Download 101 algebra problems from the training of the USA IMO team

ONE HUNDRED AND ONE
ALGEBRA PROBLEMS
FROM THE TRAINING OF
THE USA IMO TEAM
Titu Andreescu
and
Zuming Feng
Contents
Acknowledgments
iii
Abbreviations and Notations
v
Preface
vii
Introduction
ix
1 Introductory Problems
1
2 Advanced Problems
9
3 Solutions to Introductory Problems
19
4 Solutions to Advanced Problems
55
Glossary
113
Further Reading
119
ii
Acknowledgments
Thanks to Tiankai Liu who helped in proof reading and preparing solutions.
Many problems are either inspired by or fixed from mathematical contests
in different countries and from the following journals:
High-School Mathematics, China
Revista Matematicǎ Timişoara, Romania
Kvant, Russia
We did our best to cite all the original sources of the problems in the solution part. We express our deepest appreciation to the original proposers
of the problems.
iii
Abbreviations and
Notations
Abbreviations
AHSME
AIME
AMC10
AMC12
ARML
IMO
USAMO
MOSP
Putnam
St. Petersburg
American High School Mathematics Examination
American Invitational Mathematics Examination
American Mathematics Contest 10
American Mathematics Contest 12, which replaces AHSME
American Regional Mathematics League
International Mathematical Olympiad
United States of America Mathematical Olympiad
Mathematical Olympiad Summer Program
The William Lowell Putnam Mathematical Competition
St. Petersburg (Leningrad) Mathematical Olympiad
v
vi
ABBREVIATIONS and NOTATIONS
Notations for Numerical Sets and Fields
Z
Zn
N
N0
Q
Q+
Q0
Qn
R
R+
R0
Rn
C
the
the
the
the
the
the
the
the
the
the
the
the
the
set
set
set
set
set
set
set
set
set
set
set
set
set
of
of
of
of
of
of
of
of
of
of
of
of
of
integers
integers modulo n
positive integers
nonnegative integers
rational numbers
positive rational numbers
nonnegative rational numbers
n-tuples of rational numbers
real numbers
positive real numbers
nonnegative real numbers
n-tuples of real numbers
complex numbers
Notations for Sets, Logic, and Geometry
⇐⇒
=⇒
|A|
A⊂B
A⊆B
A\B
A∩B
A∪B
a∈A
if and only if
implies
the number of elements in set A
A is a proper subset of B
A is a subset of B
A without B
the intersection of sets A and B
the union of sets A and B
the element a belongs to the set A
Preface
This book contains one hundred highly selected problems used in the training and testing of the USA International Mathematical Olympiad (IMO)
team. It is not a collection of one hundred very difficult, impenetrable
questions. Instead, the book gradually builds students’ algebraic skills
and techniques. This work aims to broaden students’ view of mathematics
and better prepare them for possible participation in various mathematical competitions. It provides in-depth enrichment in important areas of
algebra by reorganizing and enhancing students’ problem-solving tactics
and strategies. The book further stimulates students’ interest for future
study of mathematics.
vii
viii
PREFACE
Introduction
In the United States of America, the selection process leading to participation in the International Mathematical Olympiad (IMO) consists of a
series of national contests called the American Mathematics Contest 10
(AMC 10), the American Mathematics Contest 12 (AMC 12), the American Invitational Mathematics Examination(AIME), and the United States
of America Mathematical Olympiad (USAMO). Participation in the AIME
and the USAMO is by invitation only, based on performance in the preceding exams of the sequence. The Mathematical Olympiad Summer Program
(MOSP) is a four-week intense training of 24-30 very promising students
who have risen to the top on the American Mathematics Competitions.
The six students representing the United States of America in the IMO
are selected on the basis of their USAMO scores and further IMO type
testing that takes place during MOSP. Throughout MOSP, full days of
classes and extensive problem sets give students thorough preparation in
several important areas of mathematics. These topics include combinatorial arguments and identities, generating functions, graph theory, recursive
relations, telescoping sums and products, probability, number theory, polynomials, theory of equations, complex numbers in geometry, algorithmic
proofs, combinatorial and advanced geometry, functional equations and
classical inequalities.
Olympiad-style exams consist of several challenging essay problems. Correct solutions often require deep analysis and careful argument. Olympiad
questions can seem impenetrable to the novice, yet most can be solved with
elementary high school mathematics techniques, cleverly applied.
Here is some advice for students who attempt the problems that follow.
• Take your time! Very few contestants can solve all the given problems.
ix
x
INTRODUCTION
• Try to make connections between problems. A very important theme
of this work is: all important techniques and ideas featured in the
book appear more than once!
• Olympiad problems don’t “crack” immediately. Be patient. Try
different approaches. Experiment with simple cases. In some cases,
working backward from the desired result is helpful.
• Even if you can solve a problem, do read the solutions. They may
contain some ideas that did not occur in your solutions, and they may
discuss strategic and tactical approaches that can be used elsewhere.
The formal solutions are also models of elegant presentation that
you should emulate, but they often obscure the torturous process
of investigation, false starts, inspiration and attention to detail that
led to them. When you read the solutions, try to reconstruct the
thinking that went into them. Ask yourself, “What were the key
ideas?” “How can I apply these ideas further?”
• Go back to the original problem later, and see if you can solve it in
a different way. Many of the problems have multiple solutions, but
not all are outlined here.
• All terms in boldface are defined in the Glossary. Use the glossary
and the reading list to further your mathematical education.
• Meaningful problem solving takes practice. Don’t get discouraged if
you have trouble at first. For additional practice, use the books on
the reading list.
Chapter 1
Introductory Problems
1. Let a, b, and c be real and positive parameters. Solve the equation
√
√
√
√
√
√
a + bx + b + cx + c + ax = b − ax + c − bx + a − cx.
2. Find the general term of the sequence defined by x0 = 3, x1 = 4 and
xn+1 = x2n−1 − nxn
for all n ∈ N.
3. Let x1 , x2 , . . . , xn be a sequence of integers such that
(i) −1 ≤ xi ≤ 2, for i = 1, 2, . . . , n;
(ii) x1 + x2 + · · · + xn = 19;
(iii) x21 + x22 + · · · + x2n = 99.
Determine the minimum and maximum possible values of
x31 + x32 + · · · + x3n .
ax + b
, where a, b, c, and d are
cx + d
nonzero real numbers, has the properties f (19) = 19, f (97) = 97,
d
and f (f (x)) = x, for all values of x, except − . Find the range of
c
f.
4. The function f , defined by f (x) =
1
2
CHAPTER 1. INTRODUCTORY PROBLEMS
5. Prove that
(a − b)2
a+b √
(a − b)2
≤
− ab ≤
8a
2
8b
for all a ≥ b > 0.
6. Several (at least two) nonzero numbers are written on a board. One
may erase any two numbers, say a and b, and then write the numbers
b
a
a + and b − instead. Prove that the set of numbers on the board,
2
2
after any number of the preceding operations, cannot coincide with
the initial set.
7. The polynomial
1 − x + x2 − x3 + · · · + x16 − x17
may be written in the form
a0 + a1 y + a2 y 2 + · · · + a16 y 16 + a17 y 17 ,
where y = x + 1 and ai ’s are constants. Find a2 .
8. Let a, b, and c be distinct nonzero real numbers such that
a+
1
1
1
=b+ =c+ .
b
c
a
Prove that |abc| = 1.
9. Find polynomials f (x),g(x), and h(x), if they exist, such that for all
x,


if x < −1
−1
|f (x)| − |g(x)| + h(x) = 3x + 2
if −1 ≤ x ≤ 0


−2x + 2 if x > 0.
10. Find all real numbers x for which
8x + 27x
7
= .
12x + 18x
6
11. Find the least positive integer m such that
µ ¶ n1
2n
<m
n
for all positive integers n.
3
12. Let a, b, c, d, and e be positive integers such that
abcde = a + b + c + d + e.
Find the maximum possible value of max{a, b, c, d, e}.
13. Evaluate
4
2001
3
+
+ ··· +
.
1! + 2! + 3! 2! + 3! + 4!
1999! + 2000! + 2001!
14. Let x =
of x.
√
a2 + a + 1 −
√
a2 − a + 1, a ∈ R. Find all possible values
15. Find all real numbers x for which
10x + 11x + 12x = 13x + 14x .
16. Let f : N × N → N be a function such that f (1, 1) = 2,
f (m + 1, n) = f (m, n) + m and f (m, n + 1) = f (m, n) − n
for all m, n ∈ N. Find all pairs (p, q) such that f (p, q) = 2001.
17. Let f be a function defined on [0, 1] such that
f (0) = f (1) = 1 and |f (a) − f (b)| < |a − b|,
for all a 6= b in the interval [0, 1]. Prove that
|f (a) − f (b)| <
1
.
2
18. Find all pairs of integers (x, y) such that
x3 + y 3 = (x + y)2 .
19. Let f (x) =
2
for real numbers x. Evaluate
+2
µ
¶
µ
¶
µ
¶
1
2
2000
f
+f
+ ··· + f
.
2001
2001
2001
4x
4
CHAPTER 1. INTRODUCTORY PROBLEMS
20. Prove that for n ≥ 6 the equation
1
1
1
+ 2 + ··· + 2 = 1
x21
x2
xn
has integer solutions.
21. Find all pairs of integers (a, b) such that the polynomial ax17 +bx16 +
1 is divisible by x2 − x − 1.
22. Given a positive integer n, let p(n) be the product of the non-zero
digits of n. (If n has only one digit, then p(n) is equal to that digit.)
Let
S = p(1) + p(2) + · · · + p(999).
What is the largest prime factor of S?
23. Let xn be a sequence of nonzero real numbers such that
xn =
xn−2 xn−1
2xn−2 − xn−1
for n = 3, 4, . . . . Establish necessary and sufficient conditions on x1
and x2 for xn to be an integer for infinitely many values of n.
24. Solve the equation
x3 − 3x =
√
x + 2.
25. For any sequence of real numbers A = {a1 , a2 , a3 , · · · }, define ∆A
to be the sequence {a2 − a1 , a3 − a2 , a4 − a3 , . . . }. Suppose that all
of the terms of the sequence ∆(∆A) are 1, and that a19 = a92 = 0.
Find a1 .
26. Find all real numbers x satisfying the equation
2x + 3x − 4x + 6x − 9x = 1.
27. Prove that
16 <
80
X
1
√ < 17.
k
k=1
28. Determine the number of ordered pairs of integers (m, n) for which
mn ≥ 0 and
m3 + n3 + 99mn = 333 .
5
29. Let a, b, and c be positive real numbers such that a + b + c ≤ 4 and
ab + bc + ca ≥ 4. Prove that at least two of the inequalities
|a − b| ≤ 2,
|b − c| ≤ 2,
|c − a| ≤ 2
are true.
30. Evaluate
n
X
k=0
31. Let 0 < a < 1. Solve
1
.
(n − k)!(n + k)!
x
a
xa = ax
for positive numbers x.
32. What is the coefficient of x2 when
(1 + x)(1 + 2x)(1 + 4x) · · · (1 + 2n x)
is expanded?
33. Let m and n be distinct positive integers. Find the minimum value
of |xm − xn |, where x is a real number in the interval (0, 1).
34. Prove that the polynomial (x − a1 )(x − a2 ) · · · (x − an ) − 1, where
a1 , a2 , · · · , an are distinct integers, cannot be written as the product
of two non-constant polynomials with integer coefficients, i.e., it is
irreducible.
35. Find all ordered pairs of real numbers (x, y) for which:
(1 + x)(1 + x2 )(1 + x4 ) = 1 + y 7
(1 + y)(1 + y 2 )(1 + y 4 ) = 1 + x7 .
36. Solve the equation
2
2(2x − 1)x2 + (2x − 2)x = 2x+1 − 2
for real numbers x.
6
CHAPTER 1. INTRODUCTORY PROBLEMS
37. Let a be an irrational number and let n be a integer greater than 1.
Prove that
³
´ n1 ³
´ n1
p
p
a + a2 − 1
+ a − a2 − 1
is an irrational number.
38. Solve the system of equations
(x1 − x2 + x3 )2
(x2 − x3 + x4 )2
(x3 − x4 + x5 )2
(x4 − x5 + x1 )2
(x5 − x1 + x2 )2
=
=
=
=
=
x2 (x4 + x5 − x2 )
x3 (x5 + x1 − x3 )
x4 (x1 + x2 − x4 )
x5 (x2 + x3 − x5 )
x1 (x3 + x4 − x1 )
for real numbers x1 , x2 , x3 , x4 , x5 .
39. Let x, y, and z be complex numbers such that x + y + z = 2, x2 +
y 2 + z 2 = 3, and xyz = 4. Evaluate
1
1
1
+
+
.
xy + z − 1 yz + x − 1 zx + y − 1
40. Mr. Fat is going to pick three non-zero real numbers and Mr. Taf is
going to arrange the three numbers as the coefficients of a quadratic
equation
x2 + x + = 0.
Mr. Fat wins the game if and only if the resulting equation has two
distinct rational solutions. Who has a winning strategy?
41. Given that the real numbers a, b, c, d, and e satisfy simultaneously
the relations
a + b + c + d + e = 8 and a2 + b2 + c2 + d2 + e2 = 16,
determine the maximum and the minimum value of a.
42. Find the real zeros of the polynomial
Pa (x) = (x2 + 1)(x − 1)2 − ax2 ,
where a is a given real number.
7
43. Prove that
1 3
2n − 1
1
· ···
<√
2 4
2n
3n
for all positive integers n.
44. Let
P (x) = a0 xn + a1 xn−1 + · · · + an
be a nonzero polynomial with integer coefficients such that P (r) =
P (s) = 0 for some integers r and s, with 0 < r < s. Prove that
ak ≤ −s for some k.
45. Let m be a given real number. Find all complex numbers x such
that
µ
¶2 µ
¶2
x
x
+
= m2 + m.
x+1
x−1
46. The sequence given by x0 = a, x1 = b, and
µ
¶
1
1
xn+1 =
xn−1 +
.
2
xn
is periodic. Prove that ab = 1.
47. Let a, b, c, and d be real numbers such that
(a2 + b2 − 1)(c2 + d2 − 1) > (ac + bd − 1)2 .
Prove that
a2 + b2 > 1 and c2 + d2 > 1.
48. Find all complex numbers z such that
(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2.
49. Let x1 , x2 , · · · , xn−1 , be the zeros different from 1 of the polynomial
P (x) = xn − 1, n ≥ 2. Prove that
1
1
1
n−1
+
+ ··· +
=
.
1 − x1
1 − x2
1 − xn−1
2
8
CHAPTER 1. INTRODUCTORY PROBLEMS
50. Let a and b be given real numbers. Solve the system of equations
p
x − y x2 − y 2
p
= a,
1 − x2 + y 2
p
y − x x2 − y 2
p
=b
1 − x2 + y 2
for real numbers x and y.
51. [Russia 2000] Prove that there exist 10 distinct real numbers a1 , a2 , . . . , a10
such that the equation
(x − a1 )(x − a2 ) · · · (x − a10 ) = (x + a1 )(x + a2 ) · · · (x + a10 )
has exactly 5 different real roots.
Chapter 2
Advanced Problems
1. Evaluate
µ
¶ µ
¶ µ
¶
µ
¶
2000
2000
2000
2000
+
+
+ ··· +
.
2
5
8
2000
2. Let x, y, z be positive real numbers such that x4 + y 4 + z 4 = 1.
Determine with proof the minimum value of
x3
y3
z3
+
+
.
8
8
1−x
1−y
1 − z8
3. Find all real solutions to the equation
2x + 3x + 6x = x2 .
4. Let {an }n≥1 be a sequence such that a1 = 2 and
an+1 =
an
1
+
2
an
for all n ∈ N. Find an explicit formula for an .
5. Let x, y, and z be positive real numbers. Prove that
x
y
p
p
+
+
x + (x + y)(x + z) y + (y + z)(y + x)
z
p
≤ 1.
z + (z + x)(z + y)
9
10
CHAPTER 2. ADVANCED PROBLEMS
6. Find, with proof, all nonzero polynomials f (z) such that
f (z 2 ) + f (z)f (z + 1) = 0.
7. Let f : N → N be a function such that f (n+1) > f (n) and f (f (n)) =
3n for all n. Evaluate f (2001).
8. Let F be the set of all polynomials f (x) with integers coefficients
such that f (x) = 1 has at least one integer root. For each integer
k > 1, find mk , the least integer greater than 1 for which there exists
f ∈ F such that the equation f (x) = mk has exactly k distinct
integer roots.
9. Let x1 = 2 and
xn+1 = x2n − xn + 1,
for n ≥ 1. Prove that
1−
1
1
1
1
1
+
+ ··· +
< 1 − 2n .
<
x1
x2
xn
2
22n−1
10. Suppose that f : R+ → R+ is a decreasing function such that for all
x, y ∈ R+ ,
f (x + y) + f (f (x) + f (y)) = f (f (x + f (y)) + f (y + f (x))).
Prove that f (f (x)) = x.
11. Find all functions f : Q → Q such that
f (x + y) + f (x − y) = 2f (x) + 2f (y)
for all x, y ∈ Q.
12. Let
3
4
< a < 1. Prove that the equation
x3 (x + 1) = (x + a)(2x + a)
has four distinct real solutions and find these solutions in explicit
form.
13. Let a, b, and c be positive real numbers such that abc = 1. Prove
that
1
1
1
+
+
≤ 1.
a+b+1 b+c+1 c+a+1
11
14. Find all functions f , defined on the set of ordered pairs of positive
integers, satisfying the following properties:
f (x, x) = x, f (x, y) = f (y, x), (x + y)f (x, y) = yf (x, x + y).
15. Consider n complex numbers zk , such that |zk | ≤ 1, k = 1, 2, . . . , n.
Prove that there exist e1 , e2 , . . . , en ∈ {−1, 1} such that, for any
m ≤ n,
|e1 z1 + e2 z2 + · · · + em zm | ≤ 2.
16. Find a triple of rational numbers (a, b, c) such that
q
√
√
√
3 √
3
3
2 − 1 = 3 a + b + 3 c.
17. Find the minimum of
µ
¶
µ
¶
µ
¶
1
1
1
logx1 x2 −
+ logx2 x3 −
+ · · · + logxn x1 −
4
4
4
where x1 , x2 , . . . , xn are real numbers in the interval ( 14 , 1).
18. Determine x2 + y 2 + z 2 + w2 if
x2
− 12
x2
42 − 12
x2
2
6 − 12
x2
82 − 12
22
y2
− 32
y2
+ 2
4 − 32
y2
+ 2
6 − 32
y2
+ 2
8 − 32
+
22
z2
− 52
z2
+ 2
4 − 52
z2
+ 2
6 − 52
z2
+ 2
8 − 52
+
22
w2
− 72
w2
+ 2
4 − 72
w2
+ 2
6 − 72
w2
+ 2
8 − 72
+
22
= 1,
= 1,
= 1,
= 1.
19. Find all functions f : R → R such that
f (xf (x) + f (y)) = (f (x))2 + y
for all x, y ∈ R.
20. The numbers 1000, 1001, · · · , 2999 have been written on a board.
Each time, one is allowed to erase two numbers, say, a and b, and
1
replace them by the number min(a, b). After 1999 such operations,
2
one obtains exactly one number c on the board. Prove that c < 1.
12
CHAPTER 2. ADVANCED PROBLEMS
21. Let a1 , a2 , . . . , an be real numbers, not all zero. Prove that the equation
√
√
√
1 + a1 x + 1 + a2 x + · · · + 1 + an x = n
has at most one nonzero real root.
22. Let {an } be the sequence of real numbers defined by a1 = t and
an+1 = 4an (1 − an ) for n ≥ 1. For how many distinct values of t do
we have a1998 = 0?
23. (a) Do there exist functions f : R → R and g : R → R such that
f (g(x)) = x2 and g(f (x)) = x3 for all x ∈ R?
(b) Do there exist functions f : R → R and g : R → R such that
f (g(x)) = x2 and g(f (x)) = x4 for all x ∈ R?
24. Let 0 < a1 ≤ a2 · · · ≤ an , 0 < b1 ≤ b2 · · · ≤ bn be real numbers such
n
n
X
X
that
ai ≥
bi . Suppose that there exists 1 ≤ k ≤ n such that
i=1
i=1
bi ≤ ai for 1 ≤ i ≤ k and bi ≥ ai for i > k. Prove that
a1 a2 · · · an ≥ b1 b2 · · · bn .
25. Given eight non-zero real numbers a1 , a2 , · · · , a8 , prove that at least
one of the following six numbers: a1 a3 + a2 a4 , a1 a5 + a2 a6 , a1 a7 +
a2 a8 , a3 a5 + a4 a6 , a3 a7 + a4 a8 , a5 a7 + a6 a8 is non-negative.
26. Let a, b and c be positive real numbers such that abc = 1. Prove
that
bc
ca
ab
+ 5
+ 5
≤ 1.
5
5
5
a + b + ab b + c + bc c + a5 + ca
27. Find all functions f : R → R such that the equality
f (f (x) + y) = f (x2 − y) + 4f (x)y
holds for all pairs of real numbers (x, y).
28. Solve the system of equations:
3x − y
=3
x2 + y 2
x + 3y
y− 2
= 0.
x + y2
x+
13
29. Mr. Fat and Mr. Taf play a game with a polynomial of degree at
least 4:
x2n + x2n−1 + x2n−2 + · · · + x + 1.
They fill in real numbers to empty spaces in turn. If the resulting
polynomial has no real root, Mr. Fat wins; otherwise, Mr. Taf wins.
If Mr. Fat goes first, who has a winning strategy?
30. Find all positive integers k for which the following statement is true:
if F (x) is a polynomial with integer coefficients satisfying the condition
0 ≤ F (c) ≤ k for c = 0, 1, . . . , k + 1,
then F (0) = F (1) = · · · = F (k + 1).
31. The Fibonacci sequence Fn is given by
F1 = F2 = 1, Fn+2 = Fn+1 + Fn
Prove that
F2n =
(n ∈ N).
3
3
F2n+2
+ F2n−2
3
− 2F2n
9
for all n ≥ 2.
32. Find all functions u : R → R for which there exists a strictly monotonic function f : R → R such that
f (x + y) = f (x)u(y) + f (y)
for all x, y ∈ R.
33. Let z1 , z2 , . . . , zn be complex numbers such that
|z1 | + |z2 | + · · · + |zn | = 1.
Prove that there exists a subset S of {z1 , z2 , . . . , zn } such that
¯
¯
¯X ¯ 1
¯
¯
z¯ ≥ .
¯
¯
¯ 6
z∈S
34. A polynomial P (x) of degree n ≥ 5 with integer coefficients and
n distinct integer roots is given. Find all integer roots of P (P (x))
given that 0 is a root of P (x).
14
CHAPTER 2. ADVANCED PROBLEMS
35. Two real sequences x1 , x2 , . . . , and y1 , y2 , . . . , are defined in the
following way:
p
√
x1 = y1 = 3, xn+1 = xn + 1 + x2n ,
and
yn+1 =
y
pn
1 + 1 + yn2
for all n ≥ 1. Prove that 2 < xn yn < 3 for all n > 1.
36. For a polynomial P (x), define the difference of P (x) on the interval
[a, b] ([a, b), (a, b), (a, b]) as P (b) − P (a). Prove that it is possible
to dissect the interval [0, 1] into a finite number of intervals and
color them red and blue alternately such that, for every quadratic
polynomial P (x), the total difference of P (x) on red intervals is equal
to that of P (x) on blue intervals. What about cubic polynomials?
37. Given a cubic equation
x3 + x2 + x +
= 0,
Mr. Fat and Mr. Taf are playing the following game. In one move,
Mr. Fat chooses a real number and Mr. Taf puts it in one of the
empty spaces. After three moves the game is over. Mr. Fat wins the
game if the final equation has three distinct integer roots. Who has
a winning strategy?
38. Let n > 2 be an integer and let f : R2 → R be a function such that
for any regular n-gon A1 A2 . . . An ,
f (A1 ) + f (A2 ) + · · · + f (An ) = 0.
Prove that f is the zero function.
39. Let p be a prime number and let f (x) be a polynomial of degree d
with integer coefficients such that:
(i) f (0) = 0, f (1) = 1;
(ii) for every positive integer n, the remainder upon division of f (n)
by p is either 0 or 1.
Prove that d ≥ p − 1.
15
40. Let n be a given positive integer. Consider the sequence a0 , a1 , · · · , an
1
with a0 = and
2
a2
ak = ak−1 + k−1 ,
n
for k = 1, 2, · · · , n. Prove that
1−
1
< an < 1.
n
41. Let a1 , a2 , . . . , an be nonnegative real numbers, not all zero.
(a) Prove that xn − a1 xn−1 − · · · − an−1 x − an = 0 has precisely
one positive real root R.
Pn
Pn
(b) Let A = j=1 aj and B = j=1 jaj . Prove that AA ≤ RB .
42. Prove that there exists a polynomial P (x, y) with real coefficients
such that P (x, y) ≥ 0 for all real numbers x and y, which cannot be
written as the sum of squares of polynomials with real coefficients.
43. For each positive integer n, show that there exists a positive integer
k such that
k = f (x)(x + 1)2n + g(x)(x2n + 1)
for some polynomials f, g with integer coefficients, and find the smallest such k as a function of n.
44. Let x be a positive real number.
(a) Prove that
∞
X
(n − 1)!
1
= .
(x + 1) · · · (x + n)
x
n=1
(b) Prove that
∞
X
∞
X
(n − 1)!
1
=
.
n(x + 1) . . . (x + n)
(x + k)2
n=1
k=1
45. Let n ≥ 3 be an integer, and let
X ⊆ S = {1, 2, . . . , n3 }
16
CHAPTER 2. ADVANCED PROBLEMS
be a set of 3n2 elements. Prove that one can find nine distinct
numbers ai , bi , ci (i = 1, 2, 3) in X such that the system
a1 x + b1 y + c1 z
a2 x + b2 y + c2 z
a3 x + b3 y + c3 z
=
=
=
0
0
0
has a solution (x0 , y0 , z0 ) in nonzero integers.
46. Let n ≥ 3 be an integer and let x1 , x2 , · · · , xn be positive real numn
X
1
bers. Suppose that
= 1. Prove that
1
+
xj
j=1
√
x1 +
√
x2 + · · · +
√
µ
xn ≥ (n − 1)
1
1
1
√ + √ + ··· + √
x1
x2
xn
¶
.
47. Let x1 , x2 , . . . , xn be distinct real numbers. Define the polynomials
P (x) = (x − x1 )(x − x2 ) · · · (x − xn )
and
µ
1
1
1
+
+ ··· +
x − x1
x − x2
x − xn
be the roots of Q. Show that
Q(x) = P (x)
Let y1 , y2 , . . . , yn−1
¶
.
min |xi − xj | < min |yi − yj |.
i6=j
i6=j
48. Show that for any positive integer n, the polynomial
n
f (x) = (x2 + x)2 + 1
cannot be written as the product of two non-constant polynomials
with integer coefficients.
49. Let f1 , f2 , f3 : R → R be functions such that
a1 f1 + a2 f2 + a3 f3
is monotonic for all a1 , a2 , a3 ∈ R. Prove that there exist c1 , c2 , c3 ∈
R, not all zero, such that
c1 f1 (x) + c2 f2 (x) + c3 f3 (x) = 0
for all x ∈ R.
17
50. Let x1 , x2 , . . . , xn be variables, and let y1 , y2 , . . . , y2n −1 be the sums
of nonempty subsets of xi . Let pk (x1 , . . . , xn ) be the k th elementary symmetric polynomial in the yi (the sum of every product
of k distinct yi ’s). For which k and n is every coefficient of pk (as a
polynomial in x1 , . . . , xn ) even?
For example, if n = 2, then y1 , y2 , y3 are x1 , x2 , x1 + x2 and
p1 = y1 + y2 + y3 = 2x1 + 2x2 ,
p2 = y1 y2 + y2 y3 + y3 y1 = x21 + x22 + 3x1 x2 ,
p3 = y1 y2 y3 = x21 x2 + x1 x22 .
18
CHAPTER 2. ADVANCED PROBLEMS
Chapter 3
Solutions to Introductory
Problems
1. [Romania 1974] Let a, b, and c be real and positive parameters. Solve
the equation
√
√
√
√
√
√
a + bx + b + cx + c + ax = b − ax + c − bx + a − cx.
Solution: It is easy to see that x = 0 is a solution. Since the right
hand side is a decreasing function of x and the left hand side is an
increasing function of x, there is at most one solution. Thus x = 0
is the only solution to the equation.
2. Find the general term of the sequence defined by x0 = 3, x1 = 4 and
xn+1 = x2n−1 − nxn
for all n ∈ N.
Solution: We shall prove by induction that xn = n + 3. The claim
is evident for n = 0, 1.
For k ≥ 1, if xk−1 = k + 2 and xk = k + 3, then
xk+1 = x2k−1 − kxk = (k + 2)2 − k(k + 3) = k + 4,
as desired. This completes the induction.
19
20
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
3. [AHSME 1999] Let x1 , x2 , . . . , xn be a sequence of integers such that
(i) −1 ≤ xi ≤ 2, for i = 1, 2, . . . , n;
(ii) x1 + x2 + · · · + xn = 19;
(iii) x21 + x22 + · · · + x2n = 99.
Determine the minimum and maximum possible values of
x31 + x32 + · · · + x3n .
Solution: Let a, b, and c denote the number of −1’s, 1’s, and 2’s
in the sequence, respectively. We need not consider the zeros. Then
a, b, c are nonnegative integers satisfying
−a + b + 2c = 19 and a + b + 4c = 99.
It follows that a = 40 − c and b = 59 − 3c, where 0 ≤ c ≤ 19 (since
b ≥ 0), so
x31 + x32 + · · · + x3n = −a + b + 8c = 19 + 6c.
When c = 0 (a = 40, b = 59), the lower bound (19) is achieved.
When c = 19 (a = 21, b = 2), the upper bound (133) is achieved.
ax + b
, where a, b, c,
cx + d
and d are nonzero real numbers, has the properties f (19) = 19,
f (97) = 97, and f (f (x)) = x, for all values of x, except −d/c. Find
the range of f .
4. [AIME 1997] The function f , defined by f (x) =
First Solution: For all x, f (f (x)) = x, i.e.,
³
´
a ax+b
cx+d + b
´
³
=x
+
d
c ax+b
cx+d
(a2 + bc)x + b(a + d)
=x
c(a + d)x + bc + d2
c(a + d)x2 + (d2 − a2 )x − b(a + d) = 0,
21
which implies that c(a + d) = 0. Since c 6= 0, we must have a = −d.
The conditions f (19) = 19 and f (97) = 97 lead to the equations
192 c = 2 · 19a + b and 972 c = 2 · 97a + b. Hence (972 − 192 )c = 2(97 −
19)a. It follows that a = 58c, which in turn leads to b = −1843c.
Therefore
58x − 1843
1521
f (x) =
= 58 +
,
x − 58
x − 58
which never has the value 58. Thus the range of f is R − {58}.
Second Solution: The statement implies that f is its own inverse.
ay + b
The inverse may be found by solving the equation x =
for y.
cy + d
dx − b
. The nonzero numbers a, b, c, and d
This yields f −1 (x) =
−cx + a
must therefore be proportional to d, −b, −c, and a, respectively; it
follows that a = −d, and the rest is the same as in the first solution.
5. Prove that
(a − b)2
a+b √
(a − b)2
≤
− ab ≤
8a
2
8b
for all a ≥ b > 0.
First Solution: Note that
Ã√
Ã√
√ !2
√ !2
a+ b
a+ b
√
√
≤1≤
2 a
2 b
√
√
√
√
√
√
√
√
√ 2
√
( a + b)2 ( a − b)2
( a + b)2 ( a − b)2
≤ ( a − b) ≤
4a
4b
√
(a − b)2
a − 2 ab + b
(a − b)2
≤
≤
8a
2
8b
from which the result follows.
Second Solution: Note that
¡ a+b ¢2
− ab
a+b √
(a − b)2
2
√
√ .
− ab = a+b
=
2
ab
2(a + b) + 4 ab
2 +
22
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
Thus the desired inequality is equivalent to
√
4a ≥ a + b + 2 ab ≥ 4b,
which is evident as a ≥ b > 0 (which implies a ≥
√
ab ≥ b).
6. [St. Petersburg 1989] Several (at least two) nonzero numbers are
written on a board. One may erase any two numbers, say a and b,
b
a
and then write the numbers a + and b − instead. Prove that
2
2
the set of numbers on the board, after any number of the preceding
operations, cannot coincide with the initial set.
Solution: Let S be the sum of the squares of the numbers on the
board. Note that S increases in the first operation and does not
decrease in any successive operation, as
µ
¶2 ³
b
a ´2
5
a+
+ b−
= (a2 + b2 ) ≥ a2 + b2
2
2
4
with equality only if a = b = 0. Therefore the answer is no.
7. [AIME 1986] The polynomial
1 − x + x2 − x3 + · · · + x16 − x17
may be written in the form
a0 + a1 y + a2 y 2 + · · · + a16 y 16 + a17 y 17 ,
where y = x + 1 and ai ’s are constants. Find a2 .
First Solution: Let f (x) denote the given expression. Then
xf (x) = x − x2 + x3 − · · · − x18
and
(1 + x)f (x) = 1 − x18 .
Hence
f (x) = f (y − 1) =
1 − (y − 1)18
1 − (y − 1)18
=
.
1 + (y − 1)
y
23
3
Therefore a2 is equal to
¡18the
¢ coefficient of y in the expansion of
18
1 − (y − 1) , i.e., a2 = 3 = 816.
Second Solution: Let f (x) denote the given expression. Then
f (x) = f (y − 1) = 1 − (y − 1) + (y − 1)2 − · · · − (y − 1)17
= 1 + (1 − y) + (1 − y)2 + · · · + (1 − y)17 .
Thus
µ ¶ µ ¶
µ ¶ µ ¶
2
3
17
18
a2 =
+
+ ··· +
=
.
2
2
2
3
Here we used the formula
µ ¶ µ
¶ µ
¶
n
n
n+1
+
=
k
k+1
k+1
and the fact that
¡2¢
2
=
¡3¢
3
= 1.
8. Let a, b, and c be distinct nonzero real numbers such that
a+
1
1
1
=b+ =c+ .
b
c
a
Prove that |abc| = 1.
Solution: From the given conditions it follows that
a−b=
b−c
c−a
a−b
, b−c=
, and c − a =
.
bc
ca
ab
Multiplying the above equations gives (abc)2 = 1, from which the
desired result follows.
9. [Putnam 1999] Find polynomials f (x),g(x),
such that for all x,


