Find two positive integers $x$ and $y$ such that:

$$
\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}-\sqrt{3}}=\frac{\sqrt{x}+\sqrt{y}}{2}
$$
Solution

$$
\begin{aligned}
\frac{1-\sqrt{2}+\sqrt{3}}{1+\sqrt{2}-\sqrt{3}} \cdot \frac{\sqrt{2}}{\sqrt{2}} & =\frac{-2+\sqrt{2}+\sqrt{6}}{2+\sqrt{2}-\sqrt{6}} \\
& =\frac{-2+\sqrt{2}+\sqrt{6}}{2+\sqrt{2}-\sqrt{6}} \cdot \frac{2 \sqrt{2}+2+2 \sqrt{3}}{2 \sqrt{2}+2+2 \sqrt{3}} \\
& =\frac{-\sqrt{2}+(1+\sqrt{3})}{(\sqrt{2}+1)-\sqrt{3}} \cdot \frac{\sqrt{2}+(1+\sqrt{3})}{(\sqrt{2}+1)+\sqrt{3}} \\
& =\frac{(1+\sqrt{3})^{2}-2}{(\sqrt{2}+1)^{2}-3} \\
& =\frac{\sqrt{2}+\sqrt{6}}{2}
\end{aligned}
$$

Therefore, $x=2$ and $y=6$ satisfy the statement.

Observation: It can be shown that these are the only integer solutions. Indeed, it was shown that $\sqrt{x}+\sqrt{y}=\sqrt{2}+\sqrt{6}$. Thus,

$$
\begin{aligned}
(\sqrt{x}+\sqrt{y})^{2} & =(\sqrt{2}+\sqrt{6})^{2} \\
x+y+2 \sqrt{x y} & =8+4 \sqrt{3} \\
2 \sqrt{x y}-4 \sqrt{3} & =8-x-y
\end{aligned}
$$

If $2 \sqrt{x y}-4 \sqrt{3} \neq 0$, it follows that:

$$
\begin{aligned}
2 \sqrt{x y}+4 \sqrt{3} & =\frac{(2 \sqrt{x y})^{2}-(4 \sqrt{3})^{2}}{2 \sqrt{x y}-4 \sqrt{3}} \\
& =\frac{4 x y-48}{8-x-y}
\end{aligned}
$$

Consequently, subtracting this result from the last equality found, we have

$$
\begin{aligned}
\left(\frac{4 x y-48}{8-x-y}\right)-(8-x-y) & =(2 \sqrt{x y}+4 \sqrt{3})-(2 \sqrt{x y}-4 \sqrt{3}) \\
& =8 \sqrt{3}
\end{aligned}
$$

This is a contradiction, since $\sqrt{3} \notin \mathbb{Q}$ and $\left(\frac{4 x y-48}{8-x-y}\right)-(8-x-y) \in \mathbb{Q}$. Therefore, $2 \sqrt{x y}-4 \sqrt{3}=0$ and $\sqrt{x y}=\sqrt{12}$. It remains to solve the system:

$$
\left\{\begin{aligned}
\sqrt{x}+\sqrt{y} & =\sqrt{2}+\sqrt{6} \\
\sqrt{x y} & =\sqrt{12}
\end{aligned}\right.
$$

Squaring the first equation and using that $\sqrt{x y}=\sqrt{12}$, we obtain that $x+y=8$ and $12=x y=x(8-x)=8 x-x^{2}$. The roots of $12=8 x-x^{2}$ are $x=2$ and $x=6$. Therefore, $(x, y)=(2,6)$ or (6,2).