1973 AHSME Problems/Problem 18

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Problem

If $p \geq 5$ is a prime number, then $24$ divides $p^2 - 1$ without remainder

$\textbf{(A)}\ \text{never} \qquad \textbf{(B)}\ \text{sometimes only} \qquad \textbf{(C)}\ \text{always} \qquad$ $\textbf{(D)}\ \text{only if } p =5 \qquad \textbf{(E)}\ \text{none of these}$

Solution

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 $p \ge 5$ 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 $p \equiv 1,3 \mod 4$, $p^2 - 1$ is a multiple of $8$. Also, since $p \equiv 1,5 \mod 6$, $p^2 - 1$ is a multiple of $3$. Thus, $p^2 -1$ is a multiple of $24$, so the answer is $\boxed{\textbf{(C)}}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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