# 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 $\boxed{\textbf{(D)}}$.

## See Also

 1960 AHSC (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byProblem 26 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 • 36 • 37 • 38 • 39 • 40 All AHSME Problems and Solutions
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