$30 \cdot 22$ If $p \geqslant 5$ is a prime, then 24 divides $p^{2}-1$
(A) impossible.
(B) possible only sometimes.
(C) always possible.
(D) possible only when $p=5$.
(E) none of these.
(24th American High School Mathematics Examination, 1973)
[Solution 1] Since $p \geqslant 5$ is a prime number, let $p=3k+1$ or $3k+2$.
If $p=3k+1$, then $p^{2}-1=(p-1)(p+1)=3k \cdot(3k+2)$
If $p=3k+2$, then $p^{2}-1=(3k+2)(3k+3)$,
it is clear that 3 divides $\left(p^{2}-1\right)$.
Similarly, by setting $p=8t+r, r=1,3,5,7$ (note that $p$ is a prime number), we can prove
$$
8 \mid \left(p^{2}-1\right) \text{. }
$$

Since 3 and 8 are coprime, it follows that $24 \mid \left(p^{2}-1\right)$.
Therefore, the answer is $(C)$.
[Solution 2] Let $p=6k \pm 1$ (noting that $p$ is a prime number not less than 5).
Then $p^{2}-1=(6k \pm 1)^{2}-1=36k^{2} \pm 12k=12k(3k \pm 1)$,
and $k(3k \pm 1)$ is even, so $24 \mid \left(p^{2}-1\right)$.
Therefore, the answer is $(C)$.