<s>[INST] Prove by induction that $3\mid n^3 - n$ 
Prove by induction that $3\mid n^3 - n$.

I'm having an argument with my professor whether my exam was right or not. Before I sign a formal complain to get a review on my exam, I'd like to be sure it's correct. My answer:
Proof by induction:
Proposition: $\forall n \in\mathbb{ N}$ (belongs to natural numbers), $3 | (n^3 - n)$;
Basic Step: $P(0) = 0^3 - 0$ is divisible by 3.
Inductive Step:

$p(n+1) = (n+1)^3 - (n+1)$
$= n^3 + 3n^2 + 3n + 1 - n - 1$
$= n^3 + 3n^2 + 2n$
$= n^3 + 3n^2 + 2n - (n^3 - n)$
$= 3n^2 + 3n$

By induction, $p(n+1)$ is true if:
     $3\mid n$ and $3\mid m$ and $3 | n - m$
    to prove that, assume that $n$ is $3p$ and $m$ is $3q$, therefore $3 | 3p - 3q$.
My professor wanted:

$= (n^3 -n) + 3n^2 + (2n + n)$

 [/INST] The trouble comes from saying that $n^3+3n^2+2n=n^3+3n^2+2n-(n^3-n)$. Instead we can consider $(n+1)^3-(n+1)=n^3+3n^2+3n+1-n-1$. We rewrite this as $(n^3-n)+(3n^2+3n)=(n^3-n)+3(n^2+n)$. The first term is our induction hypothesis which we can rewrite as $3k$ where $k\in\mathbb{Z}$ and the last term is divisible by three. So $3(k+n^2+n)$ is divisible by three for all integers $(k+n^2+n)$. Thus by the Principle of Mathematical Induction $3|(n^3-n)$ for all $n \in \mathbb{N}$.
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