AHSME 1973 Problem 18

If $p \geq 5$ is a prime number, then $24$ divides $p^2 - 1$ without remainder $\textbf{(A)}\ \text{never} \textbf{(B)}\ \text{sometimes only} \textbf{(C)}\ \text{always}$ $\textbf{(D)}\ \text{only if } p =5 \textbf{(E)}\ \text{none of these}$