\section*{Problem 1 - 241041}

Determine all ordered pairs \((x ; y)\) of integers for which \(\sqrt{x}+\sqrt{y}=\sqrt{1985}\) holds!
}

Obviously, \((0,1985)\) and \((1985,0)\) are solutions in \(\mathbb{Z}\). These are also the only solutions, as we will now show.

Let \(x \neq 0\). Equivalent squaring yields: \(x+y+2 \sqrt{x y}=1985 \quad(*)\).

Now let \(y=x t^{2}\) with \(t \in \mathbb{Q}\). Substituting into \((*)\) gives:

\[
x+x t^{2}+2 x t=1985 \Longleftrightarrow x=\frac{1985}{(t+1)^{2}}
\]

Divisors of 1985 are \(\{1 ; 5 ; 397 ; 1985\}\). Only \((t+1)^{2}=1\) provides a solution for rational \(t\). However, we have already noted these solutions above.