2. (48th Slovenian Mathematical Olympiad) Find all integer solutions of the equation $\sqrt{x}+\sqrt{y}=\sqrt{2004}$.

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2. Squaring both sides of the equation $\sqrt{x}=\sqrt{2004}-\sqrt{y}$, and rearranging, we get
$2 \sqrt{2004 y}=2004+y-x$.
From this, we know that $2 \sqrt{2004 y}=4 \sqrt{501 y}$ is an integer. Therefore, $y=501 k^{2}$, where $k$ is a non-negative integer (since $501=$ $3 \times 167$, it is reasonable to set $y=501 k^{2}$).
Substituting the expression for $y$ into the original equation, we get
$\sqrt{x}=(2-k) \sqrt{501}$, from which we can deduce that $k$ is $0,1,2$.
Thus, the possible pairs $(x, y)$ are $(2004,0),(501,501),(0,2004)$.