A short CRT proof, without first classifying primes modulo 6:
For any prime p > 3, p is odd. Write p = 2k + 1. Then
p² − 1 = 4k(k + 1).
One of k and k + 1 is even, so 8 divides p² − 1. Also p is not divisible by 3, hence p ≡ ±1 (mod 3), so 3 divides p² − 1. Since gcd(8, 3) = 1, their product 24 divides p² − 1. Therefore p² ≡ 1 (mod 24) for every prime p > 3.
A short CRT proof, without first classifying primes modulo 6:
For any prime p > 3, p is odd. Write p = 2k + 1. Then
p² − 1 = 4k(k + 1).
One of k and k + 1 is even, so 8 divides p² − 1. Also p is not divisible by 3, hence p ≡ ±1 (mod 3), so 3 divides p² − 1. Since gcd(8, 3) = 1, their product 24 divides p² − 1. Therefore p² ≡ 1 (mod 24) for every prime p > 3.