7. (1985 High School Mathematics Competition Question in Suzhou and Zhenjiang Cities, Jiangsu Province, China) 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.

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7. Since $p, q$ are positive primes, the original equation cannot have positive roots.
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Also, since the coefficient of $x^{2}$ in the known 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 numbers: $-1, -q, -q^{2}, -q^{3}$.
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 time, $p^{2}$ is even, and $p^{2} \neq 4$, so $p$ must be composite, which contradicts $p$ being 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 $p$ composite, so $q$ must be an even prime, i.e., $q=2$, thus $p=3$.

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