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

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?

• Hint: one of two consecutive evens is a multiple of $4\,$. – dxiv Jul 31 '17 at 4:08
• @dxiv I haven't studied that theorem – user444945 Jul 31 '17 at 4:09
• It's not any grand theorem. Just think about it for more than 40 sec which it took you to reply to my comment. – dxiv Jul 31 '17 at 4:10
• another grand theorem would be to say that one of two consecutive integers is a multiple of 2 – fonfonx Jul 31 '17 at 4:30

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 $8$.

• So Case 1: $k = 2j$ and Case 2: $k = 2j+1$? – user444945 Jul 31 '17 at 4:10
• Sure, that's one good way to do it. – Eric Wofsey Jul 31 '17 at 4:10

$(2k+1)(2k)(2k+2) = (2k+1)(4k)(k+1) =4(2k+1)k(k+1)$ 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 $4k(k+1)$.