Problem. Find all integral $x$ and $y$ satisfying $$2\sqrt{6} + 5\sqrt{10} = \sqrt{x} + \sqrt{y}.$$
Squaring both sides, we get $$274 - x - y = 2\sqrt{xy} - 20\sqrt{60}.$$ Squaring again, we get $$-160\sqrt{15xy} \in \mathbb{Z},$$ which gives us $xy = 15k^2$. We know $2\sqrt{6} + 5\sqrt{10} < 2(2.5) + 5(3.2) = 21$. Hence $x, y < 441$. There are still quite a lot of cases to check. Is there a better way to do this?
Edit: I would like to find all possible solutions, and prove that those constitute all of them. Clearly $(24,250)$ and $(250,24)$ are solutions, but can we prove that there are no other?
(1) Keeping in mind the linear independance of irrational quadratics we have $$x=6X^2, y=10Y^2\Rightarrow 2\sqrt6+5\sqrt{10}=X\sqrt6+Y\sqrt{10}\Rightarrow X=2,Y=5$$ Hence $$(x,y)=(24,250)$$
(2) Applying conjugates in quadratic fields one has the system $$\begin{cases}2\sqrt{6} + 5\sqrt{10} = \sqrt{x} + \sqrt{y}\\2\sqrt{6} - 5\sqrt{10} = \sqrt{x} - \sqrt{y}\end{cases}$$ From which it is immediate $$4\sqrt6=2\sqrt x\Rightarrow x=24\\10\sqrt{10}=2\sqrt y\Rightarrow y=250$$
The only solution is $$(x,y)=(24,250)$$