How many [ordered pairs](https://artofproblemsolving.com/wiki/index.php/Ordered_pair) $(m,n)$ of positive [integers](https://artofproblemsolving.com/wiki/index.php/Integer), with $m \ge n$, have the property that their squares differ by $96$? 
$\text{(A)}\ 3 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$
\[m^2 - n^2 = (m+n)(m-n) = 96 = 2^{5} \cdot 3\]
For every two factors $xy = 96$, we have $m+n=x, m-n=y \Longrightarrow m = \frac{x+y}{2}, n = \frac{x-y}{2}$. 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](https://artofproblemsolving.com/wiki/index.php/Parity). Thus we exclude the factors $(x,y) = (1,96)(3,32)$, and we are left with $4$ pairs $\mathrm{(B)}$.