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

Question

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!

Note: $m-3,m, m+3$ are not three consecutive integers. If $m=4$, say, then those are $1,4,7$, none of which is divisible by $3$. — lulu, Jan 22 '19 at 21:35
$n=m(m-1)(m-2)$ not, $m(m-3)(m+3)$. And so it boils down to prove that products of three consecutive integers are divisible by $6$. One easy way: do you know what is $m$ choose $3$? — Quang Hoang, Jan 22 '19 at 21:36
Check your factorization. But if n is the product of 3 consecutive integers one must be a multiple of 3, and one must be a multiple of 2. — Doug M, Jan 22 '19 at 21:36

score 5 · Accepted Answer · edited Jan 22 '19 at 21:42

5

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 $2\cdot 3 = 6$.

edited Jan 22 '19 at 21:42

J. W. Tanner

60,406

answered Jan 22 '19 at 21:36

jordan_glen

442

score 1 · Answer 2 · answered Jan 22 '19 at 21:42

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.

score 0 · Answer 3 · answered Jan 22 '19 at 21:36

0

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

answered Jan 22 '19 at 21:36

Dr. Sonnhard Graubner

95,283

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

3 Answers3