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.

Source : Link , Question Author : Don De Tina , Answer Author : darij grinberg

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