Prove that p2 – 1 is divisible by 24 if p is a prime number greater than 3?

Problem:-

Prove that p2 – 1 is divisible by 24 if p is a prime number greater than 3?

Solution:-

The most elementary proof , without explicitly mentioning any number theory: out of the three consecutive numbers p–1, p, p+1, one of them must be divisible by 3; also, since the neighbors of p are consecutive even numbers, one of them must be divisible by 2 and the other by 4, so their product is divisible by 3⋅2⋅4=24 — and of course, we can throw p out since it’s prime, and those factors cannot come from it.

according to this question

p^2 – 1 = 24*X let X>0

after solving this equation we will get

p = sqrtof(24*X+1)

now put the values of X

when X=1 , p=5
when X=2 , p=7
when X=5 , p=11

we will not consider X=3 and X=4 because we will not get the perfect squares…..:)