
0
•
which �s •
SOLUTIONS (97-98)
162
x-c
Now l e t
b e a factor of
( 96 . 3 )
c � -1 � 1 have
C learly must
As the l e f t side of ( 9 6 . 2 ) is a polynomial , we 2 and x-c I x -a . x-a . , for some i , l � i � r , l l
.
r
that i s
c
(96.4)
2
- a . c - a. a Q . l l
From ( 9 6 . 3 ) and ( 9 6 . 4) we obtain 2
c a. = l c+l which is a contradiction .
Hence no such rat ional funct ion
eX l s t s . •
97 .
For
n
a po s l t lve lnteger, set •
•
•
S (n)
n
L
-
.
k= Q
Prove that
For
n
�
2
,
•
(�
n+l
S (n)
Solut ion :
1
-
k
2 n+l L n+l k k=l 2
•
we have n
n+l
2 n+l
2 S (n) - n S (n- l ) =
n
L
kaQ
1
2
n n-l kaQ
1 -"-:n- l k
( )
-
-
f (x)
163
SOLUTIONS (97-98)
n-1 n+1 n 2 2 - n+1 + 2 L k=O (n+l) (�) n-1 n+l n 2 + 2 L n+1 k-O 1 n n+ 2 2 + i" (n+l ) . n+1 n 2n+1 2 = n+1 + 7(� : )"! n+:C:1'
..
•
-
1 n-1 n k
! 2k ! k ! (n-kl) ! (n-k) . n! (n+l ) !
n-l
L k! (n-k-l) ! (n-2k-l)
k-O n-1
L (k ! (n-k) ! - (k+1) ! (n-k-1) l )
k=O
n 2 (0 ." n . - n l O l ) (n+l) 1 = n+1 ' and so n+1 2 k L k ' k=3
n+1 2 S en) n+1 which gives the required result as
2 2 2 = -1 + � 2
•
98.
Let u (x) be a non-trivial solution of the differential equat�on tI u + pu = 0 , •
defined on the interval I = [ 1 ,00) , where p " p (x) �s cont�nuous on I . Prove that u has only finitely many zeros in any interval [a,b] , 1 � a < b (A zero of u (x) is a point z , 1 � z < 00 , with u(z) = 0 ) •
•
•
.
164
SOLUTIONS (98-99)
Let
So lut ion :
denote the set o f zeros of
5
1 � a
[ a. b ) .
and derive a contradiction.
.
u (x)
on the interval
We will assume that
is infinite
5
The Bo lzano-Weiers trass theorem imp l ies
that 5 ' has at least: one accumu 1at ion point . say
c . in [ a , b ) .
Hence ,
there exis t s e ither a decreas ing or increasing sequence of zero s {x : n .. 1 , 2 , 3 n have u ( c ) - O .
}
• • • •
converg1ng to •
As
c .
u
1 S cont1nuous we •
•
Apply ing the mean value theorem to
intervals with end-points
u
on the
x and x (n - 1 . 2 . 3 ) , there n+l n } with eX1s t s a s equence {y : n .. 1 , 2 , 3 ly ing between n x and u ' (y ) - O for n By the con tinand 1,2.3 n n we see that u ' (c) - 0 s ince the y ' s converge to c . uity o f n Now define • • • .
• • • •
•
•
Then
q (c )
=
,
•
•
.
•
0 and q ' (x) .. 2uu ' ( l-p)
s o tha t ( 98 . 1 )
Ipl
where
� K-l
on [ c , b ] .
From (9 8 . 1 ) we deduce that
K (x-c) q (x) � q (c ) e . However
q (x) � 0
[ c , b ) . that is
s o that
u (x) = 0
q ( c) - 0
c � x � b
imp l ies that
•
q (x) = 0
on
on [ c . b ] .
The proof wil l be completed by showing that
u (x) = 0
on [ a , c ] .
We set v ex) - u(a+c-x) and r ex) - p (a+c-x) for
a � x � c .
Then
v
is a so lut ion of the di fferential equation
SOUITIONS (98-99)
165
v" + sati sfying
v (a)
v (x) = 0
that
=
0 , v ' (a) .. O .
u (x) = 0 99 .
u (x) = 0
P . (j .. 0 , I , 2 , J
on [ a , c ] .
on [ a ,b] , for any
• • •
, n- l )
pol.nts on a circle of unit radius •
I pQ I
Solut ion :
u (x) = 0
b
>
a , and so
on [ 1 ,�) , contrary to as sumption .
