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

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Found problems: 988

Determine all positive integers$ n$ that can be written as the product of two consecutive integers and as well as the product of four consecutive integers numbers. In the formula: $n = a (a + 1) = b (b + 1) (b + 2) (b + 3)$.
Find all the integers $x, y$ and $z$ greater than or equal to $0$ such that $2^x + 9 \cdot 7^y = z^3$
Problem. The sum of two consecutive squares can be a square; for instance $3^2 + 4^2 = 5^2$. (a) Prove that the sum of $m$ consecutive squares cannot be a square for $m \in \{3, 4, 5, 6\}$. (b) Find an example of eleven consecutive squares whose sum is a square. Can anyone help me with this? Thanks.
Let $n$ be a natural number such that there are exactly$ 2017$ distinct pairs of natural numbers $(a, b)$, which the equation $$\frac{1}{a}+\frac{1}{b}=\frac{1}{n}$$ fulfilld. Show that $n$ is a perfect square . Remark: $(7, 4) \ne (4, 7)$
Solve in positive integers: $\frac{1}{x^2}+\frac{y}{xz}+\frac{1}{z^2}=\frac{1}{2013}$ .
Find the roots $ r_i \in \mathbb{R}$ of the polynomial \[ p(x) \equal{} x^n \plus{} n \cdot x^{n\minus{}1} \plus{} a_2 \cdot x^{n\minus{}2} \plus{} \ldots \plus{} a_n\] satisfying \[ \sum^{16}_{k\equal{}1} r^{16}_k \equal{} n.\]
Let $a, b$ be integers. Show that the equation $a^2 + b^2 = 26a$ has at least $12$ solutions.
Find all solutions of the following equation in integers $x,y: x^4+ y= x^3+ y^2$
Solve the following system of equations in integer numbers: $$\begin{cases} x^2 = yz + 1 \\ y^2 = zx + 1 \\ z^2 = xy + 1 \end{cases}$$
Determine all triples $(x, y, z)$ of positive integers $x > y > z > 0$, such that $x^2 = y \cdot 2^z + 1$
Determine all solutions of the diophantine equation $a^2 = b \cdot (b + 7)$ in integers $a\ge 0$ and $b \ge 0$. (W. Janous, Innsbruck)
Determine all primes $p$ and all non-negative integers $m$ and $n$, such that $$1 + p^n = m^3. $$
Let $n$ be a positive integer and let $d(n)$ denote the number of ordered pairs of positive integers $(x,y)$ such that $(x+1)^2-xy(2x-xy+2y)+(y+1)^2=n$. Find the smallest positive integer $n$ satisfying $d(n) = 61$. (Patrik Bak, Slovakia)
Prove that there exist infinitely many positive integers $n$ such that $p=nr$, where $p$ and $r$ are respectively the semi-perimeter and the inradius of a triangle with integer side lengths.
Determine all triples $(x, y, z)$ of positive integers $x > y > z > 0$, such that $x^2 = y \cdot 2^z + 1$
Prove that for each positive integer $n$ there exist odd positive integers $x_n$ and $y_n$ such that ${x_{n}}^2 +7{y_{n}}^2 =2^n$.
(a) Find a pair of integers (x,y) such that $15x^2 +y^2 = 2^{2000}$ (b) Does there exist a pair of integers $(x,y)$ such that $15x^2 + y^2 = 2^{2000}$ and $x$ is odd?
Given is a prime number $p$. Find the number of positive integer solutions $(a, b, c, d)$ of the system of equations $ac+bd = p(a+c)$ and $bc-ad = p(b-d)$.
Find all integers $m>1$ such that $m^3$ is a sum of $m$ squares of consecutive integers.
The equation \[x^5 + 5 \lambda x^4 - x^3 + (\lambda \alpha - 4)x^2 - (8 \lambda + 3)x + \lambda \alpha - 2 = 0\] is given. Determine $\alpha$ so that the given equation has exactly (i) one root or (ii) two roots, respectively, independent from $\lambda.$
Find all ordered pairs $(a, b)$ of positive integers such that $a^2 + b^2 + 25 = 15ab$ and $a^2 + ab + b^2$ is prime.
Let $n$ be a positive integer. Show that there are integers $x_1, x_2, \ldots , x_n$, [i]not all equal[/i], satisfying $$\begin{cases} x_1^2+x_2+x_3+\ldots+x_n=0 \\ x_1+x_2^2+x_3+\ldots+x_n=0 \\ x_1+x_2+x_3^2+\ldots+x_n=0 \\ \vdots \\ x_1+x_2+x_3+\ldots+x_n^2=0 \end{cases}$$ if, and only if, $2n-1$ is not prime.
The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$? $ \textbf{(A)}\ 255 \qquad\textbf{(B)}\ 256 \qquad\textbf{(C)}\ 257 \qquad\textbf{(D)}\ 258 \qquad\textbf{(E)}\ 259 $
Show that for any positive integer $n$ there exists an integer $m > 1$ such that $(\sqrt2-1)^n=\sqrt{m}-\sqrt{m-1}$.