If $p \geq 5$ is a prime number, then $24$ divides $p^2 - 1$ without remainder 
$\textbf{(A)}\ \text{never} \qquad \textbf{(B)}\ \text{sometimes only} \qquad \textbf{(C)}\ \text{always} \qquad$
$\textbf{(D)}\ \text{only if } p =5 \qquad \textbf{(E)}\ \text{none of these}$
Starting with some experimentation, substituting $p=5$ results in $24$, substituting $p=7$ results in $48$, and substituting $p=11$ results in $120$.  For these primes, the resulting numbers are multiples of $24$.
To show that all primes we devise the following proof:
$Proof:$
$p \ge 5$ result in $p^2 -1$ being a multiple of $24$, we can use modular arithmetic.  Note that $p^2 - 1 = (p+1)(p-1)$.  Since $p^2 \equiv 0,1,2,3 \mod 4$, $p^2 - 1$ is a multiple of $8$.  Also, since $p^2 \equiv 0,1,2 \mod 3$, $p^2 - 1$ is a multiple of $3$.  Thus, $p^2 -1$ is a multiple of $24$, so the answer is $\boxed{\textbf{(C)}}$.