If $p$ and $q$ are two primes greater than $3$ then prove that $24\mid p^2-q^2$ . My solution goes like this:

If $p,q$ are primes greater than $3$ then $8 \mid p^2-q^2$ as any odd perfect square is of the form $8k+1$ where $k\in\mathbb{Z}$ . Now, $p\equiv 1,2 \pmod{3}$ and $q\equiv 1,2\pmod{3}$ ($0 $ cant be a case as $p,q$ are primes). So, we see that all possible combinations in the remainders of $p,q$ will be equivalent to zero either by doing $p+q$ or by $p-q$ such that $3 \mid p^2-q^2$ . So, $24 \mid p^2-q^2$.

I just want to verify whether my solution is okay or not.