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?
 A: $$m(m^2-7)=m(m^2-1-6)=m[(m+1)(m-1)-6]=(m-1)m(m+1)-6m$$
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, $6|m(m^2-7)$
A: Consider the following cases:


*

*$m\equiv0\pmod6 \implies m(m^2−7)\equiv0\cdot(0^2−7)         \equiv0\pmod6$

*$m\equiv1\pmod6 \implies m(m^2−7)\equiv1\cdot(1^2−7)\equiv- 6\equiv0\pmod6$

*$m\equiv2\pmod6 \implies m(m^2−7)\equiv2\cdot(2^2−7)\equiv- 6\equiv0\pmod6$

*$m\equiv3\pmod6 \implies m(m^2−7)\equiv3\cdot(3^2−7)\equiv+ 6\equiv0\pmod6$

*$m\equiv4\pmod6 \implies m(m^2−7)\equiv4\cdot(4^2−7)\equiv+36\equiv0\pmod6$

*$m\equiv5\pmod6 \implies m(m^2−7)\equiv5\cdot(5^2−7)\equiv+90\equiv0\pmod6$

A: We have
$$\sum_{k=1}^m 3(k+1)(k-2) = m(m^2-7)$$
Note that $(k+1)(k-2)$ is always even. Hence, $3(k+1)(k-2)$ is a multiple of $6$. Hence, $\displaystyle \sum_{k=1}^m 3(k+1)(k-2)$ is also a multiple of $6$.
A: Another option you have is to approach via brute force case checking.  It is less elegant than Aniket's or Leg's approaches above, but perhaps easier to remember how to use and requires less intuition.
You have six cases:  $m=6k,~ m=6k+1,~ m=6k+2,~ m=6k+3,~ m=6k+4$ or $m=6k+5$ for some integer $k$.  (Equivalently, $m\equiv 0\pmod{6},~m\equiv 1\pmod{6},~\dots,~m\equiv 5\pmod{6}$)
Case 1: If $m=6k$ then $m(m^2-7) = 6k((6k)^2-7) = 6(36k^3-7)$ is a multiple of six.
Case 2: If $m=6k+1$ then $\begin{array}{l}m(m^2-7) = (6k+1)((6k+1)^2-7)  \\=6(k((6k+1)^2-7))+1((6k+1)^2-7) = 6(36k^3+12k^2+k-7k) + (36k^2+12k+1-7) \\= 6(36k^3+12k^2-6k+6k^2+2k-1)\end{array}$
is a multiple of six.
Instead of writing it in this way, it is easier if we describe this using modular arithmetic in order to make the simplification easier:
Case 3: If $m\equiv 2\pmod{6}$ then $m(m^2-7)\equiv 2(2^2-7)\equiv 2(4-7)\equiv 2(-3)\equiv -6\equiv 0\pmod{6}$ so it is a multiple of six.
Case 4: If $m\equiv 3\pmod{6}$ then $m(m^2-7)\equiv 3(9-7)\equiv 3\cdot 2\equiv 0\pmod{6}$ so it is a multiple of six.
Similarly for cases 5 and 6.
In all cases, you see that the expression is divisible by six, so it is true for all integers (or naturals) $m$.
A: With congruences:
Modulo $6$, we have $m(m^2-7)\equiv m(m^2-1)$, hence it is enough to show $m^3\equiv m\mod 2$ and $\bmod3$.
Modulo $2$: Little Fermat says $m^2\equiv m$, hence $m^3\equiv m^2\equiv m$.
Modulo $3$: $m^3\equiv m$ directly by Little Fermat.
