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:

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

 A: 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 $3$.
A: It is clear to me that your logic is correct.  However your solution is not well-written and as a result is confusing to read. For example, you have written 
$$n^3+3n^2+2n = n^3+3n^2+2n - (n^3-n)$$
(This is the equality between your 3rd and 4th lines.) This is false. It is clear to me that what you mean is that, since $p(n+1)=n^3+3n^2+2n$, then $p(n+1)-p(n)=n^3+3n^2+2n-(n^3-n)$. However, this is not what you wrote.
Thus, while the substance of your answer is essentially correct, the form in which it is being communicated leaves something to be desired. There is no objective standard regarding how such a situation should be graded; it is up to the professor.
This is not mathematical advice, but I suggest you approach your professor again rather than lodging a formal complaint, because the case is not sufficiently clear-cut to be adjudicated by outside parties. It will help you have the kind of conversation you want with them if you approach that conversation with the attitude that you know you had the right idea, but that you didn't communicate it well and would like to learn how to do it better, but that you want to make sure that they understand that you did have the right idea.
A: 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 $n \in \mathbb{N}$.
