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.
 A: If n is neither divisible by 3 nor by 2 then it is of the form $n = 6k \pm 1$. Then we get
$$(6k \pm 1)^2 - 1 = 36k^2 \pm 12 k \equiv 0 \pmod {24}$$
showing the divisibility
A: If $n$ is not divisible by either $2$ or $3$, then it must satisfy either $n \equiv 1 \pmod{6}$ or $n \equiv 5 \pmod{6}$. This can be used directly.
A: The long way to get at the same answer is to simply look at the squares modulo 24. Not divisible by 2 or 3 leaves us 1, 5, 7, 11, 13, 17, 19, 23. Obviously $1^2 \equiv 1 \pmod{24}$ and $5^2 = 25 \equiv 1 \pmod{24}$. Likewise $7^2 = 2 \times 24 + 1$.
This leaves five cases you can check easily enough, though this is not so much for your benefit but for those who come across this question through the duplicate.
Of course I don't recommend this approach if you're dealing with numbers greater than 24, even if just by a little bit.
A: $n^2-1=(n-1)(n+1)$
$n$ 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.