−1
|f (x)| − |g(x)| + h(x) = 3x + 2


−2x + 2
and h(x), if they exist,
if x < −1
if −1 ≤ x ≤ 0
if x > 0.
24
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
First Solution: Since x = −1, 0 are the two critical values of the
absolute functions, one can suppose that
F (x)
= a|x + 1| + b|x| + cx + d


(c − a − b)x + d − a if x < −1
=
(a + c − b)x + a + d if −1 ≤ x ≤ 0


(a + b + c)x + a + d if x > 0,
which implies that a = 3/2, b = −5/2, c = −1, and d = 1/2. Hence
f (x) = (3x + 3)/2, g(x) = 5x/2, and h(x) = −x + 12 .
Second Solution: Note that if r(x) and s(x) are any two functions,
then
max(r, s) = (r + s + |r − s|)/2.
Therefore, if F (x) is the given function, we have
F (x)
= max{−3x − 3, 0} − max{5x, 0} + 3x + 2
= (−3x − 3 + |3x + 3|)/2 − (5x + |5x|)/2 + 3x + 2
1
= |(3x + 3)/2| − |5x/2| − x + .
2
10. Find all real numbers x for which
7
8x + 27x
= .
12x + 18x
6
Solution: By setting 2x = a and 3x = b, the equation becomes
a3 + b3
7
=
a2 b + b2 a
6
a2 − ab + b2
7
=
ab
6
6a2 − 13ab + 6b2 = 0
(2a − 3b)(3a − 2b) = 0.
Therefore 2x+1 = 3x+1 or 2x−1 = 3x−1 , which implies that x = −1
and x = 1. It is easy to check that both x = −1 and x = 1 satisfy
the given equation.
25
11. [Romania 1990] Find the least positive integer m such that
µ ¶ n1
2n
<m
n
for all positive integers n.
Solution: Note that
µ ¶ µ ¶ µ ¶
µ ¶
2n
2n
2n
2n
<
+
+ ··· +
= (1 + 1)2n = 4n
n
0
1
2n
and for n = 5,
µ
¶
10
5
= 252 > 35 .
Thus m = 4.
12. Let a, b, c, d, and e be positive integers such that
abcde = a + b + c + d + e.
Find the maximum possible value of max{a, b, c, d, e}.
First Solution: Suppose that a ≤ b ≤ c ≤ d ≤ e. We need to
find the maximum value of e. Since e < a + b + c + d + e ≤ 5e, e <
abcde ≤ 5e or 1 < abcd ≤ 5. Hence (a, b, c, d) = (1, 1, 1, 2), (1, 1, 1, 3),
(1, 1, 1, 4), (1, 1, 2, 2), or (1, 1, 1, 5), which leads to max{e} = 5.
Second Solution: Note that
1
1
1
1
1
+
+
+
+
bcde cdea deab eabc abcd
1
1
1
1 1
3+d+e
≤
+
+
+ + =
.
de de de e d
de
1=
Therefore, de ≤ 3 + d + e or (d − 1)(e − 1) ≤ 4. If d = 1, then
a = b = c = 1 and 4 + e = e, which is impossible. Thus d − 1 ≥ 1
and e−1 ≤ 4 or e ≤ 5. It is easy to see that (1, 1, 1, 2, 5) is a solution.
Therefore max{e} = 5.
26
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
Comment:
The second solution can be used to determine the
maximum value of {x1 , x2 , . . . , xn }, when x1 , x2 , . . . , xn are positive
integers such that
x1 x2 · · · xn = x1 + x2 + · · · + xn .
13. Evaluate
4
2001
3
+
+ ··· +
.
1! + 2! + 3! 2! + 3! + 4!
1999! + 2000! + 2001!
Solution: Note that
k+2
k+2
=
k! + (k + 1)! + (k + 2)!
k![1 + k + 1 + (k + 1)(k + 2)]
=
k+1
1
=
k!(k + 2)
(k + 2)!
=
(k + 2) − 1
1
1
=
−
(k + 2)!
(k + 1)! (k + 2)!
By telescoping sum, the desired value is equal to
1
1
−
.
2 2001!
14. Let x =
of x.
√
a2 + a + 1 −
√
First Solution: Since
x= √
a2 − a + 1, a ∈ R. Find all possible values
p
a2 + |a| + 1 > |a| and
2a
√
,
a2 + a + 1 + a2 − a + 1
|x| < |2a/a| = 2.
Squaring both sides of
p
p
x + a2 − a + 1 = a2 + a + 1
27
yields
2x
p
a2 − a + 1 = 2a − x2 .
Squaring both sides of the above equation gives
4(x2 − 1)a2 = x2 (x2 − 4) or a2 =
x2 (x2 − 4)
.
4(x2 − 1)
Since a2 ≥ 0, we must have
x2 (x2 − 4)(x2 − 1) ≥ 0,
Since |x| < 2, x2 − 4 < 0 which forces x2 − 1 < 0. Therefore,
−1 < x < 1.
Conversely,
√ for every x ∈
√ (−1, 1) there exists a real number a such
that x = a2 + a + 1 − a2 − a + 1.
√
√
Second Solution: Let A = (−1/2, 3/2), B = (1/2, 3/2), and
P = (a, 0). Then P is a point on the x-axis and we are looking for
all possible values of d = P A − P B. By the Triangle inequality,
|P A−P B| < |AB| = 1. And it is clear that all the values −1 < d < 1
are indeed obtainable. In fact, for such a d, a half hyperbola of all
points Q such that QA − QB = d is well defined. (Points A and B
are foci of the hyperbola.) Since line AB is parallel to the x-axis,
this half hyperbola intersects the x- axis, i.e., P is well defined.
15. Find all real numbers x for which
10x + 11x + 12x = 13x + 14x .
Solution: It is easy to check that x = 2 is a solution. We claim
that it is the only one. In fact, dividing by 13x on both sides gives
µ ¶x µ ¶x µ ¶x
µ ¶x
10
11
12
14
+
+
=1+
.
13
13
13
13
The left hand side is a decreasing function of x and the right hand
side is an increasing function of x. Therefore their graphs can have
at most one point of intersection.
28
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
Comment: More generally,
a2 + (a + 1)2 + · · · + (a + k)2
= (a + k + 1)2 + (a + k + 2)2 + · · · + (a + 2k)2
for a = k(2k + 1), k ∈ N.
16. Let f : N × N → N be a function such that f (1, 1) = 2,
f (m + 1, n) = f (m, n) + m and f (m, n + 1) = f (m, n) − n
for all m, n ∈ N. Find all pairs (p, q) such that f (p, q) = 2001.
Solution: We have
f (p, q) =
=
=
=
=
f (p − 1, q) + p − 1
f (p − 2, q) + (p − 2) + (p − 1) = · · ·
p(p − 1)
f (1, q) +
2
p(p − 1)
f (1, q − 1) − (q − 1) +
= ···
2
q(q − 1) p(p − 1)
f (1, 1) −
+
= 2001.
2
2
Therefore
p(p − 1) q(q − 1)
−
= 1999
2
2
(p − q)(p + q − 1) = 2 · 1999.
Note that 1999 is a prime number and that p − q < p + q − 1 for
p, q ∈ N. We have the following two cases:
(a) p − q = 1 and p + q − 1 = 3998. Hence p = 2000 and q = 1999.
(b) p − q = 2 and p + q − 1 = 1999. Hence p = 1001 and q = 999.
Therefore (p, q) = (2000, 1999) or (1001, 999).
29
17. [China 1983] Let f be a function defined on [0, 1] such that
f (0) = f (1) = 1 and |f (a) − f (b)| < |a − b|,
for all a 6= b in the interval [0, 1]. Prove that
|f (a) − f (b)| <
1
.
2
Solution: We consider the following cases.
1
, as desired.
2
(b) |a − b| > 1/2. By symmetry, we may assume that a > b. Then
(a) |a − b| ≤ 1/2. Then |f (a) − f (b)| < |a − b| ≤
|f (a) − f (b)| = |f (a) − f (1) + f (0) − f (b)|
≤ |f (a) − f (1)| + |f (0) − f (b)| < |a − 1| + |0 − b|
1
= 1 − a + b − 0 = 1 − (a − b) < ,
2
as desired.
18. Find all pairs of integers (x, y) such that
x3 + y 3 = (x + y)2 .
Solution: Since x3 +y 3 = (x+y)(x2 −xy +y 2 ), all pairs of integers
(n, −n), n ∈ Z, are solutions.
Suppose that x + y 6= 0. Then the equation becomes
x2 − xy + y 2 = x + y
x2 − (y + 1)x + y 2 − y = 0.
Treated as a quadratic equation in x, we calculate the discriminant
∆ = y 2 + 2y + 1 − 4y 2 + 4y = −3y 2 + 6y + 1. Solving for ∆ ≥ 0 yields
√
√
3−2 3
3+2 3
≤y≤
.
3
3
Thus the possible values for y are 0, 1, and 2, which lead to the
solutions (1, 0), (0, 1), (1, 2), (2, 1), and (2, 2).
Therefore, the integer solutions of the equation are (x, y) = (1, 0),
(0, 1), (1, 2), (2, 1), (2, 2), and (n, −n), for all n ∈ Z.
30
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
19. Let f (x) =
2
for real numbers x. Evaluate
+2
¶
µ
¶
µ
¶
µ
2
2000
1
+f
+ ··· + f
.
f
2001
2001
2001
4x
Solution:
Note that f has a half-turn symmetry about point
(1/2, 1/2). Indeed,
f (1 − x) =
2
41−x
+2
=
2 · 4x
4x
,
= x
x
4+2·4
4 +2
from which it follows that f (x) + f (1 − x) = 1. Thus the desired
sum is equal to 1000.
20. Prove that for n ≥ 6 the equation
1
1
1
+ 2 + ··· + 2 = 1
x21
x2
xn
has integer solutions.
Solution: Note that
1
1
1
1
1
=
+
+
+
,
2
2
2
2
a
(2a)
(2a)
(2a)
(2a)2
from which it follows that if (x1 , x2 , · · · , xn ) = (a1 , a2 , · · · , an ) is an
integer solution to
1
1
1
+ 2 + · · · + 2 = 1,
x21
x2
xn
then
(x1 , x2 , · · · , xn−1 , xn , xn+1 , xn+2 , xn+3 )
= (a1 , a2 , · · · , an−1 , 2an , 2an , 2an , 2an , )
is an integer solution to
1
1
1
+ 2 + ··· + 2
= 1.
x21
x2
xn+3
31
Therefore we can construct the solutions inductively if there are
solutions for n = 6, 7, and 8.
Since x1 = 1 is a solution for n = 1, (2, 2, 2, 2) is a solution for n = 4,
and (2, 2, 2, 4, 4, 4, 4) is a solution for n = 7.
It is easy to check that (2, 2, 2, 3, 3, 6) and (2, 2, 2, 3, 4, 4, 12, 12) are
solutions for n = 6 and n = 8, respectively. This completes the
proof.
21. [AIME 1988] Find all pairs of integers (a, b) such that the polynomial
ax17 + bx16 + 1 is divisible by x2 − x − 1.
First Solution: Let p and q be the roots of x2 − x − 1 = 0. By
the Vieta’s theorem, p + q = 1 and pq = −1. Note that p and q
must also be the roots of ax17 + bx16 + 1 = 0. Thus
ap17 + bp16 = −1 and aq 17 + bq 16 = −1.
Multiplying the first of these equations by q 16 , the second one by
p16 , and using the fact that pq = −1, we find
ap + b = −q 16 and aq + b = −p16 .
Thus
a=
p16 − q 16
= (p8 + q 8 )(p4 + q 4 )(p2 + q 2 )(p + q).
p−q
Since
p + q = 1,
p2 + q 2 = (p + q)2 − 2pq = 1 + 2 = 3,
p4 + q 4 = (p2 + q 2 )2 − 2p2 q 2 = 9 − 2 = 7,
p8 + q 8 = (p4 + q 4 )2 − 2p4 q 4 = 49 − 2 = 47,
it follows that a = 1 · 3 · 7 · 47 = 987.
Likewise, eliminating a in (1) gives
−b =
p17 − q 17
= p16 + p15 q + p14 q 2 + · · · + q 16
p−q
=
(p16 + q 16 ) + pq(p14 + q 14 ) + p2 q 2 (p12 + q 12 )
+ · · · + p7 q 7 (p2 + q 2 ) + p8 q 8
=
(p16 + q 16 ) − (p14 + q 14 ) + · · · − (p2 + q 2 ) + 1.
(1)
32
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
For n ≥ 1, let k2n = p2n + q 2n . Then k2 = 3 and k4 = 7, and
k2n+4
=
=
=
p2n+4 + q 2n+4
(p2n+2 + q 2n+2 )(p2 + q 2 ) − p2 q 2 (p2n + q 2n )
3k2n+2 − k2n
for n ≥ 3. Then k6 = 18, k8 = 47, k10 = 123, k12 = 322, k14 = 843,
k16 = 2207. Hence −b = 2207−843+322−123+47−18+7−3+1 =
1597 or (a, b) = (987, −1597).
Second Solution: The other factor is of degree 15 and we write
(c15 x15 − c14 x14 + · · · + c1 x − c0 )(x2 − x − 1) = ax17 + bx16 + 1.
Comparing coefficients:
x0 :
1
x :
2
x :
c0 = 1,
c0 − c1 = 0, c1 = 1
− c0 − c1 + c2 = 0, c2 = 2,
and for 3 ≤ k ≤ 15,
xk :
− ck−2 − ck−1 + ck = 0.
It follows that for k ≤ 15, ck = Fk+1 (the Fibonacci number).
Thus a = c15 = F16 = 987 and b = −c14 − c15 = −F17 = −1597 or
(a, b) = (987, −1597).
Comment: Combining the two methods, we obtain some interesting facts about sequences k2n and F2n−1 . Since
3F2n+3 − F2n+5 = 2F2n+3 − F2n+4 = F2n+3 − F2n+2 = F2n+1 ,
it follows that F2n−1 and k2n satisfy the same recursive relation. It
is easy to check that k2 = F1 + F3 and k4 = F3 + F5 . Therefore
k2n = F2n−1 + F2n+1 and
F2n+1 = k2n − k2n−2 + k2n−4 − · · · + (−1)n−1 k2 + (−1)n .
33
22. [AIME 1994] Given a positive integer n, let p(n) be the product of
the non-zero digits of n. (If n has only one digit, then p(n) is equal
to that digit.) Let
S = p(1) + p(2) + · · · + p(999).
What is the largest prime factor of S?
Solution:
Consider each positive integer less than 1000 to be
a three-digit number by prefixing 0’s to numbers with fewer than
three digits. The sum of the products of the digits of all such positive
numbers is
(0 · 0 · 0 + 0 · 0 · 1 + · · · + 9 · 9 · 9) − 0 · 0 · 0
= (0 + 1 + · · · + 9)3 − 0.
However, p(n) is the product of non-zero digits of n. The sum of
these products can be found by replacing 0 by 1 in the above expression, since ignoring 0’s is equivalent to thinking of them as 1’s in the
products. (Note that the final 0 in the above expression becomes a
1 and compensates for the contribution of 000 after it is changed to
111.) Hence
S = 463 − 1 = (46 − 1)(462 + 46 + 1) = 33 · 5 · 7 · 103,
and the largest prime factor is 103.
23. [Putnam 1979] Let xn be a sequence of nonzero real numbers such
that
xn−2 xn−1
xn =
2xn−2 − xn−1
for n = 3, 4, . . . . Establish necessary and sufficient conditions on x1
and x2 for xn to be an integer for infinitely many values of n.
First Solution: We have
1
2xn−2 − xn−1
2
1
=
=
−
.
xn
xn−2 xn−1
xn−1
xn−2
Let yn = 1/xn . Then yn − yn−1 = yn−1 − yn−2 , i.e., yn is an
arithmetic sequence. If xn is a nonzero integer when n is in an
34
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
infinite set S, the yn ’s for n ∈ S satisfy −1 ≤ yn ≤ 1. Since an
arithmetic sequence is unbounded unless the common difference is
0, yn − yn−1 = 0 for all n, which in turn implies that x1 = x2 = m,
a nonzero integer. Clearly, this condition is also sufficient.
Second Solution: An easy induction shows that
x1 x2
x1 x2
xn =
=
,
(n − 1)x1 − (n − 2)x2
(x1 − x2 )n + (2x2 − x1 )
for n = 3, 4, . . . . In this form we see that xn will be an integer for
infinitely many values of n if and only if x1 = x2 = m for some
nonzero integer m.
24. Solve the equation
x3 − 3x =
√
x + 2.
First Solution: It is clear that x ≥ −2. We consider the following
cases.
(a) −2 ≤ x ≤ 2. Setting x = 2 cos a, 0 ≤ a ≤ π, the equation
becomes
p
8 cos3 a − 6 cos a = 2(cos a + 1).
or
r
a
4 cos2 ,
2
a
from which it follows that cos 3a = cos 2 . Then 3a − a2 = 2mπ,
m ∈ Z, or 3a + a2 = 2nπ, n ∈ Z. Since 0 ≤ a ≤ π, the solution
in this case are
4π
4π
x = 2 cos 0 = 2, x = 2 cos
, and x = 2 cos
.
5
7
2 cos 3a =
(b) x > 2. Then x3 − 4x = x(x2 − 4) > 0 and
x2 − x − 2 = (x − 2)(x + 1) > 0
or
x>
It follows that
√
x + 2.
x3 − 3x > x >
√
x + 2.
Hence there are no solutions in this case.
35
Therefore, x = 2, x = 2 cos 4π/5, and x = 2 cos 4π/7.
Second Solution:
that
For x > 2, there is a real number t > 1 such
x = t2 +
1
.
t2
The equation becomes
µ
¶3
µ
¶ r
1
1
1
2
2
t + 2
− 3 t + 2 = t2 + 2 + 2
t
t
t
1
1
t6 + 6 = t +
t
t
7
5
(t − 1)(t − 1) = 0,
which has no solutions for t > 1. Hence there are no solutions for
x > 2.
For −2 ≤ x ≤ 2, please see the first solution.
25. [AIME 1992] For any sequence of real numbers A = {a1 , a2 , a3 , · · · },
define ∆A to be the sequence {a2 −a1 , a3 −a2 , a4 −a3 , . . . }. Suppose
that all of the terms of the sequence ∆(∆A) are 1, and that a19 =
a92 = 0. Find a1 .
Solution:
Then
Suppose that the first term of the sequence ∆A is d.
∆A = {d, d + 1, d + 2, . . . }
with the n
th
term given by d + (n − 1). Hence
A = {a1 , a1 + d, a1 + d + (d + 1), a1 + d + (d + 1) + (d + 2), . . . }
with the nth term given by
1
an = a1 + (n − 1)d + (n − 1)(n − 2).
2
This shows that an is a quadratic polynomial in n with leading coefficient 1/2. Since a19 = a92 = 0, we must have
an =
1
(n − 19)(n − 92),
2
so a1 = (1 − 19)(1 − 92)/2 = 819.
36
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
26. Find all real numbers x satisfying the equation
2x + 3x − 4x + 6x − 9x = 1.
Solution: Setting 2x = a and 3x = b, the equation becomes
1 + a2 + b2 − a − b − ab = 0.
Multiplying both sides of the last equation by 2 and completing the
squares gives
(1 − a)2 + (a − b)2 + (b − 1)2 = 0.
Therefore 1 = 2x = 3x , and x = 0 is the only solution.
27. [China 1992] Prove that
80
X
1
√ < 17.
16 <
k
k=1
Solution: Note that
³√
√ ´
2
k+1− k = √
Therefore
2
1
√ <√ .
k+1+ k
k
80
80 ³
X
X
√ ´
√
1
√ >2
k + 1 − k = 16,
k
k=1
k=1
which proves the lower bound.
On the other hand,
³√
´
√
2
k− k−1 = √
k+
2
√
1
>√ .
k−1
k
Therefore
80
80 ³√
´
X
X
√
√
1
√ <1+2
k − k − 1 = 2 80 − 1 < 17,
k
k=1
k=2
which proves the upper bound. Our proof is complete.
37
28. [AHSME 1999] Determine the number of ordered pairs of integers
(m, n) for which mn ≥ 0 and
m3 + n3 + 99mn = 333 .
Solution:
Note that (m + n)3 = m3 + n3 + 3mn(m + n). If
m + n = 33, then
333 = (m + n)3 = m3 + n3 + 3mn(m + n) = m3 + n3 + 99mn.
Hence m + n − 33 is a factor of m3 + n3 + 99mn − 333 . We have
m3 + n3 + 99mn − 333
= (m + n − 33)(m2 + n2 − mn + 33m + 33n + 332 )
1
= (m + n − 33)[(m − n)2 + (m + 33)2 + (n + 33)2 ].
2
Hence there are 35 solutions altogether: (0, 33), (1, 32), · · · , (33, 0),
and (−33, −33).
Comment: More generally, we have
a3 + b3 + c3 − 3abc
1
= (a + b + c)[(a − b)2 + (b − c)2 + (c − a)2 ].
2
29. Let a, b, and c be positive real numbers such that a + b + c ≤ 4 and
ab + bc + ca ≥ 4. Prove that at least two of the inequalities
|a − b| ≤ 2,
|b − c| ≤ 2,
|c − a| ≤ 2
are true.
Solution: We have
(a + b + c)2 ≤ 16
a2 + b2 + c2 + 2(ab + bc + ca) ≤ 16
a2 + b2 + c2 ≤ 8
2
2
a + b + c2 − (ab + bc + ca) ≤ 4
(a − b)2 + (b − c)2 + (c − a)2 ≤ 8,
and the desired result follows.
38
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
30. Evaluate
n
X
k=0
1
.
(n − k)!(n + k)!
Solution: Let Sn denote the desired sum. Then
n
Sn
=
=
=
=
1 X
(2n)!
(2n)!
(n − k)!(n + k)!
k=0
µ
¶
¶
n
n µ
2n
1 X 2n
1 X
=
n−k
k
(2n)!
(2n)!
k=0
k=0
" 2n µ ¶ µ ¶#
1
1 X 2n
2n
·
+
(2n)! 2
k
n
k=0
·
µ ¶¸
1
1 2n
2n
22n−1
1
·
2 +
=
+
.
(2n)! 2
n
(2n)!
2(n!)2
31. [Romania 1983] Let 0 < a < 1. Solve
x
xa = ax
a
for positive numbers x.
Solution: Taking loga yields
ax loga x = xa .
Consider functions from R+ → R,
f (x) = ax ,
g(x) = loga x,
h(x) = xa .
Then both f and g are decreasing and h is increasing. It follows that
f (x)g(x) = h(x) has unique solution x = a.
32. What is the coefficient of x2 when
(1 + x)(1 + 2x)(1 + 4x) · · · (1 + 2n x)
is expanded?
39
Solution: Let
fn (x) = an,0 + an,1 x + · · · + an,n xn = (1 + x)(1 + 2x) · · · (1 + 2n x).
It is easy to see that an,0 = 1 and
an,1 = 1 + 2 + · · · + 2n = 2n+1 − 1.
Since
fn (x) =
=
=
fn−1 (x)(1 + 2n x)
¡
¢
1 + (2n − 1)x + an−1,2 x2 + · · · (1 + 2n x)
¢
¡
¢
¡
1 + 2n+1 − 1 x + an−1,2 + 22n − 2n x2 + · · · ,
we have
an,2
=
an−1,2 + 22n − 2n
=
=
an−2,2 + 22n−2 − 2n−1 + 22n − 2n = · · ·
¡
¢ ¡
¢
a1,2 + 24 + 26 + · · · + 22n − 22 + 23 + · · · + 2n
=
2+
=
24 (22n−2 − 1)
− 4(2n−1 − 1)
3
¡ n+1
¢¡
¢
2
− 1 2n+1 − 2
22n+2 − 3 · 2n+1 + 2
=
.
3
3
33. Let m and n be distinct positive integers. Find the minimum value
of |xm − xn |, where x is a real number in the interval (0, 1).
Solution: By symmetry, we can assume that m > n. Let y =
xm−n . Since 0 < x < 1, xm < xn and 0 < y < 1. Thus
¡
¢ 1
|xm − xn | = xn − xm = xn (1 − xm−n ) = y n (1 − y)m−n m−n .
Applying the AM-GM inequality yields
¶n µ
¶n
µ
(m − n)y
n
(1 − y)m−n
y n (1 − y)m−n =
m−n
n
!n+m−n
µ
¶n Ã
n · (m−n)y
+ (m − n)(1 − y)
n
n
≤
m−n
n+m−n
=
nn (m − n)m−n
.
mm
40
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
Therefore
µ
m
n
|x − x | ≤
nn (m − n)m−n
mm
1
¶ m−n
µ
= (m − n)
nn
mm
1
¶ m−n
.
Equality holds if and only if
(m − n)y
=1−y
n
or
x=
1
³ n ´ m−n
m
.
Comment: For m = n + 1, we have
xn − xn+1 ≤
nn
(n + 1)n+1
for real numbers 0 < x < 1. Equality holds if and only if x =
n/(n + 1).
34. Prove that the polynomial (x − a1 )(x − a2 ) · · · (x − an ) − 1, where
a1 , a2 , · · · , an are distinct integers, cannot be written as the product
of two non-constant polynomials with integer coefficients, i.e., it is
irreducible.
Solution:
For the sake of contradiction, suppose that f (x) =
(x − a1 )(x − a2 ) · · · (x − an ) − 1 is irreducible. Let f (x) = p(x)q(x)
such that p(x) and q(x) are two polynomials with integral coefficients
