If p is an odd prime, prove that $a^{2p-1} \equiv a \pmod{ 2p}$ Let $m = 2p$
If p is an odd prime, prove that $a^{2p - 1} \equiv a \pmod {2p} \iff a^{m - 1} \equiv a \pmod m$.
I have no idea on how to start. I was trying to find a form such that
$a^{m - 2} \equiv 1 \pmod m$. But I got stuck. Can someone give me a hint here?

Fermat: $p\mid a^{p}-a$  so $p\mid\left(a^{p-1}+1\right)\left(a^{p}-a\right)=a^{2p-1}-a$.
Next to that $2\mid a\iff2\mid a^{2p-1}$ so that $2\mid a^{2p-1}-a$.
So $2$ and $p$ are two distinct ($p$ is odd, so $p\neq2$) primes both dividing $a^{2p-1}-a$. 
Conclusion: $$2p\mid a^{2p-1}-a$$
