<s>[INST] Problem of Complex Numbers in Geometry using Roots of Unity 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?
 [/INST] 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 $\alpha \beta = q = 2$
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