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

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- Apr 22nd 2010, 11:19 PMZennieWorking with mod p (Euler's and Fermat's Theorem)
Prove that if $\displaystyle p$ is prime, then for any number $\displaystyle a$, divisible by $\displaystyle p$ or not, $\displaystyle a^p \equiv a (mod p)$

- Apr 22nd 2010, 11:22 PMDrexel28
Are you asking to prove Euler's theorem? Do you know group theory? Note that $\displaystyle \left(\mathbb{Z}/n\mathbb{Z}\right)^{\times}$ forms a group whose order is $\displaystyle \varphi(n)$ 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 $\displaystyle a\nmid p$ then by virtue of $\displaystyle p$'s primality $\displaystyle (a,p)=1$ in which case Fermat's theorem applies. If $\displaystyle a\mid p$ then either $\displaystyle a\equiv 0,1$ in which case the theorem is a triviality.