1960 AHSME Problems/Problem 25

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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 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Problem 26
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