Why one less than a prime's square is always a multiple of 24

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Try it: 5 squared minus 1 = 24; 7 squared minus 1 = 48; 11 squared minus 1 = 120; 13 squared minus 1 = 168. All multiples of 24.

Prove that for every prime p greater than 3, p squared minus 1 is divisible by 24. Does the proof actually need p to be prime?

Accepted Answer

Yes, and the proof only uses that p is not divisible by 2 or 3. Factor. p squared minus 1 = (p minus 1)(p plus 1), the product of the two neighbours of p. Divisibility by 8. Since p is an odd prime, both p minus 1 and p plus 1 are even. They are consecutive even numbers, so one of them is a multiple of 4. The product of a multiple of 2 and a multiple of 4 is a multiple of 8. Divisibility by 3. Among any three consecutive whole numbers, p minus 1, p, p plus 1, exactly one is a multiple of 3. It is not p, because p is a prime bigger than 3. So it is p minus 1 or p plus 1, and their product is a…

Why one less than a prime's square is always a multiple of 24

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