# 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 (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 All AHSME Problems and Solutions
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