$12 \cdot 29$ Find the pairs of positive integers $(x, y)$ that satisfy the following conditions:
(1) $x \leqslant y$;
(2) $\sqrt{x}+\sqrt{y}=\sqrt{1992}$.
(Australian Mathematics Competition, 1991)
[Solution] Since $\sqrt{1992}=2 \sqrt{498}$,
it follows that $(x, y)=(498,498)$ is clearly a positive integer solution to the equation.
If $(x, y)$ is a positive integer solution to the equation, then by the condition

we have
$$
\begin{array}{c}
x \leqslant y \\
x \leqslant 498 .
\end{array}
$$

However, from
we have
$$
\begin{array}{c}
\sqrt{x}+\sqrt{y}=\sqrt{1992} \\
y=1992+x-2 \sqrt{1992 x},
\end{array}
$$

it follows that $2 \sqrt{1992 x}=4 \sqrt{498 x}$ is an integer, because
$$
\begin{array}{c}
498=2 \cdot 3 \cdot 83 \\
x \geqslant 498 . \\
x=498 .
\end{array}
$$

is square-free, so
thus we can only have
which means the equation has only one unique positive integer solution $(498,498)$.