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 got that $p=-1$ but does not know how to go on to finding $q$. All help is appreciated!
Hint. Observe that
$$ \alpha\beta = \omega^4+\omega^5+\omega^6+3\omega^7+\omega^8+\omega^9+\omega^{10} = 2+\omega^4(1+\omega+\omega+\cdots+\omega^6) $$
method is due to Gauss. This book 1875. $x^2 + x + 2.$
gp-pari to check; note a^8 = a
? x = a + a^2 + a^4
%1 = a^4 + a^2 + a
? q = x^2 + x + 2
%2 = a^8 + 2*a^6 + 2*a^5 + 2*a^4 + 2*a^3 + 2*a^2 + a + 2
?
If we switched to one of the real numbers $$ t = \omega + \omega^6, $$ we would have a root of $$ t^3 + t^2 - 2 t - 1 $$
$$q=\alpha\beta=3+\sum_{j=1}^6w^j=3+w\frac{w^6-1}{w-1}=3+\frac{1-w}{w-1}=2$$
By what we know for quadratic equations, if we express $x^2+px+q=0$ as $(x-\alpha )\times (x-\beta)$, then we see that $p = -(\alpha+\beta)$ and $q = \alpha\beta$.
So, $ p = -(\omega + \ldots + \omega^6) = 1$ because $\omega$ is the seventh root of unity, hence $1 + \omega + \ldots + \omega^6 = 0$ is known.
On the other hand, $q = \alpha\beta = \omega^4+\omega^6 + 1 + \omega^5 + 1 + \omega^1 + 1 + \omega^2 + \omega^3 = 2$, because of the previous identity again. (and because $\omega^7 = 1$, I am removing multiples of $7$ from the exponent).
Hence $x^2+x+2 = 0$ is the polynomial whose roots are $\alpha$ and $\beta$. You can check this by seeing that the roots are $\frac{\pm(i \sqrt 7 \pm i)}{2}$.