CAT 2019 Shift 2 Paper

Shortcut:
Number of pairs
$105=3\times 5\times 7$
Number of factors $=2\times 2\times 2=8$
Hence, required number of pairs $=8∕2=4$
Detailed Explanation:
${\mathit{m}}^2+105={\mathit{n}}^2$
$\Rightarrow {\mathit{n}}^2-{\mathit{m}}^2=105$
$\Rightarrow \left(\mathit{n}-\mathit{m}\right)\left(\mathit{n}+\mathit{m}\right)=105$
Since $\mathit{m}$ and $\mathit{n}$ are positive integers,
Splitting 105 in two factors, we get
For $\left(\mathit{n}-\mathit{m}\right)=1$ and $\left(\mathit{n}+\mathit{m}\right)=105,\left(\mathit{m},\mathit{n}\right)=\left(52,53\right)$
$\Rightarrow \left(\mathit{n}-\mathit{m}\right)\left(\mathit{n}+\mathit{m}\right)=3\times 35$
For $\left(\mathit{n}-\mathit{m}\right)=3$ and $\left(\mathit{n}+\mathit{m}\right)=35,\left(\mathit{m},\mathit{n}\right)=\left(16,19\right)$
$\Rightarrow \left(\mathit{n}-\mathit{m}\right)\left(\mathit{n}+\mathit{m}\right)=5\times 21$
For $\left(\mathit{n}-\mathit{m}\right)=5$ and $\left(\mathit{n}+\mathit{m}\right)=21,\left(\mathit{m},\mathit{n}\right)=\left(8,13\right)$
$\Rightarrow \left(\mathit{n}-\mathit{m}\right)\left(\mathit{n}+\mathit{m}\right)=7\times 21$
For $\left(\mathit{n}-\mathit{m}\right)=7$ and $\left(\mathit{n}+\mathit{m}\right)=21,\left(\mathit{m},\mathit{n}\right)=\left(4,11\right)$
Hence there are four pairs.