360 Problems for Mathematical Contests [Andreescu] 9739417124 (1)

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'I'D ANDREESCD

DORIN ANDRICA

360 Problems for

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TITU ANDREESCU DORIN ANDRICA

360 Problems

for

Mathematical Contests

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360 Problems for Mathematical Contests Authors: Titu Andreescu, Dorin Andrica

GIL Publishing House

P.o. Box 44, Post Office 3, 4700, Zalau, Romania, tel. (+40) 260/616314 fax. (+40) 260/616414 e-mail: [email protected] www.gil.ro IMPRIMERIA �

ARTA

IV GRAFICA

I ��LIBRIS

Calea $erbanVodti 133,S.4,Cod 70517,BUCURE$TI Tel.: 336 ₂₉11 Fax: 337 07 35

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Contents

FOREWORD . . . 3

FROM THE AUTHORS . . . 5

Chapter 1 . _{ALGEBRA . . . 7}

Problems . . . 9

Solutions . . . 17

Chapter 2. NUMBER THEORY . . . .47

Problems . . . 49 Solutions . . . 57 Chapter 3. GEOMETRY . . . 85 Problems . . . 87 Solutions . . . 95 Chapter 4. TRIGONOMETRY . . . 137 Problems . . . 139 Solutions . . . 147

Chapter 5. MATHEMATICAL ANALYSIS . . . 179

Problems . . . 181

Solutions . . . 189

Chapter 6. COMPREHENSIVE PROBLEMS . . . 221

Problems . . . 223

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FOREWORD

I take great pleasure in recommending to all readers - Romanians or from abroad the book of professors Titu Andreescu and Dorin Andrica. This book is the fruit of a prodigious activity of the two authors, well-known creators of mathematics questions for Olympiads and other mathematical contests. They have published innumerable original problems in various mathematical journals.

The book is organized in six chapters: algebra, number theory, geometry, trigonometry, analysis and comprehensive problems. In addition, other fields of math ematics found their place in this book, for example, combinatorial problems can be found in the last chapter, and problems involving complex numbers are included in the trigonometry section. Moreover, in all chapters of this book the serious reader can find numerous challenging inequality problems. All featured problems are interesting, with an increased level of difficulty; some of them are real gems that will give great satisfaction to any math lover attempting to solve or even extend them.

Through their outstanding work as jury members of the National Mathematical Olympiad, the Balkan Mathematics Contest (BMO), and the International Math ematical Olympiad (IMO), the authors also supported the excellent results of the Romanian contestants in these competitions. A great effort was given in preparing lectures for summer and winter training camps and also for creating original problems to be used in selection tests to search for truly gifted mathematics students. To support the claim that the Romanian students selected to represent the country were really the ones to deserve such honor, we note that only two mathematicians of Romanian origin, both former IMO gold-medalists, were invited recently to give conferences at the International Mathematical Congress: Dan Voiculescu (Zurich, 1994) and Daniel Tataru (Beijing, 2002). The Romanian mathematical community unanimously recog nized this outstanding activity of professors Titu Andreescu and Dorin Andrica. As a consequence, Titu Andreescu, at that time professor at Loga Academy in Timi§oara and having students on the team participating in the IMO, was appointed to serve as deputy leader of the national team. Nowadays, Titu's potential, as with other Ro manians in different fields, has been fully realized in the United States, leading the USA team in the IMO, coordinating the training and selection of team contestants and serving as member of several national and regional mathematical contest juries.

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One more time, I strongly express my belief that the 360 mathematics problems featured in this book will reveal the beauty of mathematics to all students and it will be a guide to their teachers and professors.

Professor loan Tomescu

Department of Mathematics and Computer Science University of Bucharest

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FROM THE AUTHORS

This book is intended to help students preparing for all rounds of Mathematical Olympiads or any other significant mathematics contest. Teachers will also find this work useful in training young talented students.

Our experience as contestants was a great asset in preparing this book. To this we added our vast personal experience from the other side of the " barricade" , as creators of problems and members of numerous contest committees.

All the featured problems are supposed to be original. They are the fruit of our collaboration for the last 30 years with several elementary mathematics journals from all over the world. Many of these problems were used in contests throughout these years, from the first round to the international level. It is possible that some problems are already known, but this is not critical. The important thing is that an educated - to a certain extent - reader will find in this book problems that bring something new and will teach new ways of dealing with key mathematics concepts, a variety of methods, tactics, and strategies.

The problems are divided in chapters, although this division is not firm, for some of the problems require background in several fields of mathematics.

Besides the traditional fields: algebra, geometry, trigonometry and analysis, we devoted an entire chapter to number theory, because many contest problems require knowledge in this field.

The comprehensive problems in the last chapter are also intended to help under graduate students participating in mathematics contests hone their problem solving skills. Students and teachers can find here ideas and questions that can be interesting topics for mathematics circles.

Due to the difficulty level of the problems contained in this book, we deemed it appropriate to give a very clear and complete presentation of all solutions. In many cases, alternative solutions are provided.

As a piece of advice to all readers, we suggest that they try to find their own solutions to the problems before reading the given ones. Many problems can be solved in multiple ways and pertain to interesting extensions.

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This edition is significantly different from the 2002 Romanian edition. It features more recent problems, enhanced solutions, along with references for all published problems.

We wish to extend our gratitude to everyone who influenced in one way or another the final version of this book.

We will gladly receive any observation from the readers.

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Chapter

1

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PROBLEMS

1 . _{Let C be a set of}

_n

_characters

_{{Cl' C2,}

. . •

, cn}.

We call word a string of at

most m characters, m :::;

_n,

that does not start nor end with

_{Cl .}

How many words can be formed with the characters of the set C?

2. The numbers 1, 2,

_{. . . , 5n}

are divided into two disjoint sets. Prove that these sets contain at least

_n

pairs

_{(x, y), x}

>

_y,

such that the number

_{x - y}

is also an

element of the set which contains the pair.

3. Let

_{al , a2, .. . ,an}

be distinct numbers from the interval

_{[a, b]}

and let a be a

permutation of

_{I,

2,

_{. . . , n}.}

Define the function

f

:

[a, b]

-t

[a, b]

as follows:

f(x)

=

{

a

q(

i

) if

x

=

�i'

i =

1, n

x

otherwIse

Prove that there is a positive integer h such that

f[h](X)

fafa ... af.

'----v---"

htimes

x,

where

f[h]

4. _{Prove that if}

_{x, y, z}

_{are nonzero real numbers with}

_{x + y + z}

= 0, then

x2 + y2 y2 + Z2 Z2 + x2 x3 y3 Z3

---

₊

_{= - + - + -.}

x + y

y + Z

Z + x yz zx xy

5. Let

_{a, b, c, d}

be complex numbers with

_{a + b + C + d}

= O. Prove that

a3 + b3 + c3 + d3

=

3(abc + bcd + cda + dab).

6. Let

_{a, b, c}

be nonzero real numbers such that

_{a + b + c}

= 0 and

_{a3 + b3 + c3}

=

a5 + b5 + c5•

Prove that

7. Let

_{a, b, c, d}

be integers. Prove that

_{a + b + c + d}

divides

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10 1. ALGEBRA

8. _{Solve in complex numbers the equation}

(x + l)(x + 2)(x + 3)2(X + 4)(x + 5)

=

_360.