having degree less than n. Then g(x) = p(x) + q(x) is a polynomial
with integral coefficients having degree less than n.
Since p(ai )q(ai ) = f (ai ) = −1 and both p(ai ) and q(ai ) are integers,
|p(ai )| = |q(ai )| = 1 and p(ai ) + q(ai ) = 0. Thus g(x) has at least
n roots. But deg g < n, so g(x) = 0. Then p(x) = −q(x) and
f (x) = −p(x)2 , which implies that the leading coefficient of f (x)
must be a negative integer, which is impossible, since the leading
coefficient of f (x) is 1.
35. Find all ordered pairs of real numbers (x, y) for which:
(1 + x)(1 + x2 )(1 + x4 ) = 1 + y 7
(1 + y)(1 + y 2 )(1 + y 4 ) = 1 + x7 .
41
Solution: We consider the following cases.
(a) xy = 0. Then it is clear that x = y = 0 and (x, y) = (0, 0) is a
solution.
(b) xy < 0. By the symmetry, we can assume that x > 0 > y.
Then (1 + x)(1 + x2 )(1 + x4 ) > 1 and 1 + y 7 < 1. There are no
solutions in this case.
(c) x, y > 0 and x 6= y. By the symmetry, we can assume that
x > y > 0. Then
(1 + x)(1 + x2 )(1 + x4 ) > 1 + x7 > 1 + y 7 ,
showing that there are no solutions in this case.
(d) x, y < 0 and x 6= y. By the symmetry, we can assume that
x < y < 0. Multiplying by 1 − x and 1 − y the first and the
second equation, respectively, the system now reads
1 − x8 = (1 + y 7 )(1 − x) = 1 − x + y 7 − xy 7
1 − y 8 = (1 + x7 )(1 − y) = 1 − y + x7 − x7 y.
Subtracting the first equation from the second yields
x8 − y 8 = (x − y) + (x7 − y 7 ) − xy(x6 − y 6 ).
(1)
Since x < y < 0, x8 − y 8 > 0, x − y < 0, x7 − y 7 < 0, −xy < 0,
and x6 − y 6 > 0. Therefore, the left-hand side of (1) is positive
while the right-hand side of (1) is negative. Thus there are no
solutions in this case.
(e) x = y. Then solving
1 − x8 = 1 − x + y 7 − xy 7 = 1 − x + x7 − x8
leads to x = 0, 1, −1, which implies that (x, y) = (0, 0) or
(−1, −1).
Therefore, (x, y) = (0, 0) and (−1, −1) are the only solutions to the
system.
42
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
36. Solve the equation
2
2(2x − 1)x2 + (2x − 2)x = 2x+1 − 2
for real numbers x.
Solution: Rearranging terms by powers of 2 yields
2
2x x + 2x+1 (x2 − 1) − 2(x2 + x − 1) = 0.
(1)
Setting y = x2 − 1 and dividing by 2 on the both sides, (1) becomes
2y x + 2x y − (x + y) = 0
or
x(2y − 1) + y(2x − 1) = 0.
(2)
x
Since f (x) = 2 − 1 and x always have the same sign,
x(2y − 1) · y(2x − 1) > 0.
Hence if the terms on the left-hand side of (2) are nonzero, they
must have the same sign, which in turn implies that their sum is not
equal to 0. Therefore (2) is true if and only if x = 0 or y = 0, which
leads to solutions x = −1, 0, and 1.
37. Let a be an irrational number and let n be a integer greater than 1.
Prove that
³
´ n1 ³
´ n1
p
p
+ a − a2 − 1
a + a2 − 1
is an irrational number.
Solution: Let
³
´ n1 ³
´ n1
p
p
N = a + a2 − 1
+ a − a2 − 1 ,
and let
b = (a +
p
1
a2 − 1) n .
Then N = b + 1/b. For the sake of contradiction, assume that N is
rational. Then by using the identity
µ
¶µ
¶ µ
¶
1
1
1
1
m+1
m
m−1
b
+ m+1 = b +
b + m − b
+ m−1
b
b
b
b
43
repeatedly for m = 1, 2, . . . , we obtain that bm + 1/bm is rational for
all m ∈ N . In particular,
bn +
p
p
1
= a + a2 − 1 + a − a2 − 1 = 2a
n
b
is rational, in contradiction with the hypothesis. Therefore our assumption is wrong and N is irrational.
38. Solve the system of equations
(x1 − x2 + x3 )2
(x2 − x3 + x4 )2
(x3 − x4 + x5 )2
(x4 − x5 + x1 )2
(x5 − x1 + x2 )2
=
=
=
=
=
x2 (x4 + x5 − x2 )
x3 (x5 + x1 − x3 )
x4 (x1 + x2 − x4 )
x5 (x2 + x3 − x5 )
x1 (x3 + x4 − x1 )
for real numbers x1 , x2 , x3 , x4 , x5 .
Solution: Let xk+5 = xk . Adding the five equations gives
5
X
(3x2k − 4xk xk+1 + 2xk xk+2 ) =
k=1
5
X
(−x2k + 2xk xk+2 ).
k=1
It follows that
5
X
(x2k − xk xk+1 ) = 0.
k=1
Multiplying both sides by 2 and completing the squares yields
5
X
(xk − xk+1 )2 = 0,
k=1
from which x1 = x2 = x3 = x4 = x5 . Therefore the solutions to the
system are
(x1 , x2 , x3 , x4 , x5 ) = (a, a, a, a, a)
for a ∈ R.
44
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
39. Let x, y, and z be complex numbers such that x + y + z = 2, x2 +
y 2 + z 2 = 3, and xyz = 4. Evaluate
1
1
1
+
+
.
xy + z − 1 yz + x − 1 zx + y − 1
Solution: Let S be the desired value. Note that
xy + z − 1 = xy + 1 − x − y = (x − 1)(y − 1).
Likewise, yz + x − 1 = (y − 1)(x − 1) and zx + y − 1 = (z − 1)(x − 1).
Hence
S
=
=
=
=
1
1
1
+
+
(x − 1)(y − 1) (y − 1)(z − 1) (z − 1)(x − 1)
x+y+z−3
−1
=
(x − 1)(y − 1)(z − 1)
(x − 1)(y − 1)(z − 1)
−1
xyz − (xy + yz + zx) + x + y + z − 1
−1
.
5 − (xy + yz + zx)
But
2(xy + yz + zx) = (x + y + z)2 − (x2 + y 2 + z 2 ) = 1.
Therefore S = −2/9.
40. [USSR 1990] Mr. Fat is going to pick three non-zero real numbers
and Mr. Taf is going to arrange the three numbers as the coefficients
of a quadratic equation
x2 + x +
= 0.
Mr. Fat wins the game if and only if the resulting equation has two
distinct rational solutions. Who has a winning strategy?
Solution: Mr. Fat has the winning strategy. A set of three distinct
rational nonzero numbers a, b, and c, such that a + b + c = 0, will do
45
the trick. Let A, B, and C be any arrangement of a, b, and c, and
let f (x) = Ax2 + Bx + C. Then
f (1) = A + B + C = a + b + c = 0,
which implies that 1 is a solution. Since the product of the two
solutions is C/A, the other solution is C/A, and it is different from
1.
41. [USAMO 1978] Given that the real numbers a, b, c, d, and e satisfy
simultaneously the relations
a + b + c + d + e = 8 and a2 + b2 + c2 + d2 + e2 = 16,
determine the maximum and the minimum value of a.
First Solution: Since the total of b, c, d, and e is 8 − a, their
average is x = (8 − a)/4. Let
b = x + b1 ,
c = x + c1 ,
d = x + d1 ,
e = x + e1 .
Then b1 + c1 + d1 + e1 = 0 and
16 = a2 + 4x2 + b21 + c21 + d21 + e21 ≥ a2 + 4x2 = a2 +
or
(8 − a)2
(1)
4
0 ≥ 5a2 − 16a = a(5a − 16).
Therefore 0 ≤ a ≤ 16/5, where a = 0 if and only if b = c = d = e = 2
and a = 16/5 if and only if b = c = d = e = 6/5.
Second Solution:
from
By the RMS-AM inequality, (1) follows
(b + c + d + e)2
(8 − a)2
=
,
4
4
and the rest of the solution is the same.
b2 + c2 + d2 + e2 ≥
42. Find the real zeros of the polynomial
Pa (x) = (x2 + 1)(x − 1)2 − ax2 ,
where a is a given real number.
46
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
Solution: We have
(x2 + 1)(x2 − 2x + 1) − ax2 = 0.
Dividing by x2 yields
µ
¶µ
¶
1
1
x+
x−2+
− a = 0.
x
x
By setting y = x + 1/x, the last equation becomes
y 2 − 2y − a = 0.
It follows that
x+
√
1
= 1 ± 1 + a,
x
which in turn implies that, if a ≥ 0, then the polynomial Pa (x) has
the real zeros
p
√
√
1+ 1+a± a+2 1+a−2
x1,2 =
.
2
In addition, if a ≥ 8, then Pa (x) also has the real zeros
p
√
√
1− 1+a± a−2 1+a−2
x3,4 =
.
2
43. Prove that
1 3
2n − 1
1
· ···
<√
2 4
2n
3n
for all positive integers n.
Solution: We prove a stronger statement:
1 3
2n − 1
1
· ···
≤√
.
2 4
2n
3n + 1
We use induction.
For n = 1, the result is evident.
47
Suppose the statement is true for some positive integer k, i.e.,
1 3
2k − 1
1
.
· ···
<√
2 4
2k
3k + 1
Then
1 3
2k − 1 2k + 1
1
2k + 1
· ···
·
<√
·
.
2 4
2k
2k + 2
3k + 1 2k + 2
In order for the induction step to pass it suffices to prove that
√
2k + 1
1
1
·
<√
.
2k
+
2
3k + 1
3k + 4
This reduces to
µ
2k + 1
2k + 2
¶2
(4k 2 + 4k + 1)(3k + 4)
0
<
3k + 1
3k + 4
<
(4k 2 + 8k + 4)(3k + 1)
< k,
which is evident. Our proof is complete.
Comment: By
√ using Stirling numbers, the upper bound can be
improved to 1/ πn for sufficiently large n.
44. [USAMO Proposal, Gerald Heuer] Let
P (x) = a0 xn + a1 xn−1 + · · · + an
be a nonzero polynomial with integer coefficients such that P (r) =
P (s) = 0 for some integers r and s, with 0 < r < s. Prove that
ak ≤ −s for some k.
Solution: Write P (x) = (x − s)xc Q(x) and
Q(x) = b0 xm + b1 xm−1 + · · · + bm ,
where bm 6= 0. Since Q has a positive root, by Descartes’ rule of
signs, either there must exist some k for which bk > 0 ≥ bk+1 , or
bm > 0.
48
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
If there exists a k for which bk > 0 ≥ bk+1 , then
ak = −sbk + bk+1 ≤ −s.
If bm > 0, then am = −sbm ≤ −s.
In either case, there is a k such that ak ≤ −s, as desired.
45. Let m be a given real number. Find all complex numbers x such
that
¶2 µ
¶2
µ
x
x
+
= m2 + m.
x+1
x−1
Solution: Completing the square gives
µ
¶2
x
x
2x2
+
= 2
+ m2 + m
x+1 x−1
x −1
µ
¶2
2x2
2x2
= 2
+ m2 + m.
2
x −1
x −1
Setting y = 2x2 /(x2 − 1), the above equation becomes
y 2 − y − (m2 + m) = 0,
(y − m − 1)(y + m) = 0.
Thus
2x2
2x2
=
−m
or
= m + 1,
x2 − 1
x2 − 1
which leads to solutions
r
r
m
m+1
x=±
if m 6= −2 and x = ±
if m 6= 1.
m+2
m−1
46. The sequence given by x0 = a, x1 = b, and
µ
¶
1
1
xn+1 =
xn−1 +
.
2
xn
is periodic. Prove that ab = 1.
49
Solution: Multiplying by 2xn on both sides of the given recursive
relation yields
2xn xn+1 = xn−1 xn + 1
or
2(xn xn+1 − 1) = xn−1 xn − 1.
Let yn = xn−1 xn − 1 for n ∈ N. Since yn+1 = yn /2, {yn } is a
geometric sequence. If xn is periodic, then so is yn , which implies
that yn = 0 for all n ∈ N. Therefore
ab = x0 x1 = y1 + 1 = 1.
47. Let a, b, c, and d be real numbers such that
(a2 + b2 − 1)(c2 + d2 − 1) > (ac + bd − 1)2 .
Prove that
a2 + b2 > 1 and c2 + d2 > 1.
Solution: For the sake of the contradiction, suppose that one of
a2 + b2 or c2 + d2 is less than or equal to 1. Since (ac + bd − 1)2 ≥ 0,
a2 + b2 − 1 and c2 + d2 − 1 must have the same sign. Thus both
a2 + b2 and c2 + d2 are less than 1. Let
x = 1 − a2 − b2 and y = 1 − c2 − d2 .
Then 0 < x, y ≤ 1. Multiplying by 4 on both sides of the given
inequality gives
4xy
>
=
=
≥
(2ac + 2bd − 2)2 = (2 − 2ac − 2bd)2
(a2 + b2 + x + c2 + d2 + y − 2ac − 2bd)2
[(a − c)2 + (b − d)2 + x + y]2
(x + y)2 = x2 + 2xy + y 2 ,
or 0 > x2 − 2xy + y 2 = (x − y)2 , which is impossible. Thus our
assumption is wrong and both a2 + b2 and c2 + d2 are greater than
1.
50
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
48. Find all complex numbers z such that
(3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2.
Solution: Note that
8(3z + 1)6(4z + 1)4(6z + 1)2(12z + 1) = 768
(24z + 8)(24z + 6)(24z + 4)(24z + 2) = 768.
Setting u = 24z + 5 and w = u2 yields
(u + 3)(u + 1)(u − 1)(u − 3) = 768
(u2 − 1)(u2 − 9) = 768
w2 − 10w − 759 = 0
(w − 33)(w + 23) = 0.
Therefore the solutions to the given equation are
√
√
± 33 − 5
± 23i − 5
z=
and z =
.
24
24
49. Let x1 , x2 , · · · , xn−1 , be the zeros different from 1 of the polynomial
P (x) = xn − 1, n ≥ 2. Prove that
1
1
1
n−1
+
+ ··· +
=
.
1 − x1
1 − x2
1 − xn−1
2
First Solution: For i = 1, 2, . . . , n, let ai = 1 − xi . Let
Q(x) =
P (1 − x)
(1 − x)n − 1
=
.
x
x
Then
n n−1
Q(x) = (−1) x
+ (−1)
n−1
µ ¶
µ ¶
µ ¶
n n−2
n
n
x
+ ··· +
x−
1
2
1
and ai ’s are the nonzero roots of the polynomial Q(x), as
Q(ai ) =
(1 − ai )n − 1
xn − 1
= i
= 0.
ai
xi − 1
51
Thus the desired sum is the sum of the reciprocals of the roots of
polynomial Q(x), that is,
1
1
1
+
+ ··· +
1 − x1
1 − x2
1 − xn−1
1
1
1
=
+
+ ··· +
a1
a2
an−1
a2 a3 · · · an + a1 a3 · · · an + · · · + a1 a2 · · · an−1
=
a1 a2 · · · an
By the Vieta’s theorem, the ratio between
S = a2 · · · an + a1 a3 · · · an + · · · + a1 a2 · · · an−1
and
P = a1 · · · an
is equal to the additive inverse of the ratio between the coefficient of
x and the constant term in Q(x), i.e., the desired value is equal to
¡n¢
S
n−1
= − ¡2n¢ =
,
P
2
− 1
as desired.
Second Solution: For any polynomial R(x) of degree n−1, whose
zeros are x1 , x2 , . . . , xn−1 , the following identity holds:
1
1
1
R0 (x)
+
+ ··· +
=
.
x − x1
x − x2
x − xn−1
R(x)
xn − 1
= xn−1 + xn−2 + · · · + x + 1, R(1) = n and
x−1
n(n − 1)
R0 (1) = (n − 1) + (n − 2) + · · · + 1 =
. It follows that
2
For R(x) =
1
1
R0 (1)
n−1
1
+
+ ··· +
=
=
.
1 − x1
1 − x2
1 − xn−1
R(1)
2
52
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
50. Let a and b be given real numbers. Solve the system of equations
p
x − y x2 − y 2
p
= a,
1 − x2 + y 2
p
y − x x2 − y 2
p
=b
1 − x2 + y 2
for real numbers x and y.
Solution: Let u = x + y and v = x − y. Then
0 < x2 − y 2 = uv < 1, x =
u−v
u+v
, and y =
.
2
2
Adding the two equations and subtracting the two equations in the
original system yields the new system
√
√
u − u uv = (a + b) 1 − uv
√
√
v + v uv = (a − b) 1 − uv.
Multiplying the above two equations yields
uv(1 − uv) = (a2 − b2 )(1 − uv),
hence uv = a2 − b2 . It follows that
√
√
(a + b) 1 − a2 + b2
(a − b) 1 − a2 + b2
√
√
u=
and v =
,
1 − a2 − b2
1 + a2 − b2
which in turn implies that
Ã
!
√
√
a + b a2 − b2 b + a a2 − b2
(x, y) = √
, √
,
1 − a2 + b2
1 − a2 + b2
whenever 0 < a2 − b2 < 1.
51. [Russia 2000] Prove that there exist 10 distinct real numbers a1 , a2 , . . . ,
a10 such that the equation
(x − a1 )(x − a2 ) · · · (x − a10 ) = (x + a1 )(x + a2 ) · · · (x + a10 )
53
has exactly 5 different real roots.
Solution:
We show that {a1 , a2 , . . . , a10 } = {7, 6, . . . , −2} is a
group of numbers satisfying the conditions given in the problem.
The given equality becomes
(x − 2)(x − 1)x(x + 1)(x + 2)g(x2 ) = 0,
where
g(u)
= 2[((7 + 6 + · · · + 3)u2 +
(7 · 6 · 5 + 7 · 6 · 4 + · · · + 5 · 4 · 3)u + 7 · 6 · · · 3].
If g(u) = has no real solutions, then g(x2 ) = 0 has no real solutions.
If u1 and u2 are real solutions of g(u) = 0, then u1 + u2 < 0 and
u1 u2 > 0, that is, both u1 and u2 are negative. It follows again that
g(x2 ) = 0 has no real solutions. Our proof is complete.
54
CHAPTER 3. SOLUTIONS TO INTRODUCTORY PROBLEMS
Chapter 4
Solutions to Advanced
Problems
1. Evaluate
µ
¶ µ
¶ µ
¶
µ
¶
2000
2000
2000
2000
+
+
+ ··· +
.
2
5
8
2000
Solution: Let
f (x) = (1 + x)2000 =
Let ω = (−1 +
2000
Xµ
k=0
√
¶
2000 k
x .
k
3i)/2. Then ω 3 = 1 and ω 2 + ω + 1 = 0. Hence
µµ
¶ µ
¶
µ
¶¶
2000
2000
2000
3
+
+ ··· +
2
5
2000
= f (1) + ωf (ω) + ω 2 f (ω 2 )
= 22000 + ω(1 + ω)2000 + ω 2 (1 + ω 2 )2000
= 22000 + ω(−ω 2 )2000 + ω 2 (−ω)2000
= 22000 + ω 2 + ω = 22000 − 1.
Thus the desired value is
22000 − 1
.
3
55
56
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
2. Let x, y, z be positive real numbers such that x4 + y 4 + z 4 = 1.
Determine with proof the minimum value of
x3
y3
z3
+
+
.
1 − x8
1 − y8
1 − z8
Solution: For 0 < u < 1, let f (u) = u(1 − u8 ). Let A be a positive
real number. By the AM-GM inequality,
· 8
¸9
Au + 8(1 − u8 )
8
8
8
8
.
A(f (u)) = Au (1 − u ) · · · (1 − u ) ≤
9
Setting A = 8 in the above inequality yields
µ ¶9
8
8
8
8(f (u)) ≤
or f (u) ≤ √
.
4
9
39
It follows that
y3
z3
x3
+
+
8
8
1−x
1−y
1 − z8
x4
y4
z4
+
+
x(1 − x8 ) y(1 − y 8 ) z(1 − z 8 )
√
√
4
943
(x4 + y 4 + z 4 ) 39
=
,
≥
8
8
=
with equality if and only if
1
.
x=y=z= √
4
3
Comment: This is a simple application of the result of problem
33 in the previous chapter.
3. [Romania 1990] Find all real solutions to the equation
2x + 3x + 6x = x2 .
Solution: For x < 0, the function f (x) = 2x + 3x + 6x − x2 is
increasing, so the equation f (x) = 0 has the unique solution x = −1.
57
Assume that there is a solution s ≥ 0. Then
s2 = 2s + 3s + 6s ≥ 3,
so s ≥
√
3, and hence bsc ≥ 1. But then s ≥ bsc yields
2s ≥ 2bsc = (1 + 1)bsc ≥ 1 + bsc ≥ s,
which in turn implies that
2
6s > 4s = (2s ) ≥ s2 .
So 2s + 3s + 6s > s2 , a contradiction.
Therefore x = −1 is the only solution to the equation.
4. Let {an }n≥1 be a sequence such that a1 = 2 and
an+1 =
1
an
+
2
an
for all n ∈ N. Find an explicit formula for an .
Solution: Solving the equation
x=
x 1
+
2 x
√
leads to x = ± 2. Note that
Ã
√
√
√ !2
an+1 + 2
a2n + 2 2an + 2
an + 2
√ =
√
√
=
.
an+1 − 2
a2n − 2 2an + 2
an − 2
Therefore,
Ã
√
√ !2n−1 ³
´ 2n
√
an + 2
a1 + 2
√ =
√
=
2+1
an − 2
a1 − 2
and
i
√ h¡√
¢2n
2
2+1
+1
an =
.
¡√
¢2n
2+1
−1
58
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
5. Let x, y, and z be positive real numbers. Prove that
x
y
p
p
+
+
x + (x + y)(x + z) y + (y + z)(y + x)
z
p
≤ 1.
z + (z + x)(z + y)
Solution: Note that
p
√
√
(x + y)(x + z) ≥ xy + xz.
In fact, squaring both sides of the above inequality yields
√
x2 + yz ≥ 2x yz,
which is evident by the AM-GM inequality. Thus
√
x
x
p
√ =√
√ .
≤
√
√
x + xy + xz
x+ y+ z
x + (x + y)(x + z)
x
Likewise,
√
y
√ ,
p
≤√
√
x
+
y+ z
y + (y + z)(y + x)
y
and
z+
z
p
(z + x)(z + y)
≤√
√
z
√ .
√
x+ y+ z
Adding the last three inequalities leads to the desired result.
6. Find, with proof, all nonzero polynomials f (z) such that
f (z 2 ) + f (z)f (z + 1) = 0.
Solution: Let f (z) = az m (z − 1)n g(z), where m and n are nonnegative integers and
g(z) = (z − z1 )(z − z2 ) · · · (z − zk ),
zi 6= 0 and zi 6= 1, for i = 1, 2, . . . , k. The given condition becomes
az 2m (z − 1)n (z + 1)n (z 2 − z1 )(z 2 − z2 ) · · · (z 2 − zk )
= −a2 z m+n (z + 1)m (z − 1)n (z − z1 )(z − z2 ) · · · (z − zk )
·(z + 1 − z1 )(z + 1 − z2 ) · · · (z + 1 − zk ).
59
Thus a = −a2 , and f is nonzero, so a = −1. Since zi 6= 1, 1 − zi 6=
0. Then z 2m = z m+n , that is, m = n. Thus f is of the form
−z m (z − 1)m g(z). Dividing by z 2m (z − 1)n (z + 1)n , the last equation
becomes
g(z 2 ) = g(z)g(z + 1).
We claim that g(z) ≡ 1. Suppose not; then clearly g must have at
least one complex root r 6= 0. Now g(r2 ) = g(r)g(r + 1) = 0, g(r4 ) =
0, g(r8 ) = 0, and so on. Since g cannot have infinitely many roots,
all its roots must have absolute value 1. Now, g((r − 1)2 ) = g(r −
1)g(r) =√0, so √
|(r − 1)2 | = 1. Clearly, if |r| = |(r − 1)2 | = 1, then
1+ 3i 1− 3i
r ∈ { 2 , 2 }. But r2 is also a root of g, so the same should
√
√
be true of r2 : r2 ∈ { 1+2 3i , 1−2 3i }. This is absurd. Hence, g cannot
have any roots, and g(z) ≡ 1.
Therefore, the f (z) are all the polynomials of the form −z m (z − 1)m
for m ∈ N.
7. Let f : N → N be a function such that f (n+1) > f (n) and f (f (n)) =
3n for all n. Evaluate f (2001).
First Solution: We prove the following lemma.
Lemma For n = 0, 1, 2, . . . ,
(a) f (3n ) = 2 · 3n ; and
(b) f (2 · 3n ) = 3n+1 .
Proof:
We use induction.
For n = 0, note that f (1) 6= 1, otherwise 3 = f (f (1)) = f (1) = 1,
which is impossible. Since f : N → N, f (1) > 1. Since f (n + 1) >
f (n), f is increasing. Thus 1 < f (1) < f (f (1)) = 3 or f (1) = 2.
Hence f (2) = f (f (1)) = 3.
Suppose that for some positive integer n ≥ 1,
f (3n ) = 2 · 3n and f (2 · 3n ) = 3n+1 .
Then,
and
f (3n+1 ) = f (f (2 · 3n )) = 2 · 3n+1 ,
¡
¢
f (2 · 3n+1 ) = f f (3n+1 ) = 3n+2 ,
60
