Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$ Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$


I can not understand how to prove it. Please help me.
HINT :

$(n-1)n(n+1)(n+2)$

is "consective numbers".
The first very obvious thing to note is that $n(n^2 - 1)(n+2)$ can be written as $n(n - 1)(n+1)(n+2)$ which means that they are four consecutive integers. Obviously one of these must be divisible by $4$.
The proof:
Any number $n$ is either of the form $2m$ or $2m+1$.   

Case I:  $n = 2m$
$n(n^2 - 1)(n+2)$ = $2m(4m^2 - 1)(2m + 2) = 2m(4m^2 - 1)*2(m+1) = 4x$
Case II: $n = 2m+1$
$n(n^2 - 1)(n+2) = (2m+1)((2m+1)^2 - 1)(2m+1 + 2) = (2m+1)((4m^2 + 4m +1) - 1)(2m+3) = 4x$
Hence proved.

The link provided by @lab bhattacharjee is quite useful.