Show that $m(m^2 − 7)$ for any natural m is always divisible by 6 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?

$$m(m^2-7)=m(m^2-1-6)=m[(m+1)(m-1)-6]=(m-1)m(m+1)-6m$$
Now, product of any 3 consecutive natural numbers is always divisible by $6$ and $6m$ is always divisible by $6$ $\forall$ $m \in \mathbb{N}$.
Hence we can say that $6|[(m-1)m(m+1)-6m]$ or, $6|m(m^2-7)$
