Show there does not exist $n \geq 1$ such that $\omega^n \in \mathbb{Q}$, where $\omega$ is a root of $x^2 + x + 2$. The title question arises due to the following setup:
Consider the ring $R = \dfrac{\mathbb{Q}[A,B]}{\langle A^2 + AB + 2B^2\rangle}$, where $A$ and $B$ are indeterminates. Let $a$ and $b$ denote the images of $A$ and $B$ in $R$. I am trying to show does not exist $n \geq 1$ such that $a^n \in \langle b^n \rangle$.
To start, give $A$ and $B$ degree 1. Then $R$ is an $\mathbb{N}$-graded ring. Now, suppose such $n$ exists and try to find a contradiction: taking homogeneous components of degree $n$ of the inclusion $a^n \in \langle b^n \rangle$, we find that $a^n = c b^n$ for some $c \in \mathbb{Q}$.
The case $c = 0$ readily leads to a contradiction since $R$ is a domain. Assume $c \neq 0$. Lifting the equation $a^n = cb^n$ to $\mathbb{Q}[A,B]$, we find $A^n - cB^n \in \langle A^2 + AB + 2B^2 \rangle$.
Let $\omega$ denote a root of the polynomial $x^2 + x + 2$. Now, send $A \rightarrow \omega B$ and we find $\omega^nB^n - cB^n \in \langle \omega^2 B^2 + \omega B^2 + 2B^2 \rangle = \langle 0 \rangle$. Hence $\omega^n - c = 0$, so $w^n = c \in \mathbb{Q}$. Is this a contradiction? In other words, is it impossible to have $\omega^n \in \mathbb{Q}$? I don't know of any technique to prove this.

If $x^2 + x+ 2 = 0,$ then
$$x^3 = -x^2-2x = x+2-2x = -x+2\implies x^4 = -x^2+2x = 3x+2...$$
The point is you can express $x^n$ as a linear function of $x$ with integer coefficients. Then your conclusion is trivial since $a\omega + b$ clearly cannot be rational when $w$ is a complex solution to $x^2+x+2 = 0.$
Or more simply, you can recognize $x^2+x+2 = 0$ corresponds to a recurrence relation $a_{n+1}+a_n+2a_{n-1} = 0.$ and go from here to reach the same conclusion.
EDIT: yeah the comments below are right - more work needed. If you let $\omega^n = a_n\omega + b_n,$ each of $a_n, b_n$ satisfies the recurrence:
$$x_{n+2}+x_{n+1} + 2x_n = 0$$
with initial values $a_1 = 1, a_2 = -1$ and $b_1 = 0, b_2 = -2.$ Luckily for us, this recurrence guarantees that $a_n$'s all have the same parities and $a_1 \equiv a_2\equiv 1\pmod 2$, so $a_n$ cannot be zero.
