Question

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Answer 1

factor
\[p^2-1=(p+1)(p-1)\]

Answer 2

ok so \(p\) is prime larger than 5
now consider the numbers \(p-1,p,p+1\)
one is divisible by 3, because there are three of them
one is divisible by 2 and therefore one is divisible by 4

Answer 3

and they have be \(p-1\) or \(p+1\) because by assumption \(p\) is prime, and therefore not divisible by 3 or 4

Answer 4

therefore \(p^2-1\) must be divisible by 2, 3 and 4

Answer 5

a somewhat fancier idea is that since \(p\) is congruent to either 1, 3, 5, or 7 mod 8, then \(p^2\) is congruent to 1 mod 8 and so \(p^2-1\) is congruent to 0 mod 8, i.e. it is divisible by 8

Answer 6

and again since either \(p-1\) or \(p+1\) must be divisible by 3, then \(p^2-1\) must be divisible by \(8\times 3=24\)