9. _{Solve in real numbers the equation}

-IX

₊

VY

₊

2v'z=2

₊

.jU

₊

-IV =

_{x + y +}

z

₊

u

₊

v.

10. Find the real solutions to the equation

(x + y) 2

=

_{(X + 1) (y - 1).}

1 1 . Solve the equation

�

x +

J

4X +

V

16X +

l··

+

J4nx + 3

-..;x =

_1.

1 2 . Solve the equation

-J

_{x + a +}

-J

_{x + b +}

-J

_{x +}

c = -J

_{x + a + b -}

c,

where

_{a, b,}

c are real parameters. Discuss the equation in terms of the values of the

parameters.

13. Let

_a

and

_b

be distinct positive real numbers. Find all pairs of positive real numbers

_{(x, y),}

solutions to the system of equations

14. Solve the equation

{

x4 - y4

=

_{ax - by}

x2

_

y2

= {/

_a2

_

b2.

[

25x - 2

]

=

_{l3x + 4}

4 3 '

where

_[a]

denotes the integer part of a real number

_a.

.

1 + V5

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1. 1. PROBLEMS 11

16. _{Prove that if}

_{x, y, z}

_{are real numbers such that}

_{x S}

₊

_{y S}

₊

_{Z S}

_{f:. 0, then the}

ratio

_{2xyz - (x}

+

_y

+

_z)

x S

+

y S

+

Z S

equals

�

if and only if

_x

+

_y

+

_z

= O.

17. _{Solve in real numbers the equation}

1

vx;:-=-r +

2 .JX2

_{- 4}

+ . . . +

_{nJX n - n2}

=

_{"2 (xl}

+

_X2

+ . . . +

_xn).

18. _{Find the real solutions to the system of equations}

1

1 1

-+ - = 9

x y

1 1 1 1 -+- 1 +- 1 + - - 18

(

?Ix �

)(

?Ix

)(

�

)

-1 9. _{Solve in real numbers the system of equations}

y2

+

u2

+

v2

+

w2

=

_{4x -}

1

x2

+

u2

+

v2

+

w2

=

4y

- 1

x2

+

y2

+

v2

+

w2

=

_{4u -}

1

x2

+

y2

+

u2

+

w2

=

_{4v -}

1

x2

+

y2

+

u2

+

v2

=

_{4w -}

1

20. Let

_{aI, a2, a s, a4, a5}

be real numbers such that

_al

+

_a2

+

_{a s}

+

_a4

+

_a5

= 0 and

max

_{lai - ajl}

< 1. Prove that

_ai

+

_a�

+

_{a �}

+

_{a �}

+

_{a �}

< 10.

l�i<j�5

-

-2 1 . Let

_{a, b, c}

be positive real numbers. Prove that

1 1 1 1 1 1

- + - + -

2a 2b 2c - a

> --+

b b

+ -- ++

c c

--+

a

22. Let

_{a, b, c}

be real numbers such that the sum of any two of them is not equal

to zero. Prove that

23. Let

_{a, b, c}

be real numbers such that

_abc

= 1. Prove that at most two of the

numbers

are greater than 1.

1

2a - -b'

_{2b - -}

1

c'

1

2c

-a

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12

1.

ALGEBRA

24. _Let

a, b,

c,

_d

be real numbers. Prove that

2

�

2 1

min(a

_{- b b -}

c c

_{- a-}

d - a )

<

-,

,

_,

- 4'

25. _Let

_{aI , a2, ' .. ,an}

_{be numbers in the interval}

_{(0, 1)}

_{and let}

_{k ;:::}

_{2 be an integer}

Find the maximum value of the expression

n

L V'ai(1

- ai+1),

i=l

26. _Let

_m

_and

_n

_{be positive integers. Prove that}

xmn - 1 xn - l

---

_>

--m - x

for any positive real number

_x.

27. _{Prove that}

m! ;::: (n!)[�]

for all positive integers

m

and

n.

28. _{Prove that}

1

1 �

2

1 + - + - + ' ' ' + - > n - -

v'2 v'3 \Iii

_{n + 1}

for any integer

_n

;::: 2.

29. _{Prove that}

n (1 - 1/

vn)

+ 1 > 1 +

�

2

+

�

3 + .. . +

.!.

_n

> n

(\In

+ 1 - 1)

for any positive integer

_n.

( )

_{na1a2 ' " an}

30. _Let

_{aI, a2, .. . , an}

E

_{0, 1}

and let

_tn

= . Prove that

a1 + a2 + .. . + an

n

L

loga.

tn ;::: (n - l)n.

i=l

31. _{Prove that between n and}

_3n

_{there is at least a perfect cube for any intege}

n ;::: 10.

32. _{Compute the sum}

33. _{Compute the sums:}

n

"'" 1c(Ic+l)

Sn

=

�

(-1)

2

k.

k=l

a₎

_Sn

₌

�

_{(k + l)l(k +}

₂₎

(�);

_b₎

Tn =

�

_{(k + l)(k + 2)(k + 3)}

(�).

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1 .1.

PROBLEMS

34. _{Show that for}any positive integer n the number

C

n

;

1

)

2

_2n

+

C

n

:

1

)

_22n-2

. 3 + . . . +

C

n2

:

1

)

3

_n

is the sum of two consecutive perfect squares.

35. _{Evaluate the sums:}

36. _{Prove that}

12 (�)

+ 3

₂

(�)

+

₅₂

(�)

+ ... = n(n + 1)2

_n-3

for all integers n � 3.

37. _{Prove that}

2n

L

[log2

k

] = (n - 2)2n + n + 2

k=l

for all positive integers n.

38. _Let

_Xn

₌₂

_2n

+ 1, n = 1, 2, 3, . . . Prove that

for all positive integers n.

1 2 2

₂

_{2n -l}

1

- + - + - + . . . +-- <

Xl X2 X3

Xn

3

39. _{Let f}: C -t C be a function such that f(z)f(iz) =

_Z2

for all z E C.

Prove that

f (z) + f ( -z) =

₀

for all z E C

13

40. _{Consider a function f}:

_{(0, 00)}

-t � and a real number a >

₀

such that

f(a) = 1. Prove that if

f(x)f(y) + f

(�)

f

(�)

= 2f(xy) for all x, y E

_{(0, 00),}

then f is a constant function.

41. _{Find with proof if the function}t: _�_-t_{[- 1, 1], f(x)}₌_{sin[x] is periodical.}

n

42 . _{For all i, j}₌ _1,_n_{define S(i, j)}₌

_L

ki+i.

_{Evaluate the determinant}� =

k=l

(20)

14

43. _Let 1. ALGEBRA

{

ai

if i =

_j

xii

= 0 if i

_{i j,}

i +

_{j i}

2n + 1

bi

_if _i₊

j

= 2n + 1

where ai, bi are real numbers.

Evaluate the determinant

_�2n

=

_IXiil·

44. _a) _{Compute the determinant}

x y

z v

y x

v z z v

_{x y}

v z

_{y x}

b) Prove that if the numbers

_{abed, bade, cdab, deba}

are divisible by a prime p,

then at least one of the numbers

a

+

b

+

_e

+

_{d, a}

+

_{b - e - d, a - b}

+

_{e - d, a - b - e}

+

_d,

is divisible by p.

45. _{Consider the quadratic polynomials}

_{tl (x)}

₌

_x2

+

_{PI X}

+

_qr

and

_{t2 (x)}

=

_x2

+

P2X

+

q�,

where

PI,P2, ql, q2

are real numbers.

