1. Find all integer solutions of the equation $\sqrt{x}+\sqrt{y}=\sqrt{2004}$.
III/1. Let's square the equation $\sqrt{x}=\sqrt{2004}-\sqrt{y}$ and express $2 \sqrt{y \cdot 2004}=2004+y-x$. From this, it follows that $2 \sqrt{y \cdot 2004}=4 \sqrt{y \cdot 501}$ must be an integer, so $y=501 \cdot k^{2}$, where $k$ is a non-negative integer. If we now substitute $y$ into the original equation, we get $\sqrt{x}=(2-k) \sqrt{501}$, from which it follows that $k$ can be 0, 1, or 2. All possible pairs $(x, y)$ are thus $(2004,0),(501,501)$ and $(0,2004)$.