\section*{Problem 1 - 021041}

Determine all pairs \((x ; y)\) of positive integers \(x\) and \(y\) for which \(\sqrt{x}+\sqrt{y}=\sqrt{50}\) holds!
}

Lower bound estimation: \(x>0\) and \(y>0\); upper bound estimation: \(\sqrt{x}=\sqrt{50}-\sqrt{y}<\sqrt{y}\), thus \(x<50\); similarly for \(y\).

\[
\sqrt{x}+\sqrt{y}=\sqrt{50} \quad x=50+y-2 \sqrt{50 y} \quad x=50+y-10 \sqrt{2 y}
\]

Therefore, \(\sqrt{2 y}\) must be an integer. This is achieved by \(y=2 \sqrt{a} 2\) with \(a \in \mathbb{N}\). Within the above bounds, this means:

\[
\begin{array}{ll}
y_{1}=2 \cdot 1^{2}=2 \Rightarrow x_{1}=32 ; & y_{2}=2 \cdot 2^{2}=8 \Rightarrow x_{2}=18 \\
y_{3}=2 \cdot 3^{2}=18 \Rightarrow x_{3}=8 ; & y_{4}=2 \cdot 4^{2}=32 \Rightarrow x_{4}=2
\end{array}
\]

The verification confirms all solutions.