Proving divisibility of $a^3 - a$ by $6$ As part of a larger proof, I need to show why $a^3-a$ is always divisible by $6$. I'm having trouble getting started.

by Euler's theorem $ a^{\phi (n)} = a (mod n) $ with $\phi(n)$ the number of integers lower than n and prime with n
if n = 6 there are three integers less than 6 and first ones with 6: 1,3,5
let $a^3 = a (mod 6) $
