If $z^7 = 1$, $z \in \mathbb{C}$ and $$x = \frac{z}{1+z^2}+\frac{z^2}{1+z^4}+\frac{z^3}{1+z^6}$$ Prove that $x \in \mathbb{Z}$.
I wrote $z$ as $z = \cos{\frac{2k\pi}{7}} + i\sin{\frac{2k\pi}{7}}$, $k$ having values from $0$ to $6$, and $$\frac{z}{1+z^2} = \frac{1}{z^6 + z} = \frac{1}{2\cos(k\pi)\cos(\frac{5k\pi}{7})}$$ and $$x = \frac{1}{2\cos(k\pi)} \left(\frac{1}{\cos\frac{5k\pi}{7}} + \frac{1}{\cos\frac{3k\pi}{7}} +\frac{1}{\cos\frac{k\pi}{7}}\right)$$ But after this I haven't had any successes in trying to prove that $x$ is an integer.
