Found problems: 1513
2013 IMO, 5
Let $\mathbb Q_{>0}$ be the set of all positive rational numbers. Let $f:\mathbb Q_{>0}\to\mathbb R$ be a function satisfying the following three conditions:
(i) for all $x,y\in\mathbb Q_{>0}$, we have $f(x)f(y)\geq f(xy)$;
(ii) for all $x,y\in\mathbb Q_{>0}$, we have $f(x+y)\geq f(x)+f(y)$;
(iii) there exists a rational number $a>1$ such that $f(a)=a$.
Prove that $f(x)=x$ for all $x\in\mathbb Q_{>0}$.
[i]Proposed by Bulgaria[/i]
2018 IFYM, Sozopol, 5
Find all functions $f :[0, +\infty) \rightarrow [0, +\infty)$ for which
$f(f(x)+f(y)) = xy f (x+y)$
for every two non-negative real numbers $x$ and $y$.
2006 Germany Team Selection Test, 1
We denote by $\mathbb{R}^\plus{}$ the set of all positive real numbers.
Find all functions $f: \mathbb R^ \plus{} \rightarrow\mathbb R^ \plus{}$ which have the property:
\[f(x)f(y)\equal{}2f(x\plus{}yf(x))\]
for all positive real numbers $x$ and $y$.
[i]Proposed by Nikolai Nikolov, Bulgaria[/i]
2014 IMO Shortlist, A6
Find all functions $f : \mathbb{Z} \to\mathbb{ Z}$ such that
\[ n^2+4f(n)=f(f(n))^2 \]
for all $n\in \mathbb{Z}$.
[i]Proposed by Sahl Khan, UK[/i]
1977 IMO Shortlist, 9
For which positive integers $n$ do there exist two polynomials $f$ and $g$ with integer coefficients of $n$ variables $x_1, x_2, \ldots , x_n$ such that the following equality is satisfied:
\[\sum_{i=1}^n x_i f(x_1, x_2, \ldots , x_n) = g(x_1^2, x_2^2, \ldots , x_n^2) \ ? \]
2021 2nd Memorial "Aleksandar Blazhevski-Cane", 6
Let $\mathbb{R}^{+}$ be the set of all positive real numbers. Find all the functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that for all $x, y \in \mathbb{R}^{+}$,
\[ f(x)f(y) = f(y)f(xf(y)) + \frac{1}{xy}. \]
1991 IMO Shortlist, 23
Let $ f$ and $ g$ be two integer-valued functions defined on the set of all integers such that
[i](a)[/i] $ f(m \plus{} f(f(n))) \equal{} \minus{}f(f(m\plus{} 1) \minus{} n$ for all integers $ m$ and $ n;$
[i](b)[/i] $ g$ is a polynomial function with integer coefficients and g(n) = $ g(f(n))$ $ \forall n \in \mathbb{Z}.$
2015 Miklos Schweitzer, 8
Prove that all continuous solutions of the functional equation
$\left(f(x)-f(y)\right)\left(f\left(\frac{x+y}{2}\right)-f\left(\sqrt{xy}\right)\right)=0 \ , \ \forall x,y\in (0,+\infty)$
are the constant functions.
2005 Portugal MO, 6
Prove that there is a unique function $f: N\to N$, that verifies $$f(a + b)f(a - b) = f(a^2)$$, for any $a, b\in N$ such that $a > b$.
2015 Postal Coaching, Problem 2
Find all functions $f: \mathbb{Q} \to \mathbb{R}$ such that $f(xy)=f(x)f(y)+f(x+y)-1$ for all rationals $x,y$
2014 France Team Selection Test, 4
Let $\mathbb{Z} _{>0}$ be the set of positive integers. Find all functions $f: \mathbb{Z} _{>0}\rightarrow \mathbb{Z} _{>0}$ such that
\[ m^2 + f(n) \mid mf(m) +n \]
for all positive integers $m$ and $n$.
