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. 
 A: You can say more....
The number is divisible by 24. Why?
We have 4 consecutive numbers. So two of them have to be even and one of them should be divisible by 4. So 8 should divide the product. 
Also, since we have 4 consecutive numbers, one of them is divisible  by 3. 
Hence 24 divides the product.
A: $$\text{Here is a different way out; Recall that } \displaystyle \sum_{k=1}^m k = \dfrac{m(m+1)}2$$
$$\text{Hence, we have }1 + 2 + \cdots +(n-1) = \dfrac{n(n-1)}2 \text{ and } 1 + 2 + \cdots +(n+1)  = \dfrac{(n+1)(n+2)}2$$
$$\text{Putting the above two together, we get that}$$
$$\color{red}{\left(1 + 2 + \cdots +(n-1) \right) \left(1 + 2 + \cdots + n + (n+1)\right)} = \color{blue}{\dfrac{n(n^2-1)(n+2)}4}$$
$$\text{ Clearly, the }\color{red}{\text{left side is an integer}}\text{ and hence }\color{blue}{\text{right side is also an integer}}.$$
A: Hint: Consider two cases: $n$ is even, or $n$ is odd. Can you find two parts of the product that are both even in either case? In the second, you may find it useful to write
$$n^2 - 1 = (n - 1)(n + 1)$$
A: HINT :
$(n-1)n(n+1)(n+2)$
is "consective numbers".
A: You have $n(n^2-1)(n+2)$ which can be seen as $(n-1)n(n+1)(n+2)$
Now, you should be able to see that this is divisible by $4$
Because... any four consecutive numbers will have "....."
A: Hint 1 $\ $ Partition the $\,4\,$ factors of the product into $\rm\color{#c00}{two}$ sets, each of which contains  integers of opposite parity. Then each set contains an even number, so the product has $\rm\color{#c00}{two}$ factors of $\,2$.  
Hint 2 $\ $ Consecutive integers have opposite parity.
A: 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.
