# Answer to Puzzle #15: Prime Squared Minus 1 Multiple of 24

15. Why is it that if 'p' is a prime number bigger than 3, then p

^{2}-1 is always divisible by 24 with no remainder?This one took me a while, my first answer, although valid, was not the best. The final answer is quite tidy although there is some algebra required...

Before reading the answer can I interest you in a clue?

The solution relies on showing that p

First expand p

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 2x2x2, 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 p

* - I know 2 is prime, and even. But we are in the space greater than 3

The solution relies on showing that p

^{2}- 1 is a multiple of 2x2x2x3First expand p

^{2}- 1 to give:p^{2} - 1 = (p - 1) x (p + 1)

^{*},) so p - 1 and p + 1 must be even. 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 2x2x2, 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 p

^{2}- 1 has the factors 2, 2, 2 & 3 hence:p^{2} - 1 = 24n

* - I know 2 is prime, and even. But we are in the space greater than 3

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