Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic $x^2 + px + q = 0$ for some integers $p$ and $q$. Find the ordered pair $(p,q)$.
I am unable to find a good starting place for this problem. Should I start by working backwards using the roots?
On
From the development of the geometric series with reason $\omega$ we get $1 + \omega + \omega^2 + \dots + \omega^6 = (\omega^7 - 1) / (\omega - 1) = 0$, from which we deduce $1 + \alpha + \beta = 0$.
Using the same development again, $\alpha \beta = 1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 + 1 + 1 = 2$.
$\alpha$ and $\beta$ being roots of $x^2 + px + q$ gives $x^2 + px + q = (x - \alpha)(x - \beta) = x^2 - (\alpha + \beta) x + \alpha \beta$ from which we deduce the equations $\alpha + \beta = -p = 1$ and $\alpha \beta = q = 2$
$$w^7-1=(w-1)(w^6+w^5+w^4+\dots+w+1) $$ So $$w^6+w^5+w^4+\dots+w+1 =0 \tag{1}$$
Now use Vieta's Formula.
First note that the sum of roots is $-p$, so we have that $$-p=\alpha+\beta=w^6+w^5+w^4+w^3+w^2+w=-1$$ From $\text{(1)}$ Also, $$q=(\omega + \omega^2 + \omega^4) \times (\omega^3 + \omega^5 + \omega^6)$$ Expanding, $$q=w^4(w^{6}+w^{5}+w^{4}+w^{3}+w^{2}+w+1)+2w^7=2$$ So $p=1, q=2$.