Let
where
By the above argument we deduce
on [ a , c ] , and thus
This shows that
.. 0
rv
be
n (�2)
equa l ly spaced
Evaluate the sum
•
deno tes the distance between the points Without loss of genera l ity we may take to be the po int
I Pj Pk I
2
..
0
�
l exp ( 2rrj i/n)
j
j
and
<
k
�
n- l , we have
- exp ( 2rrki/n)
1
.. 2 - exp (2rr (k-j ) i/n) - exp (-2rr(k-j ) i /n)
S (n) "
n- l
L
L
j -O k-j+l
.. 2
n-2
L
j "' O
( 2 - exp (2rr(k-j ) i /n) - exp (-2rr (k-j ) i /n»
-
(n- l-j ) - A - A , n n
where A n and
-
A
n
n-2 ..
L
n-l
L
j =O k-j+l
exp (2rr(k-j ) i/n)
denotes the comp lex conj ugate of
A n
•
Now
•
. . •
Izl "
2
and so n-2
Q
( j .. 0 , 1 , 2 ,
on the unit c ircle
exp ( 2rrj i/n)
Then , for
in the complex p l ane .
P
P
, n- I )
1
166
SOLUTIONS (99-100)
A = n
n 2
I
j =O
exp ( 2n i!n ) - exp ( 2n i (n-j ) !n) 1 - exp ( 2ni!n)
(n- l ) e xp ( 2n i !n ) = I - exp ( 2ni /n)
_
(n- 1 ) exp ( 2n i f n) = 1 - exp (2ni !n)
_
I--, �--' -_ 7= / � 1 - exp (2ni n)
n-2 \' L
j =O
exp (- 2 n iJ' !n )
1::,_ -..,; ,,--.-..,...... 1 - exp (-2ni !n)
n exp ( 2'ITi !n) , I - exp (2'ITiln)
= and so
A + A = -n . n n -
Hence we obtain ,
S en) = 2 (n-1 )
2
(n-2) (n- 1 ) - 2 + n 2
gl.vl.ng •
•
S en) = n
100 .
be a
M
Let
the f inite field
Z
invertible? Solut ion :
Let
2
3x3
2 •
matrix with entries chosen a t random from
· {O , I }
M = (a. .) , l.J
•
1
What i s the probab i l i ty that
i , j
�
M
•
l. S
3 , so that
D = det M = where
If
(A
(a
,a
(A (a
11 ll 11
,A ,a
ll I2 12 I2
,A ,a ,A ,a
12 13 13 13
,A
)
13
)
( 0 , 0 , 0) , the number of corresponding trip l e s
such that
) )
'"'
..
D = °
is
8 .
For each triple
(0 , 0 , 0 ) , the number of corresponding triples
with
D
=
°
is
4 .
Hence the number
N
of matrices
SOLUTIONS (99-100)
M
with
D
=
1S
0
•
( 1 00 . 1 ) where A
N = Sn + 4 ( 64-n) = 4n + 256 , n
If
a 1 2 A
=
a
2
•
0 . - a
22
= A1 ll 2
there are
(a 1 , a , a , a 1 , a , a ) 22 2 3 3 3 2 33 2
is the numb er of sextup l e s
= A = A 1 ll 13 2
wi th
167
•
A1
- 0
23 3
trip les
=
there are
O .
8
(a 1 , a , a ) 3 2 33 3
with
A
(a ll
Hence we have n and so
N = 344
=
lX8 + 7X2
•
22 ,
•
The required probab i l i ty is 16 8 5 1 2 - 344 = .:,:.;1� 1 5 2 5 2
•
(a 1 ,a , a ) 32 33 3
triples
For eac h trip l e
0 . 328
•
2l
=
with
,a
A1
22
,a
23
)
�
(0 , 0 , 0) ,
= A1 = 0 . 2 3
ABBREVIATIONS ADM
Arch i v de r Mathema tik
AI Spr nger-Ver ag
I
(
by D . S .
Mi trinovi c ,
AMM
Ame rican Mathemati cal Monthly
BLMS
Bul l e t i n of the London Ma thema t i ca l Soci ety
CF
Convex F igures by I . M . Yag l om and V . G . Bol tyanski i , Ho l t , Rinehart and Winston ( 1 9 6 1 1
CM
Crux Mathema t i corum
CMB
C anadian Mathema t i cal Bul l e t i n
CN
Cou r s e Notes for Mathemat i c s 6 9 . 1 1 2 , Carl eton Universi ty ( 1 9 8 4 1
( f ormerly Eureka l
ETN
by W.
