Number Theory problem 4 by Dhaval Furia

$(n-m)(n+m)=105=1\times 3\times 5\times 7$ .

So there are $\boxed 4$ pairs of $(m, n)$ in all; $(4,11),(8,13),(16,19),(52,53)$

Similar solution as @Alak Bhattacharya 's just more details for the better understanding of some.

$\begin{aligned} m^2 + 105 & = n^2 \\ \implies n^2 - m^2 & = 105 \\ (n-m)(n+m) & = 105 \end{aligned}$

Since $m$ and $n$ are positive integers, this means that $n>m$ , and $n-m$ and $n+m$ are the factors of $105$ , with $n-m < n+m$ . For example, $105 = 1 \times 105$ , this means that $n-m = 1$ and $n+m = 105$ , $\implies n-m + n + m = 1 + 105 \implies 2n = 106 \implies n = 53 \implies m = 52$ . Then we have:

$\implies (n-m)(n+m) = \begin{cases} 1 \times 105 & \implies n = 53, \ m = 52 \\ 3 \times 35 & \implies n = 19, \ m = 16 \\ 5 \times 21 & \implies n = 13, \ m = 8 \\ 7 \times 15 & \implies n = 11, \ m = 4 \end{cases}$ . Therefore there are $\boxed 4$ solution pairs.