Find all $w \in G_{28}$ such that $1 + w^2 + w^4 + w^6 =0$

Question

We can at once discard $w=1$ since it doesn't verify the equation.

We notice that

$$1 + w^2 + w^4 + w^6 =0 \ \iff \\ w^4 + w^6 = (-1)(1+w^2)$$

hence, we can square both sides (can we?, are we losing solutions?),

$$w^8 + 2w^{10} + w^{12} = w^4 + 2w^2 + 1 \ \iff \\ (w^4)^2 (1+2w^2+w^4) = 1+2w^2+w^4$$

Since $\mathbb{C}$ is an integral domain, we know that either $1+2w^2+w^4 = 0$ or $(w^4)^2 = 1$.

We also know that

$$ 1+2w^2+w^4 = 0 \iff w^2 =-1 \\ (w^4)^2 = 1 \iff w^4 = \pm 1$$

But $w^2 =-1$ only if $w=\pm i$ and $w^4 = 1$ only if $w\in G_4 \subset G_{28}$; either way, $G_4 \setminus \{1\}$ is a solution.

On the other hand, $w^4 = -1$ only if $w^2 = \pm i$. But $w^2 \in G_{14}$ and $G_{14} \cap \{\pm i\} = \emptyset$ (since $4$ is not a divisor of $14$).

So our solution is $G_4 \setminus \{1\}$, right?

Answer 1

The systematic way is to compute the gcd of the polynomials $1 + w^2 + w^4 + w^6$ and $1-w^{28}$. This gcd is $w^2+1$ and so the solution is $w=\pm i$. This set is $G_4 \setminus G_2$.

Answer 2

Let $\zeta=\exp(2i\pi/28)$.

Then $(\zeta^n)^8=1 \iff 7|n,\;$ and $(\zeta^n)^2=1\iff14|n$.

So $1+(\zeta^n)^2+(\zeta^n)^4+(\zeta^n)^6=\dfrac{(\zeta^n)^8-1}{(\zeta^n)^2-1}=0 \iff 7|n$ and not $14|n \iff \zeta^n=\pm\, i .$

Answer 3

HINT \begin{align*} 1 + w^{2} + w^{4} + w^{6} = 0 \Longleftrightarrow (1+w^{2}) + w^{4}(1+w^{2}) = 0 \Longleftrightarrow (1+w^{4})(1+w^{2}) = 0 \end{align*}