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
as desired. This completes the induction.
n
n
2
n
There are 3 − 1 integers m such that 3 < m < 2 · 3 and there are
3n − 1 integers m0 such that
f (3n ) = 2 · 3n < m0 < 3n+1 = f (2 · 3n ) .
Since f is an increasing function,
f (3n + m) = 2 · 3n + m,
for 0 ≤ m ≤ 3n . Therefore
f (2 · 3n + m) = f (f (3n + m)) = 3 (3n + m)
for 0 ≤ m ≤ 3n . Hence
¡
¢
¡
¢
f (2001) = f 2 · 36 + 543 = 3 36 + 543 = 3816.
Second Solution: For integer n, let n(3) = a1 a2 · · · a` denote the
base 3 representation of n. Using similar inductions as in the first
solution, we can prove that
½
2a2 · · · a`
if a1 = 1,
f (n)(3) =
1a2 · · · a` 0 if a1 = 2.
Since 2001(3) = 2202010, f (2001)(3) = 12020100 or f (2001) = 1 ·
32 + 2 · 34 + 2 · 36 + 1 · 37 = 3816.
8. [China 1999] Let F be the set of all polynomials f (x) with integers
coefficients such that f (x) = 1 has at least one integer root. For
each integer k > 1, find mk , the least integer greater than 1 for
which there exists an f ∈ F such that f (x) = mk has exactly k
distinct integer roots.
Solution: Suppose that fk ∈ F satisfies the condition that fk (x) =
mk has exactly k distinct integer roots, and let a be an integer
such that fk (a) = 1. Let gk be the polynomial in F such that
gk (x) = fk (x + a) for all x. Now gk (0) = fk (a) = 1, so the constant
term of gk is 1. Now gk (x) = mk has exactly k distinct integer roots
r1 , r2 , . . . , rk , so we can write
gk (x) − mk = (x − r1 )(x − r2 ) . . . (x − rk )qk (x),
61
where qk (x) is an integer polynomial. Note that r1 r2 · · · rk divides
the constant term of gk (x)−mk , which equals 1−mk . Since mk > 1,
1 − mk cannot be 0, |1 − mk | ≥ |r1 r2 · · · rk |. Now r1 , r2 , · · · rk are
distinct integers, and none of them is 0, so
¯
¯
|r1 r2 · · · rk | ≥ ¯1 · (−1) · 2 · (−2) · 3 · · · (−1)k+1 dk/2e¯ ,
hence mk ≥ bk/2c! · dk/2e! + 1. This value of mk is attained by
gk (x) =
k−1
(−1)( 2 ) (x − 1)(x + 1)(x − 2)(x + 2)
¡
¢
· · · x + (−1)k dk/2e + bk/2c! · dk/2e! + 1.
Thus,
mk = bk/2c! · dk/2e! + 1.
9. Let x1 = 2 and
xn+1 = x2n − xn + 1,
for n ≥ 1. Prove that
1−
1
1
1
1
1
<
+
+ ··· +
< 1 − 2n .
x1
x2
xn
2
22n−1
Solution: Since x1 = 2 and
xn+1 − 1 = xn (xn − 1),
xn is increasing. Then xn − 1 6= 0. Hence
1
1
1
1
=
=
−
xn+1 − 1
xn (xn − 1)
xn − 1 xn
or
1
1
1
=
−
,
xn
xn − 1 xn+1 − 1
which implies that
1
1
1
1
+
+ ··· +
=1−
.
x1
x2
xn
xn+1 − 1
Thus it suffices to prove that, for n ∈ N,
1−
1
1
1
<1−
< 1 − 2n .
xn+1 − 1
2
22n−1
62
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
or
22
n−1
n
< xn+1 − 1 < 22 .
(1)
We use induction to prove (1).
For n = 1, x2 = x21 − x1 + 1 = 3 and (1) becomes 2 < 3 < 4, which
is true.
Now suppose that (1) is true for some positive integer n = k, i.e.,
22
k−1
k
< xk+1 − 1 < 22 .
(2)
Then for n = k + 1, the lower bound of (1) follows from
xk+2 − 1 = xk+1 (xk+1 − 1) > 22
k−1
· 22
k−1
k
= 22 .
Since xk+1 is an integer, the lower bound of (2) implies that
k
k
xk+1 ≤ 22 and xk+1 − 1 ≤ 22 − 1,
from which it follows that
³ k
´
k
k+1
xk+2 − 1 = xk+1 (xk+1 − 1) ≤ 22 · 22 − 1 < 22 ,
as desired. This finishes the induction and we are done.
10. [Iran 1997] Suppose that f : R+ → R+ is a decreasing function such
that for all x, y ∈ R+ ,
f (x + y) + f (f (x) + f (y)) = f (f (x + f (y)) + f (y + f (x))).
Prove that f (f (x)) = x.
Solution: Setting y = x gives
f (2x) + f (2f (x)) = f (2f (x + f (x))).
Replacing x with f (x) yields
f (2f (x)) + f (2f (f (x))) = f (2f (f (x) + f (f (x)))).
Subtracting these two equations gives
f (2f (f (x))) − f (2x) = f (2f (f (x) + f (f (x)))) − f (2f (x + f (x))).
63
If f (f (x)) > x, the left hand side of this equation is negative, so
f (f (x) + f (f (x)) > f (x + f (x))
and f (x) + f (f (x)) < x + f (x), a contradiction. A similar contradiction occurs if f (f (x)) < x. Thus f (f (x)) = x as desired.
Comment: In the original formulation f was meant to be a continous function. The solution above shows that this condition is not
necessary.
11. [Nordic Contest 1998] Find all functions f : Q → Q such that
f (x + y) + f (x − y) = 2f (x) + 2f (y)
for all x, y ∈ Q.
Solution: The only such functions are f (x) = kx2 for rational k.
Any such function works, since
f (x + y) + f (x − y) = k(x + y)2 + k(x − y)2
= kx2 + 2kxy + ky 2 + kx2 − 2kxy + ky 2
= 2kx2 + 2ky 2
= 2f (x) + 2f (y).
Now suppose f is any function satisfying
f (x + y) + f (x − y) = 2f (x) + 2f (y).
Then letting x = y = 0 gives 2f (0) = 4f (0), so f (0) = 0. We will
prove by induction that f (nz) = n2 f (z) for any positive integer n
and any rational number z. The claim holds for n = 0 and n = 1;
let n ≥ 2 and suppose the claim holds for n − 1 and n − 2. Then
letting x = (n − 1)z, y = z in the given equation we obtain
f (nz) + f ((n − 2)z) = f ((n − 1)z + z) + f ((n − 1)z − z)
= 2f ((n − 1)z) + 2f (z)
64
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
so
f (nz) = 2f ((n − 1)z) + 2f (z) − f ((n − 2)z)
= 2(n − 1)2 f (z) + 2f (z) − (n − 2)2 f (z)
= (2n2 − 4n + 2 + 2 − n2 + 4n − 4)f (z)
= n2 f (z)
and the claim holds by induction. Letting x = 0 in the given equation
gives
f (y) + f (−y) = 2f (0) + 2f (y) = 2f (y),
so f (−y) = f (y) for all rational y; thus f (nz) = n2 f (z) for all
integers n. Now let k = f (1); then for any rational number x = p/q,
q 2 f (x) = f (qx) = f (p) = p2 f (1) = kp2
so
f (x) = kp2 /q 2 = kx2 .
Thus the functions f (x) = kx2 , k ∈ Q, are the only solutions.
12. Let
3
4
< a < 1. Prove that the equation
x3 (x + 1) = (x + a)(2x + a)
has four distinct real solutions and find these solutions in explicit
form.
Solution:
a:
Look at the given equation as a quadratic equation in
a2 + 3xa + 2x2 − x3 − x4 = 0.
The discriminant of this equation is 9x2 −8x2 +4x3 +4x4 = (x+2x2 )2 .
Thus
a=
−3x ± (x + 2x2 )
.
2
2
The first choice a = −x+x2 yields
√the quadratic equation x −x−a =
0, whose solutions are x = (1 ± 1 + 4a)/2.
65
The second choice a = −2x − x2 yields√the quadratic equation x2 +
2x + a = 0, whose solutions are −1 ± 1 − a. The inequalities
√
√
√
√
1 + 1 + 4a
1 − 1 + 4a
−1 − 1 − a < −1 + 1 − a <
<
2
2
show that the four solutions are distinct. Indeed
√
√
1 − 1 + 4a
−1 + 1 − a <
2
√
√
√
reduces to 2 1 − a < 3− 1 + 4a which is equivalent to 6 1 + 4a <
6 + 8a, or 3a < 4a2 , which is evident.
13. [APMO 1996] Let a, b, and c be positive real numbers such that
abc = 1. Prove that
1
1
1
+
+
≤ 1.
a+b+1 b+c+1 c+a+1
First Solution: Setting x = a + b, y = b + c and z = c + a, the
inequality becomes
1
1
1
+
+
≤1
x+1 y+1 z+1
1
1
x
+
≤
y+1 z+1
x+1
y+z+2
x
≤
(y + 1)(z + 1)
x+1
xy + xz + 2x + y + z + 2 ≤ xyz + xy + xz + x
x + y + z + 2 ≤ xyz
2(a + b + c) + 2 ≤ (a + b)(b + c)(c + a)
2(a + b + c) ≤ a2 b + ab2 + b2 c + bc2 + c2 a + ca2 .
√
3
By the AM-GM inequality, (a2 b + a2 c + 1) ≥ 3 a4 bc = 3a.
Likewise, (b2 c + b2 a + 1) ≥ 3b and (c2 a + c2 b + 1) ≥ 3c. Therefore
we only need to prove that
2(a + b + c) + 3 ≤ 3(a + b + c)
3 ≤ a + b + c,
66
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
which is evident from AM-GM inequality and abc = 1.
Second Solution:
Note that
Let a = a31 , b = b31 , c = c31 . Then a1 b1 c1 = 1.
a31 + b31 − a21 b1 − a1 b21 = (a1 − b1 )(a21 − b21 ) ≥ 0,
which implies that a31 + b31 ≥ a1 b1 (a1 + b1 ). Therefore,
1
1
1
= 3
≤
3
a+b+1
a1 + b1 + a1 b1 c1
a1 b1 (a1 + b1 ) + a1 b1 c1
=
c1
a1 b1 c1
=
.
a1 b1 (a1 + b1 + c1 )
a1 + b1 + c1
Likewise,
1
a1
1
b1
≤
and
≤
.
b+c+1
a1 + b1 + c1
c+a+1
a1 + b1 + c1
Adding the three inequalities yields the desired result.
14. [AIME 1988] Find all functions f , defined on the set of ordered pairs
of positive integers, satisfying the following properties:
f (x, x) = x, f (x, y) = f (y, x), (x + y)f (x, y) = yf (x, x + y).
Solution: We claim that f (x, y) = lcm(x, y), the least common
multiple of x and y. It is clear that lcm(x, x) = x and lcm(x, y) =
lcm(y, x). Note that
lcm(x, y) =
xy
and gcd (x, y) = gcd (x, x + y),
gcd (x, y)
where gcd (u, v) denote the greatest common divisor of u and v.
Then
(x + y)lcm(x, y) = (x + y) ·
=y·
xy
gcd (x, y)
x(x + y)
= ylcm(x, x + y).
gcd (x, x + y)
67
Now we prove that there is only one function satisfying the conditions
of the problem. For the sake of contradiction, assume that there is
another function g(x, y) also satisfying the given conditions. Let S
be the set of all pairs of positive integers (x, y) such that f (x, y) 6=
g(x, y), and let (m, n) be such a pair with minimal sum m + n. It is
clear that m 6= n, otherwise f (m, n) = f (m, m) = m = g(m, m) =
g(m, n). By symmetry (f (x, y) = f (y, x)), we can assume that n −
m > 0. Note that
nf (m, n − m) = [m + (n − m)]f (m, n − m)
= (n − m)f (m, m + (n − m)) = (n − m)f (m, n)
or
f (m, n − m) =
n−m
· f (m, n).
n
Likewise,
n−m
· g(m, n).
n
Since f (m, n) 6= g(m, n), f (m, n − m) 6= g(m, n − m). Thus (m, n −
m) ∈ S. But (m, n − m) has a smaller sum m + (n − m) = n,
a contradiction. Therefore our assumption is wrong and f (x, y) =
lcm(x, y) is the only solution.
g(m, n − m) =
15. [Romania 1990] Consider n complex numbers zk , such that |zk | ≤ 1,
k = 1, 2, . . . , n. Prove that there exist e1 , e2 , . . . , en ∈ {−1, 1} such
that, for any m ≤ n,
|e1 z1 + e2 z2 + · · · + em zm | ≤ 2.
Solution: Call a finite sequence of complex numbers each with absolute value not exceeding 1 a green sequence. Call a green sequence
{zk }nk=1 happy if it has a friend sequence {ek }nk=1 of 1’s and −1’s,
satisfying the condition of the problem. We will prove by induction
on n that all green sequences are happy.
For n = 2, this claim is obviously true.
Suppose this claim is true when n equals some number m. For the
case of n = m + 1, think of the zk as points in the complex plane.
For each k, let `k be the line through the origin and the point corresponding to zk . Among the lines `1 , `2 , `3 , some two are within 60◦
68
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
of each other; suppose they are `α and `β , with the leftover one being
`γ . The fact that `α and `β are within 60◦ of each other implies that
there exists some number eβ ∈ {−1, 1} such that z 0 = zα + eβ zβ has
absolute value at most 1. Now the sequence z 0 , zγ , z4 , z5 , . . . , zk+1 is
a k-term green sequence, so, by the induction hypothesis, it must be
happy; let e0 , eγ , e4 , e5 , . . . , ek+1 be its friend. Let eα = 1. Then the
k+1
sequence {ei }k+1
i=1 is the friend of {zi }i=1 . Induction is now complete.
16. [ARML 1997] Find a triple of rational numbers (a, b, c) such that
q
√
√
√
3 √
3
3
2 − 1 = 3 a + b + 3 c.
Solution:
Let x =
√
x = 3 y − 1. Note that
p
√
3
3
2 − 1 and y =
√
3
2. Then y 3 = 2 and
1 = y 3 − 1 = (y − 1)(y 2 + y + 1),
and
y2 + y + 1 =
3y 2 + 3y + 3
y 3 + 3y 2 + 3y + 1
(y + 1)3
=
=
,
3
3
3
which implies that
x3 = y − 1 =
1
3
=
y2 + y + 1
(y + 1)3
or
√
3
x=
3
.
y+1
(1)
On the other hand,
3 = y 3 + 1 = (y + 1)(y 2 − y + 1)
from which it follows that
1
y2 − y + 1
=
.
y+1
3
Combining (1) and (2), we obtain
r
√
√
3
3
3 1
x=
( 4 − 2 + 1).
9
(2)
69
Consequently,
µ
(a, b, c) =
4 2 1
,− ,
9 9 9
¶
is a desired triple.
17. [Romania 1984] Find the minimum of
logx1
¶
µ
¶
µ
¶
µ
1
1
1
x2 −
+ logx2 x3 −
+ · · · + logxn x1 −
4
4
4
where x1 , x2 , . . . , xn are real numbers in the interval ( 14 , 1).
Solution: Since loga x is a decreasing function of x when 0 < a < 1
and, since (x − 1/2)2 ≥ 0 implies x2 ≥ x − 1/4, we have
logxk
µ
¶
1
log xk+1
xk+1 −
≥ logxk x2k+1 = 2 logxk xk+1 = 2
.
4
logxk
It follows that
µ
logx1
¶
µ
¶
µ
¶
1
1
1
+ logx2 x3 −
+ · · · + log=xn x1 −
4
4
4
µ
¶
log x2
log x3
log xn
log x1
≥2
+
+ ··· +
+
≥ 2n
log x1
log x2
log xn−1
log xn
x2 −
by the AM-GM inequality. Equalities hold if and only if x1 =
x2 = · · · = xn = 1/2.
18. [AIME 1984] Determine x2 + y 2 + z 2 + w2 if
x2
− 12
x2
42 − 12
x2
2
6 − 12
x2
82 − 12
22
y2
− 32
y2
+ 2
4 − 32
y2
+ 2
6 − 32
y2
+ 2
8 − 32
+
22
z2
− 52
z2
+ 2
4 − 52
z2
+ 2
6 − 52
z2
+ 2
8 − 52
+
22
w2
− 72
w2
+ 2
4 − 72
w2
+ 2
6 − 72
w2
+ 2
8 − 72
+
22
= 1,
= 1,
= 1,
= 1.
70
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
Solution: The claim that the given system of equations is satisfied
by x2 , y 2 , z 2 , and w2 is equivalent to claiming that
x2
y2
z2
w2
+
+
+
=1
2
2
2
t−1
t−3
t−5
t − 72
(1)
is satisfied by t = 4, 16, 36, and 64. Multiplying to clear fractions,
we find that for all values of t for which it is defined (i.e., t 6= 1, 9, 25,
and 49), (1) is equivalent to the polynomial equation
P (t) = 0,
where
P (t) = (t − 1)(t − 9)(t − 25)(t − 49)
−x2 (t − 9)(t − 25)(t − 49) − y 2 (t − 1)(t − 25)(t − 49)
−z 2 (t − 1)(t − 9)(t − 49) − w2 (t − 1)(t − 9)(t − 25).
Since deg P (t) = 4, P (t) = 0 has exactly four zeros t = 4, 16, 36, and
64, i.e.,
P (t) = (t − 4)(t − 16)(t − 36)(t − 64).
Comparing the coefficients of t3 in the two expressions of P (t) yields
1 + 9 + 25 + 49 + x2 + y 2 + z 2 + w2 = 4 + 16 + 36 + 64,
from which it follows that
x2 + y 2 + z 2 + w2 = 36.
19. [Balkan 1997] Find all functions f : R → R such that
f (xf (x) + f (y)) = (f (x))2 + y
for all x, y ∈ R.
Solution:
yields
Let f (0) = a. Setting x = 0 in the given condition
f (f (y)) = a2 + y,
71
for all y ∈ R. Since the range of a2 + y consists of all real numbers,
f must be surjective. Thus there exists b ∈ R such that f (b) = 0.
Setting x = b in the given condition yields
f (f (y)) = f (bf (b) + f (y)) = (f (b))2 + y = y,
for all y ∈ R. It follows that, for all x, y ∈ R,
(f (x))2 + y = f (xf (x) + f (y))
= f [f (f (x))f (x) + f (y)] = f [f (x)f (f (x)) + y]
= f (f (x))2 + y = x2 + y,
that is,
(f (x))2 = x2 .
(1)
It is clear that f (x) = x is a function satisfying the given condition.
Suppose that f (x) 6= x. Then there exists some nonzero real number
c such that f (c) = −c. Setting x = cf (c) + f (y) in (1) yields
[f (cf (c) + f (y))]2 = [cf (c) + f (y)]2 = [−c2 + f (y)]2 ,
for all y ∈ R, and, setting x = c in the given condition yields
f (cf (c) + f (y)) = (f (c))2 + y = c2 + y,
for all y ∈ R. Note that (f (y))2 = y 2 . It follows that
[−c2 + f (y)]2 = (c2 + y)2 ,
or
f (y) = −y,
for all y ∈ R.
Therefore the only functions satisfy the given condition are f (x) = x
or f (x) = −x, for x ∈ R.
20. The numbers 1000, 1001, · · · , 2999 have been written on a board.
Each time, one is allowed to erase two numbers, say, a and b, and
1
replace them by the number min(a, b). After 1999 such operations,
2
one obtains exactly one number c on the board. Prove that c < 1.
72
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
1
Solution: By symmetry, we may assume a ≤ b. Then min(a, b) =
2
a
. We have
2
1
1 1
+ ≤ a,
a b
2
from which it follows that the sum of the reciprocals of all the numbers on the board is nondecreasing (i.e., the sum is a monovariant).
At the beginning this sum is
S=
1
1
1
1
+
+ ··· +
≤ ,
1000 1001
2999
c
where 1/c is the sum at the end. Note that, for 1 ≤ k ≤ 999,
1
1
4000
4000
1
+
=
>
=
.
2000 − k 2000 + k
20002 − k 2
20002
1000
Rearranging terms in S yields
µ
¶ µ
¶
1
1
1
1
1
1
≥
+
+
+
+
+
c
1000
1001 2999
1002 2998
µ
¶
1
1
1
··· +
+
+
1999 2001
2000
>
1
1
× 1000 +
> 1,
1000
2000
or c < 1, as desired.
21. [Bulgaria 1998] Let a1 , a2 , . . . , an be real numbers, not all zero. Prove
that the equation
√
√
√
1 + a1 x + 1 + a2 x + · · · + 1 + an x = n
has at most one nonzero real root.
√
Solution: Notice√that fi (x) = 1 + ai x is concave. Hence f (x) =
√
1 + a1 x + · · · + 1 + an x is concave. Since f 0 (x) exists, there
can be at most one point on the curve y = f (x) with derivative 0.
Suppose there is more than one nonzero root. Since x = 0 is also a
73
root, we have three real roots x1 < x2 < x3 . Applying the MeanValue theorem to f (x) on intervals [x1 , x2 ] and [x2 , x3 ], we can find
two distinct points on the curve with derivative 0, a contradiction.
Therefore, our assumption is wrong and there can be at most one
nonzero real root for the equation f (x) = n.
22. [Turkey 1998] Let {an } be the sequence of real numbers defined by
a1 = t and an+1 = 4an (1 − an ) for n ≥ 1. For how many distinct
values of t do we have a1998 = 0?
First Solution: Let f (x) = 4x(1 − x). Observe that
f −1 (0) = {0, 1},
f −1 (1) = {1/2},
f −1 ([0, 1]) = [0, 1],
and |{y : f (y) = x}| = 2 for all x ∈ [0, 1).
Let An = {x ∈ R : f n (x) = 0}; then
An+1 = {x ∈ R : f n+1 (x) = 0}
= {x ∈ R : f n (f (x)) = 0} = {x ∈ R : f (x) ∈ An }.
We claim that for all n ≥ 1, An ⊂ [0, 1], 1 ∈ An , and |An | = 2n−1 +1.
For n = 1, we have A1 = {x ∈ R | f (x) = 0} = {0, 1}, and the
claims hold.
Now suppose n ≥ 1 and An ⊂ [0, 1], 1 ∈ An , and |An | = 2n−1 + 1.
Then x ∈ An+1 ⇒ f (x) ∈ An ⊂ [0, 1] ⇒ x ∈ [0, 1], so An+1 ⊂ [0, 1].
Since f (0) = f (1) = 0, we have f n+1 (1) = 0 for all n ≥ 1, so
1 ∈ An+1 . Now we have
X
|An+1 | = |{x : f (x) ∈ An }| =
|{x : f (x) = a}|
a∈An
= |{x : f (x) = 1}| +
= 1+
X
X
|{x : f (x) = a}|
a∈An
a∈[0,1)
2 = 1 + 2(|An | − 1)
a∈An
a∈[0,1)
= 1 + 2(2n−1 + 1 − 1) = 2n + 1.
Thus the claim holds by induction.
74
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
Finally, a1998 = 0 if and only if f 1997 (t) = 0, so there are 21996 + 1
such values of t.
Second Solution: As in the previous solution, observe that if
f (x) ∈ [0, 1] then x ∈ [0, 1], so if a1998 = 0√we must have t ∈ [0, 1].
Now choose θ ∈ [0, π/2] such that sin θ = t. Observe that for any
φ ∈ R,
f (sin2 φ) = 4 sin2 φ (1 − sin2 φ) = 4 sin2 φ cos2 φ = sin2 2φ;
since a1 = sin2 θ, it follows that
a2 = sin2 2θ, a3 = sin2 4θ, . . . , a1998 = sin2 21997 θ.
Therefore
a1998 = 0 ⇐⇒ sin 21997 θ = 0 ⇐⇒ θ =
kπ
21997
for some k ∈ Z. Thus the values of t which give a1998 = 0 are
sin2 (kπ/21997 ),
k ∈ Z, giving 21996 + 1 such values of t.
23. [IMO 1997 short list]
(a) Do there exist functions f : R → R and g : R → R such that
f (g(x)) = x2 and g(f (x)) = x3 for all x ∈ R?
(b) Do there exist functions f : R → R and g : R → R such that
f (g(x)) = x2 and g(f (x)) = x4 for all x ∈ R?
Solution:
(a) The conditions imply that f (x3 ) = f (g(f (x))) = [f (x)]2 , whence
x ∈ {−1, 0, 1} =⇒ x3 = x =⇒ f (x) = [f (x)]2 =⇒ f (x) ∈
{0, 1}. Thus, there exist different a, b ∈ {−1, 0, 1} such that
f (a) = f (b). But then a3 = g(f (a)) = g(f (b)) = b3 , a contradiction. Therefore, the desired functions f and g do not exist.
75
(b) Let

