$31 \cdot 25$ Let $a$ be any given positive integer, then the equation for $x$ and $y$
$$
x^{2}-y^{2}=a^{3}
$$
(A) has no positive integer solutions.
(B) has only positive integer solutions.
(C) has positive integer solutions only when $a$ is even.
(D) always has integer solutions.
(Jiangsu Province, China Junior High School Mathematics Competition, 1987)
[Solution] From $x^{2}-y^{2}=a^{3}$, we have $(x-y)(x+y)=a^{3}$.
If we let $\left\{\begin{array}{l}x+y=a^{2}, \\ x-y=a,\end{array}\right.$ solving gives $\left\{\begin{array}{l}x=\frac{a^{2}+a}{2}, \\ y=\frac{a^{2}-a}{2} .\end{array}\right.$
Since $a^{2} \pm a$ are both even, then $x, y$ are both integers, meaning the equation has integer solutions.
When $a=3$, we have $x^{2}-y^{2}=27$, which gives $x=6, y=3$ as one solution, thus ruling out $(A),(C)$.

Letting $a=1$, we have $x^{2}-y^{2}=1$, which gives $x=-1, y=0$ as one solution, thus ruling out $(B)$.
Therefore, the answer is $(D)$.