Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega
eq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic

Question

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 $(p,q)$.

Reiny · Accepted Answer

can 't read it, don't use codes, they don't work here

Steve · Answer

This problem is discussed here: http://math.stackexchange.com/questions/2144834/let-alpha-omega-omega2-omega4-and-beta-omega3-omega5 google is your friend. Try using it.

Anonymous · Answer

the actual answer is here (it says i cant post urls... so just add "h t t p s" to the beginning of the phrase below):

://math.stackexchange.com/questions/1253515/quadratic-using-the-roots-of-unity-where-omega7-1-omega-neq-1