ln |x|

|x|
g(x) = |x|− ln |x|


0
if |x| ≥ 1
if 0 < |x| < 1
if x = 0.
Note that g is even and |a| = |b| whenever g(a) = g(b); thus,
we are allowed to define
f (x) = y 2 , where y is such that g(±y) = x.
We claim that the functions f, g described above satisfy the
conditions of the problem. It isp
clear from the definition of f
that f (g(x)) = x2 . Now let y = f (x). Then g(y) = x and

¢4
¡
2 ln(y 2 )

= y 4 ln y = y ln y
(y )
¡
¢4
2
g(f (x)) = g(y 2 ) =
(y 2 )− ln(y ) = y − ln y


0
if y ≥ 1
if 0 < y < 1
if y = 0
= [g(y)]4 = x4 .
24. [Weichao Wu] Let 0 < a1 ≤ a2 · · · ≤ an , 0 < b1 ≤ b2 · · · ≤ bn be
n
n
X
X
real numbers such that
ai ≥
bi . Suppose that there exists
i=1
i=1
1 ≤ k ≤ n such that bi ≤ ai for 1 ≤ i ≤ k and bi ≥ ai for i > k.
Prove that
a1 a2 · · · an ≥ b1 b2 · · · bn .
First Solution: We define two new sequences. For i = 1, 2, . . . , n,
let
bi ak
a0i = ak and b0i =
.
ai
Then
a0i − b0i = ak −
bi ak
ak
=
(ai − bi )
ai
ai
or
(a0i − b0i ) − (ai − bi ) =
(ak − ai )(ai − bi )
≥ 0.
ai
76
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
Therefore nak = a01 + a02 + · · · + a0n ≥ b01 + b02 + · · · + b0n . Applying
the AM-GM inequality yields
µ
b1 b2 · · · bn ank
a1 a2 · · · an
¶ n1
1
= (b01 b02 · · · b0n ) n ≤
b01 + b02 + · · · + b0n
≤ ak
n
from which the desired result follows.
Second Solution: We define two new sequences. For i = 1, 2, . . . , n,
let
a0i = ak and b0i = bi + ak − ai > 0.
Then
b01 + b02 + · · · + b0n ≤ nak .
(1)
Note that, for cy(x − y)(y + c) ≥ 0,
x+c
x
≥
, x ≥ y and c ≥ 0;
y
y+c
Setting x = ai , y = bi , and c = ak − ai , the above inequality implies
that ai /bi ≥ a0i /b0i , for i = 1, 2, . . . , n. Thus,
a0 a0 · · · a0n
a1 a2 · · · an
≥ 10 02
b1 b2 · · · bn
b1 b2 · · · b0n
(2)
Using (1) and the AM-GM inequality yields
1
(a01 a02 · · · a0n ) n = ak ≥
or
1
b01 + b02 + · · · + b0n
≥ (b01 b02 · · · b0n ) n
n
a01 a02 · · · a0n ≥ b01 b02 · · · b0n .
(3)
It is clear that the desired result follows from (2) and (3).
25. Given eight non-zero real numbers a1 , a2 , · · · , a8 , prove that at least
one of the following six numbers: a1 a3 + a2 a4 , a1 a5 + a2 a6 , a1 a7 +
a2 a8 , a3 a5 + a4 a6 , a3 a7 + a4 a8 , a5 a7 + a6 a8 is non-negative.
Solution:
First, we introduce some basic knowledge of vector
operations. Let u = [a, b] and v = [m, n] be two vectors. Define
their dot product u · v = am + bn. It is easy to check that
77
(i) v · v = m2 + n2 = |v|2 , that is, the dot product of vector with
itself is the square of the magnitude of v and v · v ≥ 0 with
equality if and only if v = [0, 0];
(ii) u · v = v · u;
(iii) u · (v + w) = u · v + u · w, where w is a vector;
(iv) (cu) · v = c(u · v), where c is a scalar.
When vectors u and v are placed tail-by-tail at the origin O, let A
−−→
and B be the tips of u and v, respectively. Then AB = v − u. Let
∠AOB = θ. Applying the law of cosines to triangle AOB yields
|v − u|2 = AB 2 = OA2 + OB 2 − 2OA · OB cos θ
= |u|2 + |v|2 − 2|u||v| cos θ.
It follows that
(v − u) · (v − u) = u · u + v · v − 2|u||v| cos θ,
u·v
or cos θ =
. Consequently, if 0 ≤ θ ≤ 90◦ , u · v ≥ 0.
|u||v|
Consider vectors v1 = [a1 , a2 ], v2 = [a3 , a4 ], v3 = [a5 , a6 ], and v4 =
[a7 , a8 ]. Note that the numbers a1 a3 +a2 a4 , a1 a5 +a2 a6 , a1 a7 +a2 a8 ,
a3 a5 + a4 a6 , a3 a7 + a4 a8 , a5 a7 + a6 a8 are all the dot products of
distinct vectors vi and vj . Since there are four vectors, when placed
tail-by-tail at the origin, at least two of them form a non-obtuse
angle, which in turn implies the desired result.
26. [IMO 1996 short list] Let a, b and c be positive real numbers such
that abc = 1. Prove that
bc
ca
ab
+
+
≤ 1.
a5 + b5 + ab b5 + c5 + bc c5 + a5 + ca
Solution: We have a5 +b5 ≥ a2 b2 (a+b), because (a3 −b3 )(a2 −b2 ) ≥
0, with equality if and only if a = b. Hence
ab
ab
≤ 2 2
a5 + b5 + ab
a b (a + b) + ab
1
abc
=
ab(a + b) + 1
ab(a + b + c)
c
=
.
a+b+c
=
78
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
a
ca
b
bc
≤
and 5
≤
.
5
5
+ c + bc
a+b+c
c + a + ca
a+b+c
Adding the last three inequalities leads to the desired result. Equality holds if and only if a = b = c = 1.
Likewise,
b5
Comment: Please compare the solution to this problem with the
second solution of problem 13 in this chapter.
27. [Czech-Slovak match 1997] Find all functions f : R → R such that
the equality
f (f (x) + y) = f (x2 − y) + 4f (x)y
holds for all pairs of real numbers (x, y).
Solution:
Clearly, f (x) = x2 satisfies the functional equation.
Now assume that there is a nonzero value a such that f (a) 6= a2 .
x2 − f (x)
Let y =
in the functional equation to find that
2
µ
¶
µ
¶
f (x) + x2
f (x) + x2
f
=f
+ 2f (x)(x2 − f (x))
2
2
or 0 = 2f (x)(x2 − f (x)). Thus, for each x, either f (x) = 0 or
f (x) = x2 . In both cases, f (0) = 0.
Setting x = a, it follows from above that either f (a) = 0 or f (a) = 0
or f (a) = a2 . The latter is false, so f (a) = 0. Now, let x = 0 and
then x = a in the functional equation to find that
f (y) = f (−y),
f (y) = f (a2 − y)
and so f (y) = f (−y) = f (a2 + y); that is, the function is periodic with nonzero period a2 . Let y = a2 in the original functional
equation to obtain
f (f (x)) = f (f (x) + a2 ) = f (x2 − a2 ) + 4a2 f (x) = f (x2 ) + 4a2 f (x).
However, putting y = 0 in the functional equation gives f (f (x)) =
f (x2 ) for all x. Thus, 4a2 f (x) = 0 for all x. Since a is nonzero,
f (x) = 0 for all x. Therefore, either f (x) = x2 or f (x) = 0.
79
28. Solve the system of equations:
3x − y
=3
x2 + y 2
x + 3y
y− 2
= 0.
x + y2
x+
First Solution: Multiplying the second equation by i and adding
it to the first equation yields
x + yi +
(3x − y) − (x + 3y)i
= 3,
x2 + y 2
or
3(x − yi) i(x − yi)
− 2
= 3.
x2 + y 2
x + y2
1
x − yi
Let z = x + yi. Then = 2
. Thus the last equation becomes
z
x + y2
x + yi +
z+
or
3−i
= 3,
z
z 2 − 3z + (3 − i) = 0.
Hence
3±
√
−3 + 4i
3 ± (1 + 2i)
=
,
2
2
that is, (x, y) = (2, 1) or (x, y) = (1, −1).
z=
Second Solution: Multiplying the first equation by y, the second
by x, and adding up yields
2xy +
(3x − y)y − (x + 3y)x
= 3y,
x2 + y 2
3y + 1
. Substituting
2y
this into the second equation of the given system gives
# µ
"µ
¶2
¶
3y + 1
3y + 1
2
y
+y −
− 3y = 0,
2y
2y
or 2xy − 1 = 3y. It follows that y 6= 0 and x =
80
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
or
4y 4 − 3y 2 − 1 = 0.
It follows that y 2 = 1 and that the solutions to the system are (2, 1)
and (1, −1).
29. [China 1995] Mr. Fat and Mr. Taf play a game with a polynomial of
degree at least 4:
x2n + x2n−1 + x2n−2 + · · · + x + 1.
They fill in real numbers to empty spaces in turn. If the resulting
polynomial has no real root, Mr. Fat wins; otherwise, Mr. Taf wins.
If Mr. Fat goes first, who has a winning strategy?
Solution: Mr. Taf has a winning strategy.
We say a space odd (even) if it coefficient of an odd (even) power of
x. First Mr. Taf will fill in arbitrary real numbers into one of the
remaining even spaces, if there are any. Since there are only n − 1
even spaces, there will be at least one odd spaces left after 2n − 3
plays, that is, the given polynomial becomes
p(x) = q(x) + xs + x2t−1 ,
where s and 2t − 1 are distinct positive integers and q(x) is a fixed
polynomial. We claim that there is a real number a such that
p(x) = q(x) + axs + x2t−1
will always has a real root regardless of the coefficient of x2t−1 . Then
Mr. Taf can simply fill in a in front of xs and win the game.
Now we prove our claim. Let b be the coefficient of x2t−1 in p(x).
Note that
1
p(2) + p(−1)
22t−1
µ
¶
1
s−2t+1
=
q(2)
+
2
a
+
b
+ [q(−1) + (−1)s a − b]
22t−1
µ
¶
1
=
q(2) + q(−1) + a[2s−2t+1 + (−1)s ].
22t−1
81
Since s 6= 2t − 1, 2s−2t+1 + (−1)s 6= 0. Thus
1
2t−1 q(2) + q(−1)
a = − 2 s−2t+1
2
+ (−1)s
is well defined such that a is independent of b and
1
p(2) + p(−1) = 0.
22t−1
It follows that either p(−1) = p(2) = 0 or p(−1) and p(2) have
different signs, which implies that there is a real root of p(x) in
between −1 and 2. In either case, p(x) has a real root regardless the
coefficient of x2t−1 , as claimed. Our proof is thus complete.
30. [IMO 1997 short list] Find all positive integers k for which the following statement is true: if F (x) is a polynomial with integer coefficients
satisfying the condition
0 ≤ F (c) ≤ k
for c = 0, 1, . . . , k + 1,
then F (0) = F (1) = · · · = F (k + 1).
Solution: The statement is true if and only if k ≥ 4. We start by
proving that it does hold for each k ≥ 4.
Consider any polynomial F (x) with integer coefficients satisfying the
inequality 0 ≤ F (c) ≤ k for each c ∈ {0, 1, . . . , k + 1}. Note first that
F (k + 1) = F (0), since F (k + 1) − F (0) is a multiple of k + 1 not
exceeding k in absolute value. Hence
F (x) − F (0) = x(x − k − 1)G(x),
where G(x) is a polynomial with integer coefficients. Consequently,
k ≥ |F (c) − F (0)| = c(k + 1 − c)|G(c)|
(1)
for each c ∈ {1, 2, . . . , k}. The equality c(k + 1 − c) > k holds for
each c ∈ {2, 3, . . . , k − 1}, as it is equivalent to (c − 1)(k − c) > 0.
Note that the set {2, 3, . . . , k − 1} is not empty if k ≥ 3, and for any
c in this set, (1) implies that |G(c)| < 1. Since G(c) is an integer,
G(c) = 0. Thus
F (x) − F (0) = x(x − 2)(x − 3) · · · (x − k + 1)(x − k − 1)H(x), (2)
82
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
where H(x) is a polynomial with integer coefficients. To complete
the proof of our claim, it remains to show that H(1) = H(k) = 0.
Note that for c = 1 and c = k, (2) implies that
k ≥ |F (c) − F (0)| = (k − 2)! · k · |H(c)|.
For k ≥ 4, (k − 2)! > 1. Hence H(c) = 0.
We established that the statement in the question holds for any
k ≥ 4. But the proof also provides information for the smaller values
of k as well. More exactly, if F (x) satisfies the given condition then
0 and k + 1 are roots of F (x) and F (0) for any k ≥ 1; and if k ≥ 3
then 2 must also be a root of F (x) − F (0). Taking this into account,
it is not hard to find the following counterexamples:
F (x) = x(2 − x)
F (x) = x(3 − x)
for k = 1,
for k = 2,
F (x) = x(4 − x)(x − 2)2
for k = 3.
31. The Fibonacci sequence Fn is given by
F1 = F2 = 1, Fn+2 = Fn+1 + Fn
Prove that
F2n =
(n ∈ N).
3
3
+ F2n−2
F2n+2
3
− 2F2n
9
for all n ≥ 2.
Solution: Note that
F2n+2 − 3F2n = F2n+1 − 2F2n = F2n−1 − F2n = −F2n−2 ,
whence
3F2n − F2n+2 − F2n−2 = 0
(1)
for all n ≥ 2. Setting a = 3F2n , b = −F2n+2 , and c = −F2n−2 in the
algebraic identity
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)
gives
3
3
3
27F2n
− F2n+2
− F2n−2
− 9F2n+2 F2n F2n−2 = 0.
83
Applying (1) twice gives
2
2
F2n+2 F2n−2 − F2n
= (3F2n − F2n−2 )F2n−2 − F2n
2
2
= F2n (3F2n−2 − F2n ) − F2n−2
= F2n F2n−4 − F2n−2
2
= · · · = F6 F2 − F4 = −1.
The desired result follows from
3
2
9F2n+2 F2n F2n−2 − 9F2n
= 9F2n (F2n+2 F2n−2 − F2n
) = −9F2n .
32. [Romania 1998] Find all functions u : R → R for which there exists
a strictly monotonic function f : R → R such that
f (x + y) = f (x)u(y) + f (y)
for all x, y ∈ R.
Solution: The solutions are u(x) = ax , a ∈ R+ . To see that these
work, take f (x) = x for a = 1. If a 6= 1, take f (x) = ax − 1; then
f (x + y) = ax+y − 1 = (ax − 1)ay + ay − 1 = f (x)u(y) + f (y)
for all x, y ∈ R.
Now suppose u : R → R, f : R → R are functions for which f is
strictly monotonic and f (x + y) = f (x)u(y) + f (y) for all x, y ∈ R.
We must show that u is of the form u(x) = ax for some a ∈ R+ .
First, letting y = 0, we obtain f (x) = f (x)u(0) + f (0) for all x ∈ R.
Thus, u(0) 6= 1 would imply f (x) = f (0)/(1 − u(0)) for all x, which
would contradict the fact that f is strictly monotonic, so we must
have u(0) = 1 and f (0) = 0. Then f (x) 6= 0 for all x 6= 0. Next, we
have
f (x)u(y) + f (y) = f (x + y) = f (x) + f (y)u(x),
or
f (x)(u(y) − 1) = f (y)(u(x) − 1)
for all x, y ∈ R. That is,
u(x) − 1
u(y) − 1
=
f (x)
f (y)
84
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
for all xy 6= 0. It follows that there exists C ∈ R such that
u(x) − 1
=C
f (x)
for all x 6= 0. Thus, u(x) = 1 + Cf (x) for x 6= 0; since u(0) = 1,
f (0) = 0, this equation also holds for x = 0. If C = 0, then u(x) = 1
for all x, and we are done. Otherwise, observe
u(x + y) = 1 + Cf (x + y) = 1 + Cf (x)u(y) + Cf (y)
= u(y) + Cf (x)u(y) = u(x)u(y)
for all x, y ∈ R. Thus u(nx) = u(x)n for all n ∈ Z, x ∈ R. Since
u(x) = 1 + Cf (x) for all x, u is strictly monotonic, and u(−x) =
1/u(x) for all x, so u(x) > 0 for all x as u(0) = 1. Let a = u(1) > 0;
then u(n) = an for all n ∈ N, and u(p/q) = (u(p))1/q = ap/q for all
p ∈ Z, q ∈ N, so u(x) = ax for all x ∈ Q. Since u is monotonic and
the rationals are dense in R, we have u(x) = ax for all x ∈ R. Thus
all solutions are of the form u(x) = ax , a ∈ R+ .
33. [China 1986] Let z1 , z2 , . . . , zn be complex numbers such that
|z1 | + |z2 | + · · · + |zn | = 1.
Prove that there exists a subset S of {z1 , z2 , . . . , zn } such that
¯
¯
¯X ¯ 1
¯
¯
z¯ ≥ .
¯
¯
¯ 6
z∈S
First Solution: Let `1 , `2 , and `3 be three rays from origin that
form angles of 60◦ , 180◦ , and 300◦ , respectively, with the positive
x-axis. For i = 1, 2, 3, let Ri denote the region between `i and `i+1
(here `4 = `1 ), including the ray `i . Then
1=
X
zk ∈R1
|zk | +
X
zk ∈R1
|zk | +
X
|zk |.
zk ∈R1
By the Pigeonhole Principle, at least one of the above sums is not less
than 1/3. Say it’s R3 (otherwise, we apply a rotation, which does
85
not effect the magnitude of a complex number). Let zk = xk + iyk .
Then for zk ∈ R3 , xk = |xk | ≥ |zk |/2. Consequently,
¯
¯ ¯
¯
¯ X
¯ ¯ X
¯ 1 X
1 1
1
¯
¯ ¯
¯
zk ¯ ≥ ¯
xk ¯ ≥
|zk | ≥ · = ,
¯
¯
¯ ¯
¯ 2
2 3
6
zk ∈R3
zk ∈R3
zk ∈R3
as desired.
Second Solution: We prove a stronger statement: there is subset
S of {z1 , z2 , . . . , zn } such that
¯
¯
¯X ¯ 1
¯
¯
z¯ ≥ .
¯
¯ 4
¯
z∈S
For 1 ≤ k ≤ n, let zk = xk + iyk . Then
1
=
≤
=
|z1 | + |z2 | + · · · + |zn |
(|x1 | + |y1 |) + (|x2 | + |y2 |) + · · · + (|xn | + |yn |)
X
X
X
X
|xk | +
|xk | +
|yk | +
|yk |.
xk ≥0
xk <0
yk ≥0
yk <0
By the Pigeonhole Principle, at least one of the above sums is not
less than 1/4. By symmetry, we may assume that
¯
¯
¯ X ¯
X
1
¯
¯
≤
|xk | = ¯
xk ¯.
¯
¯
4
xk ≥0
Consequently,
xk ≥0