Prove that if polynomials

_tl

and

_t2

have zeros of the same nature, then the polynomial

has real zeros.

46. _Let

_{a, b, e}

_{be real numbers with}

_a

_>_{0 such that the quadratic polynomial}

T (x)

=

_ax2

+

_bex

+

_{b S}

+

_{e S - 4abe}

has nonreal zeros.

Prove that exactly one of the polynomials

_{TI (x)}

=

_ax2

+

_bx

+

_e

and

_{T2 (x)}

ax2

+

ex

+

b

has only positive values.

47. _{Consider the polynomials with complex coefficients}

P(x)

=

_xn

+

_alxn-l

+ . . . +

_an

and

_Q(x)

=

_xn

+

_blxn-l

+ . . . +

_bn

having zeros

_{Xl , X2, .. . ,Xn}

and

_{xi, x�, .. . ,x;}

respectively.

Prove that if

_al

+

_{a s}

+

_a5

+ . . . and

_a2

+

_a4

+

_a6

+ . . . are real numbers, then

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1.1.

BROBLEMS

48. _Let

_P(x)

_{be a polynomial of degree n. If}

k

P(k)

=

-k-

+ 1 for

k

= 0, 1, . . . ,n evaluate

_P(m),

where

_m

> n.

49. _{Find all polynomials}

_P(x)

_{with integral coefficients such that}

for all real numbers

_x.

15

50. _{Consider the polynomials Pi, i}₌ _{1, 2, . . . , n with degrees at least 1. Prove}

that if the polynomial

P(x)

=

_{P1 (Xn+1)}

+

_{XP2 (xn+1)}

+ . . . +

_{xn-1pn (xn+1),}

is divisible by

_{xn +xn-1}

+ . . ·+ x + 1, then all polynomials Pi(x), i = 1, n, are divisible

by x - 1.

51 . _{Let P be a prime number and let}

P(x)

=

_aoxn

+

_a1xn-1

+ . . . +

_an

be a polynomial with integral coefficients such that

_an

=1= 0 (mod

_p).

Prove that if there are n + 1 integers

_{0'1, 0'2,}

• • •

, an+ 1

such that

P ( ar)

== 0 (mod

_p)

for all r = 1, 2, . . . ,n + 1, then there exist i,j with i i-j such that

_a

i ==

_aj

(mod

_p).

52. _{Determine all polynomials}

_P

_{with real coefficients such that}

_pn(x)

₌

_P(xn)

for all real numbers x, where n > 1 is a given integer.

53. _Let

P(x)

=

_aoxn

+

_a1xn-1

+ . . . +

_{an, an}

i-0,

be a polynomial with complex coefficients such that there is an integer

_m

with

Prove that the polynomial

_P

has at least a zero with the absolute value less than 1.

54. _{Find all polynomials}

_P

_{of degree n having only real zeros}

_{Xl , X2, .. . , Xn}

_such

that

n

1 n2

�

P(x) - Xi

=

_{XP I(X) '}

for all nonzero real numbers x.

55. _{Consider the polynomial with real coefficients}

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16 1. ALGEBRA

and an

_f:.

O.

Prove that if the equation P(x) = 0 has all of its roots real and distinct, then the

equation

x2 PII(x) + 3xP' (x) + P(x) = 0

has the same property.

56. _Let_R

�?

_and_R

f�

₎_{be the sets of polynomials with real coefficients having}

no multiple zeros and having multiple zeros of order n respectively. Prove that if

P(x) E R

�?

and P(Q(x)) E R

�?

, then Q' (x) E R

�

]-1).

57. _LetP(x) be a polynomial with real coefficients of degree at least 2. Prove that if there is a real number a such that

P(a)plI(a) > (P' (a))2 ,

then P has at least two nonreal zeros.

58. _{Consider the equation}

aoxn + a1Xn-1 + . . . + an = 0

with real coefficients ai . Prove that if the equation has all of its roots real, then

(n - l)ar 2:: 2naOa2 . Is the reciprocal true?

59. _{Solve the equation}

X4 - (2m + 1)x3 + (m - 1)x2 + (2m2 + l)x + m = 0,

where m is a real parameter.

60. _{Solve the equation}

x2n + a1 x2n-1 + . . . + a2n_2X2 - 2nx + 1 = 0,

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SOLUTIONS

1 . _{Let Nk be the number of words having exactly k characters from the set C,}

1 :s; k :s; m. Clearly,

_N1

= n - 1. The number that we seek is

_N1

+

_N

2 + . . . +

_Nm·

Let Ak =

_{I,

2, . . . , k}, 1 :s; k :s; m. We need to find out the number of functions

f : Ak ---+ A, k = 2, n with the properties

f(l)

_{f:. a1}

and f(k)

_{f:. a1}

For f(l) and f(k) there are n - 1 possibilities of choosing a character from

C2, • . • , Cn and for f(i), 1 < i < k there are n such possibilities. Therefore the number

of strings f(l)f(2) . . . f(k - l)f(k) is

N

k = (n - 1)2nk-2

It follows that

N1

+

N

2 + . .. +

Nm

= (n - 1) + (n - 1)2nO + (n - 1)2n

₁

+ . . . + (n -1)2n

_m

-2 =

(Dorin Andrica)

2. _{Suppose, for the sake of contradiction, that there are two sets}A and B such that Au B =

_{I,

2, .. . ,5n}, An B = 0 and the sets contain together less than n pairs

(x,

_y),

x >

_y,

with the desired property.

Let k be a given number, k = 1, n. If k and 2k are in the same set - A or B

-the same can be said about -the difference 2k - k = k. The same argument is applied

for 4k and 2k. Consider the case when k and 4k are elements of A and 2k is an

element of B. If 3k is an element of A, then 4k - 3k = k E A, so let 3k E B. Now if 5k E A, then 5k - 4k = k E A and if 5k E B, then 5k - 3k = 2k E B; so among the numbers k, 2k, 3k, 4k, 5k there is at least a pair with the desired property. Because k = 1, 2, . . . , n, it follows that there are at least n pairs with the desired property.

(Dorin Andrica,

Revista Matematica Timi§oara (RMT), No. 2(1978), pp. 75,

Problem 3698)

3. _{Note that}

(1)

(24)

18

_1.

ALGEBRA

and furthermore

(2)

for all integers

_{ml , m2}

2::

_1.

Suppose that for all integers k 2::

₁

we have

_I[k](x)

f=.

_x.

Because there are n ! permutations, it follows that for k > n ! there are distinct

positive integers

_nl

>

_n2

such that

(3) Let _h= nl

_{- n2}

> 0 and observe that for all k the functions

_I[k]

are injective,

since numbers

_{ai, i}

=

_1,

n are distinct. From relation (3) we derive that

or

l[n2+h] (x)

=

_{l[n2] (x), x}

_E

[a, b],

(10

_{l[n2+h-l])(x)}

= (10

_{l[n2-1 ])(x), X}

E

_{[a, b].}

Because

_I

is injective, we obtain

l[n2+h-l] (x)

=

_{l[n2-1] (x), x}

E

_{[a, b]}

and in the same manner or

l[h+l](X)

=

_{I(x), x}

E

_{[a, b]}

l[h](X)

=

_{x, x}

E

_{[a, b].}

Alternative solution.

