Question

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

$3$

$4$

$6$

$9$

$12$

Write $m^2 - n^2 = (m + n)(m - n) = 96$

$m + n$ and $m - n$ have the same parity, so both must be even

The factors $m + n$ and $m - n$ have the same parity. Since their product is $96,$ they cannot both be odd, so both must be even.

The even factor pairs are $(48, 2),$ $(24, 4),$ $(16, 6),$ and $(12, 8),$ giving $(m, n) = (25, 23),$ $(14, 10),$ $(11, 5),$ and $(10, 2).$

So there are $4$ ordered pairs.

Thus, the correct answer is B.