$31 \cdot 39$ The number of integer solutions $(x, y)$ for the equation $\sqrt{x}+\sqrt{y}=\sqrt{336}$ is
(A) 2.
(B) 3.
(C) 4.
(D) 5.
(China Beijing Junior High School Mathematics Competition, 1993)
[Solution] According to the problem, we have $\sqrt{x}+\sqrt{y}=4 \sqrt{21}$, thus the integer solutions $(x, y)$ that satisfy the equation are
\begin{tabular}{c|c|c|c|c|c}
\hline$x$ & 0 & 21 & $2^{2} \cdot 21$ & $3^{2} \cdot 21$ & $4^{2} \cdot 21$ \\
\hline$y$ & $4^{2} \cdot 21$ & $3^{2} \cdot 21$ & $2^{2} \cdot 21$ & 21 & 0 \\
\hline
\end{tabular}

There are 5 solutions in total.
Therefore, the answer is $(D)$.