$31 \cdot 36$ satisfies $0 \leqslant x \leqslant y$ and $\sqrt{1992}=\sqrt{x}+\sqrt{y}$ for the number of different integer pairs $(x, y)$ is
(A) 0.
(B) 1.
(C) 2.
(D) 4.
(Chinese Zhejiang Province Junior High School Mathematics Competition, 1993)
[Solution] Since $1992=2^{2} \cdot 498$ (498 is not a perfect square). Then the equation becomes
$$
\sqrt{x}+\sqrt{y}=2 \sqrt{498} \text {, }
$$

The integer solutions of the equation are $(1992,0),(0,1992)$ or $(498,498)$.
Given the condition $y \geqslant x \geqslant 0$, only two sets are suitable.
Therefore, the answer is $(C)$.