Các đề toán học từ các kỳ thi toán học trên toàn thế giới

Problem 1. Find all functions f : R → R satisfying the following conditions : a) f(x + y) − f(x) − f(y) ∈ {0, 1} for all x, y ∈ R b) bf(x)c = bxc for all real x. ( Korea Winter, 2018 ) Problem 2. Find all functions f : R → R satisfying the following conditions : f(x 3 ) + f(y 3 ) = (x + y)(f(x

Pham Quoc Sang The art of Mathematics

Mathematical problems from mathematical examinations around

the world

Functional Equation

Ho Chi Minh City - 2018

1 Problem.

Problem 1 Find all functions f : R → R satisfying the following conditions :

a) f (x + y) − f (x) − f (y) ∈ {0, 1} for all x, y ∈ R

b) bf (x)c = bxc for all real x

( Korea Winter, 2018 ) Problem 2 Find all functions f : R → R satisfying the following conditions :

f (x3) + f (y3) = (x + y)(f (x2) + f (y2) − f (xy))

( British MO 2009 ) Problem 3 Define a function f (n) from the positive integers to the positive integers such that f (f (n)) is the number of positive integer divisors of n Prove that if p is a prime, then f (p) is prime

( China MO 2017 ) Problem 4 Find all functions f : R → R such that

f (f (x)2+ f (y)) = xf (x) + y, ∀x, y ∈ R

( Kyrgyzstan MO 2012 ) Problem 5 Functions f, g : Z → Z satisfy

f (g(x) + y) = g(f (y) + x), ∀x, y ∈ Z If f is bounded, prove that g is periodic

( China TST 2018 ) Problem 6 Find all functions f : R → R suc that

f (f (x) + y) = f (x2− y) + 4f (x)y

( Iran MO 99 ) Problem 7 Find all functions f : R → R such that

f (x2+ y2f (x)) = xf (y)2− f (x)2, ∀x, y ∈ R

( Brazilian Undergrad MO 2016 ) Problem 8 Let f be a function from the set of integers to the set of positive integers Suppose that, for any two integers m and n, the difference f (m) − f (n) is divisible by

f (m − n) Prove that, for all integers m and n with f (m) ≤ f (n), the number f (n) is divisible by f (m)

( IMO 2011 ) Problem 9 Let k be a positive integer Find all functions f : N → N satisfying the following two conditions:

• For infinitely many prime numbers p there exists a positve integer c such that

f (c) = pk

• For all positive integers m and n, f (m) + f (n) divides f (m + n)

( Iran MO 3rd round 2017 )

Problem 10 Find all functions f : R → R such that

f (x3+ y3+ xy) = x2f (x) + y2f (y) + f (xy), ∀x, y ∈ R

( IZHO 2015 )

Problem 11 Find all the functions f : R → R such that

f (x2) + 4y2f (y) = (f (x − y) + y2)(f (x + y) + f (y)), ∀x, y ∈ R

( Turkey TST 2015 )

Problem 12 Determine all functions f : Z → Z satisfying

f f (m) + n
+ f (m) = f (n) + f (3m) + 2014 for all integers m and n ( IMO shortlist 2014 )

Problem 13 a is a real number, for every x, y are real numbers

f (xy + f (y)) = f (x)y + a Find the solution of functions depends on a

( Turkey TST 2017 )

Problem 14 Find all functions f : R → R, such that

f (x.f (y)) + f (f (x) + f (y)) = y.f (x) + f (x + f (y))

holds for all x, y ∈ R, where R denotes the set of real numbers.

