<s>[INST] Prove that for all $n\in \mathbb{N}$ we have that $6$ divides $n^3-n$ I'd like to know if my proof of the following is valid and if there's a shorter proof:

Prove that for all $n\in \mathbb{N}$ we have that $6$ divides $n^3-n$

Proof:
We proceed by induction...
$P(1)=(1)^3-(1)=0$ which $6$ divides
Suppose the proposition is true for $n=k$, that is, there exists $j\in \mathbb{N}$ s.t. $k^3-k=6j$.
We show that $P(k+1)=(k+1)^3-(k+1)=6i$ for some $i$:
$$(k+1)^3-(k+1)=k^3+3k^2+2k$$
$$=(k^3-k)+3k^2+3k$$
$$=\color{red}{6j+3k(k+1)}$$

Now we show that $6$ divides $3k(k+1)$. Either $3k(k+1)$ is even (case 1), or odd (case 2).
(case 1) There exists $l$ s.t. $k=2l\implies3k(k+1)=6l(2l+1)$
(case 2) There exists $m$ s.t. $k=2m+1\implies3k(k+1)=3(2m+1)(2m+1+1)=6(2m^2+3m+1)$
Now, (case 1) and (case 2) $\implies 6$ divides $3k(k+1)\implies 3k(k+1)=\color{blue}{6q}$ for some $q\in\mathbb{N}$

Next, $$P(k+1)=\color{red}{6j+3k(k+1)}=6j+\color{blue}{6q}=6(i)$$
Therefore, by induction $6$ divides $n^3-n$ for all $n\in \mathbb{N}$
 [/INST] Since $n^3 \equiv n \pmod{2}$, we know that $2$ divides $n^3 - n$.
Now use induction to show that $n^3 - n$ is also divisible by $3$:
Base case: Certainly $3$ divides $n^3 - n$ when $n = 0$.
Inductive step:
Now suppose $3$ divides $n^3 - n$.
Algebra allows us to write
$\quad (n+1)^3 - (n+1) = n^3 + 3n^2 + 2n = n^3 - n + 3n^2 + 3n$
But then
$\quad [(n+1)^3 - (n+1)] \pmod{3}\equiv$
$\quad\quad [n^3 - n] \pmod{3} + [3n^2 + 3n]\pmod{3}\equiv$
$\quad\quad 0 \pmod{3} + 0 \pmod{3}\equiv 0\pmod{3}$
We conclude that $3$ also divides $n^3 -n$.
Since both $2$ and $3$ are prime factors, the product $6$ must also divide $n^3 -n$.
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