A number theory problem by Shanthanu Rai

$\begin{aligned} \color{#3D99F6}{\text{Let}} \quad \quad n^2 + 192 & = (n+\color{#3D99F6}{m})^2 \quad \quad \quad \quad \color{#3D99F6}{\text{where } m \text{ is a positive integer.}} \\ & = n^2 + 2mn + m^2 \\ \Rightarrow 2mn + \color{#3D99F6}{m^2} & = 192 \quad \quad \quad \quad \quad \quad \space \space \color{#3D99F6}{\text{We note that } m^2 \text{ and hence } m \text{ must be even.}} \\ n & = \frac{192-m^2}{2m} \\ & = \frac{96}{m} - \frac{m}{2} \end{aligned}$

$\color{#3D99F6}{\text{The possible }m \text{ and } n \text{ are:}}$

$\begin{cases} m = 2 & \Rightarrow n = 48- 1 & = 47 \\ m = 4 & \Rightarrow n = 24 - 2 & = 22 \\ m = 6 & \Rightarrow n = 16 - 3 & = 13 \\ m = 8 & \Rightarrow n = 12 - 4 & = 8 \\ m = 12 & \Rightarrow n = 8 - 6 & = 2 \end{cases}$

$\color{#3D99F6}{\text{The sum of all possible } n} = 47+22+13+8+2 = \boxed{92}$