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.

Consider modulo $24$.. We have $i\mod 24$ for $0\leq i\leq 23$
We have to neglect $a\mod 24$ if $a$ is a multiple of $2$ or  a multiple of $3$.. We are left with 


*

*$1\mod 24$

*$5\mod 24$

*$7\mod 24$

*$11\mod 24$

*$13\mod 24\equiv -11\mod 24$

*$17\mod 24\equiv -7\mod 24$

*$19\mod 24\equiv -19\mod 24$

*$23\mod 24\equiv -1\mod 24$


considering squares, we have 


*

*$1^2\mod 24\equiv 1\mod 24$

*$5^2\mod 24\equiv 1\mod 24$

*$7^2\mod 24\equiv 1\mod 24$

*$11^2\mod 24\equiv 1\mod 24$


For same reason as above, squares of below congruence classes are equal to $1\mod 24$.


*

*$13\mod 24\equiv -11\mod 24$

*$17\mod 24\equiv -7\mod 24$

*$19\mod 24\equiv -19\mod 24$

*$23\mod 24\equiv -1\mod 24$


So, $p^2\equiv 1\mod 24$.
