Let $a, b$ be positive integers and $p, q$ be prime numbers for which $p \nmid q - 1$ and $q \mid a^p - b^p$. Prove that $q \mid a - b$.

A positive interger number $k$ is called “$t-m$”-property if forall positive interger number $a$, there exists a positive integer number $n$ such that ${{1}^{k}}+{{2}^{k}}+{{3}^{k}}+...+{{n}^{k}} \equiv a (\bmod m).$ a) Find all positive integer numbers $k$ which has $t-20$-property. b) Find smallest positive integer number $k$ which has $t-{{20}^{15}}$-property.

The sequence of real numbers $a_1, a_2, a_3, ...$ is defined as follows: $a_1 = 2019$, $a_2 = 2020$, $a_3 = 2021$ and for all $n \ge 1$ $$a_{n+3} = 5a^6_{n+2} + 3a^3_{n+1} + a^2_n.$$ Show that this sequence does not contain numbers of the form $m^6$ where $m$ is a positive integer.

Let $p$ be an odd prime number, and let $\mathbb{F}_p$ denote the field of integers modulo $p$. Let $\mathbb{F}_p[x]$ be the ring of polynomials over $\mathbb{F}_p$, and let $q(x) \in \mathbb{F}_p[x]$ be given by $q(x) = \sum_{k=1}^{p-1} a_k x^k$ where $a_k = k^{(p-1)/2}$ mod $p$. Find the greatest nonnegative integer $n$ such that $(x-1)^n$ divides $q(x)$ in $\mathbb{F}_p[x]$.

Let $p$ and $a$ be positive integer numbers having no common divisors except of $1$. Prove that $p$ is prime if and only if all the coefficients of the polynomial \[ F(x) = (x-a)^p - (x^p - a) \] are divisible by $p$.

What is the smallest positive integer $n$ such that $20 \equiv n^{15} \pmod{29}$?

Show that there do not exist natural numbers $a_1, a_2, \dots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \] are all powers of $5$ [i]Proposed by Tejaswi Navilarekallu[/i]

Let $ A$ be the subset of the set of positive integers, having the following $ 2$ properties: 1) If $ a$ belong to $ A$,than all of the divisors of $ a$ also belong to $ A$; 2) If $ a$ and $ b$, $ 1 < a < b$, belong to $ A$, than $ 1 \plus{} ab$ is also in $ A$; Prove that if $ A$ contains at least $ 3$ positive integers, than $ A$ contains all positive integers.

For any set $A=\{a_1,a_2,\cdots,a_m\}$, let $P(A)=a_1a_2\cdots a_m$. Let $n={2010\choose99}$, and let $A_1, A_2,\cdots,A_n$ be all $99$-element subsets of $\{1,2,\cdots,2010\}$. Prove that $2010|\sum^{n}_{i=1}P(A_i)$.

Say a positive integer $n>1$ is $d$-coverable if for each non-empty subset $S\subseteq \{0, 1, \ldots, n-1\}$, there exists a polynomial $P$ with integer coefficients and degree at most $d$ such that $S$ is exactly the set of residues modulo $n$ that $P$ attains as it ranges over the integers. For each $n$, find the smallest $d$ such that $n$ is $d$-coverable, or prove no such $d$ exists. [i]Proposed by Carl Schildkraut[/i]

Jerry's favorite number is $97$. He knows all kinds of interesting facts about $97$: [list][*]$97$ is the largest two-digit prime. [*]Reversing the order of its digits results in another prime. [*]There is only one way in which $97$ can be written as a difference of two perfect squares. [*]There is only one way in which $97$ can be written as a sum of two perfect squares. [*]$\tfrac1{97}$ has exactly $96$ digits in the [smallest] repeating block of its decimal expansion. [*]Jerry blames the sock gnomes for the theft of exactly $97$ of his socks.[/list] A repunit is a natural number whose digits are all $1$. For instance, \begin{align*}&1,\\&11,\\&111,\\&1111,\\&\vdots\end{align*} are the four smallest repunits. How many digits are there in the smallest repunit that is divisible by $97?$

Let $\phi(n)$ be the number of positive integers less than $n$ that are relatively prime to $n$, where $n$ is a positive integer. Find all pairs of positive integers $(m,n)$ such that \[2^n + (n-\phi(n)-1)! = n^m+1.\]

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 the three-digit numbers $\overline{abc}$ such that the $6003$-digit number $\overline{abcabc\ldots abc}$ is divisible by $91$.

Find all functions $f : Z \to Z$ satisfying $\bullet$ $ f(p) > 0$ for all prime numbers $p$, $\bullet$ $p| (f(x) + f(p))^{f(p)}- x$ for all $x \in Z$ and all prime numbers $p$.