S i erpirtski ,
Warsaw ( 1 9 4 1 GCEA
Oxf ord and Cambr idge Schools Exami na t i o n Board , General Cer t i f i cate Exami na tion , Scho larship Leve l , Mathema t i c s and H igher Ma thema tics
GCEB
Oxford and C ambr idge Scho o l s Examinat ion Boar d , General Cert i f ica te Exami na tion , Scho l a r sh i p Leve l , Mat hema tics f o r Sci ence
HCM
Oxford a n d C ambr idge Schoo l s Exami na t i on Board, Higher Cer t i f i cate Mathemat ics ( G roup 1 1 1 1
IMO
I n terna t i onal Ma thema t i cal Olympiad
JUM
Journa l of Undergraduate Mathema tics
MM
Mathema ti cs Maga z i ne
NMT
Nordi sk Ma tema ti sk Ti dskr i f t
PMA PSM TN
WLP
by W . Rudi n , Publ i caci ons , Secci6 de Ma tematique s , Uni ve rsi ta t Autonoma de Barcelona by G . B . Ma thews , Chel sea N . Y . Wi l l i am Lowel l Putnam Mathema t i cal Compe t i tion 169
REFERENCES P roblem 01:
sugge st ed by Problem 3 ,
IMO ( 1 9 7 0 ) .
02 :
see P r oblem 3 8 2 ,
CM 4 ( 1 97 8 ) , p . 2 5 0 .
03:
see P roblem 3 0 4 ,
CM 4 ( 1 97 8 ) , p . l l .
04 :
see P r oblem 1 3 8 , CM 2 ( 1 97 6 ) , p . 6 8 .
05 :
sugge s ted by Problem 1 6 1 ,
06:
see P roblem 1 9 ,
08 :
sugge s ted by Problem 1 6 2 CM 2 ( 1 9 7 6 ) , p . 1 3 5 .
09:
sugge s ted by Problem 2 0 7 , CM 3 ( 1 97 7 ) , p . 1 0 .
10 :
sugges ted by Problem 5 , GCEA
11 :
see Prob lem 1 , HCM
13 :
see Problem 3 ,
15:
see P roblem 4 , GCEA ( 1 9 5 2 ) , Paper
16:
see P r oblem 1 , GCEA
( 1 9 5 2 ) , Paper
17 :
see Problem 9 ,
GCEA
( 1 9 5 5 ) , Pape r
18 :
see Problem 4 ,
GCEA ( 1 9 5 5 ) , Paper V .
19 :
sugge s ted by Problem 2 , HCM ( 1 9 4 9 ) , Paper 7 .
20 :
ba sed on P roblem 6 , GCEA
21:
see P roblem 2 ( i ) , GCES ( 1 9 5 5 ) , paper V .
23 :
sugge s t ed by P roblem 1 8 , TN,
24:
sugges ted by Problem J- 2 ,
26 :
sugge s t ed by P r oblem A- 3 , WLP ( 1 9 8 0 ) .
27 :
see P r ob l em 3 , HCM
28 :
sugge sted by P r oblem 2 ( i i ) , GCES ( 1 9 5 5 ) ,
29:
sugge s ted by Problem 9 1 6 , CM 10
30 :
see AMM 7 6
31 :
see P roblem 5 2 8 , CM 7
32:
based on Problem 4 ,
33:
see P roblem 3 , HCM
36 :
sugge sted by Problem 1 2 4 , CM 2 ( 1 9 7 6 ) , p . 1 1 9 .
37 :
ba sed on Problem 8 ,
38 :
ba sed on P r oblem 4 , HCM ( 1 9 4 3 ) , paper 5 .
40:
see P roblem 1 0 ( i i ) , GCEA ( 1 9 5 4 ) ,
41:
ba sed on P r oblem 5 , HCM ( 1 9 4 4 ) ,
CM 2
( 1 97 6 ) , p . 1 3 5 .
PMA , p . 1 2 9 .
( 1947 ) ,
( 19 5 4 ) ,
Paper I .
Paper 5 .
IMO ( 1 9 8 3 ) . v.
v.
v.
( 1 9 5 2 ) , Paper V .
( 1947 ) ,
p . 31 8 .
CM 6
( 19 8 0 ) , p . 1 4 5 .
paper 5 . paper V .
( 198 4 ) , p . 5 4 .
( 1 96 9 ) , pp . 1 l 3 0 -1 1 3 l . ( 1 9 8 1 ) , p . 90 .
GCEA ( 1 9 5 3 ) , Pape r VI I . ( 1948 ) ,
paper 7 .
GCEA ( 1 9 5 2 ) ,
171
Paper V . Paper
VI .
paper 5 .
