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

Let $$p>3$$ be a prime. Prove that $$24 \mid p^2-1$$.

I know this is very basic and old hat to many, but I love this question and I am interested in seeing whether there are any proofs beyond the two I already know.

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.