<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] Note: the following "equality" in your argument is not valid (third to fourth line):
$${\bf n^3 + 3n^2 + 2n} = {\bf n^3 + 3n^2 + 2n} \color{red}{\bf - (n^3 - n)}$$
First, your missing the inductive hypothesis:
Assume that for $n = k$, we have $p(k) = k^3 - k$ is divisible by $3$.
Then, we show based on that assumption, that $p(k+1)$ is true.
Here is what your professor is arguing:
$$\begin{align} p(k+1) & = (k+1)^3 - (k+1) \\ \\ 
& = k^3 + 3k^2 + 3k + 1 - k - 1\\ \\
& = k^3 + 3k^2 + 2k \\ \\
& = k^3 - k + k + 3k^2 + 2k \\ \\
& = (\underbrace{k^3 - k}_{\large \text{IH}}) + 3(k^2 + k)\end{align}$$
By the inductive hypothesis (IH), we have that the first term is divisible by $3$, and we see that the second term is also divisible by three.
Hence $p(k + 1)$ is true.
Therefore, by induction, $n^3 - n$ is divisible by $3$.
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