For any prime $p > 3$, why is $p^2-1$ always divisible by 24?

The most elementary proof I can think of, without explicitly mentioning any number theory: out of the three consecutive numbers $p - 1$, $p$, $p + 1$, one of them must be divisible by $3$; also, since the neighbours of p are consecutive even numbers, one of them must be divisible by $2$ and the other by $4$, so their product is divisible by $3 · 2 · 4 = 24$ — and of course, we can throw $p$ out since it's prime, and those factors cannot come from it.

$p$ must be congruent either to 1,3,5,7 modulo 8. Then $p^2$ is congruent to $1$ modulo $8$ in either case. So $8$ divides $p^2-1$.

Now, $p$ is not a multiple of 3, so either $p-1$ or $p+1$ is a multiple of three. So $3$ divides $p^2-1$.

Together, it follows that 24 divides $p^2 -1 $.