The number of distinct pairs of [integers](https://artofproblemsolving.com/wiki/index.php/Integers) $(x, y)$ such that $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ } 7$
We can simplify $\sqrt{1984}$ to $8\sqrt{31}$. Therefore, the only solutions are $a\sqrt{31}+b\sqrt{31}$ such that $a+b=8$ and $0<a<b$. The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$. Each of these gives distinct pairs of $(x, y)$, so there are $3$ pairs, $\boxed{\text{C}}$.