1. Find all pairs of integers $x$ and $y$ for which

$$
\sqrt{x \sqrt{5}}-\sqrt{y \sqrt{5}}=\sqrt{6 \sqrt{5}-10} .
$$
1. From the form of the given equation, it immediately follows that $x>y \geq 0$ (since $6 \sqrt{5}-10>0$). For such $x, y$, we can square both (positive) sides of the equation and perform further equivalent transformations:

$$
\begin{aligned}
x \sqrt{5}-2 \sqrt{5 x y}+y \sqrt{5} & =6 \sqrt{5}-10 \\
x-2 \sqrt{x y}+y & =6-2 \sqrt{5} \\
x+y-6 & =2(\sqrt{x y}-\sqrt{5}) .
\end{aligned}
$$

By squaring and further manipulation, we get that for the sought integers $x, y$, the following must hold:

$$
\begin{aligned}
(x+y-6)^{2} & =4(x y-2 \sqrt{5 x y}+5) \\
8 \sqrt{5 x y} & =4(x y+5)-(x+y-6)^{2}
\end{aligned}
$$

From the last equation, it follows that the value $\sqrt{5 x y}$ is rational, and thus an integer, ${ }^{1}$ so $5 x y$ is a square of a non-negative integer, which is clearly divisible by five. ${ }^{2}$ Therefore, $5 x y = (5 k)^{2}$ or $x y = 5 k^{2}$, where $k$ is a non-negative integer. At this point, it is advantageous to substitute not into equation (2), but directly into equation (1). We then get the equation

$$
x+y-6=2\left(\sqrt{5 k^{2}}-\sqrt{5}\right) \quad \text { or } \quad x+y-6=2(k-1) \sqrt{5} \text {, }
$$

from which, due to the irrationality of the number $\sqrt{5}$, it follows that to satisfy equation (1), both equations $k=1$ and $x+y-6=0$ must hold. From the system of equations

$$
x y=5 k=5, \quad x+y=6
$$

we easily find that $\{x, y\}=\{5,1\}$, i.e., $x=5$ and $y=1$, since $x>y$ according to the initial consideration.

The sought pair $(x, y)$ is unique, and it is $(x, y)=(5,1)$.

For a complete solution, award 6 points, with 1 point for the first and 2 points for the second of the two squarings. Known facts about square roots and squares mentioned in both footnotes can be used by solvers directly, without stating (or even proving) them as rules (i.e., in general form). If, in an otherwise complete solution, the pair $(x, y)=(1,5)$ is not excluded or the condition $x>y$ is not mentioned, and there is no check in the subsequent manipulation after squaring (which is not needed in the presented solution), award only 5 points.