Roots inside the Roots

$\begin{aligned} \sqrt{50 + \sqrt{n}} + \sqrt{50 - \sqrt{n}} & = m \\ \left( \sqrt{50 + \sqrt{n}} + \sqrt{50 - \sqrt{n}} \right)^2 & = m^2 \\ 50 + \sqrt{n} + 50 - \sqrt{n} + 2\sqrt{(50+n)(50-n)} & = m^2 \\ 100 + 2\sqrt{2500-n} & = m^2 \\ \sqrt{2500-n} & = \dfrac{m^2-100}{2} \\ n & = 2500 - \left( \dfrac{m^2 -100}{2} \right)^2 \end{aligned}$

For $n$ to be positive real integer, then $\left( \dfrac{m^2 - 100}{2} \right)^2 < 2500$ and $m^2 \geq 100$ , or $m \geq 10$ and $m^2 <200$ .

Then, we have $m = \big\{ 10, 12, 14 \big\}$ .

For $m = 10,$ we have $n = 2500$ .

For $m = 12,$ we have $n = 2016$ .

For $m = 14,$ we have $n = 196$ .

Then, $n = ( 196, 2016, 2500 )$ . Hence, the answer is $4712$ .