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.

AND:
OR:
NO:

Found problems: 451

1962 IMO Shortlist, 4

Solve the equation $\cos^2{x}+\cos^2{2x}+\cos^2{3x}=1$

2013 Dutch Mathematical Olympiad, 4

For a positive integer n the number $P(n)$ is the product of the positive divisors of $n$. For example, $P(20) = 8000$, as the positive divisors of $20$ are $1, 2, 4, 5, 10$ and $20$, whose product is $1 \cdot 2 \cdot 4 \cdot 5 \cdot 10 \cdot 20 = 8000$. (a) Find all positive integers $n$ satisfying $P(n) = 15n$. (b) Show that there exists no positive integer $n$ such that $P(n) = 15n^2$.

1950 Moscow Mathematical Olympiad, 180

Solve the equation $\sqrt {x + 3 - 4 \sqrt{x -1}} +\sqrt{x + 8 - 6 \sqrt{x - 1}}= 1$.

2002 IMO Shortlist, 3

Let $P$ be a cubic polynomial given by $P(x)=ax^3+bx^2+cx+d$, where $a,b,c,d$ are integers and $a\ne0$. Suppose that $xP(x)=yP(y)$ for infinitely many pairs $x,y$ of integers with $x\ne y$. Prove that the equation $P(x)=0$ has an integer root.

2006 Cezar Ivănescu, 1

Solve the equation [b]a)[/b] $ \log_2^2 +(x-1)\log_2 x =6-2x $ in $ \mathbb{R} . $ [b]b)[/b] $ 2^{x+1}+3^{x+1} +2^{1/x^2}+3^{1/x^2}=18 $ in $ (0,\infty ) . $ [i]Cristinel Mortici[/i]

2011 India IMO Training Camp, 2

Find all pairs $(m,n)$ of nonnegative integers for which \[m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).\] [i]Proposed by Angelo Di Pasquale, Australia[/i]

2005 Gheorghe Vranceanu, 1

Tags: equation , algebra
Solve in the real numbers the equation $ 3^{x+1}=(x-1)(x-3). $

2002 Croatia National Olympiad, Problem 1

Tags: equation , algebra
Solve the equation $$\left(x^2+3x-4\right)^3+\left(2x^2-5x+3\right)^3=\left(3x^2-2x-1\right)^3.$$

2020 Iran MO (3rd Round), 4

We call a polynomial $P(x)$ intresting if there are $1398$ distinct positive integers $n_1,...,n_{1398}$ such that $$P(x)=\sum_{}{x^{n_i}}+1$$ Does there exist infinitly many polynomials $P_1(x),P_2(x),...$ such that for each distinct $i,j$ the polynomial $P_i(x)P_j(x)$ is interesting.

1997 Akdeniz University MO, 1

Prove that, $$15x^2-7y^2=9$$ equation has any solutions in integers.

2015 Germany Team Selection Test, 1

Determine all pairs $(x, y)$ of positive integers such that \[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\] [i]Proposed by Titu Andreescu, USA[/i]

1984 IMO Longlists, 25

Prove that the product of five consecutive positive integers cannot be the square of an integer.

2016 Middle European Mathematical Olympiad, 8

For a positive integer $n$, the equation $a^2 + b^2 + c^2 + n = abc$ is given in the positive integers. Prove that: 1. There does not exist a solution $(a, b, c)$ for $n = 2017$. 2. For $n = 2016$, $a$ is divisible by $3$ for all solutions $(a, b, c)$. 3. There are infinitely many solutions $(a, b, c)$ for $n = 2016$.

2012 IMO Shortlist, N7

Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, \ldots, a_n$ such that \[ \frac{1}{2^{a_1}} + \frac{1}{2^{a_2}} + \cdots + \frac{1}{2^{a_n}} = \frac{1}{3^{a_1}} + \frac{2}{3^{a_2}} + \cdots + \frac{n}{3^{a_n}} = 1. \] [i]Proposed by Dusan Djukic, Serbia[/i]

2013 Taiwan TST Round 1, 5

An integer $a$ is called friendly if the equation $(m^2+n)(n^2+m)=a(m-n)^3$ has a solution over the positive integers. [b]a)[/b] Prove that there are at least $500$ friendly integers in the set $\{ 1,2,\ldots ,2012\}$. [b]b)[/b] Decide whether $a=2$ is friendly.

