Let's determine

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

the value of $m$ such that the roots of the equation form an arithmetic progression.
The roots of the given equations:

$$
\begin{aligned}
& x_{1}=+\sqrt{\frac{3 m+2+\sqrt{5 m^{2}+12 m+4}}{2}} \\
& x_{2}=+\sqrt{\frac{3 m+2-\sqrt{5 m^{2}+12 m+4}}{2}} \\
& x_{3}=-\sqrt{\frac{3 m+2+\sqrt{5 m^{2}+12 m+4}}{2}} \\
& x_{4}=-\sqrt{\frac{3 m+2-\sqrt{5 m^{2}+12 m+4}}{2}}
\end{aligned}
$$

If these roots are ordered by magnitude, then:

$$
x_{1}>x_{2}>x_{4}>x_{3}
$$

The arithmetic progression of the consecutive terms is thus:

$$
x_{1}, x_{2}, x_{4}, x_{3}
$$

so

$$
x_{1}-x_{2}=x_{2}-x_{4}
$$

and since

$$
x_{4}=-x_{2}
$$

therefore

$$
x_{1}=3 x_{2}
$$

or

$$
x_{1}^{2}=9 x_{2}^{2}
$$

Substituting the given roots into this equation:

$$
5 \sqrt{5 m^{2}+12 m+4}=12 m+8
$$

or

$$
19 m^{2}-108 m-36=0
$$

from which

$$
m_{1}=6, m_{2}=-\frac{6}{19}
$$

Substituting these values, the progressions are:

$$
3 \sqrt{2}, \sqrt{2},-\sqrt{2},-3 \sqrt{2}
$$

and

$$
3 \sqrt{\frac{2}{19}}, \sqrt{\frac{2}{19}},-\sqrt{\frac{2}{19}},-3 \sqrt{\frac{2}{19}}
$$

(Mihály Devecis.)

The problem was also solved by: Baumgarten Sz., Bella I., Bojedain F., Détshy K., Erdős A., Fekete J., Freibauer E., Groffits G., Kertész L., Koós A., Laczkó E., Lövi I., Magyar S., Makk I., Manheim E., Perl Gy., Posgay B., Prohászka J., Raab L., Spitzer Ö., Szabó I., Szabó K., Weisz H., Weisz J.