Let

Sn

be the symmetric group of order n and

Hn

the cyclic

subgroup generated by

_a.

It is clear that

_Hn

is a finite group and therefore there is integer h such that

_a[h]

is identical permutation.

Notice that

I[k] (x)

=

{

aq[kJ (i)

if

x

=

�i' i

=

1,

n

x

otherwIse

Then

_I[h](x)

=

_x

and the solution is complete.

(Dorin Andrica,

Revista Matematica Timi§oara (RMT), No.

2(1978),

pp. 53,

Problem 3540)

4. Because

_x

+

_y

+

_z

= 0, we obtain

x

+

y

=

_{-z, y}

+

_z

=

_{-x, z}

+

_x

=

_-y,

or, by squaring and rearranging,

x2

+

y2

=

_Z2

_

2xy, y2

+

Z2

=

_x2

_

2yz, Z2

+

x2

=

_{y2 - 2zx.}

The given equality is equivalent to

Z2 - 2xy x2 - 2y Z y2 - 2zx x3 y3 Z3

--- + + = - + - + -,

(25)

1.2. SOLUTIONS

and consequently to

(

Xy yz zx

)

X3 y3 Z3

- (x + y + z) + 2 - + - + -

=

-

_{+ - +}

-

.

z x Y

yz zx xy

The last equality is equivalent to

2(X2y2 + y2z2 + Z2X2)

=

_{x4 + y4 + Z4.}

On the other side, from

_{x + y + z}

=

₀

we obtain

_{(x + y + Z)2}

=

₀

or

x2 + y2 + Z2

=

_{-2( xy + y z + zx).}

Squaring yields

X4 + y4 + Z4 + 2(X2y2 + y2z2 + Z2X2)

=

_{4(X2y2 + y2z2 + Z2X2) + 8xyz(x + y + z)}

or

as desired.

19

( Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No.

3(1971),

pp. 25,

Problem

₄₈₃

; Gazeta Matematica (GM-B), No.

_12(1977),

pp.

_501,

Problem

₆₀₉₀₎

5. _{We assume that numbers}

_{a, b, c, d}

_{are different from zero. Consider the equation}

x4

- (2:

a

)

x3 +

(2:

ab

)

x2

- (2:

abc

)

x + abed

=

₀

with roots

_{a, b, c, d.}

Substituting

_x

with

_{a, b, e}

and

_d

and simplifying by

_{a, b, c, d f:. 0,}

after summing up we obtain

Because

2:

_a

=

_0,

it follows that

If one of the numbers is zero, say

_a,

then

b + c + d = 0,

or

_{b + c}

=

_-d.

It is left to prove that

_{b3 + c3 + d3}

=

_3bcd.

Now

b3 + c3 + d3

=

_{b3 + c3 -}

(

_{b + c)3}

=

_-3bc

(

_{b + c)}

=

_3bcd

as desired.

(Dorin Andrica,

1-2(1979),

pp.

47,

Problem

₃₈₀₃₎

6. _Because

_{a + b + c}

₌

_0,

_{we obtain}

(26)

20

or

1. ALGEBRA

The given relation becomes

3abc

=

_{5abc(a2 + b2 + c2 + ab + bc + ca)}

3 a2 + b2 + c2 + ab + bc +

ca = 5'

since

_{a, b, c}

are nonzero numbers. It follows that

1

3 - [(a + b + C)2 + a2 + b2 + c2]

₂

= -

5

and, using again the relation

_{a + b + c}

=

_0,

we obtain

a2 + b2 + c2

=

�

5 '

as desired.

(Titu Andreescu,

2(1977),

pp.

59,

Problem

₃₀₁₆₎

7. _{Consider the equation}

x4

-

(L:

a

)

x3 +

(L:

ab

)

x2

-

(L:

abc

)

x + abed

=

_0,

with roots

_{a, b, e, d.}

Substituting

_x

with

_{a, b, e}

and

_d,

respectively, we obtain after summation that

L:

a4 +

(

L:

ab

) L:

a2 + 4abed

is divisible by

L:

_a.

Taking into account that we deduce that

is divisible by

L:

_a.

Hence

2(a4 + b4 + e4 + d4) - (a2 + b2 + e2 + d2)2 + 8abed

is divisible by

_{a + b + e + d,}

as desired.

(Dorin Andrica)

8. _{The equation is equivalent to}

(x2 + 6x + 5)(x2 + 6x + 8)(x2 + 6x + 9)

=

_360.

Setting

_{x2 + 6x}

=

_y

yields

(27)

1.2.

SOLUTIONS

21

or

y3 +

22y2

+

157y = 0,

with solutions

Yl

= 0,

Y2

= -11

+

6i,

Y3

_{= -11 - 6i.}

Turning back to the substitution, we obtain a first equation,

X2 +

_{6x = 0,}

with solutions Xl

_{= 0, X2 = -6.}

The equation

X2 +

_{6x = -11}

+

_6i

is equivalent to

(

x

+

_{3)2 = -2}

+

_6i.

Setting

X

+

_{3 = u}

+

_{iv, u, v}

E lR, we obtain the system

{ U2 -

v2 = -2

2uv = 6

It follows that

_(u2

+ V2)2

_{= (u2}

-V2)2 + (2UV)2

_{= 4}

+

_{36 = 40.}

Therefore

{

U2

u2 +V2

- v2 = -2

= 2v'W

and

_{u2 = v'W}

- 1,

_{v2 =}

_y'lO

+

1, yielding the solutions

X3,4

_{= -3}

±

V

V16

_-

1 ±

i

V

V16

+

₁

where the signs

+

and - correspond.

The equation

X2 +

_{6x = -11 - 6i}

can be solved in a similar way and it has the solutions

X5

_{= -3}

+

J

y'lO

_{- 1 -}

i

V

y'lO

+

_1,

X6

_{= -3}

- V.Jill

_{- 1}

+

i

V

y'lO

+

l.

( Titu Andreescu,

3(1972),

pp.

26,

Problem

₁₂₅₅₎

or

9. _{The equation is equivalent to}

X

_{- ..;x}

+

Y

_-

VY + z

-

_2vz-=-2

+

_{u -}

Vu

+

_{v - ..;v = 0,}

(

vx

_-

D2 +

(

v'Y

_-

D2 + (vz-=-2_1)2 +

+ (.ru-D\ (VV-D2

= 0

Because x, y, Z,

U,

_v

are real numbers, it follows that

Hence

..;x =

VY

=

Vu

= ..;v =

�

and vz-=-2

= 1.

1

X

₌

Y

₌

U

_{= v =}

4' Z

_{= 3.}

( Titu Andreescu,

2(1974),

pp.

47,

(28)

22

or

1. ALGEBRA

10. _Setting

X

=

_{x + 1}

and Y = y

_{- 1}

yields

(X

+

y)2

=

Xy

�[X2

+

y2

+

(X

+

y)2]

=

_0.

2

Hence

X

= Y =

_0,

so the solution is

_x

=

_-1

and y =

_1.

( Titu Andreescu,

Revista Matematidl Timi§oara (RMT) , No.

1(1977),

pp.

40,

Problem

₂₈₁₁₎

1 1. _{The equation is equivalent to}

�

x +

V

4x +

V

16x +

V/

. .

