2007 AMC 10A Problems/Problem 23

Problem

How many ordered pairs $(m,n)$ of positive integers, with $m \ge n$ , have the property that their squares differ by $96$ ?

For every two factors $xy = 96$ , we have $m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}$ . Since $m \ge n > 0$ , $x+y \ge x-y > 0$ , from which it follows that the number of ordered pairs $(m,n)$ is given by the number of ordered pairs $(x,y): xy=96, x > y > 0$ . There are $(5+1)(1+1) = 12$ factors of $96$ , which give us six pairs $(x,y)$ . However, since $m,n$ are positive integers, we also need that $\frac{x+y}{2}, \frac{x-y}{2}$ are positive integers, so $x$ and $y$ must have the same parity. Thus we exclude the factors $(x,y) = (1,96)(3,32)$ , and we are left with four pairs $\mathrm{(B)}$ .