This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

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Found problems: 10

2023 Dutch IMO TST, 4

Find all functions $f: \mathbb{Q^+} \rightarrow \mathbb{Q}$ satisfying $f(x)+f(y)= \left(f(x+y)+\frac{1}{x+y} \right) (1-xy+f(xy))$ for all $x, y \in \mathbb{Q^+}$.

2020 Latvia Baltic Way TST, 2

Determine all functions $f:\mathbb R\to\mathbb R$ that satisfy equation: $$ f(x^3+y^3) =f(x^3) + 3x^2f(x)f(y) + 3f(x)f(y)^2 + y^6f(y) $$ for all reals $x,y$

2022 Indonesia MO, 1

Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any $x,y \in \mathbb{R}$ we have \[ f(f(f(x)) + f(y)) = f(y) - f(x) \]

2008 Bulgarian Autumn Math Competition, Problem 12.3

Find all continuous functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that \[(f(x)f(y)-1)f(x+y)=2f(x)f(y)-f(x)-f(y)\quad \forall x,y\in \mathbb{R}\]

2020 Final Mathematical Cup, 1

Find all such functions $f:\mathbb{R} \to \mathbb{R}$ that for any real $x,y$ the following equation is true. $$f(f(x)+y)+1=f(x^2+y)+2f(x)+2y$$

2023 Olympic Revenge, 1

Find all $f:\mathbb{R}\rightarrow \mathbb{R}$ continuous functions such that $\lim_{x\rightarrow \infty} f(x) =\infty$ and $\forall x,y\in \mathbb{R}, |x-y|>\varphi, \exists n<\varphi^{2023}, n\in \mathbb{N}$ such that $$f^n(x)+f^n(y)=x+y$$

2023 Dutch IMO TST, 4

Find all functions $f: \mathbb{Q^+} \rightarrow \mathbb{Q}$ satisfying $f(x)+f(y)= \left(f(x+y)+\frac{1}{x+y} \right) (1-xy+f(xy))$ for all $x, y \in \mathbb{Q^+}$.

2023 District Olympiad, P4

Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that any real numbers $x{}$ and $y{}$ satisfy \[f(xf(x)+f(y))=f(f(x^2))+y.\]

2017 239 Open Mathematical Olympiad, 3

Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real number $x,y$, $$(y+1)f(yf(x))=yf(x(y+1)).$$

2022 Turkey Team Selection Test, 2

Find all functions $f: \mathbb{Q^+} \rightarrow \mathbb{Q}$ satisfying $f(x)+f(y)= \left(f(x+y)+\frac{1}{x+y} \right) (1-xy+f(xy))$ for all $x, y \in \mathbb{Q^+}$.