I have come across this proof that for prime p >3 p^2-1 is always a multiple of 24:

p^2 – 1 = (p – 1)(p + 1) and given that p is prime, it must be odd.

So p – 1 and p + 1 are both even, but for two adjacentevennumbersone of them must be divisible by four.

So (p – 1)(p + 1) must be divisible by eight.

Now if p is not 3 and is prime then p can't be divisible by three, so one of (p – 1) or (p + 1) must be divisible by three since every third number is divisible by three.So (p – 1)(p + 1) is divisible by 3, and by 8, so it is divisible by 24.

The two bits i bold, I am not sure about? Can anyone help me see why they are true?

Thanks