Working with mod p (Euler's and Fermat's Theorem)

Quote:

Originally Posted by Zennie

Prove that if $p$ is prime, then for any number $a$ , divisible by $p$ or not, $a^p \equiv a (mod p)$

Are you asking to prove Euler's theorem? Do you know group theory? Note that $\left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$ forms a group whose order is $\varphi(n)$ and since any element of a group to it's order is the identity element the conclusion follows.

If you're just trying to apply it merely notice that if $a\nmid p$ then by virtue of $p$ 's primality $(a,p)=1$ in which case Fermat's theorem applies. If $a\mid p$ then either $a\equiv 0,1$ in which case the theorem is a triviality.