Find parameter so that the equation has roots in arithmetic progression

Question

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 1

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

Answer 2

Let $0 < a < b$ be the roots of $y^2 - m y + m^4$ (note that $0 < m < 1/2$ is needed for these to be real and positive). Then the real roots of $x^8 - m x^4 + m^4$ are (in order) $-b^{1/4}, -a^{1/4}, a^{1/4}, b^{1/4}$. These are in a.p. iff $2 a^{1/4} = b^{1/4} - a^{1/4}$, i.e. $b = 3^4 a$. That turns out to be true for $m = 9/82$.