Post #2331374
2026-03-28 02:49 UTC
Informal proof that for a prime p >=5, p²-1 must be a multiple of 24.
p²-1 = (p-1)(p+1)
p-1, p, p+1 are three consecutive integers. One of them must be divisible by 3 - and it can't be p, because p is prime. So either p-1 or p+1 is a multiple of 3.
Also, p is odd, so p-1 and p+1 are both even - and one or the other must be divisible by 4. One is a multiple of 2, the other of 4.
So the product of p-1 and p+1 has factors of 2, 3 and 4, and must be a multiple of 2*3*4 =24.
Replies (4)
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@shayman@cosocial.ca 2026-03-28 03:13
This has been another exciting episode of Friday Night Math.
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@inecas@mstdn.social 2026-03-28 07:53
@shayman@cosocial.ca so the real condition is the p number is “every number not divisible by 2 and 3”, the “prime” part now feels like a clickbait.
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@dearlove@mathstodon.xyz 2026-03-28 07:56
@shayman@cosocial.ca A prime ≥ 5 must be of the form 6n±1. (6n±1)²-1 = 36n²±12n = 12n(3n±1) which is obviously a multiple of 24 if n is even, but if n is odd 3n±1 is even, so it is a multiple of 24 again. So it's really a property of numbers 6n±1 rather than just primes.
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@geomannie@mastodon.scot 2026-03-28 10:02
@shayman@cosocial.ca 👏Lovely logic!