1973 AHSME Problems/Problem 18

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

1973 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1973 AHSME Problems/Problem 18

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