Why is that if p is a prime number bigger than 3, then p^2-1 is always divisible by 24 with no remainder?
The solution relies on showing that p^2 – 1 is a multiple of 2×2×2×3
First expand p^2 – 1 to give:
p^2 – 1 = (p – 1) x (p + 1)
Then consider the terms on the right hand side, firstly since we know that p must be odd p -1 and p + 1 must be even. so we have two of the factors we require.
Additionally since p – 1 and p + 1 effectively form 2 consecutive even numbers one of them must be a multiple of 4 thus we have another of our factors of 2. So far we have 2×2×2, now to get the factor of 3
p – 1, p & p + 1 form three consecutive numbers. in any three consecutive numbers one will be a multiple of 3, we know it is not p which is a multiple of 3, as this is prime, hence either p – 1 or p + 1 is a multiple. Therefore p2 – 1 has the factors 2, 2, 2 & 3 hence:
p^2 – 1 = 24n
Why must p be greater than 3? Because 3 is the only number which is both a multiple of 3 and prime.