$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement. "Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$"
Resolution:
$$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)$$and$$8\mid n(n^{2}-1)(3n+2)?$$$$$$
Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you.
 A: We know from here or here, the product $n$  consecutive integers is divisible by $n!$ for integer  $n>0$
As $24=4!$and we already have $n(n^2-1)=(n-1)n(n+1)$ the product of $3$ consecutive integers
So, if we can arrange the next or previous term as multiplier, we shall have the product of $4$ consecutive integers, hence divisible by $4!=24$
If we consider the previous term $(n-2),$ 
$3n+2=3(n-2)+4$
$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n-2)+4\}$$
$$=3\underbrace{(n+1)n(n-1)(n-2)}_{\text{ the product of } 4 \text{ consecutive integers }} +4 \underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$
If we consider the next term $(n+2),$ 
$3n+2=3(n+2)-4$
$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n+2)-4\}$$
$$=3\underbrace{(n+2)(n+1)n(n-1)}_{\text{ the product of } 4 \text{ consecutive integers }}-4\underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$
A: Hint:
$(n-1)n(n+1)$ is product of three consecutive numbers. So they are always divisible by $8$ and $3$ if $n$ is odd. You don't even have to bother about $3n+2$. Now think, what happens to $3n+2$ if $n$ is even.
A: To prove $8|n(n^2-1)(3n+2)$ you can simply break the problem in three cases:
Case 1 $n$ odd.
Then $n^2-1=(n-1)(n+1)$ is the product of two consecutive even numbers. Thus one must be divisible by $4$ and the other by $2$.
Case 2 $n$ multiple of 4.
Then $4|n$ and $2|3n+2$.
Case 3 $n$ even but not divisible by 4.
Then $2|n$ and $n-2$ is divisible by $4$. Hence 
$$4|3n+2=4n-(n-2) \,.$$
A: Another approach, perhaps more elementary (but lengthier):
$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)$$
(1) Clearly, exactly one of $\,n-1\,,\,n\,,\,n+1\;$ is divisible by $\,3\,$
(2) If $\,n\,$ is odd, then $\,n-1\,,\,n+1\;$ are even with exactly one of them divisible by $\,4\,\implies (n-1)n(n+1)\,$ is divisible by $\,2\cdot 4=8\;$
(3) If $\,n\,$ is even, then also $\;3n+2\;$ is even, and exactly one of them is divisible by $\,4\,$ , since
$$n=4k+2\implies 3n+2=12k+8=4(3k+2) $$
and again $\;n(3n+2)\;$ is divisible by $\;4\cdot 2=8\;$.
The above cover all the possible cases and in each one the given expression is divisible by $\;3\;$ and by $\;8\;$ , so also by $\,3\cdot 8 =24\;$
