5. 74 Find all real numbers $p, q$ such that the polynomial $x^{4}+p x^{2}+q$ has 4 real roots forming an arithmetic progression.

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[Solution] If and only if the polynomial $y^{2}+p y+q$ (where $y=x^{2}$) has two non-negative real roots, i.e., $p$ and $q$ satisfy $p^{2} \geqslant 4 q, q \geqslant 0, p \leqslant 0$, then the polynomial $x^{4}+p x^{2}+q$ has 4 real roots. If the polynomial has 4 real roots (denoted as $-x_{1},-x_{2}, x_{1}, x_{2}$, assuming $x_{1} \geqslant$ $x_{2} \geqslant 0$), then
$$
\left\{\begin{array}{l}
-2 x_{2}=-x_{1}+x_{2} \\
x_{1}^{2}+x_{2}^{2}=-p \\
x_{1}^{2} x_{2}^{2}=q
\end{array}\right.
$$

From (1) $x_{1}=3 x_{2}$
Substituting into (2) and (3) gives $\quad\left\{\begin{array}{l}10 x_{2}^{2}=-p \\ 9 x_{2}^{4}=q\end{array}\right.$
From (4) $\quad 100 x_{2}^{4}=p^{2}$
From (5) and (6) we get $\frac{p^{2}}{100}=\frac{q}{9}$,
i.e., $q=0.09 p^{2}$, they form an arithmetic sequence. Therefore, the required pairs $p, q$ should satisfy $p \leqslant 0, q=0.09 p^{2}$ (from the latter equation, we can obtain $p^{2} \geqslant 4 q$ and $q \geqslant 0$).