Školjka

%V0 A natural number $n$ is said to have the property $P,$ if, for all $a, n^2$ divides $a^n - 1$ whenever $n$ divides $a^n - 1.$ a.) Show that every prime number $n$ has property $P.$ b.) Show that there are infinitely many composite numbers $n$ that possess property $P.$

%V0 Let $a,b,n$ be positive integers, $b > 1$ and $b^n-1|a.$ Show that the representation of the number $a$ in the base $b$ contains at least $n$ digits different from zero.

%V0 Find all positive integers $n$ such that there exists a unique integer $a$ such that $0\leq a < n!$ with the following property: $$n!\mid a^n + 1$$

%V0 Denote by $d(n)$ the number of divisors of the positive integer $n$. A positive integer $n$ is called highly divisible if $d(n) > d(m)$ for all positive integers $m < n$. Two highly divisible integers $m$ and $n$ with $m < n$ are called consecutive if there exists no highly divisible integer $s$ satisfying $m < s < n$. (a) Show that there are only finitely many pairs of consecutive highly divisible integers of the form $(a, b)$ with $a\mid b$. (b) Show that for every prime number $p$ there exist infinitely many positive highly divisible integers $r$ such that $pr$ is also highly divisible.

%V0 Let $a$, $b$ be positive integers such that $b^n+n$ is a multiple of $a^n+n$ for all positive integers $n$. Prove that $a=b$.

%V0 For every integer $k \geq 2,$ prove that $2^{3k}$ divides the number $$\binom{2^{k + 1}}{2^{k}} - \binom{2^{k}}{2^{k - 1}}$$ but $2^{3k + 1}$ does not. Author: unknown author, Poland