1960 AHSME Problems/Problem 25

Problem

Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$ . The largest integer which divides all possible numbers of the form $m^2-n^2$ is:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution

First, factor the difference of squares. $(m+n)(m-n)$ Since $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$ , where $a$ and $b$ can be any integer. $(2a+2b+2)(2a-2b)$ Factor the resulting expression. $4(a+b+1)(a-b)$ If $a$ and $b$ are both even, then $a-b$ is even. If $a$ and $b$ are both odd, then $a-b$ is even as well. If $a$ is odd and $b$ is even (or vise versa), then $a+b+1$ is even. Therefore, in all cases, $8$ can be divided into all numbers with the form $m^2-n^2$ .

This can be confirmed by setting $m=3$ and $n=1$ , making $m^2-n^2=9-1=8$ . Since $8$ is not a multiple of $3$ and is less than $16$ , we can confirm that the answer is $\boxed{\textbf{(D)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1960 AHSME Problems/Problem 25

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