$30 \cdot 21$ Let $p$ be a prime number. If the equation $x^{2}-p x-580 p=0$ has two integer roots, then
(A) $0<p<10$.
(B) $10<p<20$.
(C) $20<p<30$.
(D) $30<p<40$.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1987)
[Solution] Suppose the roots of the equation are
$$
x_{1,2}=\frac{p \pm \sqrt{p^{2}+4 \cdot 580 p}}{2}=\frac{p \pm \sqrt{p(p+4 \cdot 580)}}{2} \text {, }
$$

Since they are integers, $p(p+4 \cdot 580)$ must be a perfect square. Also, $p$ is a prime number.
Then $p+4 \cdot 580=p \cdot m^{2} \quad(m \in \mathbb{N})$,
which means $\left(m^{2}-1\right) p=4 \cdot 580=4^{2} \cdot 5 \cdot 29$,
and since $x_{1}+x_{2}=p, x_{1} x_{2}=-580 p$, we know that one of $x_{1}, x_{2}$ must be even, and $p$ must be odd. Since $p$ is a prime number, we have
$$
p=5 \text { or } 29 .
$$

If $p=5$, then $m^{2}-1=4 \cdot 4 \cdot 29$, which is not valid;
If $p=29$, then we have $m^{2}-1=4 \cdot 4 \cdot 5=80$, giving $m=9$.
Therefore, the answer is $(C)$.