# 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, $6|m(m^2-7)$

• Just to add ...the fact that the in general the product of $r$ consecutive natural numbers is always divisible by $r!$(which is a much harder question!!)... So as here $r=3$ so it is always divisible by $3!=6$... – Freelancer Nov 19 '15 at 12:08
• @Freelancer Was the comment meant for me or the OP? you should mention that... – SchrodingersCat Nov 19 '15 at 12:11
• It was for everyone who was using this fact (divisbilty by 6)..here.. And even trying to prove it... – Freelancer Nov 19 '15 at 12:14

Consider the following cases:

• $m\equiv0\pmod6 \implies m(m^2−7)\equiv0\cdot(0^2−7) \equiv0\pmod6$
• $m\equiv1\pmod6 \implies m(m^2−7)\equiv1\cdot(1^2−7)\equiv- 6\equiv0\pmod6$
• $m\equiv2\pmod6 \implies m(m^2−7)\equiv2\cdot(2^2−7)\equiv- 6\equiv0\pmod6$
• $m\equiv3\pmod6 \implies m(m^2−7)\equiv3\cdot(3^2−7)\equiv+ 6\equiv0\pmod6$
• $m\equiv4\pmod6 \implies m(m^2−7)\equiv4\cdot(4^2−7)\equiv+36\equiv0\pmod6$
• $m\equiv5\pmod6 \implies m(m^2−7)\equiv5\cdot(5^2−7)\equiv+90\equiv0\pmod6$

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 $6$.

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.

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) $m$.