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: 1513

2021 Science ON all problems, 4

Find all functions $f:\mathbb{Z}_{\ge 1}\to \mathbb{R}_{>0}$ such that for all positive integers $n$ the following relation holds: $$\sum_{d|n} f(d)^3=\left (\sum_{d|n} f(d) \right )^2,$$ where both sums are taken over the positive divisors of $n$. [i] (Vlad Robu) [/i]

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 $x, y \in \mathbb{Q^+}$.

2016 Thailand Mathematical Olympiad, 9

A real number $a \ne 0$ is given. Determine all functions $f : R \to R$ satisfying $f(x)f(y) + f(x + y) = axy$ for all real numbers $x, y$.

2015 Switzerland Team Selection Test, 7

Find all finite and non-empty sets $A$ of functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that for all $f_1, f_2 \in A$, there exists $g \in A$ such that for all $x, y \in \mathbb{R}$ $$f_1 \left(f_2 (y)-x\right)+2x=g(x+y)$$

2009 IMO Shortlist, 3

Determine all functions $ f$ from the set of positive integers to the set of positive integers such that, for all positive integers $ a$ and $ b$, there exists a non-degenerate triangle with sides of lengths \[ a, f(b) \text{ and } f(b \plus{} f(a) \minus{} 1).\] (A triangle is non-degenerate if its vertices are not collinear.) [i]Proposed by Bruno Le Floch, France[/i]

1986 Traian Lălescu, 2.1

Let be a nonnegative integer $ n. $ Find all continuous functions $ f:\mathbb{R}_{\ge 0}\longrightarrow\mathbb{R} $ for which the following equation holds: $$ (1+n)\int_0^x f(t) dt =nxf(x) ,\quad\forall x>0. $$

1973 Bulgaria National Olympiad, Problem 4

Find all functions $f(x)$ defined in the range $\left(-\frac\pi2,\frac\pi2\right)$ that are differentiable at $0$ and satisfy $$f(x)=\frac12\left(1+\frac1{\cos x}\right)f\left(\frac x2\right)$$ for every $x$ in the range $\left(-\frac\pi2,\frac\pi2\right)$. [i]L. Davidov[/i]

2014 Contests, 1

Determine all functions $f:\mathbb{R} \to \mathbb{R}$ such that \[ xf(y) + f(xf(y)) - xf(f(y)) - f(xy) = 2x + f(y) - f(x+y)\] holds for all $x,y \in \mathbb{R}$.

1997 Nordic, 4

Let f be a function defined in the set $\{0, 1, 2,...\}$ of non-negative integers, satisfying $f(2x) = 2f(x), f(4x+1) = 4f(x) + 3$, and $f(4x-1) = 2f(2x - 1) -1$. Show that $f $ is an injection, i.e. if $f(x) = f(y)$, then $x = y$.

2020 Serbian Mathematical Olympiad, Problem 5

For a natural number $n$, with $v_2(n)$ we denote the largest integer $k\geq0$ such that $2^k|n$. Let us assume that the function $f\colon\mathbb{N}\to\mathbb{N}$ meets the conditions: $(i)$ $f(x)\leq3x$ for all natural numbers $x\in\mathbb{N}$. $(ii)$ $v_2(f(x)+f(y))=v_2(x+y)$ for all natural numbers $x,y\in\mathbb{N}$. Prove that for every natural number $a$ there exists exactly one natural number $x$ such that $f(x)=3a$.

2005 Taiwan TST Round 3, 1

Find all functions $ f: \mathbb{N^{*}}\to \mathbb{N^{*}}$ satisfying \[ \left(f^{2}\left(m\right)+f\left(n\right)\right) \mid \left(m^{2}+n\right)^{2}\] for any two positive integers $ m$ and $ n$. [i]Remark.[/i] The abbreviation $ \mathbb{N^{*}}$ stands for the set of all positive integers: $ \mathbb{N^{*}}=\left\{1,2,3,...\right\}$. By $ f^{2}\left(m\right)$, we mean $ \left(f\left(m\right)\right)^{2}$ (and not $ f\left(f\left(m\right)\right)$). [i]Proposed by Mohsen Jamali, Iran[/i]

1990 IMO Longlists, 6

Let function $f : \mathbb Z_{\geq 0}^0 \to \mathbb N$ satisfy the following conditions: (i) $ f(0, 0, 0) = 1;$ (ii) $f(x, y, z) = f(x - 1, y, z) + f(x, y - 1, z) + f(x, y, z - 1);$ (iii) when applying above relation iteratively, if any of $x', y', z$' is negative, then $f(x', y', z') = 0.$ Prove that if $x, y, z$ are the side lengths of a triangle, then $\frac{\left(f(x,y,z) \right) ^k}{ f(mx ,my, mz)}$ is not an integer for any integers $k, m > 1.$

2010 Belarus Team Selection Test, 2.4

Find all functions $f, g : Q \to Q$ satisfying the following equality $f(x + g(y)) = g(x) + 2 y + f(y)$ for all $x, y \in Q$. (I. Voronovich)