4. find the largest natural number $n$ that is a divisor of $a^{25}-a$ for every integer $a$.
## 1st solution:

For a prime number $p$, $p^{25}-p$ is not divisible by $p^{2}$, therefore $n$ cannot be divisible by a square $>1$. Now let $p$ be a prime number that is a divisor of $a^{25}-a$ for every integer $a$. Then $a^{25}-a must always be \equiv 0$ modulo $p$. If $a$ is not divisible by $p$, then you can reduce this congruence with $a$ and obtain $a^{24} \equiv 1$. Let $d$ be the order of $a$ modulo $p$, i.e. the smallest natural number with $a^{d} \equiv 1(p)$. Then on the one hand $d \mid \varphi(p)=p-1$ and on the other hand $d \mid 24$. We now choose $a$ as the primitive root modulo $p$, i.e. so that $d=p-1$. Now $p-1 \mid 24$ follows, i.e. $p=2,3,5,7$ or 13 . With the help of Fermat's theorem, it is easy to show that $a^{25}-a$ is always divisible by these five prime numbers, so $n=2 \cdot 3 \cdot 5 \cdot 7 \cdot 13=2730$.