Polynomial equations in roots of unity

Question

I am interested in finding all solutions (for $x$ and $n$) of the equation $$w^{2x+2}-w^{2x+1}+w^{2x}-w^{x+2}-w^x+w^2-w+1=0$$ where $w$ is an $n$-th root of unity. The solution is straightforward for $n$ prime (using cyclotomic polynomials) but how can one proceed for general $n$?

Answer 1

If we take $w$ a complex $n$-th root of unity, then this equation has a explicit solution $x \in \mathbb{C}$ for almost every $n$. Let's rewrite the equation as:

$$(w^2 - w + 1)/(w^2 + 1) = w^x/(w^{2x} + 1)$$

Assuming, of course, that $w^2 \neq -1$ and that $w^{2x} \neq - 1$ (those cases can be dealt with later). Let's set $a = (w^2 - w + 1)/(w^2 + 1)$ and $y = w^x$. Then, we have:

$$a = y/(y^2 + 1)$$

The solution to this is

$$y = \frac{1 \pm \sqrt{1 - 4a^2}}{2a}$$

From it, we recover $x = \log y/\log w$. Now, there are some things that we need to be careful. Given that $w \neq 0$, we must have that $y \neq 0$. Now, given that $a$ is non zero (because $w^2 - w + 1$ isn't zero for a $n$-th root of unity), it follows that $y \neq 0$ as it was required. We actually found infinite values of $x$ satisfying the equation for given $n$, corresponding to the infinite branches of the complex logarithm.

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If $w^2 = -1$, the equation becomes:

$$(-1)^{x+1} = w/(w - 1)$$

This, once again, has infinite complex solutions $x$ corresponding to different logarithm branches. I guess that the case $w^{2x} = -1$ works out similarly.