$\text{ Number of positive integers }n\text{ for which }{n}^2+96\text{ is a perfect square is}$

$\text{ Let }m\text{ be a positive integer for which }{n}^2+96=m^2$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{m}}^2-{\mathrm{n}}^2=96\Rightarrow (\mathrm{m}+\mathrm{n})(\mathrm{m}-\mathrm{n})=96\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{m}+\mathrm{n})\left\{\right(\mathrm{m}+\mathrm{n})-2\mathrm{n}\}=96\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{m}+\mathrm{n}\text{ and }\mathrm{m}-\mathrm{n}\text{ must be both even }\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}96=2\times 48\text{ or }4\times 24\text{ or }6\times 16\text{ or }8\times 12\end{array}$

Number of solution = 4