+

v' 4nx + 3

=

_{..jii + 1}

Squaring the equation yields

V

4X +

V

16X +

V/

. . +

v'4nx

+ 3

=

_{2..jii + 1}

Squaring again implies

V

16X +

V/

. .

+

v'4nx

+ 3

=

_{4..jii + 1}

Continuing this procedure yields

4nx + 3

=

_{4nx + 2 · 2n..;x + 1}

1

and

_{2 . 2n}

Vx =

_2.

Hence

_x

=

_{4n '}

(Titu Andreescu,

Revista Matematidl Timi§oara (RMT), No.

4-5(1972),

pp.

43,

Problem

₁₃₈₅₎

1 2 . _{We distinguish two cases:}

1) b

=

_c.

The equation becomes

Jx

+ a +

Jx

+ b

=

Jx

_{+ a,}

so

_{x;:= -b.}

2) b f:. c.

The equation is equivalent to

Jx

+ b +

Jx

+ c

=

Jx

_{+ a + b - c -}

Jx

_{+ a,}

or

_{b - c}

Jx + b - Jx + c - Jx + a + b - c + Jx + a'

so

Jx

+ b -

Jx

+ c

=

Jx

_{+ a + b - c +}

Jx

_{+ a.}

(1)

(2)

(29)

1.2.

SOLUTIONS

Summing up relations

₍₁₎

and (2) we obtain

v'

x + b =

v'

x + a + b -

c,

and then

_{a =}

c.

To conclude, we have found that

(i) If

_{b =}

c, then the equation has the solution

_{x = -b.}

(ii) If

_b

f=. c and

_a

f=. c, there is no solution.

(iii) If

_b

f=. c and

_{a =}

c, then

_{x = -a}

is the only solution.

23 ( Titu Andreescu,

2(1978),

pp.

26,

Problem

₃₀₁₇₎

13. _Because

a

and

b

are distinct numbers,

x

and y are distinct as well. The second

equation could be written as

and the system could be solved in terms of

_a

and

_b.

We have

a2b2 = b2y2 + 2by(x4 _ y4) + (X4 _ y4)2

a2x2 = b2x2 + X2(X2 _ y2)3.

Subtracting the first equation from the second yields

which reduces to

Solving the quadratic equation in

_b

yields

_{b = y3 + 3x2y}

₍and

_{a = x3 + 3xy2)}

or

_{b = y3 - x2y}

(and

_{a = x3 - xy2).}

The second alternative is not possible because

a = x(x2 _y2)

and

b = y(y2 _X2)

cannot be both positive. It follows that

a = x3+3xy2

and

_{b = 3x2y + y3.}

Hence

_{a + b = (x + y)3}

and

_{a - b = (x - y)3.}

The system now becomes

x + y = �a + b

x - y = {!a - b

and its unique solution is

_{x =}

(�a

_{+ b +}

�a

_{- b)/2, y =}

(�a

_{+ b -}

�a

_{- b)/2.}

( Titu A ndreescu,

Korean Mathematics Com petitions,

2001)

13x + 4

14. _Let

_{3 = y,}

Y

E Z. It follows that

3y - 4

x = 13

(30)

24

1. ALGEBRA

and the equation is equivalent to

or

[

�

(3Y

:

4) -

2 ]

_{= Y,}

[

75Y -

52

126 ]

=

Y

Using that for any real number

_{a, [a] ::; a < [a] + 1,}

we obtain

75y - 126

1 Y ::; 52 < y + ,

126

178

or

_{126 ::; 23y < 178,}

so 23

_{::; y <}

23·

Note that

_y

E Z, therefore

_{y = 6}

or

_{y = 7,}

thus

14

17 Xl

= 13

and

X2

= 13'

are the desired solutions.

( Titu Andreescu,

3(1972),

pp.

_25,

Problem

₁₅₅₂₎

1 + V5

.

1

15. _From

_a

_�

--

_-

_{we obtam}

_{a2 - a - I}

_{� 0, or}

_a

_�a

_{+ 1.}

_{We have}

₂

Hence That is because and

[

1 + na2

_a

]

= - + na - a,

_a

1 O::;a < 1.

[

1 +

[

1 :

ana2

]

=

[

1 +

� :

na - a

]

=

[(

1 +

�

- a

)

�

+ n

]

= n, n

�

O. (

1 + - - a -::; (a - a)- = 1 - - < 1

_a

1 )

1 _a

_a

1 a

_a

(

1 +

�

- a

)

�

>

�

>

_O.

a

a a2

( Titu Andreescu,

2(1978),

pp.

_45,

Problem

₃₄₇₉₎

16. _{First we consider the case when}

X

_{+ y + z = O.}

_Then

_{x3 + y3 + Z3 = 3xyz}

_and

(31)

Conversely, if then

and so

Using the formula

1.2.

SOLUTIONS

2xy z -(x + y + z) 2

_{x3 + y3 + Z3}

-"3'

2(x3 + y3 + Z3 -3xyz) + 3(x + y + z)

=

_O.

x3 + y3 + Z3 -3xyz

=

(x + y + z)(x2 + y2 + Z2 -xy -yz -zx),

we obtain by factorization that and so

(x + y + z)[(x -y)2 + (y -Z)2 + (z -X)2 + 3]

=

_O.

Because the second factor is positive, it follows that

x + y + z

= 0, as desired.

25 ( Titu Andreescu,

1(1 973),

pp. 30,

Problem

1513)

17. _{We write the equation}_as

Xl -2JXI -1 + X2 -2· 2VX2 -22 + ... + Xn -2nvxn -n2

= 0,

or

(JXI -1 -1)2 + ( VX2 -22 -2) 2 + ... + (V Xn -n2

_

n) 2

= O.

Because the numbers

Xi,

_i

=

1, n

are real, it follows that

( Titu Andreescu,

1(1977),

pp.

14,

Problem

2243)

18. _{Using the identity}

(a +

b

+ c)3

=

a3 + b3 +

_c3

+ 3(a +

_b)(b

+

_c

)(

_c

+ a)

we obtain

(� + � +1)3

_{{IX flY} =

�

X

+

�

+1+3 (� +�) (1+ _1_) (1+ _1_)

=

Y

{IX flY {IX flY

=

9+ 1 + 54

=

64

(32)

26

1. ALGEBRA

and so

1

1 - + - = 3.

_{fIX flY}

The system is now reduced to

1

�+� = 9

x y

1

fIX

₊

flY

_{= 3,}

which is a symmetric system, having the solution

1

1 x =

'8'

y = 1

and

x = 1,

Y

=

'8'

( Titu Andreescu,

4-5(1972

)

,

pp.

43,

Problem

₁₃₈₆₎

19. _{By summing up the equations of the system we obtain}

It follows that

(4x2 - 4x + 1) + (4y2 - 4y + 1) + (4u2 - 4u + 1)+

+(4v2 - 4v + 1) + (4w2 - 4w + 1) =

O.

Due to the fact that

_{x, y, u, v, w}

are real numbers, we obtain

1 x = y = u = v = w = -2

(Dorin Andrica,

Gazeta Matematica (GM-B), No.

8(1977

)

,

pp.

321,

Problem

16782

)

20. _{From the triangle inequality we deduce}

Hence

So

lai - ajl ::; lai - ai+l l

_{+ .. . +}

laj-l - ajl

_::;

< (J'

_{- i)}

max la· - a"1 < J"

_{- i 1}

< _

_i

< J" < _

₅

- l Si<jS5 � J -

_,

I:

(ai - aj)2::;

I:

_{(i - j)2 =}

lSi<jS5 l Si<jS5 5 5

4

_�_a�

-

2 � a"a" <

₅₀

� l � � J-i=l i , i =l i>Fi

(33)

1.2.

