
Problem 1. Find the number of all integers $n>1$, for which the number $a^{25}-a$ is divisible by $n$ for every integer $a$.


Solution. Let $n$ be with the required property. Then $p^{2}$ ( $p$ prime) does not divide $n$ since $p^{2}$ does not divide $p^{25}-p$. Hence $n$ is a product of pairwise different prime numbers. On the other hand $2^{25}-2=2.3^{2} .5 \cdot 7 \cdot 13$.17.241. But $n$ is not divisible by 17 and 241 because $3^{25} \equiv-3$

$(\bmod 17)$ and $3^{25} \equiv 32(\bmod 241)$. The Fermat theorem implies that $a^{25} \equiv a(\bmod p)$ when $p=2,3,5,7,13$. Thus $n$ should be equal to the divisor of $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13$, different from 1 . Therefore the number we are looking for is $2^{5}-1=31$.
