2
$\begingroup$

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?

  • 3
    As 3 doesn't divide $p$, it has to divide either $(p-1)$ or $(p+1)$2012-11-06

1 Answers 1