Using proof by cases -- stuck 
Let $n$ be an integer.
  If $3$ does not divide $n$, then $3$ divides $n^2 - 1$.

I'm trying to prove this using a "proof by cases". However, I'm lost as to how to start. I thought proof by cases started by making a logical expression, but I can't seem to find the right one.
Thanks

Look at the possible remainders when $3$ divides an integer. The set of possible remainders when a number is divided by $3$ is $\{0,1,2\}$. You are given that $3$ does not divide $n$. Hence, the possible remainders when $3$ divides $n$ are $1$ and $2$. Hence $n=3k+1$ or $n=3k+2$. Now look at the square of the two cases.
If $n=3k+1$, then $n^2 = 9k^2 + 6k + 1 = 3(3k^2+2k) + 1 \Rightarrow (n^2-1) = 3(3k^2+2k)$
If $n=3k+2$, then $n^2 = 9k^2 + 12k + 4 = 3(3k^2+4k+1) + 1 \Rightarrow (n^2-1) = 3(3k^2+4k+1)$
Hence in both the cases $3|(n^2-1)$.
