TITLE: Proving the remainder is $1$ if the square of a prime is divided by $12$
QUESTION [6 upvotes]: Given, $p$ is a prime number and $p>3$. How do we prove that the remainder $r$ is always $1$ if $p^2$ is divided by $12$?

REPLY [5 votes]: $ p^2 -1 =(p-1)(p+1) $. Since $p$ is odd, these are two even factors, so the product is divisible by $4$. Also, one of $p-1$, $p$, $p+1$ must be divisible by $3$, but since $p$ is prime, it is not divisible by $3$. Thus one of $p-1$, or $p+1$ must be divisible by three.