Prime Squared

Since $p^{2}$ has a remainder of $1$ when divided by $x$ , then $x | p^{2}-1$ . $p^{2}-1$ = $(p-1)*(p+1)$ . $p$ is prime, so it is odd, given that the only even prime number is 2 and $p > 3$ , so $p-1$ and $p+1$ are even, and because they're consecutive even numbers, one of them is a multiple of $4$ . Also, $p$ , $p-1$ and $p+1$ are $3$ consecutive integers, so one of them is a multiple of $3$ and it can't be $p$ because it is prime and greater than $3$ . Therefore ${2*4*3}|p^{2}-1$ . So the answer is $24$