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 $q \mid a - b$.

Let $c,d \geq 2$ be naturals. Let $\{a_n\}$ be the sequence satisfying $a_1 = c, a_{n+1} = a_n^d + c$ for $n = 1,2,\cdots$. Prove that for any $n \geq 2$, there exists a prime number $p$ such that $p|a_n$ and $p \not | a_i$ for $i = 1,2,\cdots n-1$.

Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.

Find all prime numbers $p$ for which the number $p^2+11$ has less than $11$ divisors.

For positive integers $a>b>1$, define \[x_n = \frac {a^n-1}{b^n-1}\] Find the least $d$ such that for any $a,b$, the sequence $x_n$ does not contain $d$ consecutive prime numbers. [i]V. Senderov[/i]

For any integer $d > 0,$ let $f(d)$ be the smallest possible integer that has exactly $d$ positive divisors (so for example we have $f(1)=1, f(5)=16,$ and $f(6)=12$). Prove that for every integer $k \geq 0$ the number $f\left(2^k\right)$ divides $f\left(2^{k+1}\right).$ [i]Proposed by Suhaimi Ramly, Malaysia[/i]

Call a number "prime-looking" if it is composite but not divisible by 2, 3, or 5. The three smallest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000? $ \textbf{(A)}\ 100 \qquad \textbf{(B)}\ 102 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 106 \qquad \textbf{(E)}\ 108$

Let $p_1$, $p_2$ be two odd prime numbers and $\alpha $, $n$ be positive integers with $\alpha >1$, $n>1$. Prove that if the equation $\left (\frac{p_2 -1}{2} \right )^{p_1} + \left (\frac{p_2 +1}{2} \right )^{p_1} = \alpha^n$ does not have integer solutions for both $p_1 =p_2$ and $p_1 \neq p_2$.

Let $P(x)$ be a polynomial with integer coefficients. We will denote the set of all prime numbers by $\mathbb P$. Show that the set $\mathbb S := \{p\in\mathbb P : \exists\text{ }n \text{ s.t. }p\mid P(n)\}$ is finite if and only if $P(x)$ is a non-zero constant polynomial.

Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them $S$- and $Z$-tetrominoes, respectively. Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove that no matter how we tile $P$ using only $S$- and $Z$-tetrominoes, we always use an even number of $Z$-tetrominoes. [i]Proposed by Tamas Fleiner and Peter Pal Pach, Hungary[/i]

Compute the prime factorization of $1007021035035021007001$. (You should write your answer in the form $p_1^{e_1}p_2^{e_2}\ldots p_k^{e_k}$ where $p_1,\ldots,p_k$ are distinct prime numbers and $e_1,\ldots,e_k$ are positive integers.)

If $p>2$ is prime number and $m$ and $n$ are positive integers such that $$\frac{m}{n}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{p-1}$$ Prove that $p$ divides $m$

An ordered triple $(p, q, r)$ of prime numbers is called [i]parcera[/i] if $p$ divides $q^2-4$, $q$ divides $r^2-4$ and $r$ divides $p^2-4$. Find all parcera triples.

Let $p$ and $q$ be prime numbers such that $p^2+pq+q^2$ is perfect square. Prove that $p^2-pq+q^2$ is prime

Let $p$ be an odd prime number. Around a circular table, $p$ students sit. We give $p$ pieces of candy to those students in the following manner. The first candy we give to an arbitrary student. Then, going around clockwise, we skip two students and give the next student a piece of candy, then we skip 4 students and give another piece of candy to the next student... In general in the $k-$th turn we skip $2k$ students and give the next student a piece of candy. We do this until we don't give out all $p$ pieces of candy. $a)$ How many students won't get any pieces of candy? $b)$ How many pairs of neighboring students (those students who sit next to each other on the table) both got at least a piece of candy?

Let $p \geq 7$ be a prime number and $$S = \bigg\{jp+1 : 1 \leq j \leq \frac{p-5}{2}\bigg\}.$$ Prove that at least one element of $S$ can be written as $x^2+y^2$, where $x, y$ are integers.

Le $S$ be the set of positive integers greater than or equal to $2$. A function $f: S\rightarrow S$ is italian if $f$ satifies all the following three conditions: 1) $f$ is surjective 2) $f$ is increasing in the prime numbers(that is, if $p_1<p_2$ are prime numbers, then $f(p_1)<f(p_2)$) 3) For every $n\in S$ the number $f(n)$ is the product of $f(p)$, where $p$ varies among all the primes which divide $n$ (For instance, $f(360)=f(2^3\cdot 3^2\cdot 5)=f(2)\cdot f(3)\cdot f(5)$). Determine the maximum and the minimum possible value of $f(2020)$, when $f$ varies among all italian functions.

Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? [asy]path card=((0,0)--(0,3)--(2,3)--(2,0)--cycle); draw(card, linewidth(1)); draw(shift(2.5,0)*card, linewidth(1)); draw(shift(5,0)*card, linewidth(1)); label("$44$", (1,1.5)); label("$59$", shift(2.5,0)*(1,1.5)); label("$38$", shift(5,0)*(1,1.5));[/asy] $ \textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$

Show that, for any fixed integer $\,n \geq 1,\,$ the sequence \[ 2, \; 2^2, \; 2^{2^2}, \; 2^{2^{2^2}}, \ldots (\mbox{mod} \; n) \] is eventually constant. [The tower of exponents is defined by $a_1 = 2, \; a_{i+1} = 2^{a_i}$. Also $a_i \; (\mbox{mod} \; n)$ means the remainder which results from dividing $a_i$ by $n$.]

Let $p>$ be a prime number and $S$ be a set of $p+1$ integers. Prove that there exist pairwise distinct numbers $a_1,a_2,...,a_{p-1}\in S$ that $$ a_1+2a_2+3a_3+...+(p-1)a_{p-1}$$ is divisible by $p$.

Determine all functions $ f$ defined on the natural numbers that take values among the natural numbers for which \[ (f(n))^p \equiv n\quad {\rm mod}\; f(p) \] for all $ n \in {\bf N}$ and all prime numbers $ p$.

Find the total number of primes $p<100$ such that $\lfloor (2+\sqrt{5})^p \rfloor-2^{p+1}$ is divisible by $p$. Here $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.

is there a function $f:\mathbb{N}\rightarrow \mathbb{N}$ such that $i) \exists n\in \mathbb{N}:f(n)\neq n$ $ii)$ the number of divisors of $m$ is $f(n)$ if and only if the number of divisors of $f(m)$ is $n$

Let $m,n$ be positive integers and $p$ be a prime number. Show that if $\frac{7^m + p \cdot 2^n}{7^m - p \cdot 2^n}$ is an integer, then it is a prime number.

Given a positive integrer number $n$ ($n\ne 0$), let $f(n)$ be the average of all the positive divisors of $n$. For example, $f(3)=\frac{1+3}{2}=2$, and $f(12)=\frac{1+2+3+4+6+12}{6}=\frac{14}{3}$. [b]a[/b] Prove that $\frac{n+1}{2} \ge f(n)\ge \sqrt{n}$. [b]b[/b] Find all $n$ such that $f(n)=\frac{91}{9}$.