Squares of Primes

Prime numbers $(p>3)$ can be obtained by $6k\pm 1$ where $k\in\mathbb N$ . Now $p^2-1 = \,(6k\pm 1)^2-1 = \,(36k^2 \pm 12k +1 )-1 = 12k(3k \pm 1)$ Notice that $\text{gcd}\,(k,3k\pm 1) =1$ which implies that $k(3k\pm 1) = 2m$ for some positive integer $m$ . Plugging back we have $p^2-1 = 12\cdot 2m = 24m$ which is always divisible by $8$ and $6$ . Thus the answer is True .