Task B-4.4. Determine all values of the real parameter $p$ for which the equation $x^{4}-(3 p+2) x^{2}+p^{2}=0$ has four solutions that form four consecutive terms of an arithmetic sequence.
Solution. Since the equation is biquadratic, its solutions are $x_{1}, -x_{1}, x_{2} \text{ and } -x_{2}$ (symmetric with respect to the origin). Let $x_{1}$ and $x_{2}$ be positive real numbers and let $x_{1} < x_{2}$. Then the order of the terms of the sequence (from the smallest) is $-x_{2}, -x_{1}, x_{1}, x_{2}$.

It follows that $x_{1} = -x_{1} + d$ or $x_{1} = \frac{1}{2} d$.

Then $x_{2} = x_{1} + d = \frac{3}{2} d$, and $-x_{2} = -\frac{3}{2} d$.

The given sequence is of the form $-\frac{3}{2} d, -\frac{1}{2} d, \frac{1}{2} d, \frac{3}{2} d$.

If we let $t = x^{2}$ in the given equation, we get $t^{2} - (3p + 2)t + p^{2} = 0$.

From the previous considerations, we conclude that its solutions are $t_{1} = \frac{1}{4} d^{2}, t_{2} = \frac{9}{4} d^{2}$.

Using Vieta's formulas, the following equalities follow

$$
\begin{array}{rr}
\frac{1}{4} d^{2} + \frac{9}{4} d^{2} = 3p + 2, \quad \text{or} \quad \frac{5}{2} d^{2} = 3p + 2 \\
\frac{1}{4} d^{2} \cdot \frac{9}{4} d^{2} = p^{2}, \quad \text{or} \quad \frac{9}{16} d^{4} = p^{2}
\end{array}
$$

From the second equality,

$$
d^{2} = \frac{4}{3}|p|
$$

which, with the first equality, gives the equation

$$
\frac{10}{3}|p| = 3p + 2
$$

Then

$$
\begin{gathered}
\text{for } p \geq 0 \text{, } 10p = 9p + 6 \text{, and } p = 6 \\
\text{for } p < 0 \text{, } -10p = 9p + 6 \text{ and } p = -\frac{6}{19}
\end{gathered}
$$