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]

For any odd prime $p$ and any integer $n,$ let $d_p (n) \in \{ 0,1, \dots, p-1 \}$ denote the remainder when $n$ is divided by $p.$ We say that $(a_0, a_1, a_2, \dots)$ is a [i]p-sequence[/i], if $a_0$ is a positive integer coprime to $p,$ and $a_{n+1} =a_n + d_p (a_n)$ for $n \geqslant 0.$ (a) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $(a_0, a_1, a_2, \dots)$ and $(b_0, b_1, b_2, \dots)$ such that $a_n >b_n$ for infinitely many $n,$ and $b_n > a_n$ for infinitely many $n?$ (b) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $(a_0, a_1, a_2, \dots)$ and $(b_0, b_1, b_2, \dots)$ such that $a_0 <b_0,$ but $a_n >b_n$ for all $n \geqslant 1?$ [I]United Kingdom[/i]

Suppose that $A$ is a set of positive integers less than $N$ and that no two distinct elements of $A$ sum to a perfect square. That is, if $a_1, a_2 \in A$ and $a_1 \neq a_2$ then $|a_1+a_2|$ is not a square of an integer. Prove that the maximum number of elements in $A$ is at least $\left\lfloor\frac{11}{32}N\right\rfloor$ .

Let $p$ be a prime number and $f$ be a bijection from $\left\{0,1,\ldots,p-1\right\}$ to itself. Suppose that for integers $a,b \in \left\{0,1,\ldots,p-1\right\}$, $|f(a) - f(b)|\leqslant 2024$ if $p \mid a^2 - b$. Prove that there exists infinite many $p$ such that there exists such an $f$ and there also exists infinite many $p$ such that there doesn't exist such an $f$.

Find all pairs of integers $(m,n)$ such that $m^6 = n^{n+1} + n -1$.

Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4 + n^2 + 1$ is equal to the largest prime divisor of $(n+1)^4 + (n+1)^2 +1$.

Do there exist real numbers $b$ and $c$ such that each of the equations $x^2+bx+c = 0$ and $2x^2+(b+1)x+c+1 = 0$ have two integer roots? [i]N. Agakhanov[/i]

Four positive integers $x,y,z$ and $t$ satisfy the relations \[ xy - zt = x + y = z + t. \] Is it possible that both $xy$ and $zt$ are perfect squares?

Let $N=4^KL$ where $L\equiv\ 7\pmod 8$. Prove that $N$ cannot be written as a sum of 3 squares

Suppose we have distinct positive integers $a, b, c, d$, and an odd prime $p$ not dividing any of them, and an integer $M$ such that if one considers the infinite sequence \begin{align*} ca &- db \\ ca^2 &- db^2 \\ ca^3 &- db^3 \\ ca^4 &- db^4 \\ &\vdots \end{align*} and looks at the highest power of $p$ that divides each of them, these powers are not all zero, and are all at most $M$. Prove that there exists some $T$ (which may depend on $a,b,c,d,p,M$) such that whenever $p$ divides an element of this sequence, the maximum power of $p$ that divides that element is exactly $p^T$.

Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4 + n^2 + 1$ is equal to the largest prime divisor of $(n+1)^4 + (n+1)^2 +1$.

Five squirrels together have a supply of 2022 nuts. On the first day 2 nuts are added, on the second day 4 nuts, on the third day 6 nuts and so on, i.e. on each further day 2 nuts more are added than on the day before. At the end of any day the squirrels divide the stock among themselves. Is it possible that they all receive the same number of nuts and that no nut is left over?

Let $p$ be a prime number of the form $4k+1$. Show that \[\sum^{p-1}_{i=1}\left( \left \lfloor \frac{2i^{2}}{p}\right \rfloor-2\left \lfloor \frac{i^{2}}{p}\right \rfloor \right) = \frac{p-1}{2}.\]

Prove that $5^n-3^n$ is not divisible by $2^n+65$ for any positive integer $n$.