2007 AMC 10A Problem 23

The condition is $m^2 - n^2 = 96$, which factors as
$$(m+n)(m-n) = 96.$$
Write $x = m + n$ and $y = m - n$. Then $x > y \ge 0$, $xy = 96$, and recovering the pair requires $m = \frac{x+y}{2}$, $n = \frac{x-y}{2}$ to be positive integers: $x$ and $y$ must have the same parity, and $y > 0$ since $n > 0$ forces $x \ne y$ (indeed $96$ is not a square).

Because $96$ is even, $x$ and $y$ cannot both be odd, so both must be even. Factor pairs of $96$ with $x > y$:
$$(96,1),\ (48,2),\ (32,3),\ (24,4),\ (16,6),\ (12,8).$$
Discard $(96,1)$ and $(32,3)$ for mixed parity. The remaining four give
$$(m,n) = (25,23),\ (14,10),\ (11,5),\ (10,2),$$
and each checks: $625 - 529 = 196 - 100 = 121 - 25 = 100 - 4 = 96$.

The answer is $\boxed{\textbf{(B)}\ 4}$.

A difference of squares turns a quadratic Diophantine equation into a factoring problem, and the parity link between $m+n$ and $m-n$ is the filter that removes spurious factor pairs. Equivalently, both factors even means $\frac{x}{2} \cdot \frac{y}{2} = 24$, and $24$ has $4$ factor pairs with the first larger, which is a quick cross-check.

• Counting all six factor pairs of $96$ and answering $6$, forgetting that the pairs $1 \cdot 96$ and $3 \cdot 32$ have mixed parity and give non-integer $m$ and $n$.
• Counting both orders of each factor pair (answering $12$ or $8$) even though $m \ge n$ forces $m + n \ge m - n$.