Why primes above 3 give p squared minus 1 divisible by 24

Let $p$ be any prime larger than $3$. Explain why $24$ always divides $p^2-1$.

Factor $p^2-1=(p-1)(p+1)$. Since $p$ is odd, $p-1$ and $p+1$ are consecutive even numbers, one of them a multiple of $4$, so $8$ divides the product. Among $p-1,p,p+1$ one is a multiple of $3$ and it is not $p$, so $3$ divides the product too; hence $24$ does.