( MEMO 2009 )

Problem 15 Find all functions f : R+ → R+ such that for all positive real numbers

x, y :

f (y)f (x + f (y)) = f (x)f (xy)

( Iran MO 3rd round 2016 finals )

Problem 16 Find all functions f : R −→ R such that

f (x2+ yf (z)) = xf (x) + zf (y), ∀x, y, z ∈ R

( INMO 2005 )

Problem 17 Does there exist a function f : R → R satisfying the following conditions: i) For each real y there is a real x such that f (x) = y

ii) f (f (x)) = (x − 1)f (x) + 2 for all real x

( International Zhautykov olympiad 2014 )

Problem 18 A function f : N → N is called nice if fa(b) = f (a + b − 1), where fa(b) denotes a times applied function f

Let g be a nice function, and an integer A exists such that g(A + 2018) = g(A) + 1 a) Prove that g(n + 20172017) = g(n) for all n ≥ A + 2

b) If g(A + 1) 6= g(A + 1 + 20172017) find g(n) for n < A

( Serbia TST 2017 )

Problem 19 Show that there is no continuous function f : R → R such that for every real number x

f (x − f (x)) = x

2.

( Turkey TST 2001 )

Problem 20 Determine all functions f : R → R such that for all x, y ∈ R hold:

f (xf (y) − yf (x)) = f (xy) − xy

( SMO 2014, Dusan Djukic )

Problem 21 Suppose that f : N → N is a function for which the expression

af (a) + bf (b) + 2ab for all a, b ∈ N is always a perfect square

Prove that f (a) = a for all a ∈ N

( Iran TST 2011 )

Problem 22 Find all functions f : R → R such that

f (x + yf (x)) = f (x) + xf (y)

( Motivation behind HongKong 2004 )

Problem 23 Find all functions f : R → R satisfying relation :

f (xf (y) − f (x)) = 2f (x) + xy, ∀x, y ∈ R

( Viet Nam National Olympiad 2017 )

Problem 24 Find all functions f : Q → Q satisfying these conditions:

f (1) = 2 , ∀x, y ∈ Q : f (xy) = f (x).f (y) − f (x + y) + 1

( Luxembourg 1980 )

Problem 25 let R+ be the set of all positive real number Determine all function f :

R+→ R+ that satisfy

f (xf (y) + 1) = yf (x + y), ∀x, y ∈ R

( KTOM November 2017 )

Problem 26 Find all functions f : R → R such that

(x + y2)f (yf (x)) = xyf (y2+ f (x))x, y ∈ R+

( IZHO 2017 )

Problem 27 Determine all functions f(n) from the positive integers to the positive in-tegers satisfying the following condition: whenever a,b,c are positive inin-tegers such that

a +

1

b =

1

c, then

1

f (a) +

1

f (b) =

1

f (c)

( British Maths Olympiad )

Problem 28 Find all functions f : N → N satisfying

f (mn) = lcm(m, n) · gcd(f (m), f (n)) for all positive integer m, n ( Korea National 2013 )

Problem 29 Find all functions f : R 7→ R such that

(f (x) + y) (f (x − y) + 1) = f (f (xf (x + 1)) − yf (y − 1)) for all x, y ∈ R

( Swiss IMO TST 2016 )

Problem 30 Determine all functions f : Z → Z with the property that

f (x − f (y)) = f (f (x)) − f (y) − 1 holds for all x, y ∈ Z

( IMO Shortlist 2015 )

Problem 31 Let R denote the set of all real numbers Find all functions f : R → R such that

f x2+ f (y)
= y + (f (x))2

for all x, y ∈ R

( IMO 1992 )

Problem 32 Find all real numbers c such that there exists a function f : R≥0 → R which satisfies the following

For all nonnegative reals x, y, f (x + y2) ≥ cf (x) + y

Here R≥0 is the set of all nonnegative reals

( KMO 2017 )

Problem 33 Let g(x) be a polynomial of degree at least 2 with all of its coefficients positive Find all functions f : R+−→ R+ such that

f (f (x) + g(x) + 2y) = f (x) + g(x) + 2f (y) ∀x, y ∈ R+

( Iran TST 2012 )

Problem 34 Find all the functions f : N → N such that the inequality

f (x) + y.f (f (x)) 6 x (1 + f (y))