¯
¯ ¯
¯
¯X ¯ ¯X ¯ 1
¯
¯ ¯
¯
zk ¯ ≥ ¯
xk ¯ ≥ .
¯
¯
¯ ¯
¯ 4
xk ≥0
xk ≥0
Comment: Using advanced mathematics, the lower bound can be
further improved to 1/π.
34. [Czech-Slovak Match 1998] A polynomial P (x) of degree n ≥ 5 with
integer coefficients and n distinct integer roots is given. Find all
integer roots of P (P (x)) given that 0 is a root of P (x).
86
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
Solution: The roots of P (x) are clearly integer roots of P (P (x));
we claim there are no other integer roots. We prove our claim by
contradiction. Suppose, on the contrary, that P (P (k)) = 0 for some
integer k such that P (k) 6= 0. Let
P (x) = a(x − r1 )(x − r2 )(x − r3 ) · · · (x − rn ),
where a, r1 , r2 , . . . , rn are integers,
r1 = 0 ≤ |r2 | ≤ |r3 | ≤ · · · ≤ |rn |.
Since P (k) 6= 0, we must have |k − ri | ≥ 1 for all i. Since the ri are
all distinct, at most two of |k − r2 |, |k − r3 |, |k − r4 | equal 1, so
|a(k − r2 ) · · · (k − rn−1 )| ≥ |a||k − r2 ||k − r3 ||k − r4 | ≥ 2,
and |P (k)| ≥ 2|k(k − rn )|. Also note that P (k) = ri0 for some i0 , so
|P (k)| ≤ |rn |. Now we consider the following two cases:
(a) |k| ≥ |rn |.
diction.
Then |P (k)| ≥ 2|k(k − rn )| ≥ 2|k| > |rn |, a contra-
(b) |k| < |rn |, that is, 1 ≤ |k| ≤ |rn | − 1. Let a, b, c be real
numbers, a ≤ b. For x ∈ [a, b], the function
f (x) = x(c − x)
reaches its minimum value at an endpoint x = a or x = b, or at
both endpoints. Thus
|k(k − rn )| = |k||rn − k| ≥ |k|(|rn | − |k|) ≥ |rn | − 1.
It follows that
|rn | ≥ |P (k)| ≥ 2|k(k − rn )| ≥ 2(|rn | − 1),
which implies that |rn | ≥ 2. Since n ≥ 5, this is only possible if
P (x) = (x + 2)(x + 1)x(x − 1)(x − 2).
But then it is impossible to have k 6= ri and |k| ≤ |rn |, a
contradiction.
87
Thus our assumption was incorrect, and the integer roots of P (P (x))
are exactly all the integer roots of P (x).
35. [Belarus 1999] Two real sequences x1 , x2 , . . . , and y1 , y2 , . . . , are
defined in the following way:
p
√
x1 = y1 = 3, xn+1 = xn + 1 + x2n ,
and
yn+1 =
y
pn
1 + 1 + yn2
for all n ≥ 1. Prove that 2 < xn yn < 3 for all n > 1.
First Solution: Let zn = 1/yn and note that the recursion for yn
is equivalent to
p
zn+1 = zn + 1 + zn2 .
√
Also note that z2 = 3 = x1 ; since the xi ’s and zi ’s satisfy the same
recursion, this means that zn = xn−1 for all n > 1. Thus,
xn yn =
xn
xn
=
.
zn
xn−1
q
Note that 1 + x2n−1 > xn−1 . Thus xn > 2xn−1 and xn yn > 2,
which is the lower bound of the desired inequality.
Since xn ’s are increasing for n > 1, we have
x2n−1 ≥ x21 = 3 >
which implies that
1
,
3
q
2xn−1 >
1 + x2n−1 .
Thus 3xn−1 > xn , which leads to the upper bound of the desired
inequality.
Second Solution: Setting xn = cot θn for 0 < θn < 90◦ yields
µ ¶
p
θn
2
xn+1 = cot θn + 1 + cot θn = cot θn + csc θn = cot
.
2
88
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
30◦
Since θ1 = 30◦ , we have in general θn = n−1 . Similar calculation
2
shows that
2 tan θn
.
yn = tan(2θn ) =
1 − tan2 θn
It follows that
2
.
xn yn =
1 − tan2 θn
Since tan θn 6= 0, tan2 θn is positive and xn yn > 2. And since for
n > 1 we have θn < 30◦ , we also have tan2 θn < 13 so that xn yn < 3.
Comment: From the closed forms for xn and yn in the second
1
solution, we can see the relationship yn = xn−1
used in the first
solution.
36. [China 1995] For a polynomial P (x), define the difference of P (x)
on the interval [a, b] ([a, b), (a, b), (a, b]) as P (b) − P (a). Prove that
it is possible to dissect the interval [0, 1] into a finite number of intervals and color them red and blue alternately such that, for every
quadratic polynomial P (x), the total difference of P (x) on red intervals is equal to that of P (x) on blue intervals. What about cubic
polynomials?
Solution: For an interval i, let ∆i P denote the difference of polynomial P on i. For a positive real number c and a set S ⊆ R, let
S + c denote the set obtained by shifting S in the positive direction
by c.
We prove a more general result.
Lemma Let ` be a positive real number, and let k be
a positive integer. It is always possible to dissect interval
Ik = [0, 2k `] into a finite number of intervals and color
them red and blue alternatively such that, for every polynomial P (x) with deg P ≤ k, the total difference of P (x)
on the red intervals is equal to that on the blue intervals.
Proof:
We induct on k.
For k = 1, we can just use intervals [0, `] and (`, 2`]. It is easy to see
that a linear or constant polynomial has the same difference on the
two intervals.
89
Suppose that the statement is true for k = n, where n is a positive
integer; that is, there exists a set Rn of red disjoint intervals and a
set Bn of blue disjoint intervals such that Rn ∩Bn = ∅, Rn ∪Bn = In ,
and, for any polynomials P (x) with deg P ≤ n, the total differences
of P on Rn is equal to that of P on Bn .
Now consider polynomial f (x) with deg f ≤ n + 1. Define
g(x) = f (x + 2n `) and h(x) = f (x) − g(x).
Then deg h ≤ n. By the induction hypothesis,
X
X
∆r h,
∆b h =
r∈Rn
b∈Bn
or
X
b∈Bn
It follows that
∆b f +
X
∆r g =
r∈Rn
X
X
∆r f +
r∈Rn
∆b f =
0
b∈Bn+1
X
X
∆b g.
r∈Bn
∆r f,
0
r∈Rn+1
where
0
= Rn ∪ (Bn + 2n `),
Rn+1
0
= Bn ∪ (Rn + 2n `).
Bn+1
0
0
both contain the number 2n `, that number may
and Bn+1
(If Rn+1
0
0
form
and Rn+1
be removed from one of them.) It is clear that Bn+1
a dissection of In+1 and, for any polynomial f with deg f ≤ n + 1,
0
0
.
is equal to that of f on Rn+1
the total difference of f on Bn+1
0
0
might
∪Rn+1
The only possible trouble left is that the colors in Bn+1
not be alternating (which can happen at the end of the In and the
beginning of In + 2n `). But note that if intervals i1 = [a1 , b1 ] and
i2 = [b1 , c1 ] are in the same color, then ∆i1 f + ∆i2 f = ∆i3 f, where
0
0
, we can simply put consecutive
∪ Rn+1
i3 = [a1 , c1 ]. Thus, in Bn+1
same color intervals into one bigger interval in the same-color. Thus,
there exists a dissection In+1 = Bn+1 ∪ Rn+1 such that, for every
polynomial f (x) with deg f ≤ n + 1,
X
X
∆b f =
∆r f.
b∈Bn+1
r∈Rn+1
90
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
This completes the induction and the proof of the lemma.
1
4
2
1
8
Setting first ` = and then ` = in the lemma, it is clear that the
answer to each of the given questions is “yes.”
37. [USSR 1990] Given a cubic equation
x3 + x2 + x +
= 0,
Mr. Fat and Mr. Taf are playing the following game. In one move,
Mr. Fat chooses a real number and Mr. Taf puts it in one of the
empty spaces. After three moves the game is over. Mr. Fat wins the
game if the final equation has three distinct integer roots. Who has
a winning strategy?
Solution: Mr. Fat has a winning strategy. Let the polynomial be
x3 + ax2 + bx + c. Mr. Fat can pick 0 first. We consider the following
cases:
(a) Mr. Taf chooses a = 0, yielding the polynomial equation x3 +
bx + c = 0. Mr. Fat then picks the number −(mnp)2 , where
m, n, and p are three positive integers such that m2 + n2 = p2 .
If Mr. Taf chooses b = −(mnp)2 , then Mr. Fat will choose c = 0.
The given polynomial becomes x(x − mnp)(x + mnp).
If Mr. Taf chooses c = −(mnp)2 , then Mr. Fat will choose
b = m2 n2 − n2 p2 − p2 m2 . The given polynomial becomes
(x + m2 )(x + n2 )(x − p2 ).
(b) Mr. Taf chooses b = 0, yielding the equation x3 + ax2 + c = 0.
Mr. Fat then picks the number m2 (m+1)2 (m2 +m+1)3 , where
m is an integer greater than 1.
If Mr. Taf chooses a = m2 (m + 1)2 (m2 + m + 1)3 , then Mr. Fat
can choose c = −m8 (m + 1)8 (m2 + m + 1)6 . The polynomial
becomes (x − mp)[x + (m + 1)p][x + m(m + 1)p], where p =
m2 (m + 1)2 (m2 + m + 1)2 .
If Mr. Taf chooses c = m2 (m + 1)2 (m2 + m + 1)3 , then Mr.
Fat can choose a = −(m2 + m + 1)2 . The polynomial becomes
(x + mq)[x − (m + 1)q][x − m(m + 1)q], where q = m2 + m + 1.
(c) Mr. Taf chooses c = 0. Then the problem reduces to problem 40
of the previous chapter. Mr. Fat needs only to pick two integers
91
a and b such that ab(a − 1)(b − 1) 6= 0 and a + b = −1. The
polynomial becomes either x(x − 1)(x − a) or x(x − 1)(x − b).
Our proof is complete.
Below is an example of what Mr. Fat and Mr. Taf could do:
F
0
”
”
T
a
”
b
F
−3600
”
4 · 9 · 73
T
b
c
a
F
0
−481
−28 · 38 · 76
”
”
”
”
c
”
”
2
”
c
a
b
−49
−3
−3
Roots
−60, 0, 60
−16, −9, 25
−8 · 27 · 49,
−4 · 27 · 49,
8 · 9 · 49
−14, 21, 42
−3, 0, 1
0, 1, 2
38. [Romania 1996] Let n > 2 be an integer and let f : R2 → R be a
function such that for any regular n-gon A1 A2 . . . An ,
f (A1 ) + f (A2 ) + · · · + f (An ) = 0.
Prove that f is the zero function.
Solution: We identify R2 with the complex plane and let ζ =
e2πi/n . Then the condition is that for any z ∈ C and any positive
real t,
n
X
f (z + tζ j ) = 0.
j=1
In particular, for each of k = 1, . . . , n, we obtain
n
X
f (z − ζ k + ζ j ) = 0.
j=1
Summing over k, we have
n X
n
X
f (z − (1 − ζ m )ζ k ) = 0.
m=1 k=1
For m = n the inner sum is nf (z); for other m, the inner sum again
runs over a regular polygon, hence is 0. Thus f (z) = 0 for all z ∈ C.
92
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
39. [IMO 1997 short list] Let p be a prime number and let f (x) be a
polynomial of degree d with integer coefficients such that:
(i) f (0) = 0, f (1) = 1;
(ii) for every positive integer n, the remainder upon division of f (n)
by p is either 0 or 1.
Prove that d ≥ p − 1.
First Solution: For the sake of the contradiction, assume that d ≤
p−2. Then by Lagrange’s interpolation formula the polynomial
f (x) is determined by its values at 0, 1, . . . , p − 2; that is,
p−2
X
f (x) =
f (k)
k=0
=
p−2
X
f (k)
k=0
x · · · (x − k + 1)(x − k − 1) · · · (x − p + 2)
k · · · 1 · (−1) · · · (k − p + 2)
x · · · (x − k + 1) (x − k − 1) · · · (x − p + 2)
.
k!(−1)p−k
(p − k − 2)!
Setting x = p − 1 gives
f (p − 1) =
p−2
X
k=0
≡
p−2
X
k=0
f (k)
(p − 1)(p − 2) · · · (p − k)
(−1)p−k k!
p−2
f (k)
X
(−1)k k!
f (k)
≡ (−1)p
p−k
(−1)
k!
(mod p).
k=0
It follows that
S(f ) := f (0) + f (1) + · · · + f (p − 1) ≡ 0 (mod p).
(1)
On the other hand, (ii) implies that S(f ) ≡ j (mod p), where j
denotes the number of those k ∈ {0, 1, . . . , p − 1} for which f (k) ≡ 1
(mod p). But (i) implies that 1 ≤ j ≤ p − 1. So S(f ) 6≡ 0 (mod p),
which contradicts (1). Thus our original assumption was wrong, and
our proof is complete.
Second Solution: Again, we approach the problem indirectly.
Assume that d ≤ p − 2, and let
f (x) = ap−2 xp−2 + · · · + a1 x + a0 .
93
Then
S(f ) =
p−1
X
f (k) =
p−1
X
i
ai k =
k=0 i=0
k=0
where Si =
p−1 X
p−2
X
p−2
X
ai
i=0
p−1
X
i
k =
k=0
p−2
X
ai Si ,
i=0
ki .
k=0
We claim that Si ≡ 0 (mod p) for all i = 0, 1, . . . , p − 2. We use
strong induction on i to prove our claim. The statement is true for
i = 0 as S0 = p. Now suppose that S0 ≡ S1 ≡ · · · ≡ Si−1 ≡ 0
(mod p) for some 1 ≤ i ≤ p − 2. Note that
0 ≡ pi+1 =
p
X
k i+1 −
k=1
=
p−1
X
k i+1 =
k=0
p−1
X
[(k + 1)i+1 − k i+1 ]
k=0
p−1 X
i µ
X
¶
¶
i−1 µ
X
i+1 j
i+1
k = (i + 1)Si +
Sj
j
j
j=0
k=0 j=0
≡ (i + 1)Si
(mod p)
Since 0 < i + 1 < p, it follows that Si ≡ 0 (mod p). This completes
the induction and the proof of the claim. Therefore,
S(f ) =
p−2
X
ai Si ≡ 0 (mod p).
i=0
The rest is the same as in the first solution.
40. Let n be a given positive integer. Consider the sequence a0 , a1 , · · · , an
1
with a0 = and
2
a2
ak = ak−1 + k−1 ,
n
for k = 1, 2, · · · , n. Prove that
1−
1
< an < 1.
n
First Solution: We prove a stronger statement: For k = 1, 2, . . . , n,
n
n+1
< ak <
.
2n − k + 2
2n − k
(1)
94
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
We use induction to prove both inequalities.
We first prove the upper bound. For k = 1, it is easy to check that
a1 =
1
1
2n + 1
n
+
=
<
.
2 4n
4n
2n − 1
Suppose that
ak <
n
,
2n − k
for some positive integer k < n. Then
ak+1
=
1
ak
=
(n + ak ) <
n
2n − k
µ
n+
n
2n − k
¶
n(2n − k + 1)
n
,
<
(2n − k)2
2n − k − 1
as (2n − k + 1)(2n − k − 1) = (2n − k)2 − 1 < (2n − k)2 . Thus our
induction step is complete. In particular, for k = n − 1,
an = ak+1 <
n
= 1,
2n − (n − 1) − 1
as desired.
Now we prove the upper bound. For k = 1, it is easy to check that
a1 =
2n + 1
n+1
>
.
4n
2n + 1
Suppose that
ak >
n+1
,
2n − k + 2
for some positive integer k < n. Then
ak+1 = ak +
a2k
n+1
(n + 1)2
>
+
.
n
2n − k + 2 n(2n − k + 2)2
95
It follows that
n+1
2n − k + 1
n+1
(n + 1)2
≥−
+
(2n − k + 1)(2n − k + 2) n(2n − k + 2)2
µ
¶
n + 1 2n − k + 2
n+1
−
=
2n − k + 2)2
n
2n − k + 1
µ
¶
1
n+1
1
−
=
> 0.
2n − k + 2)2 n 2n − k + 1
ak+1 −
This complete the induction step. In particular, for n = k − 1, we
obtain
an = ak+1 >
n+1
n+1
1
1
=
=1−
>1− ,
2n − (n − 1) + 1
n+2
n+2
n
as desired.
Second Solution: Rewriting the given condition as
1
1
=
ak
ak−1 +
a2k−1
n
yields
n
1
1
=
−
ak−1 (n + ak−1 )
ak−1
n + ak−1
=
1
ak−1
−
1
1
=
,
ak
n + ak−1
for k = 1, 2, . . . , n.
It is clear that ak ’s are increasing. Thus
an > an−1 > · · · > a0 =
1
.
2
Thus (2) implies that
1
1
1
−
<
ak−1
ak
n
for k = 1, 2, . . . , n. Telescoping summation gives
1
1
−
<1
a0
an
(2)
96
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
or
1
1
>
− 1 = 2 − 1 = 1,
an
a0
that is, an < 1, which gives the desired upper bound.
1
Since an < 1, and, since ak ’s are increasing, = a0 < ak ≤ an < 1
2
for k = 1, 2, . . . , n. Then (2) implies
1
1
1
1
−
=
>
,
ak−1
ak
n + ak−1
n+1
for k = 1, 2, . . . , n. Telescoping sum gives
1
1
n
−
>
a0
an
n+1
or
1
1
n
n+2
=
,
<
−
an
a0
n+1
n+1
that is,
n+1
1
1
=1−
>1− ,
n+2
n+2
n
which is the desired lower bound.
an >
41. [IMO 1996 short list] Let a1 , a2 , . . . , an be nonnegative real numbers,
not all zero.
(a) Prove that xn − a1 xn−1 − · · · − an−1 x − an = 0 has precisely
one positive real root R.
Pn
Pn
(b) Let A = j=1 aj and B = j=1 jaj . Prove that AA ≤ RB .
Solution:
(a) Consider the function
f (x) =
a1
a2
an
+ 2 + ··· + n.
x
x
x
Note that f decreases from ∞ to 0 as x increases from 0 to ∞.
Hence there is a unique real number R such that f (R) = 1,
that is, there exists a unique positive real root R of the given
polynomial.
97
P
(b) Let cj = aj /A. Then cj ’s are non-negative and
cj = 1. Since
− ln x is a convex function on the interval (0, ∞), by Jensen’s
inequality,


