$31 \cdot 31$ The number of integer solutions to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{7}$ is
(A) 1 .
(B) 3 .
(C) 5 .
(D) 6 .
(Changchun, Jilin Province, China Junior High School Mathematics Competition, 1990)
［Solution］The given equation can be transformed into $x y-7 x-7 y=0$,
$$
\begin{array}{l}
\because \quad x y-7 x-7 y=(x-7)(y-7)-49, \\
\therefore x y-7 x-7 y=0 \quad \text{ becomes } \\
(x-7)(y-7)=49=7 \cdot 7=1 \cdot 49. \\
\therefore \quad\left\{\begin{array}{l}
x-7=1, \quad-1, \quad 7, \quad-7, \quad 49,-49, \\
y-7=49,-49, \quad 7, \quad-7, \quad 1, \quad-1 .
\end{array}\right. \\
\end{array}
$$

Among them, $x=0, y=0$ are not valid, so they are discarded, leaving five sets of integer solutions. Therefore, the answer is $(C)$.