SOLUTIONS

and consequently

5

t,

_{a� -}

(

t,

_a;

r

� 50

5

Note that

_{L a}

i = 0 and so

_{L ar}

::; 10, as claimed.

i=l i=l

27 ( Titu Andreescu,

Romanian Mathematica Olympiad - second round 1979; Revista

Matematica Timi§oara (RMT), No. 1-2(1980), pp. 61, Problem 4094)

2 1 . _{The inequality}

_{(a + b}

₎

2

_�

_4ab

_yields

1 1 4

- +

- >

--a b - --a + b

1 1 4 1 1 4 and similarly

_{- + -}

>

--- + ---

> --.

,

' b e - b + e' e a - e + a

Summing up these inequalities yields

1 1 1 1 1 1

2a + 2b +

2

e

�

a + b + b + e + e +

a'

as desired.

(Dorin Andrica,

Gazeta Matematica (GM-B), No. 8(1977), Problem 5966)

22. _{Using the identities}

a5 + b5 + e5

=

_{(a + b + e)5 - 5(a + b)(b + e)(e +}

a

)(

a2

_{+ b2 + e2 + ab + be +}

00)

and we obtain

or

a5 + b5 + e5 - (a + b + e)5

_{3 b3}

_c3 (

_{b )3}

= -3 5

_{(a + b + e + ab + be +}

2 2 2

ca)

a + + - a + + e

It suffices now to prove that

5 10

3

_{(a2 + b2 + e2 + ab + be +}

00) � 9 (

a

_{+ b + e)2}

3(a2 + b2 + e2 + ab + be +

ca) �

2(a2 + b2 + e2 + 2ab + 2be +

2ca).

The last inequality is equivalent to

a2 + b2 + e2

�

a

_{b + be +}

_00, which is clearly true.

( Titu Andreescu,

Problem 4295; Gazeta Matematica (GM-B), No. 6(1980), pp. 280, Problem 0-148; No. 11(1982), pp. 422, Problem 19450)

(34)

28

1.

ALGEBRA

23. _{Assume by contradiction that all numbers}

2a -

�,

b c a

2b -

�,

2c -

�

_{are greater}

than 1. Then

and

(2a-

D (2b-D (2C-D >

1

(2a-

D + (2b-D + (2C-D >3.

From the relation (1) and using

abc

= 1 we obtain

3>2(a+b+c)-G+�+D.

On the other hand, relation

(2)

gives

2(a+b+c)-G+�+D >3

which is a contradiction. The proof is complete.

(1)

(2)

(3)

( Titu Andreescu,

Revista Matematica Timi§oara (RMT) , No. 2(1986), pp. 72,

Problem 5982)

24. _{Assume by contradiction that all numbers are greater than 1/4. Then}

hence

2 2

�J2

2

1 1 1 1

a -b + b -c + c -a-+ d -a > -+ -+ -+ -

4 4 4 4

0> G-ar + G-br + G-cr + G-dr

This is a contradiction so the claim holds.

( Titu Andreescu,

Problem 5479)

25. _Setting

a

i

= sin2

ai

for

_i

= 1, n, where

aI, a2, ... , an

are real numbers, the

expression becomes

n

E =

2: \!

sin2

ai

cos2

ai+b an+l

=

al·

i=1

Using the AM-GM inequality yields 1

k

-2:

bf ;::: b1 b

2

• • .

bk, bi > 0,

i

=

_U.

k

i=1

For

b1

= sinf

ai,

b

2

= cosf

ai+l

and

b3

=

b4

= . . . =

bk

= k

{/2

�

we obtain

1

(

.

2

2 k

-2)

1

_\I'

2 2

(35)

1.2. SOLUTIONS

Summing up these relations for

_k

=

_{1, 2, .. . , n}

yields

and so

�

(

_{n(k - 2)}

)

_>

_l_ E

k n

+

2 - /0

_2-/0

-2 ,

-n · 2/ok2 -n · 21-�

E <

----

2

{14

n

Hence the maximum value of

_E

is

;;

and it is reached if and only if

1 al

=

_a2

= . . . =

_an

= 2"'

29 (Dorin Andrica,

Revista Matematidl Timi§oara (RMT) , No.

1(1978),

pp. 63,

Problem 3266)

26. _Because

_x

_and_m_{are positive, we have to prove that}

x(xmn

-

_{1) - m(xn - 1)}

� 0,

or

(xn - l)[(xn)m-Ix

+

_(xn)m-2x

+ . . . +

_x

- m

_]

� 0.

Define

_E(x)

=

_(xn)m-Ix

+

_(xn)m-2x

+ . . . +

_x

- m and note that if

_x

�

_1,

then

xn

�

₁

and

_E(x)

� 0, so the inequality holds. In the other case, when

_{x < 1,}

we have

xn < 1

_{( Titu Andreescu,}

and

E(x) <

° and again the inequality holds, as claimed. Revista Matematica Timi§oara (RMT) , No.

_2(1978),

pp. 45, Problem 3480)

27. _For_m_:::;

_n

_{the inequality is clearly true, so consider}_{m >}

_n

_{and define}

p =

[:].

This implies that m = pn + q with q E

_{{O, 1, .. . ,n - I}}

and the inequality

can be written as

(pn + q)! �

_(n!)P.

We have

(pn + q)! � (pn)! =

=

_{(1 · 2 .. . n)(n}

+

_{1) .. . (2n)}

... ((P

_{- l)n}

+

₁₎

. . . (pn) �

_(n!)P,

and we are done.

( Titu Andreescu,

2(1977),

pp. 61,

Problem 3034)

28. _{We will use the inequality}

Xml

+

_x2m

+ . . . +

_xm

_{n - nm-l}

>

_{_1_}

(Xl

+

X2

+ . . . +

X )m

_{n ,}

(36)

30 1. ALGEBRA

Now set

_Xl

=

1,

_X2

=

2,

_{. . . , Xn}

=

n and m

=

_{_.!..}

We obtain

n

₁

1 1 1 1

[

_n

(

_n

+ 1)

]-

n

�2

1 +

-

_{v'2 y'3}+

-

+ · · · +

-

_\Iii>

--

= n

-n-�-l

2 n + 1 '

as desired.

(Titu Andreescu,

Problem 2035)

or

29. _{From the AM-GM inequality we deduce}

.!.

t

_i

�

1

_�

n

IT

i

�

1

=

\in

+ 1,

n i=l

�

i=l

� 1

_n

1 1 +

-

2:

-;-

_�

\in

+ 1,

n i=l

� and so 1 +

!

_{2 3}+ ! + . . . + .!. >

_{n (}

\in

+ 1 - 1

)

n

'

as desired.

Ob serve t at t e mequa Ity IS stnct ecause t e num ers h h · 1· ·

_.

b h b

_i

-.

+ 1 . � = -1 -

_{, n,}

are -,

�

distinct.

or

In order to prove the first inequality we apply the AM-GM inequality in the form

Therefore 1 2

_{n - 1}

1 + - + - + · · · + --_{2 3}

if!

