(10) How many integer solutions $(x, y)$ does the equation $\sqrt{x}+\sqrt{y}=\sqrt{1968}$ have?
(1968 Kyiv Mathematical Olympiad)
Given the equation can be transformed as
$$
\begin{aligned}
\sqrt{x}+\sqrt{y} & =\sqrt{1968}=4 \sqrt{123}=0+4 \sqrt{123} \\
& =\sqrt{123}+3 \sqrt{123}=2 \sqrt{123}+2 \sqrt{123} \\
& =3 \sqrt{123}+\sqrt{123}=4 \sqrt{123}+0 .
\end{aligned}
$$

The corresponding $(x, y)$ pairs are $(0,1968),(123,1107),(492,492),(1107,123)$,
$$
(1968,0) \text {. }
$$