( 6th european mathematical cup, 2017 )

Problem 35 Find all non-decreasing functions from real numbers to itself such that for all real numbers x, y f (f (x2) + y + f (y)) = x2+ 2f (y) holds

( Turkey NMO 2012 )

Problem 36 Let f (x) be a polynomial function of degree n with integer coefficients and satisfying the given conditions: f (0) = 39, and f (xi) = 2017 for every i = 1, 2, 3 , n with all such xi distinct Find all possible values of n

( Indonesia MO )

Problem 37 Let N = {0, 1, 2, } Determine all functions f : N → N such that

xf (y) + yf (x) = (x + y)f (x2+ y2) x, y ∈ N

( Canada MO 2002 )

Problem 38 Let N = {0, 1, 2, } Determine all functions f : N → N such that

xf (y) + yf (x) = (x + y)f (x2+ y2) x, y ∈ N

( Canada MO 2002 )

Problem 39 Find all functions f : Z2 → [0, 1] such that for any integers x and y,

f (x, y) = f (x − 1, y) + f (x, y − 1)

( USA Winter Team Selection Test 1 for IMO 2018, Yang Liu and Michael Kura )

Problem 40 Find all the functions f : R → R such that

|f (x) − f (y)| = 2|x − y|

( ISI Entrance 2015 )

Problem 41 Determine all surjective functions f : Z → Z such that

f (xyz + xf (y) + yf (z) + zf (x)) = f (x) f (y) f (z) for all x, y, z in Z

(Taiwan TST Round 2, 2017 )

Problem 42 Determine all integer solutions (x, y) of the equation

(x − 1)x(x + 1) + (y − 1)y(y + 1) = 24 − 9xy

(Austrian Regional Competition for Advanced Students, 2012 )

Problem 43 Solve the function f : R → R:

f (x3) + f (y3) = (x + y)(f (x2) + f (y2) − f (xy))

(Albania IMO Selection )

Problem 44 Let f be a non-constant function from the set of positive integers into the set of positive integer, such that a − b divides f (a) − f (b) for all distinct positive integers

a, b Prove that there exist infinitely many primes p such that p divides f (c) for some positive integer c

( IMO Shortlist 2009 )

Problem 45 Find all functions f : R → R such that for all real numbers x and y,

(f (x) + xy) · f (x − 3y) + (f (y) + xy) · f (3x − y) = (f (x + y))2

( USAJMO 2016 )

Problem 46 The function f satisfies for all reals x, y the following inequality

f (x2+ 2y) ≥ f (x2+ 3y)

It is known that f (100) = 100 Find all possible values of f (200)

( Saint Petersburg 2016 )

Problem 47 Determine all functions f : R → R such that

xf (xy) + xyf (x) ≥ f (x2)f (y) + x2y holds for all x, y ∈ R

( Middle European Mathematical Olympiad T-2 )

Problem 48 Find all functions f : N∗ → N∗ satisfying

f2(m) + f (n)
| m2 + n
2 for any two positive integers m and n

Remark The abbreviation N∗ stands for the set of all positive integers:

N∗ = {1, 2, 3, }

By f2(m), we mean (f (m))2 (and not f (f (m)))

( IMO ShortList 2004 )

Problem 49 Let R+ be the set of all positive real numbers

Find all functions f : R+ → R+ that satisfy the following conditions:

• f (xyz) + f (x) + f (y) + f (z) = f (√xy)f (√

yz)f (√

zx) for all x, y, z ∈ R+

• f (x) < f (y) for all 1 ≤ x < y

( IMO ShortList 2003, Hojoo Lee, Korea )

Problem 50 Find all functions f : R → R such that

f (f (xy − x)) + f (x + y) = yf (x) + f (y) for all real numbers x and y

( Hong Kong TST, 2018 )

