$31 \cdot 35$ satisfies $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ for the number of different integer pairs $(x, y)$ is
(A) 0.
(B) 1.
(C) 3.
(D) 4.
(E) 7.
(35th American High School Mathematics Examination, 1984)
[Solution 1] Given $1984=2^{6} \cdot 31$. Also, when $04$, $y<x$.
Therefore, the original equation has only three sets of solutions that satisfy the conditions.
Thus, the answer is $(C)$.
[Solution 2] Given $1984=2^{6} \cdot 31$, then
$$
\begin{aligned}
\sqrt{1984} & =8 \sqrt{31}=\sqrt{31}+7 \sqrt{31}=2 \sqrt{31}+6 \sqrt{31} \\
& =3 \sqrt{31}+5 \sqrt{31} . \\
2 \sqrt{31}= & \sqrt{124}, \quad 3 \sqrt{31}=\sqrt{279}, \quad 5 \sqrt{31}=\sqrt{775}, \\
6 \sqrt{31}= & \sqrt{1116}, \quad 7 \sqrt{31}=\sqrt{1519} .
\end{aligned}
$$

Therefore, the equation has three sets of solutions.
Thus, the answer is $(C)$.