1

n

> n _

=

_.

n

\Iii

n

(

1 - _\Iii 1_

)

+

1

> 1 +

�

2 + . . . +

.!..

_n

(Dorin Andrica,

Problem 3037)

30. _{Because the numbers}

_{al , a2, .. . ,an}

_{are positive, from the AM-GM inequality}

al

+

a2

+ . . . +

an

---

_n

� -tlala2 .. . an

we deduce that

_{tn ::; (ala2 .. . an)}

n;;:l

.

Using that numbers

ai

are less than 1 we obtain

n - 1

logai

_tn

�

--

logai

_{(ala2 .. . an).}

(37)

1.2. SOLUTIONS

Summing up these inequalities yields

n

1

_n

2:

loga;

_tn

2:: n -

2:

loga; (

_{a1 a2 .. . an)}

=

i=1

n

i=1

n - 1

=

-n-[n +

(

logal

a2

+ loga2

a1)

+ . . . +

(

logal

an

+ logan

a1)

+ . . . + +

₍

logan

_an-1

+ logan_l

_an)]·

1

Note that

_a

+ - > 2 for all

a

> 0, so

a

-�

n - 1 L.J logai

_tn

2:: --[n + 2(n - 1) + 2(n - 2) + . . . +

_2]

= (n - l)n,

i=1

n as claimed. 3

₁

(Dorin Andrica,

Revista Matematica Timi§oara

(

RMT

)

, No. 2(1977), pp. 62,

Problem 3038)

31 . _{We begin with the following lemma.}

Lemma.

_{Let a}

>

_{b be two positive integers such that}

� - � > 1.

Then between numbers a and b there is at least a perfect cube.

Proof.

Suppose, for the sake of contradiction, that there is no perfect cube between

a

and

b.

Then there is an integer c such that

c3 �

_b

<

_a

� (c + 1) 3 .

This means

c � � < � � c + 1, so

which is false. 0

Now we can easily check that for n = 10, 11, 12, 13, 14, 15 the statement holds.

If n 2:: 16, then or Hence 3 1 1 n > (2, 5) = (1, 4 - 1)3 >

(�_

1)3 ' aC 1 yn >

�_(

{I3n -_-rn> 1, and using the above lemma the problem is solved.

( Titu Andreescu,

(

RMT

)

, No. 1-2(1990), pp. 59,

(38)

32

1.

ALGEBRA

k(k

+

1) . .

32 . _{Note that the number}

2

IS odd for

k =

_{4p +}

1 or

k =

_{4p +}

2

and IS

even for

k =

_{4p + 3}

or

k =

_4p,

where

_p

is a positive integer. We have the following cases:

i) if

n =

_4m,

then

n

m-l

"" k(k+l) ""

Sn=

L...,.(-1) 2

k=

L..J

(-4p - 1

- 4p -

2 + 4p + 3 + 4p + 4) =

4m.

k=1 p=O ii) if

n =

_{4m +}

1, then

Sn =

4m - (4m +

1)

=

- 1. iii) if

n =

_{4m +}

2,

then

Sn =

4m - (4m +

1)

- (4m +

2) =

-(4m + 3).

iv) if

n =

_{4m + 3}

then

Sn =

4m - (4m +

1)

- (4m +

2)

+ (4m + 3)

=

O.

Hence

{

4m

if

n=4m

- 1 if

n=

_{4m +}

1

Sn=

�(4m + 3)

if

n=4m+2

if

n =

_{4m + 3.}

(Dorin Andrica,

Revista Matematidl Timi§oara (RMT) , No. 1(1981), pp. 50,

Problem

₄₃₀₃₎

33. _{a) Summing up the identities}

(n

+

2)

(n

+

2)(n

+

1)

(n)

k - (n

+

2 - k)(n

+

1 -

k) k

for

k =

₀

to

k = n

yields

Sn= (n

+

2)1(n

+

1)

(�(n;2) - (�:�) - (�:�)) =

1

[2n+2 ( 2)

1] _

2n+2 - (n

+ 3)

(n

+

2)(n

+

1)

- n

+

-

- (n

+

2)(n

+

1) .

b) Summing up the identities

(n

+

3)

(n

+

3)(n

+

2)(n

+

1)

(n)

k - (n

+ 3

-

k)(n

+

2 - k)(n

+

1 -

k) k

for

k =

₀

to

k = n

yields

(39)

1.2.

SOLUTIONS

. (�

(

n + 3

)

_

(

n + 3

)

_

(

n + 3

)

_

(

n + 3

)

= k=O k n + 1 n + 2 n + 3 = 1

(

_2n

+

₃_ �(n2 + 3n + 2)

)

= (n + 3) (n + 2) (n + 1) 2 _ 2n+

4

- (n2 + 3n + 2) - 2(n + 3) (n

₊

2) (n + 1) . 33

(Dorin Andrica,

Problem 2116)

34. _{Let Sn be the number in the statement.}

It is not difficult to see that

Sn =

�

[

(

2 + v'3

t<+

l +

(

2 - v'3

t

H

]

.

The required property says: there exists k > 0 such that Sn = (k - 1)2 + k2 , or,

equi val entl y,

2k2 - 2k + 1 - Sn =

_O.

The discriminant of this equation is Ll = 4(2Sn - 1), and, after usual

computa-tions, we obtain

� =

(

(1 +

v3)

2n+l

;

(1 -

v3)

2nH

)

2 Solving the equation, we find

k = 2n

+

1 + (1 + v'.3

)

2n+l + (1 - v'3) 2n+l

2n+2

Therefore, it is sufficient to prove that k is an integer. Let us denote

_Em

(1 + .J3)

_m

+ (1 -

.J3)

_m,

where m is a positive integer. Clearly,

_Em

is an integer for

all m. We will prove that 2[

_�]

divides

_Em,

m = 1, 2, 3, . . . Moreover, the numbers

Em

satisfy the relation

Em

= 2

_Em

-1 +

_2Em-2•

The property now follows by induction.

(Dorin Andrica,

Romanian IMO Selection Test, 1999)

35. _{Differentiating the identity}

yields where

sin nx = sinn X

((�)

cotn-1 X -

(�)

cotn-3 x +

_(�)

cotn-5 X -• • •

)

n cos nx = n sinn-1 x cos xP(cot x) - sinn X

�

PI (cot X),

(40)

34

and

1. ALGEBRA

For

_{x =}

�

we obtain Because 7f

(0.)

n

n cos n "4

₌

2

(nP(l) - 2P'(1)).

nP(l) = n (�) - n (;) + n (�) - . . .

-2P'(1) = -2(n - 1) (�) + 2(n - 3) (;) - 2(n -

5)

(�) + .. . ,

we have

nP(1) - 2P'(1) = - [(n - 2) (�) - (n - 6) (;) + (n - 10) (�) - . . . J =

= -n ( (�) - (;) + (�) - . . . ) + 28n·

To conclude, use that

hence

n

(0.)

n

(

n

7f

_n

7f

)

S =

_{cos - +}

sin-n

₂

4 4

(Dorin Andrica,

2(1977),

pp.

Problem

₃₂₀₀₎

36. _{Differentiating with respect to}

_x

_{the identities}

(x + l)n =

(�)

+ (�) X + .. . + (�) xn

and

(x - l)n = (�) xn - (n

�

1) xn-1 + . . . + (-1)n

(�)

yields and

n(x _ 1)n-1 = n (�) xn-1 - (n - 1) (n

�

1) xn-2 + .. . + (_1)n-1

(�).

