$12 \cdot 56$ Equation $\sqrt{x}+\sqrt{y}=\sqrt{1968}$ has how many integer solutions $(x, y)$ ?
(Kiev Mathematical Olympiad, 1968)
[Solution 1]. The given equation can be transformed into
$$
\sqrt{x}=\sqrt{1968}-\sqrt{y},
$$

Squaring both sides of the equation, we get
$$
x=y+1968-8 \sqrt{123 y} \text {. }
$$

Thus, the number $123 y$ must be a perfect square. Let

Similarly,
$$
\begin{array}{ll}
123 v^{2}=y, & v \in \mathbb{Z} . \\
123 u^{2}=x, & u \in \mathbb{Z} .
\end{array}
$$

From this, we can derive the original equation as
$$
|u|+|v|=4 .
$$

The non-negative integer solutions to this equation are
$$
(0,4),(1,3),(2,2),(3,1),(4,0) \text {. }
$$

Thus, the integer solutions $(x, y)$ are
$$
(0,1968),(123,1107),(492,492),(1107,123),(1968,0) \text {. }
$$
[Solution 2] The given equation can be transformed into
$$
\begin{aligned}
\sqrt{x}+\sqrt{y} & =\sqrt{1968}=4 \sqrt{123}=0+4 \sqrt{123} \\
& =\sqrt{123}+3 \sqrt{123}=2 \sqrt{123}+2 \sqrt{123} \\
& =3 \sqrt{123}+\sqrt{123}=4 \sqrt{123}+0 .
\end{aligned}
$$

The corresponding $(x, y)$ are $(0,1968),(123,1107),(492,492),(1107,123)$, $(1968,0)$.