42 :
see P roblem 8 5 3 , CM 9 ( 1 9 8 3 ) , p . 1 7 8 .
43 :
see Problem 3 ( i i ) , HCM ( 1 9 4 9 ) , paper 5 .
44:
see P roblem 7 6 7 ,
45:
see Probl em 5 ,
48 :
sugge sted by problem A- 6 , WLP ( 1 9 7 3 ) .
49 :
ba sed on Problem B-6 , WLP
50:
sugge sted by Problem 4 , HCM ( 1 9 4 9 ) , Pape r 7 .
53 :
based on
54:
see Problem 6 0 2 , eM 8 ( 1 9 8 2 ) , p . 1 7 .
55:
based on ETN, pp . 1 2 4 -1 2 5 .
57 :
see Problem 7 , HCM
58 :
sugge sted by Problem 6 , eM 7
59:
see Problem 5 , HCM
61:
sugge st ed by Problem B-2 , WLP ( 1 9 8 4 ) .
63 :
sugge sted by Problem A- 3 , WLP ( 1 9 78 ) .
64:
ba sed on Problem A- 5 , WLP ( 1 9 8 4 ) .
65 :
sugge st ed by Problem B-1 , WLP ( 1 9 7 6 ) .
66:
sugge sted by Problem 3 2 6 , CM 4 ( 19 7 8 ) , p . 6 6 .
67:
sugge sted by Problem B- 5 , WLP ( 1 9 6 8 ) .
69:
sugge s ted by Problem 3 , eM 7 ( 1 9 8 1 ) , p . 26 8 .
71 :
see AI , p . 2 2 1 .
72 :
for a related result see
73:
see P roblem 4 ( Af te rnoon Session ) , WLP ( 1 9 6 1 ) .
74:
sugge sted by Problem 1 , IMO ( 1 9 7 5 ) .
75:
suggested by Problem B- 2 , WLP ( 1 9 7 1 ) .
76:
for mo re ge neral results see BLMS 1 1 ( 1 9 7 9 ) ,
MM
CM 9 ( 1 9 8 3 ) , p . 2 8 1 .
GCEA ( 1 9 5 5 ) , pape r V .
44
( 19 7 4 ) .
( 1 9 7 1 ) , pp . 9-10 .
( 1 9 48 ) ,
Pape r 5 . ( 19 81 ) , p . 2 3 6 .
( 1 9 4 3 ) , Pape r 7 .
AMM
76
( 19 6 9 ) , p . 1 1 2 5 .
pp . 26 8 -27 2 . 77:
based on ADM 2 5 ( 1 9 7 4 ) , pp . 41 - 4 4 .
78 :
see Probl em E 2 0 3 2 ,
79:
see AI , pp . 3 4 6 - 3 4 7 .
80 :
see P roblem 2 1 , CN , p . A7 2 * 3 /4 .
81:
sugge s ted by Problem A- 6 , WLP ( 1 9 7 2 ) .
82:
for related resul ts see JUM 1 5 ( 1 9 8 3 ) , pp . 4 9 - 5 2 .
83:
based on
84:
based on
85:
ba sed on
86 :
suggested by problem 2 ( Mo rn i ng Sessi on ) , WLP
AMM
AMM
AMM
AMM
7 5 ( 19 6 8 ) , p . 11 2 4 .
73
( 1 96 6 ) , pp . 47 7 - 4 8 3 .
57
( 1 9 5 0 ) , pp . 26 - 2 8 .
69
( 19 6 2 ) , pp .
172
215-21 7 . ( 19 6 4 ) .
BB .
see cr ,
Chapte r 2 .
90 .
ba •• d on PSM ,
91 .
ba sed o n P roblem 83 , eMB 7
( 1 96 4 ) , p . 3 0 6 .
92:
sugge s ted by Problem 1 0 6 0 ,
MM
93:
see Problem 1 0 5 , CMB 9
95 :
see P robl em 1 7 1 , CMB 1 5
96 :
see P robl em 1 7 8 ,
97 :
see NMT 2 9
99 :
sugge s ted by Problem 1 1 0 4 ,
2B
( 1 98 4 ) ,
CMB 1 5
( 1 9 47 ) ,
pp . 7 5 - B O . 5 2 ( 1 97 9 ) , p . 46 .
( 1966 ) , p . 53 2 . ( 1 97 2 ) , p . 3 1 3 . ( 1 97 2 ) , pp . 6 1 4 -61 5 .
pp . 97 - 1 0 3 .
173
MM
53
( 19 8 0 ) , p . 2 4 4 .
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