µ
¶
n
n
X
X
A
A
cj − ln j ≥ − ln 
cj j  = − ln (f (R)) = 0.
R
R
j=1
j=1
It follows that
n
X
cj (− ln A + j ln R) ≥ 0
j=1
or
n
X
j=1
cj ln A ≤
n
X
jcj ln R.
j=1
Substituting cj = aj /A, we obtain the desired inequality.
Comment: Please compare the solution of (a) with that of the
problem 15 in the last chapter.
42. Prove that there exists a polynomial P (x, y) with real coefficients
such that P (x, y) ≥ 0 for all real numbers x and y, which cannot be
written as the sum of squares of polynomials with real coefficients.
Solution: We claim that
P (x, y) = (x2 + y 2 − 1)x2 y 2 +
1
27
is a polynomial satisfying the given conditions.
First we prove that P (x, y) ≥ 0 for all real numbers x and y. If
x2 + y 2 − 1 ≥ 0, then it clear that P (x, y) > 0; if x2 + y 2 − 1 < 0,
then applying the AM-GM inequality gives
µ
¶3
1 − x2 − y 2 + x2 + y 2
1
2
2 2 2
(1 − x − y )x y ≤
=
,
3
27
or
(x2 + y 2 − 1)x2 y 2 ≥ −
1
.
27
98
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
It follows that P (x, y) ≥ 0.
We are left to prove that P (x, y) cannot be written as the sum of
squares of polynomials with real coefficients. For the sake of conn
X
tradiction, assume that P (x, y) =
Qi (x, y)2 . Since deg P = 6,
i=1
deg Qi ≤ 3. Thus
Qi (x, y) =
Ai x3 + Bi x2 y + Ci xy 2 + Di y 3
+Ei x2 + Fi xy + Gi y 2 + Hi x + Ii y + Ji .
Pn
Comparing the coefficients, in P (x, y) and i=1 Qi (x, y)2 , of terms
n
n
P 2
P 2
x6 and y 6 gives
Ai =
Di = 0, or Ai = Di = 0 for all i.
i=1
i=1
Then, comparing those of x4 and y 4 gives
n
P
i=1
Ei2 =
n
P
i=1
2
G2i = 0, or
Ei = Gi = 0 for all i. Next, comparing those of x and y 2 gives
n
n
P
P
Hi2 =
Ii2 = 0, or Hi = Ii = 0 for all i. Thus,
i=1
i=1
Qi (x, y) = Bi x2 y + Ci xy 2 + Fi xy + Ji .
But, finally, comparing the coefficients of the term x2 y 2 , we have
n
P
Fi2 = −1, which is impossible for real numbers Fi . Thus our
i=1
assumption is wrong, and our proof is complete.
43. [IMO 1996 short list] For each positive integer n, show that there
exists a positive integer k such that
k = f (x)(x + 1)2n + g(x)(x2n + 1)
for some polynomials f, g with integer coefficients, and find the smallest such k as a function of n.
Solution: First we show that such a k exists. Note that x + 1
divides 1 − x2n . Then for some polynomial a(x) with integer coefficients, we have
(1 + x)a(x) = 1 − x2n = 2 − (1 + x2n ),
99
or
2 = (1 + x)a(x) + (1 + x2n ).
Raising both sides to the (2n)th power, we obtain
22n = (1 + x)2n (a(x))2n + (1 + x2n )b(x),
where b(x) is a polynomial with integer coefficients. This shows that
a k satisfying the condition of the problem exists. Let k0 be the
minimum such k.
Let 2n = 2r · q, where r is a positive integer and q is an odd integer.
We claim that k0 = 2q .
First we prove that 2q divides k0 . Let t = 2r . Note that x2n + 1 =
(xt + 1)Q(x), where
Q(x) = xt(q−1) − xt(q−2) + · · · − xt + 1.
The roots of xt + 1 are
µ
¶
µ
¶
(2m − 1)π
(2m − 1)π
ωm = cos
+ i sin
,
t
t
m = 1, 2, . . . , t;
that is,
R(x) = xt + 1 = (x − ω1 )(x − ω2 ) · · · (x − ωt ).
Let f (x) and g(x) be polynomials with integer coefficients such that
k0 = f (x)(x + 1)2n + g(x)(x2n + 1)
= f (x)(x + 1)2n + g(x)Q(x)(xt + 1).
It follows that
f (ωm )(ωm + 1)2n = k0 ,
1 ≤ m ≤ t.
(1)
Since r is positive, t is even. So
2 = R(−1) = (1 + ω1 )(1 + ω2 ) · · · (1 + ωt ).
Since f (ω1 )f (ω2 ) · · · f (ωt ) is a symmetric polynomial in ω1 , ω2 , . . . , ωt
with integer coefficients, it can be expressed as a polynomial with
integer coefficients in the elementary symmetric functions in
100
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
ω1 , ω2 , . . . , ωt and therefore F = f (ω1 )f (ω2 ) · · · f (ωt ) is an integer.
Taking the product over m = 1, 2, . . . , t, (1) gives 22n F = k0t or
r
r
22 ·q F = k02 . It follows that 2q divides k0 .
It now suffices to prove that k0 ≤ 2q . Note that Q(−1) = 1. It
follows that Q(x) = (x + 1)c(x) + 1, where c(x) is a polynomial with
integer coefficients. Hence
(x + 1)2n (c(x))2n = (Q(x) − 1)2n = Q(x)d(x) + 1,
(2)
for some polynomial d(x) with integer coefficients. Also observe that,
for any fixed m,
2j−1
{ωm
: j = 1, 2, . . . , t} = {ω1 , ω2 , . . . , ωt }.
Thus
3
2t−1
(1 + ωm )(1 + ωm
) · · · (1 + ωm
) = R(−1) = 2,
and writing
2j−1
2
2j−2
1 + ωm
= (1 + ωm )(1 − ωm + ωm
− · · · + ωm
),
we find that for some polynomial h(x), independent of m, with integer coefficients such that
(1 + ωm )t h(ωm ) = 2.
But then (x + 1)t h(x) − 2 is divisible by xt + 1 and hence we can
write
(x + 1)t h(x) = 2 + (xt + 1)u(x),
for some polynomial u(x) with integer coefficients. Raising both
sides to the power q we obtain
(x + 1)2n (h(x))q = 2q + (xt + 1)v(x),
(3)
where v(x) is a polynomial with integer coefficients. Using (2) and
(3) we obtain
(x + 1)2n (c(x))2n (xt + 1)v(x)
= Q(x)d(x)(xt + 1)v(x) + (xt + 1)v(x)
= Q(x)d(x)(xt + 1)v(x) + (x + 1)2n (h(x))q − 2q ,
101
that is,
2q = f1 (x)(x + 1)2n + g1 (x)(x2n + 1),
where f1 (x) and g1 (x) are polynomials with integer coefficients.
Hence k0 ≤ 2q , as desired. Our proof is thus complete.
44. [USAMO 1998 proposal, Kiran Kedlaya] Let x be a positive real
number.
(a) Prove that
∞
X
1
(n − 1)!
= .
(x
+
1)
·
·
·
(x
+
n)
x
n=1
(b) Prove that
∞
X
∞
X
(n − 1)!
1
=
.
n(x
+
1)
.
.
.
(x
+
n)
(x
+
k)2
n=1
k=1
Solution: We use infinite telescoping sums to solve the problem.
(a) Equivalently, we have to show that
∞
X
n!x
= 1.
n(x
+
1)
· · · (x + n)
n=1
Note that
x
1
1
= −
.
n(x + n)
n x+n
It follows that
n!x
n(x + 1) · · · (x + n)
(n − 1)!
n!
=
−
,
(x + 1) · · · (x + n − 1) (x + 1) · · · (x + n)
and this telescoping summation yields the desired result.
(b) Let
f (x) =
∞
X
(n − 1)!
.
n(x + 1) . . . (x + n)
n=1
102
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
1
Then, by (a), f (x) < . In particular, f (x) converges to 0 as
x
x approaches ∞, so we can write f as an infinite telescoping
series
∞
X
f (x) =
[f (x + k − 1) − f (x + k)].
(1)
k=1
On the other hand, the result in (a) gives
f (x − 1) − f (x)
∞
X
µ
1
1
−
x x+n
=
(n − 1)!
n(x
+
1)
. . . (x + n − 1)
n=1
=
∞
1X
1
(n − 1)!
= 2.
x n=1 (x + 1) · · · (x + n)
x
¶
Substituting the last equation to (1) gives
f (x) =
∞
X
k=1
1
,
(x + k)2
as desired.
45. [Romania 1996] Let n ≥ 3 be an integer, and let
X ⊆ S = {1, 2, . . . , n3 }
be a set of 3n2 elements. Prove that one can find nine distinct
numbers ai , bi , ci (i = 1, 2, 3) in X such that the system
a1 x + b1 y + c1 z
=
0
a2 x + b2 y + c2 z
a3 x + b3 y + c3 z
=
=
0
0
has a solution (x0 , y0 , z0 ) in nonzero integers.
Solution: Label the elements of X in increasing order x1 < · · · <
x3n2 , and put
X1
X2
X3
=
=
=
{x1 , . . . , xn2 },
{xn2 +1 , . . . , x2n2 },
{x2n2 +1 , . . . , x3n2 }.
103
Define the function f : X1 × X2 × X3 → S × S as follows:
f (a, b, c) = (b − a, c − b).
The domain of f contains n6 elements. The range of f , on the other
hand, is contained in the subset of S × S of pairs whose sum is at
most n3 , a set of cardinality
3
nX
−1
k=1
k=
n6
n3 (n3 − 1)
<
.
2
2
By the Pigeonhole Principle, some three triples (ai , bi , ci ) (i =
1, 2, 3) map to the same pair, in which case x = b1 − c1 , y = c1 −
a1 , z = a1 − b1 is a solution in nonzero integers. Note that ai cannot
equal bj since X1 and X2 are disjoint, and that a1 = a2 implies that
the triples (a1 , b1 , c1 ) and (a2 , b2 , c2 ) are identical, a contradiction.
Hence the nine numbers chosen are indeed distinct.
46. [Xuanguo Huang] Let n ≥ 3 be an integer and let x1 , x2 , · · · , xn be
n
X
1
positive real numbers. Suppose that
= 1. Prove that
1 + xj
j=1
√
x1 +
√
x2 + · · · +
√
µ
xn ≥ (n − 1)
1
1
1
√ + √ + ··· + √
x1
x2
xn
¶
.
Solution: By symmetry, we may assume that x1 ≤ x2 ≤ · · · ≤ xn .
We have the following lemma.
Lemma For 1 ≤ i < j ≤ n,
√
√
xj
xi
≥
.
1 + xi
1 + xj
Proof:
Since n ≥ 3, and, since
n
X
i=1
1>
1
= 1,
1 + xi
1
1
2 + xi + xj
+
=
1 + xi
1 + xj
(1 + xi )(1 + xj )
104
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
or
1 + xi + xj + xi xj > 2 + xi + xj .
It follows that xi xj > 1. Thus
√
√
√
√
xj
xi (1 + xj ) − xj (1 + xi )
xi
−
=
1 + xi
1 + xj
(1 + xi )(1 + xj )
√
√
√
( xi − xj )(1 − xi xj )
=
≥ 0,
(1 + xi )(1 + xj )
as desired.
2
By the lemma, we have
√
√
√
x1
x2
xn
≥
≥ ··· ≥
,
1 + x1
1 + x2
1 + xn
and, since
1
1
1
√ ≥ √ ≥ ··· ≥ √ ,
x1
x2
xn
it follows by the Chebyshev inequality
√
√ ¶ X
n
n
n µ
n
X
xi
xi
1X 1 X
1
1
≤
=
= 1. (1)
√
√ ·
n i=1 xi i=1 1 + xi
x
1
+
x
1
+
xi
i
i
i=1
i=1
By the Cauchy-Schwartz inequality, we have
√
n
n
X
xi X 1 + xi
≥ n2 ,
√
1
+
x
x
i
i
i=1
i=1
or
!
à n
√
n
n
X
X 1
X
xi
√
xi ≥ n2 .
√ +
1
+
x
x
i
i
i=1
i=1
i=1
n
(2)
1X 1
Multiplying by
√ on both sides of (2) and applying (1) gives
n i=1 xi
n
n
n
X
X
X
√
1
1
xi ≥ n
√ +
√ ,
x
xi
i
i=1
i=1
i=1
which in turn implies the desired inequality.
105
47. Let x1 , x2 , . . . , xn be distinct real numbers. Define the polynomials
P (x) = (x − x1 )(x − x2 ) · · · (x − xn )
and
µ
Q(x) = P (x)
1
1
1
+
+ ··· +
x − x1
x − x2
x − xn
¶
.
Let y1 , y2 , . . . , yn−1 be the roots of Q. Show that
min |xi − xj | < min |yi − yj |.
i6=j
Solution:
i6=j
By symmetry, we may assume that d = min |yi − yj | =
i6=j
y2 − y1 . Let sk = y1 − xk , for k = 1, 2, . . . , n. By symmetry, we may
also assume that s1 < s2 < · · · < sn , i.e., x1 > x2 > · · · > xn .
For the sake of contradiction, assume that
d ≤ min |xi − xj | = min xi − xj = min sj − si .
i<j
i6=j
i<j
(1)
Since P has no double roots, it shares non with Q. Then
µ
¶
1
1
1
P (yi )
+
+ ··· +
= Q(y1 ) = 0,
yi − x1
yi − x2
yi − xn
or
1
1
1
+
+ ··· +
= 0.
yi − x1
yi − x2
yi − xn
In particular, setting i = 1 and i = 2 gives
n
n
X
X
1
1
=
= 0.
sk
sk + d
k=1
(2)
k=1
We claim that there is a k such that sk (sk + d) < 0, otherwise, we
n
X
1
1
1
have
<
for all k, which in turn implies that
<
sk + d
sk
sk + d
k=1
n
X
1
, which contradicts (2).
sk
k=1
106
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
Let j be the number of k’s such that sk (sk + d) < 0, that is, sk <
0 < sk + d. A simple but critical fact is that sk + d and sk+j have
the same sign. In fact, suppose that
s1 < · · · < si < si+1 < · · · < si+j < 0 < si+j+1 < · · · < sn ;
s1 + d < · · · < si + d < 0 < si+1 + d < · · · < sn + d.
Then sk+j > 0 if and only if k > i + 1, that is sk + d > 0.
From (1), we obtain sk + d ≤ sk+j , and, since sk + d and sk+j have
the same sign, we obtain
1
1
≥
,
sk + d
sk+j
for all k = 1, 2, · · · , n − j. Therefore,
n−j
X
k=1
or
1
sk+j
n
X
k=j+1
Also note that
≤
1
sk+j
j
X
1
<0<
sk
k=1
n−j
X
k=1
≤
1
,
sk + d
n−j
X
k=1
1
.
sk + d
n
X
k=n−j+1
1
.
sk + d
(3)
(4)
Adding (3) and (4) yields
n
n
X
X
1
1
<
,
sk
sk + d
k=1
k=1
which contradicts (2). Thus our assumption is wrong and our proof
is complete.
48. [Romania 1998] Show that for any positive integer n, the polynomial
n
f (x) = (x2 + x)2 + 1
cannot be written as the product of two non-constant polynomials
with integer coefficients.
107
Solution: Note that f (x) = g(h(x)), where h(x) = x2 + x and
n
g(y) = y 2 + 1. Since
g(y + 1) = (y + 1)
2n
+1=y
2n
Ã2n −1 µ ¶ !
X 2n
+
y k + 2,
k
k=1
¡ n¢
and 2k is even for 1 ≤ k ≤ 2n − 1, g is irreducible, by the Eisenstein’s criterion. Now let p be a non-constant factor of f , and let r
be a root of p. Then g(h(r)) = f (r) = 0, so s := h(r) is a root of g.
Since s = r2 + r ∈ Q(r), we have Q(s) ⊂ Q(r), so
deg p ≥ deg(Q(r)/Q) ≥ deg(Q(s)/Q) = deg g = 2n .
Thus every factor of f has degree at least 2n . Therefore, if f is
reducible, we can write f (x) = A(x)B(x) where A and B have degree
2n .
Next, observe that
f (x) ≡
≡
n
(x2 + x)2 + 1
x2
n+1
n
+ x2 + 1 ≡ (x2 + x + 1)2
n
(mod 2).
Since x2 + x + 1 is irreducible in Z2 [x], by unique factorization we
must have
A(x) ≡ B(x) ≡ (x2 + x + 1)2
n−1
n
≡ x2 + x2
n−1
+1
(mod 2).
Thus, if we write
n
A(x) = a2n x2 + · · · + a0 ,
n
B(x) = b2n x2 + · · · + b0 ,
then a2n , a2n−1 , a0 , b2n , b2n−1 , b0 are odd and all the other coefficients are even. Since f is monic, we may assume without loss of
generality that a2n = b2n = 1; also, a0 b0 = f (0) = 1, but a0 > 0,
108
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
b0 > 0 as f has no real roots, so a0 = b0 = 1. Therefore,
([x2
n
+2n−1
Ã
≡
] + [x2
n−1
])(g(x)h(x))

