My question is: Show that $m(m^2 − 7)$ for any natural m is always divisible by 6
So i know we have to use fermat's little theorem which says that if $p$ is a prime number, then $n^p-n$ is divisible by $p$ for all $n$.
Since $6=1\times 2\times 3$, what we need to do is to check that 1, 2, and 3 divide $m(m^2-7$, no matter what $n$ is.
That 3 does is direct from Fermat's little theorem.
The theorem also ensures that 2 divides $m(m^2-7)$. Now: $$m(m^2-7)=m^2(m-7)$$
I get confused when trying to apply the application. I was wondering is my process correct so far? I feel as if i'm wrong and whats throwing me off is the m outside of the brackets, who i distribute it in?