For what $n$ is $\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ an integer?
I tried setting $a=\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}$ , $b=\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}$ and playing with their squares/fourth powers, but couldn't find anything.
If you set the expression equal to $x$ and square the equation, you get $$x^2=25+2\sqrt{n}.$$ This is equivalent to $$n=\left(\frac{x^2-5^2}{2}\right)^2.$$ So just pick $x$ equal to an odd integer and then this above expression is $n.$
As Crostul points out, in order for the original expression to be well-defined, we need $n$ to be less than or equal to $\frac{625}{4}=156.25.$ And the fact that $n$ is a square means that it must be non-negative.