Problem 51 Let f : Z → Z be a function such that for all integers x and y, the following holds:

f (f (x) − y) = f (y) − f (f (x))

Show that f is bounded

( Baltic Way 2011 )

Problem 52 Find all functions f from the set of real numbers into the set of real numbers which satisfy for all x, y the identity

f (xf (x + y)) = f (yf (x)) + x2

( IMO Shortlist 2009 )

Problem 53 Find all functions f : R → R such that

f (y + x · f (x + y)) + f (x + y · f (x + y)) = f (x + f (y)) · f (y + f (x)) + f (f (x)) + y for all x, y ∈ R

( The IMO team is selected from the advanced class, Israel )

Problem 54 Find all continuous functions f : R → R such, that

f (xy) = f x2+ y2

2

 + (x − y)2 for any real numbers x and y

( Moldova TST, 2017 )

Problem 55 Find all functions f : R → R such that

f (yf (x + y) + f (x)) = 4x + 2yf (x + y)

for all x, y ∈ R

( European Girls’ Mathematical Olympiad, 2012 )

Problem 56 Determine all functions f defined in the set of rational numbers and taking their values in the same set such that the equation f (x + y) + f (x − y) = 2f (x) + 2f (y) holds for all rational numbers x and y

( Nordic Mathematical Contest, 1998 )

Problem 57 Find all functions f : R+ → R+ satisfying the following condition: for any three distinct real numbers a, b, c, a triangle can be formed with side lengths a, b, c, if and only if a triangle can be formed with side lengths f (a), f (b), f (c)

(China Team Selection Test 2016 )

Problem 58 Let P be the set of positive rational numbers and let f : P → P be such that

f (x) + f 1

x



= 1 and

f (2x) = 2f (f (x)) for all x ∈ P

Find, with proof, an explicit expression for f (x) for all x ∈ P

( Iran MO 1991)

Problem 59 A function f is defined on the natural numbers N and satisfies the following rules:

a) f (1) = 1;

b) f (2n) = f (n) and f (2n + 1) = f (2n) + 1 for all n ∈ N

Calculate the maximum value m of the set {f (n) : n ∈ N, 1 ≤ n ≤ 1989}, and determine the number of natural numbers n, with 1 ≤ n ≤ 1989, that satisfy the equation f (n) = m

( Iran MO 1989)

Problem 60 Find all continuous functions f : R → R such that for some polynomial

P (x, y) ∈ R[x, y] satisfies

f (x + y) = P (f (x), f (y)) for every x, y ∈ R

( AMM, Harry tamvakis)

2 Solution.

Problem 1 Find all functions f : R → R satisfying the following conditions :

a) f (x + y) − f (x) − f (y) ∈ {0, 1} for all x, y ∈ R

b) bf (x)c = bxc for all real x

( Korea Winter, 2018 ) Solution

Let P (x, y) be the assertion P (0; 0) gives f (0) ∈ {0; −1} and since bf (0)c = 0 we get

f (0) = 0

Let f (x) = g(x) + bxc we find 1 > g(x) ≥ 0

We have

g(x + y) − g(x) − g(y) + bx + yc − bxc − byc ∈ {0, 1}

we know that

bx + yc − bxc − byc = b{x} + {y}c If

b{x} + {y}c = 0

We find that g(x + y) = g(x) + g(y)

Its easy to see g(1) = 0 and g(x + 1) = g(x) for all real x.For all 0 < x < 1, 0 < y < 1 and

0 < x + y < 1 we have g(x + y) = g(x) + g(y) where 1 > g(x) ≥ 0.Thus we have g(x) = cx for all 0 < x < 1 and for some 1 ≥ c ≥ 0.But other hand we have g(x + 1) = g(x) for all real x.Thus we have g(x) = c {x} and f (x) = c {x} + bxc for all real numbers x.Its easy

to find c = 1 or c = 0.Thus solutions are

f (x) = {x} + bxc and

f (x) = bxc