$31 \cdot 16$ When $a \in Z$, the number of integer solutions for the equation $a x^{2}+6=0$ with respect to $x$ is
(A) 0 or 2.
(B) 0 or 1 or 2.
(C) 2.
(D) 1 or 2.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] When $a=-6$, the original equation becomes $x^{2}-1=0$, yielding $x= \pm 1$. When $a \neq-6$, and $a \in \mathbb{Z}$, the original equation has no integer roots. Therefore, the answer is $(A)$.