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