I have to prove that $\displaystyle a^5-a$ is dividable with 30 for every integer value of $\displaystyle a$.

My proof is:

$\displaystyle a^5-a=a(a^2-1)(a^2+1)$ and $\displaystyle 30 = 2(2^2-1)(2^2+1)$.

So if we put $\displaystyle a$ instead of 2 (or 2 instead of a) the proof is proven.

I have found in book much complicated solution but this seems ok to me.

Am I right?

P.S. Can someone tell me is there some guide to writing math text in messages on this forum because I couldn't find it.