2007 AMC 10A Problem 23

1. Factor the difference of squares.
$$m^2 - n^2 = (m + n)(m - n) = 96.$$
So $m + n$ and $m - n$ are two whole numbers that multiply to $96$, with $m + n$ the bigger one.
2. They are both even or both odd. The two numbers differ by $2n$, an even number. For example, if $m = 7$ and $n = 3$, they are $10$ and $4$. Both odd is impossible, because two odd numbers multiply to an odd number, not $96$. So both are even.
3. List the factor pairs of $96$ (bigger first):
$$96 \times 1, \quad 48 \times 2, \quad 32 \times 3, \quad 24 \times 4, \quad 16 \times 6, \quad 12 \times 8.$$
Cross out $96 \times 1$ and $32 \times 3$, which mix odd and even. Four pairs are left.
4. Turn each pair into $m$ and $n$. Adding the two factors gives $2m$; subtracting gives $2n$.

There are $4$ pairs.

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

A difference of two squares always factors as (sum) times (difference). That turns the problem into listing factor pairs, which is easy. The one filter to remember is that the sum and the difference of two whole numbers are both even or both odd.

Since both factors are even, write them as $2p$ and $2q$. Then $4pq = 96$, so $pq = 24$. Pairs with $p > q$: $24 \times 1$, $12 \times 2$, $8 \times 3$, $6 \times 4$. That is $4$ pairs, and each one gives $m = p + q$ and $n = p - q$.