Show that if $n$ is not divisible by $2$ or by $3$, then $n^2-1$ is divisible by $24$. 
Show that if $n$ is not divisible by $2$ or by $3$, then $n^2-1$ is divisible by $24$.

I thought I would do the following ... As $n$ is not divisible by $2$ and $3$ then $$n=2k+1\;\text{or}\\n=3k+1\;\text{or}\\n=3k+2\;\;\;\;$$for some $n\in\mathbb{N}$.
And then make induction over $k$ in each case.$$24\mid (2k+1)^2-1\;\text{or}\\24\mid (3k+1)^2-1\;\text{or}\\24\mid (3k+2)^2-1\;\;\;\;$$This question is in a text of Euclidean Division I'm reviewing, and I wonder if there is a simpler, faster, or direct this way.

$n^2-1=(n-1)(n+1)$
$n$ is not even so $n-1$ and $n+1$ are even.
Also $n=4t+1$ or $4t+3$, this means at least one of $n-1$ or $n+1$ is divisible by 4.
$n$ is not $3k$ so at least one of $n-1$ or $n+1$ must be divisible by 3.
So $n^2-1$ has factors of 4, 2(distinct from the 4) and 3 so $24|n^2-1$ 
Edit: I updated my post after arbautjc's correction in his comment.