PEN K Problems, 30
Find all functions $f: \mathbb{N}\to \mathbb{N}$ such that for all $n\in \mathbb{N}$: \[f(f(f(n)))+f(f(n))+f(n)=3n.\]
2024 Dutch IMO TST, 2
Find all functions $f:\mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ such that for all positive integers $m,n$ and $a$ we have
a) $f(f(m)f(n))=mn$ and
b) $f(2024a+1)=2024a+1$.
2017 Estonia Team Selection Test, 11
For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geq 2016$, the integer $P(n)$ is positive and $$S(P(n)) = P(S(n)).$$
[i]Proposed by Warut Suksompong, Thailand[/i]
2005 Romania Team Selection Test, 3
Let $n\geq 0$ be an integer and let $p \equiv 7 \pmod 8$ be a prime number. Prove that
\[ \sum^{p-1}_{k=1} \left \{ \frac {k^{2^n}}p - \frac 12 \right\} = \frac {p-1}2 . \]
[i]Călin Popescu[/i]
2017 IMO, 2
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that, for any real numbers $x$ and $y$, \[ f(f(x)f(y)) + f(x+y) = f(xy). \]
[i]Proposed by Dorlir Ahmeti, Albania[/i]
1995 Singapore Team Selection Test, 1
Let $N =\{1, 2, 3, ...\}$ be the set of all natural numbers and $f : N\to N$ be a function.
Suppose $f(1) = 1$, $f(2n) = f(n)$ and $f(2n + 1) = f(2n) + 1$ for all natural numbers $n$.
(i) Calculate the maximum value $M$ of $f(n)$ for $n \in N$ with $1 \le n \le 1994$.
(ii) Find all $n \in N$, with 1 \le n \le 1994, such that $f(n) = M$.
2023 IFYM, Sozopol, 4
Find all real numbers $a$ for which there exist functions $f,g: \mathbb{R} \to \mathbb{R}$, where $g$ is strictly increasing, such that $f(1) = 1$, $f(2) = a$, and
\[
f(x) - f(y) \leq (x-y)(g(x) - g(y))
\]
for all real numbers $x$ and $y$.
2019 ELMO Shortlist, A2
Find all functions $f:\mathbb Z\to \mathbb Z$ such that for all surjective functions $g:\mathbb Z\to \mathbb Z$, $f+g$ is also surjective. (A function $g$ is surjective over $\mathbb Z$ if for all integers $y$, there exists an integer $x$ such that $g(x)=y$.)
[i]Proposed by Sean Li[/i]
2019 Belarusian National Olympiad, 11.7
Find all functions $f:\mathbb R\to\mathbb R$ satisfying the equality
$$
f(f(x)+f(y))=(x+y)f(x+y)
$$
for all real $x$ and $y$.
[i](B. Serankou)[/i]
1985 IMO Shortlist, 12
A sequence of polynomials $P_m(x, y, z), m = 0, 1, 2, \cdots$, in $x, y$, and $z$ is defined by $P_0(x, y, z) = 1$ and by
\[P_m(x, y, z) = (x + z)(y + z)P_{m-1}(x, y, z + 1) - z^2P_{m-1}(x, y, z)\]
for $m > 0$. Prove that each $P_m(x, y, z)$ is symmetric, in other words, is unaltered by any permutation of $x, y, z.$
2024 CAPS Match, 5
Let $\alpha\neq0$ be a real number. Determine all functions $f:\mathbb R\to\mathbb R$ such that \[f\left(x^2+y^2\right)=f(x-y)f(x+y)+\alpha yf(y)\] holds for all $x, y\in\mathbb R.$
2004 Switzerland Team Selection Test, 11
Find all injective functions $f : R \to R$ such that for all real $x \ne y$ , $f\left(\frac{x+y}{x-y}\right) = \frac{f(x)+ f(y)}{f(x)- f(y)}$
1977 IMO Longlists, 2
Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition:
\[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]
2019 Latvia Baltic Way TST, 2
Let $\mathbb R$ be set of real numbers. Determine all functions $f:\mathbb R\to \mathbb R$ such that
$$f(y^2 - f(x)) = yf(x)^2+f(x^2y+y)$$
holds for all real numbers $x; y$