$31 \cdot 32$ The number of integer solutions (pairs) of the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{1987}$ is
(A) 0.
(B) 3.
(C) 5.
(D) None of the above conclusions is correct.
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1987)
［Solution］The original equation can be transformed into

or
$$
\begin{array}{c}
x y-1987 x-1987 y=0, \\
(x-1987)(y-1987)=1987^{2} .
\end{array}
$$
$\because 1987$ is a prime number,
$\therefore$ the above equation is equivalent to the following 6 systems of equations:
(1) $\left\{\begin{array}{l}x-1987=1, \\ y-1987=1987^{2}\end{array}\right.$
(2) $\left\{\begin{array}{l}x-1987=1987^{2} \\ y-1987=1 ;\end{array}\right.$
(3) $\left\{\begin{array}{l}x-1987=-1, \\ y-1987=-1987^{2}\end{array}\right.$;
(4) $\left\{\begin{array}{l}x-1987=-1987^{2} \\ y-1987=-1\end{array}\right.$,
(5) $\left\{\begin{array}{l}x-1987=1987, \\ y-1987=1987 ;\end{array}\right.$
(6) $\left\{\begin{array}{l}x-1987=-1987 \\ y-1987=-1987 .\end{array}\right.$

Solving these, we get:
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = 1 9 8 8 , } \\
{ y = 1 9 8 7 \cdot 1 9 8 8 ; }
\end{array} \quad \left\{\begin{array}{l}
x=1987 \cdot 1988, \\
y=1988 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x = 1 9 8 6 , } \\
{ y = - 1 9 8 7 \cdot 1 9 8 6 ; }
\end{array} \quad \left\{\begin{array}{l}
x=-1987 \cdot 1986, \\
y=1986 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ x = 2 \cdot 1 9 8 7 , } \\
{ y = 2 \cdot 1 9 8 7 ; }
\end{array} \quad \left\{\begin{array}{l}
x=0, \\
y=0 .
\end{array}\right.\right. \text { (discard) } \\
\end{array}
$$
$\therefore$ The equation has 5 integer solutions (sets).
Hence, the answer is $(C)$.