! 2n−1
2n
X
X
ai b2n +2n−1 −i + 
ai b2n−1 −i 
i=0
i=2n−1
≡ a2n b2n−1 + a2n−1 b2n + a0 b2n−1 + a2n−1 b0
≡ 2(a2n−1 + b2n−1 ) ≡ 0
as a2n−1 + b2n−1 is even. But
([x
and
¡
2n +2n−1
2n −1
2n−1 −1
2n−1
] + [x
])(f (x)) =
(mod 4)
µ
2n
2n−1
¶
µ
¶
2n − 1
= 2 n−1
,
2
−1
¢
is odd by Lucas’s theorem, so
¡ 2n +2n−1
¢
n−1 ¢¡
[x
] + [x2 ] f (x) ≡ 2 (mod 4),
a contradiction. Hence f is irreducible.
49. [Iran 1998] Let f1 , f2 , f3 : R → R be functions such that
a1 f1 + a2 f2 + a3 f3
is monotonic for all a1 , a2 , a3 ∈ R. Prove that there exist c1 , c2 , c3 ∈
R, not all zero, such that
c1 f1 (x) + c2 f2 (x) + c3 f3 (x) = 0
for all x ∈ R.
First Solution: We establish the following lemma.
Lemma Let f, g : R → R be functions such that f is
nonconstant and af + bg is monotonic for all a, b ∈ R.
Then there exists c ∈ R such that g − cf is a constant
function.
Proof: Let s, t be two real numbers such that f (s) 6= f (t). Let
u = (g(s) − g(t))/(f (s) − f (t)). Let h1 = g − d1 f for some d1 ∈ R.
Then h1 is monotonic. But
h1 (s) − h1 (t) = g(s) − g(t) − d1 (f (s) − f (t)) = (f (s) − f (t))(u − d1 ).
109
Since f (s) − f (t) 6= 0 is fixed, the monotonicity of h1 depends only
on the sign of u − d1 .
Since f is nonconstant, there exist x1 , x2 ∈ R such that f (x1 ) 6=
f (x2 ). Let
g(x1 ) − g(x2 )
c=
f (x1 ) − f (x2 )
and h = g − cf . Then r = h(x1 ) = h(x2 ) and the monotonicity of
h1 = g − d1 f , for each d1 , depends only on the sign of c − d1 . We
claim that h = g − cf is a constant function.
We prove our claim by contradiction. Suppose, on the contrary, that
there exists x3 ∈ R such that h(x3 ) 6= r. Since f (x1 ) 6= f (x2 ), at
least one of f (x1 ) 6= f (x3 ) and f (x2 ) 6= f (x3 ) is true. Without loss
of generality, suppose that f (x1 ) 6= f (x3 ). Let
c0 =
g(x1 ) − g(x3 )
.
f (x1 ) − f (x3 )
Then the monotonicity of h1 also depends only on the sign of c0 − d1 .
Since h(x3 ) 6= r = h(x1 ),
c 6=
g(x1 ) − g(x3 )
= c0 ;
f (x1 ) − f (x3 )
hence c − d1 6= c0 − d1 . So there exists some d1 such that h1 is both
strictly increasing and decreasing, which is impossible. Therefore
our assumption is false and h is a constant function.
2
Now we prove our main result. If f1 , f2 , f3 are all constant functions,
the result is trivial. Without loss of generality, suppose that f1 is
nonconstant. For a3 = 0, we apply the lemma to f1 and f2 , so
f2 = cf1 + d; for a2 = 0, we apply the lemma to f1 and f3 , so
f3 = c0 f1 + d0 . Here c, c0 , d, d0 are constant. We have
(c0 d − cd0 )f1 + d0 f2 − df3
= (c0 d − cd0 )f1 + d0 (cf1 + d) − d(c0 f1 + d0 ) = 0.
If (c0 d − cd0 , d0 , −d) 6= (0, 0, 0), then let
(c1 , c2 , c3 ) = (c0 d − cd0 , d0 , −d)
110
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
and we are done. Otherwise, d = d0 = 0 and f2 , f3 are constant
multiples of f1 . Then the problem is again trivial.
Second Solution: Define the vector
v(x) = (f1 (x), f2 (x), f3 (x))
for x ∈ R. If the v(x) span a proper subspace of R3 , we can find
a vector (c1 , c2 , c3 ) orthogonal to that subspace, and then c1 f1 (x) +
c2 f2 (x) + c3 f3 (x) = 0 for all x ∈ R.
So suppose the v(x) span all of R3 . Then there exist x1 < x2 < x3 ∈
R such that v(x1 ), v(x2 ), v(x3 ) are linearly independent, and so the
3 × 3 matrix A with Aij = fj (xi ) has linearly independent rows. But
then A is invertible, andP
its columns also span R3 . This means we can
3
find c1 , c2 , c3 such that i=1 ci (fi (x1 ), fi (x2 ), fi (x3 )) = (0, 1, 0), and
the function c1 f1 +c2 f2 +c3 f3 is then not monotonic, a contradiction.
50. [USAMO 1999 proposal, Richard Stong] Let x1 , x2 , . . . , xn be variables, and let y1 , y2 , . . . , y2n −1 be the sums of nonempty subsets of
xi . Let pk (x1 , . . . , xn ) be the k th elementary symmetric polynomial in the yi (the sum of every product of k distinct yi ’s). For
which k and n is every coefficient of pk (as a polynomial in x1 , . . . , xn )
even?
For example, if n = 2, then y1 , y2 , y3 are x1 , x2 , x1 + x2 and
p1 = y1 + y2 + y3 = 2x1 + 2x2 ,
p2 = y1 y2 + y2 y3 + y3 y1 = x21 + x22 + 3x1 x2 ,
p3 = y1 y2 y3 = x21 x2 + x1 x22 .
Solution: We say a polynomial pk is even if every coefficient of
pk is even. Otherwise, we say pk is not even. For any fixed positive
integer n, we say a nonnegative integer k is bad for n if k = 2n − 2j
for some nonnegative integer j. We will show by induction on n that
pk (x1 , x2 , · · · , xn ) is not even if and only if k is bad for n.
For n = 1, p1 (x1 ) = x1 is not even and k = 1 is bad for n = 1 as
k = 1 = 21 − 20 = 2n − 20 .
111
Suppose that the claim is true for a certain n. We now consider
pk (x1 , x2 , . . . , xn+1 ). Let σk (y1 , y2 , . . . , ys ) be the k th elementary symmetric polynomial. We have the following useful, but easy to prove,
facts:
(a) σk (y1 , y2 , · · · , ys , 0) = σk (y1 , y2 , · · · , ys );
(b) For all 1 ≤ r ≤ s,
σk (y1 , . . . , ys ) =
X
[σi (y1 , · · · , yr )σj (yr+1 , · · · , ys )];
i+j=k
(c)
σk (x + y1 , x + y2 , . . . , x + ys )
X
=
(x + yi1 )(x + yi2 ) · · · (x + yik )
i1 <i2 <···<ik
X
=
k
X
i1 <i2 <···<ik r=0
=
X
ys1 ys2 · · · ysr xk−r
s1 <s2 <···<sr
{s1 ,··· ,sr }⊆{i1 ,··· ,ik }
¶
k µ
X
s−r
σr (y1 , · · · , ys )xk−r .
k
−
r
r=0
Hence
pk (x1 , x2 , · · · , xn+1 )
X
[pi (x1 , · · · , xn ) ·
=
i+j=k
σj (xn+1 , x1 + xn+1 , · · · , x1 + x2 + · · · + xn+1 )]
¶
j µ n
X X
2 −r
=
pi (x1 , · · · , xn )pr (x1 , · · · , xn )xj−r
n+1 .
j
−
r
r=0
i+j=k
By the induction hypothesis, every term of pr (x1 , x2 · · · , xn ) is even
unless r = 2n − 2t , for some 0 ≤ t ≤ n. For such r, note that
µ n
¶ µ t ¶
2 −r
2
=
j−r
j−r
112
CHAPTER 4. SOLUTIONS TO ADVANCED PROBLEMS
is even unless j − r = 0 or j − r = 2t . Therefore, taking coefficients
modulo 2,
pk (x1 , x2 , · · · , xn+1 )
X
≡
pi (x1 , x2 , · · · , xn )pj (x1 , x2 , · · · xn )
i+j=k
+
n
X
t
pk−2n (x1 , x2 , · · · , xn )p2n −2t (x1 , x2 , · · · , xn )x2 .
t=0
By the induction hypothesis, the terms in the first sum are even
unless k − 2n = 2n − 2u for some 0 ≤ u ≤ n, that is k = 2n+1 −
2u . In the second sum, every term appears twice except the term
pk/2 (x1 , x2 , · · · , xn )2 , for k even. By the induction hypothesis, this
term is even unless k/2 = 2n − 2v , for some 0 ≤ v ≤ n, that is
k = 2n+1 − 2v+1 . It follows that pk (x1 , x2 , · · · xn+1 ) is even unless
k = 2n+1 − 2w for some 0 ≤ w ≤ n + 1, i.e., k is bad for n + 1.
Furthermore, note that the odd coefficients in pk (x1 , x2 , · · · , xn+1 )
occur for different powers of xn+1 . Therefore, the condition that k
is bad for n + 1 is also sufficient for pk (x1 , x2 , · · · , xn+1 ) to be odd.
Our induction is complete.
Glossary
Arithmetic-Geometric Mean Inequality (AM–GM Inequality)
If a1 , a2 , . . . , an are n nonnegative numbers, then
1
1
(a1 + a2 + · · · + an ) ≥ (a1 a2 · · · an ) n
n
with equality if and only if a1 = a2 = · · · = an .
Binomial Coefficient
µ ¶
n
n!
=
,
k
k!(n − k)!
the coefficient of xk in the expansion of (x + 1)n .
Cauchy-Schwarz Inequality
For any real numbers a1 , a2 , . . . , an , and b1 , b2 , . . . , bn
(a21 + a22 + · · · + a2n )(b21 + b22 + · · · + b2n )
≥ (a1 b1 + a2 b2 + · · · + an bn )2
with equality if and only if ai and bi are proportional, i = 1, 2, . . . , n.
113
114
100 Algebra Problems from USA IMO Training
Chebyshev Inequality
1. Let x1 , x2 . . . , xn and y1 , y2 , . . . , yn be two sequences of real numbers,
such that x1 ≤ x2 ≤ · · · ≤ xn and y1 ≤ y2 ≤ · · · ≤ yn . Then
1
(x1 + x2 + · · · + xn )(y1 + y2 + · · · + yn ) ≤ x1 y1 + x2 y2 + · · · + xn yn .
n
2. Let x1 , x2 . . . , xn and y1 , y2 , . . . , yn be two sequences of real numbers,
such that x1 ≥ x2 ≥ · · · ≥ xn and y1 ≥ y2 ≥ · · · ≥ yn . Then
1
(x1 + x2 + · · · + xn )(y1 + y2 + · · · + yn ) ≥ x1 y1 + x2 y2 + · · · + xn yn .
n
De Moivre’s Formula
For any angle α and for any integer n,
(cos α + i sin α)n = cos nα + i sin nα.
Elementary Symmetric Polynomials (Functions)
Given indeterminates x1 , . . . , xn , the elementary symmetric functions
s1 , . . . , sn are defined by the relation (in another indeterminate t)
(t + x1 ) · · · (t + xn ) = tn + s1 tn−1 + · · · + sn−1 t + sn .
That is, sk is the sum of the products of the xi taken k at a time. It
is a basic result that every symmetric polynomial in x1 , . . . , xn can be
(uniquely) expressed as a polynomial in the si , and vice versa.
Fibonacci Numbers
Sequence defined recursively by F1 = F2 = 1, Fn+2 = Fn+1 + Fn , for all
n ∈ N.
Jensen’s Inequality
If f is concave up on an interval [a, b] and λ1 , λ2 , . . ., λn are nonnegative
numbers with sum equal to 1, then
λ1 f (x1 ) = λ2 f (x2 ) + · · · + λn f (xn ) ≥ f (λ1 x1 + λ2 x2 + · · · + λn xn )
for any x1 , x2 , . . . , xn in the interval [a, b]. If the function is concave down,
the inequality is reversed.
Glossary
115
Lagrange’s Interpolation Formula
Let x0 , x1 , . . . , xn be distinct real numbers, and let y0 , y1 , . . . , yn be arbitrary real numbers. Then there exists a unique polynomial P (x) of degree
at most n such that P (xi ) = yi , i = 0, 1, . . . , n. This is the polynomial
given by
P (x) =
n
X
i=0
yi
(x − x0 ) · · · (x − xi−1 )(x − xi+1 ) · · · (x − xn )
.
(xi − x0 ) · · · (xi − xi−1 )(xi − xi+1 ) · · · (xi − xn )
Law of Cosines
Let ABC be a triangle. Then
BC 2 = AB 2 + AC 2 − 2AB · AC cos C.
Lucas’ Theorem
Let p be a prime; let a and b be two positive integers such that
a = ak pk + ak−1 pk−1 + · · · a1 p + a0 , b = bk pk + bk−1 pk−1 + · · · b1 p + b0 ,
where 0 ≤ ai , bi < p are integers for i = 0, 1, . . . , k. Then
µ ¶ µ ¶µ
¶ µ ¶µ ¶
a
ak
ak−1
a1
a0
≡
···
(mod p).
b
bk
bk−1
b1
b0
Pigeonhole Principle
If n objects are distributed among k < n boxes, some box contains at least
two objects.
Root Mean Square-Arithmetic Mean Inequality (RMS–
AM Inequality)
For positive numbers x1 , x2 , . . . , xn ,
r
x21 + x22 + · · · + x2k
x1 + x2 + · · · + xk
≥
.
n
n
116
100 Algebra Problems from USA IMO Training
More generally, let a1 , a2 , . . . , an be any positive numbers for which a1 +
a2 + · · · + an = 1. For positive numbers x1 , x2 , . . . , xn we define
M−∞ = min{x1 , x2 , . . . , xk },
M∞ = max{x1 , x2 , . . . , xk },
M0 = xa1 1 xa2 2 · · · xann ,
Mt = (a1 xt1 + a2 xt2 + · · · + ak xtk )1/t ,
where t is a non-zero real number. Then
M−∞ ≤ Ms ≤ Mt ≤ M∞
for s ≤ t.
Triangle Inequality
Let z = a + bi be a complex number. Define the absolute value of z to be
p
|z| = a2 + b2 .
Let α and β be two complex numbers. The inequality
|α + β| ≤ |α| + |β|
is called the triangle inequality.
Let α = α1 + α2 i and β = β1 + β2 i, where α1 , α2 , β1 , β2 are real numbers.
Then α + β = (α1 + β1 ) + (α2 + β2 )i. Vectors u = [α1 , α2 ], v = [β1 , β2 ],
and w = [α1 + β1 , α2 + β2 ] form a triangle with sides lengths |α|, |β|, and
|α + β|. The triangle inequality restates the fact the the length of any side
of a triangle is less than the sum of the lengths of the other two sides.
Vieta’s Theorem
Let x1 , x2 , . . . , xn be the roots of polynomial
P (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 .
where an 6= 0 and a0 , a1 , . . . , an ∈ C. Let sk be the sum of the products
of the xi taken k at a time. Then
sk = (−1)k
an−k
,
an
Glossary
117
that is,
x1 + x2 + · · · + xn = −
an−1
;
an
x1 x2 + · · · + xi xj + xn−1 xn =
...
a0
x1 x2 · · · xn = (−1)n .
an
an−2
;
an
Trigonometric Identities
sin2 a + cos2 a = 1,
sin a
tan x =
,
cos a
1
cot x =
tan a
addition and subtraction formulas:
sin(a ± b) = sin a cos b ± cos a sin b,
cos(a ± b) = cos a cos b ∓ sin a sin b,
tan a ± tan b
tan(a ± b) =
;
1 ∓ tan a tan b
double-angle formulas:
sin 2a = 2 sin a cos a,
cos 2a = cos2 a − sin2 a = 2 cos2 a − 1 = 1 − 2 sin2 a,
2 tan a
tan 2a =
,
1 − tan2 a
triple-angle formulas:
sin 3a = 3 sin a − 4 sin3 a,
cos 3a = 4 cos3 a − 3 cos a,
3 tan a − tan3 a
tan 3a =
;
1 − 3 tan2 a
118
100 Algebra Problems from USA IMO Training
half-angle formulas:
sin a =
2 tan a2
,
1 + tan2 a2
1 − tan2 a2
,
1 + tan2 a2
2 tan a2
tan a =
;
1 − tan2 a2
cos a =
sum-to-product formulas:
a+b
a−b
cos
,
2
2
a+b
a−b
cos a + cos b = 2 cos
cos
,
2
2
sin(a + b)
tan a + tan b =
;
cos a cos b
sin a + sin b = 2 sin
difference-to-product formulas:
a−b
a+b
cos
,
2
2
a−b
a+b
cos a − cos b = −2 sin
sin
,
2
2
sin(a − b)
tan a − tan b =
;
cos a cos b
sin a − sin b = 2 sin
product-to-sum formulas:
2 sin a cos b = sin(a + b) + sin(a − b),
2 cos a cos b = cos(a + b) + cos(a − b),
2 sin a sin b = − cos(a + b) + cos(a − b).
Further Reading
1. Andreescu, T.; Kedlaya, K.; Zeitz, P., Mathematical Contests 19951996: Olympiad Problems from around the World, with Solutions,
American Mathematics Competitions, 1997.
2. Andreescu, T.; Kedlaya, K.; Zeitz, P., Mathematical Contests 19961997: Olympiad Problems from around the World, with Solutions,
American Mathematics Competitions, 1998.
3. Andreescu, T.; Kedlaya, K.; Zeitz, P., Mathematical Contests 19971998: Olympiad Problems from around the World, with Solutions,
American Mathematics Competitions, 1999.
4. Andreescu, T.; Feng, Z., Mathematical Olympiads: Problems and
Solutions from around the World, 1998-1999 , Mathematical Association of America, 2000.
5. Andreescu, T.; Gelca, R., Mathematical Olympiad Challenges, Birkhäuser, 2000.
6. Barbeau, E., Polynomials, Springer-Verlag, 1989.
7. Beckenbach, E. F.; Bellman, R., An Introduction to Inequalities,
New Mathematical Library, Vol. 3, Mathematical Association of
America, 1961.
8. Cofman, J., What to Solve? , Oxford Science Publications, 1990.
9. Coxeter, H. S. M.; Greitzer, S. L., Geometry Revisited , New Mathematical Library, Vol. 19, Mathematical Association of America,
1967.
119
120
100 Algebra Problems from USA IMO Training
10. Engel, A., Problem-Solving Strategies, Problem Books in Mathematics, Springer, 1998.
11. Fomin, D.; Kirichenko, A., Leningrad Mathematical Olympiads 19871991 , MathPro Press, 1994.
12. Graham, Ronald L.; Knuth, Donald E.; Patashnik, Oren, Concrete
Mathematics, Addison-Wesley, 1989.
13. Greitzer, S. L., International Mathematical Olympiads, 1959-1977 ,
New Mathematical Library, Vol. 27, Mathematical Association of
America, 1978.
14. Klamkin, M., International Mathematical Olympiads, 1978-1985 ,
New Mathematical Library, Vol. 31, Mathematical Association of
America, 1986.
15. Klamkin, M., USA Mathematical Olympiads, 1972-1986 , New Mathematical Library, Vol. 33, Mathematical Association of America,
1988.
16. Kurschak, Hungarian Problem Book, volumes I & II , New Mathematical Library, Vols. 11 & 12, Mathematical Association of America, 1967.
17. Kuczma, M., 144 problems of the Austrian-Polish Mathematics Competition 1978-1993 , The Academic Distribution Center, 1994.
18. Larson, L. C., Problem-Solving Through Problems, Springer-Verlag,
1983.
19. Lozansky, E.; Rousseau, C. Winning Solutions, Springer, 1996.
20. Liu, A., Chinese Mathematics Competitions and Olympiads 19811993 , Australian Mathematics Trust, 1998.
21. Shklarsky, D. O; Chentzov, N.N; Yaglom, I. M., The USSR Olympiad
Problem Book , Freeman, 1962.
22. Slinko, A., USSR Mathematical Olympiads 1989-1992 , Australian
Mathematics Trust, 1997.
23. Stanley, R. P., Enumerative Combinatorics, Cambridge University
Press, 1997.
Further Reading
121
24. Taylor P. J., Tournament of Towns 1980-1994 , Australian Mathematics Trust, 1993.
25. Taylor P. J., Tournament of Towns 1984-1989 , Australian Mathematics Trust, 1992.
26. Taylor P. J., Tournament of Towns 1989-1993 , Australian Mathematics Trust, 1994.
27. Taylor P. J.; Storozhev, A., Tournament of Towns 1993-1997 , Australian Mathematics Trust, 1998.
28. Tomescu, I., Problems in Combinatorics and Graph Theory, Wiley,
1985.
29. Vanden Eynden, C., Elementary Number Theory, McGraw-Hill, 1987.
30. Wilf, H. S., Generatingfunctionology, Academic Press, 1994.
31. Zeitz, P., The Art and Craft of Problem Solving, John Wiley & Sons,
1999.