polynomials - Find parameter so that the equation has roots in arithmetic progression

Find the parameter $m$ so that the equation

$$x^8 - mx^4 + m^4 = 0$$
has four distinct real roots in arithmetic progression.

I tried the substitution $x^4 = y$, so the equation becomes

$$y^2 -my + m^4 = 0$$

I don't know what condition should I put now or if this is a correct approach.

I have also tried to use Viete's by the notation

$$2x_2=x_1+x_2, 2x_3=x_2 +x_4$$ but I didn't get much out of it.

Answer

Zero can't be a root, else $m=0$, in which case all the roots would be zero.

If $r$ is any root, so is $ri$, hence there must be exactly $4$ real roots, and $4$ pure imaginary roots.

Also, if $r$ is a root, so is $-r$, hence the real roots sum to zero.

Ordering the real roots in ascending order, let $d > 0$ be the common difference.

Then the four real roots are

$$-\frac{3}{2}d,\;-\frac{1}{2}d,\;\frac{1}{2}d,\;\frac{3}{2}d$$

and the other four roots are

$$-\frac{3}{2}di,\;-\frac{1}{2}di,\;\frac{1}{2}di,\;\frac{3}{2}di$$

Since the $4$-th powers of the roots satisfy the quadratic

$$y^2 - my + m^4 = 0$$

Vieta's formulas yields the equations

$$\left(\frac{1}{2}d\right)^4+\left(\frac{3}{2}d\right)^4 = m$$

$$\left(\frac{1}{2}d\right)^4\left(\frac{3}{2}d\right)^4 = m^4$$

$$\text{or, in simpler form}$$

$$\frac{41}{8}d^4 = m\tag{eq1}$$

$$\frac{81}{256}d^8 = m^4\tag{eq2}$$

Solving $(\text{eq}1)$ for $d^4$, substituting the result into $(\text{eq}2)$, and then solving for $m$ yields

$$m = \frac{9}{82}$$