prove that $[5, 7, 11, 13, 17, 19, 23]$ are the only possible variants of the remainders (read context) when dividing the prime number $p$ by 24

Question

The problem is following:

Given that $p$ is a prime number, $p > 3$. Prove that $(p^2 - 1)$ is divisible by $24$.

I started writing down the possible remainders of dividing $p$ by $24$ and got the following row: ${5; 7; 11; 13; 17; 19; 23}$

But am i even right at this point? If i am how do i prove that these are the only possible remainders?

There is a much more efficient proof. Note that if $p-1$ is a multiple of $3$ then $p+1$ is, and if $p-1$ is not a multiple of $4$, $p+1$ is. — ajotatxe, Oct 14 '19 at 18:42
Possible duplicate of For any prime $p>3$, why is $p^2-1$ always divisible by 24? — Jyrki Lahtonen, Oct 14 '19 at 18:55

José Carlos Santos · Accepted Answer · 2019-10-14T18:46:19.077

0

You have missed that $1$ can be a remainder too (take $p=73$ or $p=97$, for instance). All other elements of $\{0,1,2,\ldots,23\}$ are muliples of $2$ or of $3$. Since $p$ is prime and $24$ is a multiple of both $2$ and $3$, no such number can be the remainder of the division of $p$ by $24$, since that would make $p$ a multiple of $2$ or of $3$.

edited Oct 14 '19 at 18:46

answered Oct 14 '19 at 18:40

José Carlos Santos

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prove that $[5, 7, 11, 13, 17, 19, 23]$ are the only possible variants of the remainders (read context) when dividing the prime number $p$ by 24

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