$31 \cdot 22$ solutions of the equation $2 x^{2}+x y-y^{2}=14$ in integers are
(A) 2 pairs.
(B) 4 pairs.
(C) 8 pairs.
(D) 16 pairs.
(Shenyang City, Liaoning Province, "Yucai Cup" Junior High School Mathematics Competition, 1991)
[Solution] Given $2 x^{2}+x y-y^{2}=(2 x-y)(x+y)$, and

then $\left\{\begin{array}{l}2 x-y=1,14,-1,-14,2,7,-2,-7, \\ x+y=14,1,-14,-1,7,2,-7,-2 .\end{array}\right.$
It yields 8 sets of solutions.
Therefore, the correct choice is $(C)$.