Let $n$ be an odd positive integer. Show that $24 \vert(n^3-n).$
So since $n^3-n=(n-1)n(n+1)$, we have that $3\vert(n^3-n).$
Also since we have that $n$ is odd we can say that $n=2k+1$, for some $k \in \mathbb{Z^+}.$
This implies that $(n+1) = 2k +2 = 2(k+1)$, hence $2\vert(n+1).$ Similar argument can be made for $(n-1)$.
Now $24 = 2 \cdot 3 \cdot 4$ so essentially I'm only missing the part where I would have to show that $4$ divides some of the terms. How can I find that?
If $n$ is odd, then both $n-1$ and $n+1$ are even.
Either $n-1$ or $n+1$ must be a multiple of $4$ since for every pair of consecutive even numbers, one of them is a multiple of $4$.
Hence $8$ divides $n^3-n$.
Only remaining to show $8\mid (n-1)n(n+1)$.
Observe, for an odd $n$, one of $n-1$ and $n+1$ is divisible by $2$ and another one is divisible by $4$.
So $(n-1)(n+1)$ is divisible by $8$, if $n$ is odd.
Let n be an odd positive integer, so $n=2k+1$ so $$n^3-n=(n-1)n(n+1)\\ =(2k+1-1)(2k+1)(2k+2)\\=2k(2k+1)(2k+2)\\=4k(k+1)(2k+1)\\=4\underbrace{k(k+1)}_{even=2q}(2k+1)\\=4(2q)(2k+1)$$
