Show that for all natural $a$, $2008\mid a^{251}-a$. How to show, that for all natural $a$ coprime to 2008 the following occurs:  $2008\mid a^{251}-a$?
This means, that $a_{251} \equiv_{{}\bmod 2008} a$, right?
It's obvious if $a\mid 2008$.
In the other case I'm totally at a loss.
I thought about using Euler totient function but that's obviously doesn't apply here, since $2008$ is not prime.

Note that $2008 = 8\cdot 251$. By the Chinese Remainder Theorem it is necessary and sufficient to show $a^{251}\equiv a\pmod 8$ and $a^{251}\equiv a\pmod {251}$. The second is true for all $a$ by Fermat's little theorem.
The first is true for odd $a$ or if $8\mid a$, but otherwise it is not!
