3.19 Let $p, q$ be positive prime numbers. Does the equation $x^{2}+p^{2} x+q^{3}=0$ have rational roots? If not, provide a proof; if so, find them.
(Chinese High School Mathematics Competition, Jiangsu Province, Suzhou and Zhenjiang Cities, 1985)
[Solution] Since $p, q$ are positive primes, the original equation cannot have positive roots.
Also, since the coefficient of $x^{2}$ in the given equation is 1, if the equation has a rational root, it can only be an integer root, and it must be a divisor of $q^{3}$.

Therefore, if the original equation has a rational root, it can only take the following values: $-1, -q, -q^{2}$, $-q^{3}$ as roots.
If the two roots of the equation are $-q, -q^{2}$, then by Vieta's formulas,
$$
p^{2}=q+q^{2}=q(q+1) \geqslant 6.
$$

At this point, $p^{2}$ is an even number, and $p^{2} \neq 4$, so $p$ must be a composite number, which contradicts the fact that $p$ is a prime. Therefore, $-q, -q^{2}$ are not roots of the original equation.
If the equation has a rational root, it can only be -1 and $-q^{3}$, in which case we have
$$
p^{2}=1+q^{3},
$$

Clearly, $q$ cannot be an odd prime, otherwise $p^{2}$ would be an even number greater than 4, making $p$ a composite number. Thus, $q$ must be an even prime, i.e., $q=2$, which leads to $p=3$.

Therefore, the original equation $x^{2}+p^{2} x+q^{3}=0$ i.e., $x^{2}+9 x+8=0$ has rational roots: $x_{1}=-1, \quad x_{2}=-8$ if and only if $p=3, q=2$.