Find all triples $(a,b,c)$ of real numbers such that the following system holds: $$\begin{cases} a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\a^2+b^2+c^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\end{cases}$$ Proposed by Dorlir Ahmeti, Albania

Consider the sequence $a_1, a_2, a_3, ...$ defined by $a_1 = 9$ and $a_{n + 1} = \frac{(n + 5)a_n + 22}{n + 3}$ for $n \ge 1$. Find all natural numbers $n$ for which $a_n$ is a perfect square of an integer.

Find all triples of positive real numbers $(a, b, c)$ so that the expression $M = \frac{(a + b)(b + c)(a + b + c)}{abc}$ gets its least value.

Alice and Bob play the following game: starting with the number $2$ written on a blackboard, each player in turn changes the current number $n$ to a number $n + p$, where $p$ is a prime divisor of $n$. Alice goes first and the players alternate in turn. The game is lost by the one who is forced to write a number greater than $\underbrace{22...2}_{2020}$. Assuming perfect play, who will win the game.

Viktor and Natalia bought $2020$ buckets of ice-cream and want to organize a degustation schedule with $2020$ rounds such that: - In every round, both of them try $1$ ice-cream, and those $2$ ice-creams tried in a single round are different from each other. - At the end of the $2020$ rounds, both of them have tried each ice-cream exactly once. We will call a degustation schedule fair if the number of ice-creams that were tried by Viktor before Natalia is equal to the number of ice creams tried by Natalia before Viktor. Prove that the number of fair schedules is strictly larger than $2020!(2^{1010} + (1010!)^2)$. Proposed by Viktor Simjanoski, Macedonia

Alice and Bob play the following game: Alice picks a set $A = \{1, 2, ..., n \}$ for some natural number $n \ge 2$. Then, starting from Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen at each step should be distinct from all the already chosen numbers and should differ by $1$ from an already chosen number. The game ends when all numbers from the set $A$ are chosen. Alice wins if the sum of all the numbers that she has chosen is composite. Otherwise Bob wins. Decide which player has a winning strategy. Proposed by Demetres Christofides, Cyprus

Let $\triangle ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$. Let $E$ be the midpoint of $AD$ and $\omega$ the the circle with center $E$ and radius equal to $AE$. The line $BE$ intersects $\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $AD$ and the line $CE$ intersects $\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $AD$. If both of the intersection points of the circumcircles of $\triangle BDX$ and $\triangle CDY$ lie on the line $AD$, prove that $AB = AC$.

Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$, and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ and $(c_1)$ be the circumcircles of the triangles $\triangle AEZ$ and $\triangle BEZ$, respectively. Let $(c_2)$ be an arbitrary circle passing through the points $A$ and $E$. Suppose $(c_1)$ meets the line $CZ$ again at the point $F$, and meets $(c_2)$ again at the point $N$. If $P$ is the other point of intersection of $(c_2)$ with $AF$, prove that the points $N$, $B$, $P$ are collinear.

Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$ and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \ne A$ be a point on the line $AB$ with $AB = BZ$. Let $(c)$ be the circumcircle of the triangle $\triangle AEZ$. Let $D$ be the second point of intersection of $(c)$ with $ZC$ and let $F$ be the antidiametric point of $D$ with respect to $(c)$. Let $P$ be the point of intersection of the lines $FE$ and $CZ$. If the tangent to $(c)$ at $Z$ meets $PA$ at $T$, prove that the points $T$, $E$, $B$, $Z$ are concyclic. Proposed by Theoklitos Parayiou, Cyprus

Determine whether there is a natural number $n$ for which $8^n + 47$ is prime.

Find all positive integers $a$, $b$, $c$, and $p$, where $p$ is a prime number, such that $73p^2 + 6 = 9a^2 + 17b^2 + 17c^2$.

Find the largest integer $k$ ($k \ge 2$), for which there exists an integer $n$ ($n \ge k$) such that from any collection of $n$ consecutive positive integers one can always choose $k$ numbers, which verify the following conditions: 1. each chosen number is not divisible by $6$, by $7$, nor by $8$; 2. the positive difference of any two distinct chosen numbers is not divisible by at least one of the numbers $6$, $7$, and $8$.

Find all prime numbers $p$ such that $(x + y)^{19} - x^{19} - y^{19}$ is a multiple of $p$ for any positive integers $x$, $y$.

The positive integer $k$ and the set $A$ of distinct integers from $1$ to $3k$ inclusively are such that there are no distinct $a$, $b$, $c$ in $A$ satisfying $2b = a + c$. The numbers from $A$ in the interval $[1, k]$ will be called small; those in $[k + 1, 2k]$ - medium and those in $[2k + 1, 3k]$ - large. It is always true that there are no positive integers $x$ and $d$ such that if $x$, $x + d$, and $x + 2d$ are divided by $3k$ then the remainders belong to $A$ and those of $x$ and $x + d$ are different and are: a) small? $\hspace{1.5px}$ b) medium? $\hspace{1.5px}$ c) large? (In this problem we assume that if a multiple of $3k$ is divided by $3k$ then the remainder is $3k$ rather than $0$.)

Are there any positive integers $m$ and $n$ satisfying the equation $m^3 = 9n^4 + 170n^2 + 289$ ?

Prove that there doesn’t exist any prime $p$ such that every power of $p$ is a palindrome (a palindrome is a number that is read the same from the left as it is from the right; in particular, a number that ends in one or more zeros cannot be a palindrome).

Find all prime numbers $p$ and $q$ such that $$1 + \frac{p^q - q^p}{p + q}$$is a prime number. Proposed by Dorlir Ahmeti, Albania