prove that $p^2-1$ is divisible by $24$ if $p$ is a prime greater than $3$

Question

How to prove that $p^2-1$ is divisible by $24$ if $p$ is a prime number greater than $3$?

Consider the problem modulo $3$ and $8$. The primeness enters only via simple congruence considerations. Chances are this is a duplicate. — quid, Jan 28 '15 at 20:42

score 2 · Accepted Answer · answered Jan 28 '15 at 20:41

If $p$ is a prime greater than $3$ then either $p = 6k - 1$ or $p = 6k + 1$. In the first case you have $$p^2 - 1 = (p-1)(p+1) = (6k - 2)(6k) = 36k^2 - 12k = 12k(3k - 1).$$ If $k$ is even then $12k$ is a multiple of $24$, and if $k$ is odd then $12(3k-1)$ is a multiple of $24$.

The proof in case $p = 6k + 1$ is nearly identical.

prove that $p^2-1$ is divisible by $24$ if $p$ is a prime greater than $3$

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