AMC 10 A

Notice that $m \ge 0$, and if $(m, n)$ is a solution, then so is $(m, -n)$. Assume $n \ge 0$. Squaring both sides of the given equation gives $n^2 - 49 = m^2$, so $n^2 - m^2 = (n - m)(n + m) = 49$. Because $n - m$ and $n + m$ are positive integers, either $n - m = n + m = 7$, or $n - m = 1$ and $n + m = 49$. The first case gives $(0, 7)$ as a solution, and the second case gives $(24, 25)$ as a solution. Because $n$ is squared in the given equation, the corresponding negative values for $n$ also give solutions: $(0, -7)$ and $(24, -25)$. There are 4 solutions in all.