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

Pham Quoc Sang
The art of Mathematics

Mathematical problems from
mathematical examinations around
the world
Functional Equation

Ho Chi Minh City - 2018


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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) f (x) = x for all real x.
( Korea Winter, 2018 )
Problem 2. Find all functions f : R → R satisfying the following conditions :
f (x3 ) + f (y 3 ) = (x + y)(f (x2 ) + f (y 2 ) − 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 + y 2 f (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
1


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 + y 3 + xy) = x2 f (x) + y 2 f (y) + f (xy), ∀x, y ∈ R
( IZHO 2015 )
Problem 11. Find all the functions f : R → R such that
f (x2 ) + 4y 2 f (y) = (f (x − y) + y 2 )(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))
2


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 f a (b) = f (a + b − 1), where f a (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) = 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
3


( 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 + y 2 )f (yf (x)) = xyf (y 2 + f (x))x, y ∈ R+
( IZHO 2017 )
Problem 27. Determine all functions f(n) from the positive integers to the positive integers satisfying the following condition: whenever a,b,c are positive integers such that
4


1 1
1
+ = , then
a b
c

1
1
1
+
=
f (a) f (b)
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 → 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 + y 2 ) ≥ 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+ .
5


( Iran TST 2012 )
Problem 34. Find all the functions f : N → N such that the inequality

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

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 + y 2 ) 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 + y 2 ) 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)
.
2

( 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|

6

( 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 (y 3 ) = (x + y)(f (x2 ) + f (y 2 ) − 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 )

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

xf (xy) + xyf (x) ≥ f (x2 )f (y) + x2 y
holds for all x, y ∈ R.
( Middle European Mathematical Olympiad T-2 )
Problem 48. Find all functions f : N∗ → N∗ satisfying
f 2 (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 f 2 (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 )

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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
x2 + y 2
2

f (xy) = f

+ (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 )

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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) f (x) = x for all real x.
( Korea Winter, 2018 )
Solution.
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Let P (x, y) be the assertion. P (0; 0) gives f (0) ∈ {0; −1} and since f (0) = 0 we get
f (0) = 0.
Let f (x) = g(x) + x we find 1 > g(x) ≥ 0.
We have
g(x + y) − g(x) − g(y) + x + y − x − y ∈ {0, 1}
we know that
x + y − x − y = {x} + {y}
If
{x} + {y} = 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} + x for all real numbers x.Its easy

to find c = 1 or c = 0.Thus solutions are
f (x) = {x} + x
and
f (x) = x
.

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