Pairs satisfying difference of squares

Rearrange as $n^2 - m^2 = 105 \implies (n-m)(n+m) = 105$.

Since $105 = 3 \times 5 \times 7$, the factor pairs $(a, b)$ of 105 where $a < b$ are:

1. $(1, 105)$
2. $(3, 35)$
3. $(5, 21)$
4. $(7, 15)$

For each pair $(a, b)$, $n - m = a$ and $n + m = b \implies n = \frac{a+b}{2}, m = \frac{b-a}{2}$. Since both $a$ and $b$ are odd, $\frac{a+b}{2}$ and $\frac{b-a}{2}$ are always integers.

Thus, there are 4 positive integer pairs $(m, n)$.