$31 \cdot 20$ Let the equation $x^{2}-y^{2}=1988$ have $n$ sets of integer solutions. Then $n$ equals
(A) 4.
(B) 6.
(C) 8.
(D) 12.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1988)
【Solution】Given $(x-y)(x+y)=2^{2} \cdot 7 \cdot 71$, when $x, y$ are both odd or both even, $x+y, x-y$ are both even; when $x, y$ are one odd and one even, $x+y, \quad x-y$ are both odd.
Since 1988 is even, $x+y$ and $x-y$ can only both be even. Thus $\left\{\begin{array}{l}x+y= \pm 2 \cdot 7, \\ x-y= \pm 2 \cdot 71 ;\end{array} \quad\left\{\begin{array}{l}x+y= \pm 2 \cdot 71 \\ x-y= \pm 2 \cdot 7 ;\end{array}\right.\right.$ $\left\{\begin{array}{l}x+y= \pm 2, \\ x-y= \pm 2 \cdot 7 \cdot 71 ;\end{array} \quad\left\{\begin{array}{l}x+y= \pm 2 \cdot 7 \cdot 71 \\ x-y= \pm 2 .\end{array}\right.\right.$

Here, $x+y, x-y$ both take + or both take - .
Thus, the original equation has 8 sets of integer solutions.
Therefore, the answer is $(C)$.