For all $(n\in\Bbb N*)(\exists k \in\Bbb N)$ $ show $ : $n²(n²-1)=12k $ We had this this exercise on the exam and I unfortunately I couldn't do it right but at least I tried. Here is what I did :
$$ (\forall n\in\Bbb N*)(\exists k \in\Bbb N) :  n^2(n^2-1)=12k$$
 The first thing I did was Disjunction of cases 
If $n$ is pair then :
$$ n=2k$$
$$ n²=4k²$$
$$ n²(n²-1)= 4k²(4k²-1)$$
$$ n²(n²-1)= 16k^4-4k²$$
 and that's where I got stuck 
 So I thought I would try Proof by Induction 
For $n=1$
$$ 1²(1²-1) = 0 $$ $0$ is divisible by $12$ 
 Which means the proposition is right for $n=1$ 
Let's suppose that :  $$ n^2(n^2-1)=12k$$
Let's prove that : $$ (n+1)^2((n+1)^2-1)=12h$$
 $$ (n+1)^2((n+1)^2-1) = (n+1)^2(n^2+2n)$$
 $$  =(n^2+2n+1)(n^2+2n)$$
$$ = n^4+2n^3+2n^3+4n^2+n^2+2n $$
$$ = n^4+4n^3 +5n^2 +2n $$
 And this is where I got stuck again. I know we already passed the test but I want to know the answer. At least I tried, please don't down vote. 
 A: You only have to see that $n^2(n^2-1)$ is always divisible by $4$ and $3$ (because $4$ and $3$ are relative primes).
To check the $3$, only notice that
$$n^2(n^2-1)=n\cdot n(n-1)(n+1)$$
so you have three consecutive numbers, that means that at least one will be divisible by $3$
To check the divisibility by $4$, you have two cases.

*

*If $n$ is even, $n=2k$, so $n^2=4k^2$, letting your factor $4$.


*If $n$ is odd, $n=2k+1$, so $n^2=4k^2+4k+1$, so $n^2-1=4(k^2+k)$, letting your factor $4$ too.
A: Hints:
$$
2^2\mid n^2(n-1)(n+1).
$$
and 
$$
3\mid n^2(n-1)(n+1).
$$
A: It is much more insightful to prove a slight generalization (OP is special case $\,p=3)$
Theorem $\ $ If $\,p\,$ is an odd prime then $\,4p\mid n^{\large 2}(n^{\large p-1}-1) =: a$
Proof $\ $ If $\,2\mid n\,$ then $\,4\mid n^{\large 2}.\,$ Else $\,n\,$ is odd so $\,{\rm mod}\ 4\!:\ n\equiv \pm1\,\Rightarrow\, \color{#c00}{n^{\large 2}\equiv 1}\,$ therefore $\ n^{\large p-1}\! = (\color{#c00}{n^{\large 2}})^{\large (p-1)/2}\equiv \color{#c00}1^{\large (p-1)/2}\equiv 1\,$ so $\,4\mid n^{\large p-1}-1.\ $ Thus in all cases $\,4\mid a.\ $
By litlle Fermat $\,p\mid n^{\large p}-n \ $ so $\ p\mid a = n(n^{\large p}-n).\ $
Since both $\,4,p\mid a\,$ so does their lcm = product $= 4p.\ \ $ QED

Remark $\ $ A similar proof leads to the following generalization of the Euler-Fermat Theorem.
Theorem $\  $ Suppose that $\rm\ n\in \mathbb N\ $ has the prime factorization $\rm\:n = p_1^{e_{\:1}}\cdots\:p_k^{e_k}\ $ and suppose  that for all $\rm\,i,\,$ $\rm\ e\ge e_i\ $ and $\rm\ \phi(p_i^{e_{\:i}})\mid f.\ $ Then $\rm\ n\mid (ab)^e\,(a^f-b^f)\ $ for all $\rm\: a,b\in \mathbb Z.$
Proof $\ $ See this answer.
