21.2. (PRB, 61). Find all pairs of real numbers $p, q$, for which the polynomial $x^{4}+p x^{2}+q$ has 4 real roots forming an arithmetic progression.
21.2. The polynomial $x^{4}+p x^{2}+q$ has 4 real roots if and only if the polynomial $y^{2}+p y+q$ (relative to

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-317.jpg?height=54&width=978&top_left_y=1674&top_left_x=63)
satisfies the conditions $p^{2} \geqslant 4 q, q \geqslant 0, p \leqslant 0$. If the original polynomial has 4 real roots (namely: $-x_{1},-x_{2}, x_{2}, x_{1}$, where without loss of generality we assume that $x_{1} \geqslant x_{2} \geqslant 0$), then they form an arithmetic progression if and only if the system

$$
-2 x_{2}=-x_{1}+x_{2}, x_{1}^{2}+x_{2}^{2}=-p, x_{1}^{2} x_{2}^{2}=q
$$

(is satisfied (see Vieta's theorem and its converse)), i.e., when $q=0.09 p^{2}$. Thus, all the required pairs of numbers $p, q$ are described by the conditions

$$
p \leqslant 0, q=0.09 p^{2}
$$

(the inequalities $p^{2} \geqslant 4 q$ and $q \geqslant 0$ follow from the last equality).