Proving the Divisibility of p^2 - q^2 by 24 for Primes p and q

[PsychonautQQ]

Homework Statement

If p and q are both greater than or equal to 5, prove that 24|p^2 - q^2

Homework Equations

none

The Attempt at a Solution

24 = 2^3 * 3.
If p = q = 5, then 24|0.
If p = 7, q = 5, then 24|24.
Any other combination, p^2 - q^2 will be greater than 24. I want to show that p^2 - q^2 will always have a prime factor of either two or three, hence it will be divisible by 24.

If p and q are both odd, p^2 - q^2 will always be even, hence have a two in it's prime factorization.
A similar situation occurs when p and q are both even.

However, when p is odd and q is even, p^2 - q^2 is odd. I want to show that in this case, p^2 - q^2 has a three in it's prime factorization. I checked a few examples on wolfram and they all worked out, can't think of a way to prove this though. Anyone have any gentle guidance :D?

 

[Svein]

p² - q² = (p + q)*(p - q)

 

[PsychonautQQ]

Thanks... not to mention p and q can never be even since they are primes greater than 5... >.< lol thanks though.