[Math] Prove if $3$ does not divide $n$, then $n^2=1+3k$ for some integer $k$

Question

I am proving by cases but am getting confused. I am not sure if this leads to a contradiction or not.

Here's what I have so far:

Direct Proof.
Suppose $3$ does not divide $n$.

Case 1: remainder upon division is $1$
then $n=3k+1$, for some integer $k$
and $n^2=(3k+1)^2$ … this is where I am getting confused.

Case 2: remainder upon division is $2$
then $n=3k+2$ for some integer $k$
and $n^2=(3k+2)^2$ … and here.

Because the problem says "If $3$ does not divide $n$, then $n^2=1+3k$" but this is not what I am getting. So does this result in a contradiction or not? And so is the proof false and I need to provide a counter example?

A little lost, thanks!

Answer 1

It's as simple as:

If $n \nmid 3$, then $n \equiv 1$ or $2 \pmod 3$.

So $n^2 \equiv 1^2 = 1$ or $2^2 = 4 \equiv 1 \pmod 3$, i.e. $n^2 \equiv 1 \pmod 3$, giving the required result.