well because

2 | 6k
2 | 6k ± 2
3 | 6k + 3

cowboy credits fine, thx

fuck, it's supposed to be all primes > 3, not >= 3, but anyways

All primes >= 3 are in the form of 6k ± 1.

If we square, we get p^2 = 36k^2 ± 12k + 1

In modulo 24, we get -> 12k^2 ± 12k + 1 = 12k * (k ± 1)

k * (k ± 1) is always even, therefore it is always a multiple of 24.

Hence, we're left with the +1 at the end.

p_squared is always 1 modulo 24 :)