1. Determine all real numbers $m$ for which the equation

$$
x^{4}-(3 m+2) x^{2}+m^{2}=0
$$

has 4 real roots that are consecutive terms of an arithmetic progression.
Solution. Let $a_{1}, a_{2}, a_{3}, a_{4}$ be consecutive terms of an arithmetic progression, roots of the equation

$$
x^{4}-(3 m+2) x^{2}+m^{2}=0
$$

If $2 b$ is the difference of the progression and $s=\frac{a_{2}+a_{3}}{2}$ then $a_{1}=s-3 b, a_{2}=s-b$, $a_{3}=s+b, a_{4}=s+3 b$.

Using Vieta's formulas, from $a_{1}+a_{2}+a_{3}+a_{4}=0$ it follows that $s=0$ and $a_{1}=-3 b, a_{2}=-b, a_{3}=b, a_{4}=3 b$.

Since $a_{1}, a_{2}, a_{3}, a_{4}$ are roots of the initial equation, it follows that

$$
(x+3 b)(x+b)(x-b)(x-3 b)=x^{4}-(3 m+2) x^{2}+m^{2}
$$

That is,

$$
x^{4}-10 b^{2} x^{2}+9 b^{4}=x^{4}-(3 m+2) x^{2}+m^{2}
$$

from which the system of equations is obtained

$$
\left\{\begin{array}{c}
3 m+2=10 b^{2} \\
m^{2}=9 b^{4}
\end{array}\right.
$$

Solving the system yields $m=6$ or $m=-\frac{6}{19}$.