2021 Brazil National Olympiad, 5

Find all triples of non-negative integers \((a, b, c)\) such that \[a^{2}+b^{2}+c^{2} = a b c+1.\]

2004 Gheorghe Vranceanu, 3

Consider the function $ f:(-\infty,1]\longrightarrow\mathbb{R} $ defined as $$ f(x)=\left\{ \begin{matrix} \frac{5}{2} +2^x-\frac{1}{2^x} ,& \quad x<-1 \\ 3^{\sqrt{1-x^2}} ,& \quad x\in [-1,1] \end{matrix} \right. . $$ [b]a)[/b] For a fixed parameter, find the roots of $ f-m. $ [b]b)[/b] Study the inversability of the restrictions of $ f $ to $ (-\infty,-1] $ and $ [-1,1] $ and find the inverses of these that admit them. [i]D. Zaharia[/i]

2010 JBMO Shortlist, 2

[b]Determine all four digit numbers [/b]$\bar{a}\bar{b}\bar{c}\bar{d}$[b] such that[/b] $$a(a+b+c+d)(a^{2}+b^{2}+c^{2}+d^{2})(a^{6}+2b^{6}+3c^{6}+4d^{6})=\bar{a}\bar{b}\bar{c}\bar{d}$$

2005 Croatia National Olympiad, 1

Let $a \not = 0, b, c$ be real numbers. If $x_{1}$ is a root of the equation $ax^{2}+bx+c = 0$ and $x_{2}$ a root of $-ax^{2}+bx+c = 0$, show that there is a root $x_{3}$ of $\frac{a}{2}\cdot x^{2}+bx+c = 0$ between $x_{1}$ and $x_{2}$.

2015 Belarus Team Selection Test, 2

Determine all pairs $(x, y)$ of positive integers such that \[\sqrt[3]{7x^2-13xy+7y^2}=|x-y|+1.\] [i]Proposed by Titu Andreescu, USA[/i]

2014 Korea National Olympiad, 4

Prove that there exists a function $f : \mathbb{N} \rightarrow \mathbb{N}$ that satisfies the following (1) $\{f(n) : n\in\mathbb{N}\}$ is a finite set; and (2) For nonzero integers $x_1, x_2, \ldots, x_{1000}$ that satisfy $f(\left|x_1\right|)=f(\left|x_2\right|)=\cdots=f(\left|x_{1000}\right|)$, then $x_1+2x_2+2^2x_3+2^3x_4+2^4x_5+\cdots+2^{999}x_{1000}\ne 0$.

EGMO 2017, 5

Let $n\geq2$ be an integer. An $n$-tuple $(a_1,a_2,\dots,a_n)$ of not necessarily different positive integers is [i]expensive[/i] if there exists a positive integer $k$ such that $$(a_1+a_2)(a_2+a_3)\dots(a_{n-1}+a_n)(a_n+a_1)=2^{2k-1}.$$ a) Find all integers $n\geq2$ for which there exists an expensive $n$-tuple. b) Prove that for every odd positive integer $m$ there exists an integer $n\geq2$ such that $m$ belongs to an expensive $n$-tuple. [i]There are exactly $n$ factors in the product on the left hand side.[/i]

2009 Swedish Mathematical Competition, 2

Tags: equation , algebra
Find all real solutions of the equation \[ \left(1+x^2\right)\left(1+x^3\right)\left(1+x^5\right)=8x^5 \]

2011 Bosnia And Herzegovina - Regional Olympiad, 2

For positive integers $a$ and $b$ holds $a^3+4a=b^2$. Prove that $a=2t^2$ for some positive integer $t$

1989 IMO Longlists, 7

Prove that $ \forall n > 1, n \in \mathbb{N}$ the equation \[ \sum^n_{k\equal{}1} \frac{x^k}{k!} \plus{} 1 \equal{} 0\] has no rational roots.