Prove for every natural $n$ that if $n$ is odd then $ n^{3} - n$ is divisible by $8$.

Question

Question:

Prove for every natural $n$ that if $n$ is odd then $ n^{3} - n$ is divisible by $8$.

Hint: $ n^{3} - n = n(n-1)(n+1)$

My attempt:

If $n$ is odd then $ n = 2k+1$ for some integer $k$. Then, $ (2k+1)(2k)(2k+2) = (2k+1)(4k)(k+1)$

From here how do I show that the expression can be multiplied by 8? Should I break it into cases when $k$ is even and when $k$ is odd?

Answer 1

Hint: Consider separately the cases where $k$ is even and when $k$ is odd.

A full answer is hidden below.

If $k$ is even, then $4k$ is divisible by $8$, and hence so is $(2k+1)(4k)(k+1)$. If $k$ is odd, then $k+1$ is even and thus together with the factor of $4$ in $4k$ we again find that $(2k+1)(4k)(k+1)$ is divisible by $8$.

Answer 2

$(2k+1)(2k)(2k+2) = (2k+1)(4k)(k+1) =4(2k+1)k(k+1) $ so you know that it is divisible by at least 4.

Looking at $k(k+1)$, this is the product of two consecutive integers, so one of them has to be even (and the other odd, but that does not matter here).

Therefore $2$ divides $k(k+1)$, so $8$ divides $4k(k+1)$.