Prove $n=m^3-3m^2+2m$, for any integer $m$, then $n$ is a multiple of $6$.

Question

Prove $n=m^3-3m^2+2m$, for any integer $m$, then $n$ is a multiple of $6$.

So far I have that $n=m(m-3)(m+3)$, which are $3$ consecutive integers so at least one must be a multiple of 2. I am not sure how to get any further? Maybe use 2 cases?

Thank you!

Answer 1

Hint: $$n=m^3-3m^2+2m = m(m^2-3m + 2) = m(m-1)(m-2)$$

There are three consecutive integers in the product. One is divisible by $2$, and one by $3$, hence $n$ is divisible by $2\cdot 3 = 6$.

Answer 2

Note thatg $$m^3-3m^2+2m=(m-2)(m-1)m$$

Answer 3

This can be proved by induction. If $m=1$, then your number is $0$, which is a multiple of $6$. And if $m^3-3m^2+2m$ is a multiple of $6$; then, since\begin{align}(m+1)^3-(m+1)^2+2(m+1)-(m^3-3m^2+2m)&=3m^2-3m\\&=3m(m-1),\end{align}which is clearly a multiple of $6$, $(m+1)^3-(m+1)^2+2(m+1)$ is a multiple of $6$ too.