Determine all positive integers $x$, $y$ and $z$ such that $x^5 + 4^y = 2013^z$. ([i]Serbia[/i])

Compute the number of positive integers $n$ between $2017$ and $2017^2$ such that $n^n \equiv 1$ (mod $2017$). ($2017$ is prime.)

We say that a set $S$ of integers is [i]rootiful[/i] if, for any positive integer $n$ and any $a_0, a_1, \cdots, a_n \in S$, all integer roots of the polynomial $a_0+a_1x+\cdots+a_nx^n$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^a - 2^b$ for positive integers $a$ and $b$.

Let $p$ be a prime number. Let $\mathbb F_p$ denote the integers modulo $p$, and let $\mathbb F_p[x]$ be the set of polynomials with coefficients in $\mathbb F_p$. Define $\Psi : \mathbb F_p[x] \to \mathbb F_p[x]$ by \[ \Psi\left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=0}^n a_i x^{p^i}. \] Prove that for nonzero polynomials $F,G \in \mathbb F_p[x]$, \[ \Psi(\gcd(F,G)) = \gcd(\Psi(F), \Psi(G)). \] Here, a polynomial $Q$ divides $P$ if there exists $R \in \mathbb F_p[x]$ such that $P(x) - Q(x) R(x)$ is the polynomial with all coefficients $0$ (with all addition and multiplication in the coefficients taken modulo $p$), and the gcd of two polynomials is the highest degree polynomial with leading coefficient $1$ which divides both of them. A non-zero polynomial is a polynomial with not all coefficients $0$. As an example of multiplication, $(x+1)(x+2)(x+3) = x^3+x^2+x+1$ in $\mathbb F_5[x]$. [i]Proposed by Mark Sellke[/i]

Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed: $$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$. The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. [i]Proposed by Amine Natik, Morocco[/i]

Find all pairs of positive integers $(m, n)$ such that $m^2-mn+n^2+1$ divides both numbers $3^{m+n}+(m+n)!$ and $3^{m^3+n^3}+m+n$. [i]Proposed by Dorlir Ahmeti[/i]

For $a_1 = 3$, define the sequence $a_1, a_2, a_3, \ldots$ for $n \geq 1$ as $$na_{n+1}=2(n+1)a_n-n-2.$$ Prove that for any odd prime $p$, there exist positive integer $m,$ such that $p|a_m$ and $p|a_{m+1}.$

Find the remainder when $2^{1990}$ is divided by $1990.$

Let $p$ be a prime number. Prove that the determinant of the matrix \[ \begin{bmatrix}x & y & z\\ x^p & y^p & z^p \\ x^{p^2} & y^{p^2} & z^{p^2} \end{bmatrix} \] is congruent modulo $p$ to a product of polynomials of the form $ax+by+cz$, where $a$, $b$, and $c$ are integers. (We say two integer polynomials are congruent modulo $p$ if corresponding coefficients are congruent modulo $p$.)

Find the smallest $x \in\mathbb{N}$ for which $\frac{7x^{25}-10}{83}$ is an integer.

Let $P$ be a non-constant polynomial with integer coefficients such that if $n$ is a perfect power, so is $P(n)$. Prove that $P(x) = x$ or $P$ is a perfect power of a polynomial with integer coefficients. A perfect power is an integer $n^k$, where $n \in \mathbb Z$ and $k \ge 2$. A perfect power of a polynomial is a polynomial $P(x)^k$, where $P$ has integer coefficients and $k \ge 2$.

Prove that the function $f : \mathbb{N}\longrightarrow \mathbb{Z}$ defined by $f(n) = n^{2007}-n!$, is injective.

Let $a_0 = 2$, $a_1 = 5$, and $a_2 = 8$, and for $n>2$ define $a_n$ recursively to be the remainder when $4(a_{n-1} + a_{n-2} + a_{n-3})$ is divided by $11$. Find $a_{2018}\cdot a_{2020}\cdot a_{2022}$.

Prove that there exists infinetly many natural number $n$ such that at least one of the numbers $2^{2^n}+1$ and $2018^{2^n}+1$ is not a prime.