$30 \cdot 31$ makes $n^{5}-5 n^{3}+4 n+7$ a perfect square for natural numbers $n$ with the value
(A) There is exactly one.
(B) There are finitely many, but more than one.
(C) There are infinitely many.
(D) None exist.
(China Fujian Province Fuzhou City High School Mathematics Competition, 1990)
$$
\begin{aligned}
\text { [Solution] From } & n^{5}-5 n^{3}+4 n+7 \\
= & n^{3}\left(n^{2}-1\right)-4 n\left(n^{2}-1\right)+7 \\
= & n\left(n^{2}-1\right)\left(n^{2}-4\right)+7 \\
= & (n-2)(n-1) n(n+1)(n+2)+7, \\
\because \quad 10 \text { | } & (n-2)(n-1) n(n+1)(n+2),
\end{aligned}
$$
$\therefore$ for any natural number $n$, the units digit of $n^{5}-5 n^{3}+4 n+7$ is 7, and it is not a perfect square.
Therefore, the answer is $(D)$.
$$