Can you prove that p² ≡ 1 (mod 24) for any prime p ≥ 5?

You can start with the fact that every prime greater than 4 can be expressed as either 6n+1 or 6n+5 for some nonnegative integer n. This is because a number of the form 6n, 6n+2, 6n+3, or 6n+4 would be composite.

When you square either type of prime, you get 36n^2+12n+1 or 36n^2+60n+1. These are both equivalent to 12n^2+12n+1 = 12(n+1)n+1 (mod 24). Because either n+1 or n is even 12(n+1)n is a multiple of 24, and 12(n+1)n+1 = 1 (mod 24).