MUltiplying by

_x

gives

nx(x + 1)n-1 = (�) x + 2 (;) X2 + .. . + n (�) xn

and

nx(x _ 1)n-1 = n (�) x" - (n

- 1)

(n

�

l) x"-' + .. . + (_1),,-1 (�) x.

(41)

and

1.2. SOLUTIONS

Differentiating again we obtain

n(x + l)n-l +

n

(

n

_{- l)x(x + 1)n-2}

=

(�)

_{+ 22}

(�)

_{x + .. . +}

n

_{2 (�) xn-1}

n(x - l)n-l +

n(n

_{- l)x(x}

-1)n-2

= n

_{2 (�) xn-1}

-

₍

n

_{- 1)2}

(

n

�

₁

)

xn-2+

+ .. . + (_W-l

(�)

Setting

_x

=

₁

yields

12 (�)

+ 22

(�)

+ .. . +

n

2 (�)

= n(n

+ 1)2n-2

Summing up the last two identities gives

Sn

=

₁₂

(�)

_{+ 32}

(�)

₊

. . . = n(n

_{+ 1)2n-a,}

as desired.

35 (Dorin Andrica,

1(1978),

pp.

90,

Problem

₃₄₃₈₎

37. _{Note that}

2n

_{n_1 2i+1_1}

2:

[log2

_k]

=

2: 2:

[log2

_{k] +}

[log2

_2n],

k=l

i=O k=2i

and [log2

_k]

=

_i

for

_2i

_::;

_k

<

_{2i+1 .}

Hence

as claimed.

2n

_n-1

2:

�

Og

_{2 k]}

=

2:

_i

.

_{2i +}

[log2

_2n]

= (n

_{- 2)2n +}

n

_{+ 2}

k=l

i=O

(Dorin Andrica,

2(1981),

pp.

63,

Problem

_4585;

Gazeta Matematica (GM-B), No.

_2-3(1982),

pp.

_83,

Problem

₁₉₁₁₃₎

38. _Let

_Yn

=

_22n

-

₁

for all positive integers n. Then

�

_{_ _}

_{2_}

1

2 (22n )2 - 2 . 22n + 1

Yn Yn+1 22n

-

1 22n+1 - 1

(22n - 1)(22n+1 - 1)

and therefore

_

(22n _ 1)2

_

22n

-

1 _ 1 _ 1

- (22n _ 1)(22n+1 - 1) - (22n)2 - 1 - 22n + 1 - xn

1 1 2

=

(42)

-36

1. ALGEBRA

Xn

Yn Yn+l

Summing up these relations yields

1 2 22

2n-1 1 2n

1 - + - + - +

. .

.

+

--

=

--

<

Xl X2 X3

Xn YI Yn+l YI

for all positive integers n, as desired.

(Dorin Andrica,

1-2(1980),

pp. 67,

Problem

₄₁₃₅₎

39. Substituting

_z

with

_iz

in the relation

J(z)J(iz)

=

_Z2

yields

J(iz)J( -z)

=

_{_Z2.}

Summing up gives

J(iz)(J(z) + J( -z))

=

_0,

so

_J(iz)

=

₀

or

_{J(z) + J( -z)}

= o.

From the relation

_J(z)J(iz)

=

_Z2

we deduce that

_J(z)

=

₀

if and only if

_z

=

O.

Hence if

z

-:f.

0,

then

J(iz)

-:f.

0

and so

J(z) + J( -z)

=

₀

and, if

_z

=

_0,

then

J(z) + J( -z)

=

_2J(0)

=

_O.

Clearly,

_{J(z) + J( -z)}

=

₀

for all numbers

_z

E C, as

desired.

Remark.

A function

J

: C -+ C satisfying the relation

J(z)J(iz)

=

Z2

is

J(z)

=

(

-

�

+ i

�

)

z.

(Titu Andreescu,

2(19

76

),

pp.

56,

Problem

₂₅₈₃₎

40. _Setting

_X

=

_Y

=

₁

yields

J2 (1) + J2 (

a) =

_{2 J (1 )}

and

_{.(J(l) - 1}

)

₂

=

₀

so

_J(l)

=

_1.

Substituting

_Y

=

₁

gives

or

J(x)J(l) + J

(�)

J(a)

=

_2J(x)

a

Take now

_Y

= - and observe that

(43)

1.2. SOLUTIONS

37

Consequently,

f(

x

)f

(;)

= 1,

therefore

_{f2 (}

x

_{) = 1,}

x > o.

Now set x

₌

y

₌

0, that gives

12(,ji) + 12

(�)

= 2/(t)

and because the left-hand side is positive, it follows that

_f

is positive and

_f(

x

_{) = 1}

for all x. Then

_f

is a constant function, as claimed.

( Titu Andreescu,

(

RMT

)

, No.

12(1977),

pp.

45,

Problem

_2849;

Gazeta Matematica (GM-B

₎

, No.

_10(1980),

pp.

_439,

Problem

₁₈₄₅₅₎

41 . _{The function is not periodical. Suppose, by way of contradiction, that there}

is a number

_T

>

₀

such that

f(

x +

T) =

f(x

)

or sin

[

x

+ T] =

sin

[

x

]

, for all x E JR.

It follows that

[

x

+ T]

-[

x

] =

2k

(

x

)

7r,

X

E 1R,

where k : JR -+ Z is a function. Because 7r is irrational, we deduce that

_{k(x) = 0}

for

all x E JR and therefore

[

x

] = [

x

+ T]

for all x E JR

which is false, since the greatest integer function is not periodical.

(Dorin Andrica,

(

RMT

)

, No.

1(1978),

pp.

89,

Pro blem

₃₄₃₀₎

42 . _{Considering the determinant}

1 2 3

n

8 = 12 22 32

n

₂

In 2n 3n

n

_n

we have

1 12 1

3

In

�

_{= IS(i,j)1 = 8 · 21 22 2}

3

2n

= 82

, n1 n

₂

n3 n

_n

(44)

38

1. ALGEBRA

On the other hand,

1 1 1 1

8 = n! 1 2 3 n

1n-1

2

_n-1

3n-1

n

_n-l

(here we used the known result on Vandermonde determinants). Therefore

(Dorin Andrica,

Problem 3862) 43. The determinant is

al

0 0 0 0 0

_a2

0 0

b2

0 0

_a3

b3

0

�2n

= 0 0

b2n-2

_a2n-2

0 0

b2n-l

0 0

_a2n-l

b2n

₀ ₀ ₀ ₀

Expanding along the first and then the last row we obtain

which gives

n

�2n

=

II (aka2n-k+l - bkb2n-k+d

k=l

b1

0 0 0 0

a2n

(Dorin Andrica,

Problem 3201; Gazeta Matematica (GM-B) , No. 8(1977), pp. 325, Problem 16808)

44. a) Adding the last three columns to the first one yields that

_{x + y}

+

_{z + v}

divides the determinant.

Adding the first and second columns and subtracting the last two columns implies that

_{x + y - z - v}

divides the determinant.

Analogously we can check that

_x-y+z-v

and

_x-y-z+v

divide the determinant, and taking into account that it has degree

₄

in each of the variables, the determinant equals

..\

(

x + y + z + v)(x + y - z - v)(x - y + z - v)(x - y - z + v),