Claim: $$12\mid(p^{2}-1) \space \forall\text{ primes }p>3$$
Attempt at proof:
$$p>3\space\Rightarrow\space p\text{ is odd} $$ $$p^2-1=(p-1)(p+1)\space\Rightarrow2^2\mid(p^{2}-1)$$
How do I go on to show that $3\mid(p^2-1)$ which would complete the proof?