Let $ n$ be a positive integer and $ a_{1}, \ldots, a_{n}$ be arbitrary integers. Suppose that a function $ f: \mathbb{Z}\to \mathbb{R}$ satisfies $ \sum_{i=1}^{n}f(k+a_{i}l) = 0$ whenever $ k$ and $ l$ are integers and $ l \ne 0$. Prove that $ f = 0$.

For a given real number $a$ and a positive integer $n$, prove that: i) there exists exactly one sequence of real numbers $x_0,x_1,\ldots,x_n,x_{n+1}$ such that \[\begin{cases} x_0=x_{n+1}=0,\\ \frac{1}{2}(x_i+x_{i+1})=x_i+x_i^3-a^3,\ i=1,2,\ldots,n.\end{cases}\] ii) the sequence $x_0,x_1,\ldots,x_n,x_{n+1}$ in i) satisfies $|x_i|\le |a|$ where $i=0,1,\ldots,n+1$. [i]Liang Yengde[/i]

Let $f$ be a non-constant function from the set of positive integers into the set of positive integer, such that $a-b$ divides $f(a)-f(b)$ for all distinct positive integers $a$, $b$. Prove that there exist infinitely many primes $p$ such that $p$ divides $f(c)$ for some positive integer $c$. [i]Proposed by Juhan Aru, Estonia[/i]

Let $t$ and $n$ be fixed integers each at least $2$. Find the largest positive integer $m$ for which there exists a polynomial $P$, of degree $n$ and with rational coefficients, such that the following property holds: exactly one of \[ \frac{P(k)}{t^k} \text{ and } \frac{P(k)}{t^{k+1}} \] is an integer for each $k = 0,1, ..., m$. [i]Proposed by Michael Kural[/i]

There are 2019 students in a school, and some of these students are members of different student clubs. Each student club has an advisory board consisting of 12 students who are members of that particular club. An {\em advisory meeting} (for a particular club) can be realized only when each participant is a member of that club, and moreover, each of the 12 students forming the advisory board are present among the participants. It is known that each subset of at least 12 students in this school can realize an advisory meeting for exactly one student club. Determine all possible numbers of different student clubs with exactly 27 members.

Let $n$ points be given inside a rectangle $R$ such that no two of them lie on a line parallel to one of the sides of $R$. The rectangle $R$ is to be dissected into smaller rectangles with sides parallel to the sides of $R$ in such a way that none of these rectangles contains any of the given points in its interior. Prove that we have to dissect $R$ into at least $n + 1$ smaller rectangles. [i]Proposed by Serbia[/i]

There are two given different polynomials $P(x),Q(x)$ with real coefficients such that $P(Q(x))=Q(P(x))$. Prove that $\forall n\in \mathbb{Z_{+}}$ polynomial: \[\underbrace{P(P(\ldots P(P}_{n}(x))\ldots))- \underbrace{Q(Q(\ldots Q(Q}_{n}(x))\ldots))\] is divisible by $P(x)-Q(x)$.

Let $\mathbb{Z}^+$ be the set of positive integers. Find all functions $f:\mathbb{Z}^+ \rightarrow\mathbb{Z}^+$ such that the following conditions both hold: (i) $f(n!)=f(n)!$ for every positive integer $n$, (ii) $m-n$ divides $f(m)-f(n)$ whenever $m$ and $n$ are different positive integers.

Let $F(0)=0$, $F(1)=\frac32$, and $F(n)=\frac{5}{2}F(n-1)-F(n-2)$ for $n\ge2$. Determine whether or not $\displaystyle{\sum_{n=0}^{\infty}\, \frac{1}{F(2^n)}}$ is a rational number. (Proposed by Gerhard Woeginger, Eindhoven University of Technology)

Let $n$ be a positive integer. Tasty and Stacy are given a circular necklace with $3n$ sapphire beads and $3n$ turquoise beads, such that no three consecutive beads have the same color. They play a cooperative game where they alternate turns removing three consecutive beads, subject to the following conditions: [list] [*]Tasty must remove three consecutive beads which are turquoise, sapphire, and turquoise, in that order, on each of his turns. [*]Stacy must remove three consecutive beads which are sapphire, turquoise, and sapphire, in that order, on each of her turns. [/list] They win if all the beads are removed in $2n$ turns. Prove that if they can win with Tasty going first, they can also win with Stacy going first. [i]Yannick Yao[/i]

The sequence of real numbers $a_0,a_1,a_2,\ldots$ is defined recursively by \[a_0=-1,\qquad\sum_{k=0}^n\dfrac{a_{n-k}}{k+1}=0\quad\text{for}\quad n\geq 1.\]Show that $ a_{n} > 0$ for all $ n\geq 1$. [i]Proposed by Mariusz Skalba, Poland[/i]

Prove that for $N>1$ that $(N^{2})^{2014} - (N^{11})^{106}$ is divisible by $N^6 + N^3 +1$ Is this just a proof by induction or is there a more elegant method? I don't think calculating $N = 2$ was expected.

Let $x_1$ and $x_2$ be roots of the equation $x^2 - 6x + 1 = 0$. Prove that for any integer $n \ge 1$ the number $x_1^n + x_2^n$ is integer and is not divisible by $5$.

A class has $25$ students. The teacher wants to stock $N$ candies, hold the Olympics and give away all $N$ candies for success in it (those who solve equally tasks should get equally, those who solve less get less, including, possibly, zero candies). At what smallest $N$ this will be possible, regardless of the number of tasks on Olympiad and the student successes?

A family of sets $F$ is called perfect if the following condition holds: For every triple of sets $X_1, X_2, X_3\in F$, at least one of the sets $$ (X_1\setminus X_2)\cap X_3,$$ $$(X_2\setminus X_1)\cap X_3$$ is empty. Show that if $F$ is a perfect family consisting of some subsets of a given finite set $U$, then $\left\lvert F\right\rvert\le\left\lvert U\right\rvert+1$. [i]Proposed by Michał Pilipczuk[/i]