IMC 2008 · Day 2 · P7

Let $f(x) = x^{2n} + x^n + 1$, $g(x) = x^{2k} - x^k + 1$, $h(x) = x^{2k} + x^k + 1$. The complex number $x_1 = \cos\left( \frac{\pi}{3k} \right) + i \sin\left( \frac{\pi}{3k} \right)$ is a root of $g(x)$.

Let $\alpha = \frac{\pi n}{3k}$. Since $g(x)$ divides $f(x)$, $f(x_1) = g(x_1) = 0$. So, $$0 = x_1^{2n} + x_1^n + 1 = (\cos(2\alpha) + i \sin(2\alpha)) + (\cos\alpha + i \sin\alpha) + 1 = 0,$$ and $(2\cos\alpha + 1)(\cos\alpha + i \sin\alpha) = 0$. Hence $2\cos\alpha + 1 = 0$, i.e.\ $\alpha = \pm\frac{2\pi}{3} + 2\pi c$, where $c \in \mathbb{Z}$.

Let $x_2$ be a root of the polynomial $h(x)$. Since $h(x) = \frac{x^{3k} - 1}{x^k - 1}$, the roots of the polynomial $h(x)$ are distinct and they are $x_2 = \cos\left( \frac{2\pi s}{3k} \right) + i \sin\left( \frac{2\pi s}{3k} \right)$, where $s = 3a \pm 1$, $a \in \mathbb{Z}$. It is enough to prove that $f(x_2) = 0$. We have $$f(x_2) = x_2^{2n} + x_2^n + 1 = (\cos(4s\alpha) + i \sin(4s\alpha)) + (\cos(2s\alpha) + i \sin(2s\alpha)) + 1 = (2\cos(2s\alpha) + 1)(\cos(2s\alpha) + i \sin(2s\alpha)) = 0$$ (since $2\cos\left( \frac{4\pi}{3} s \right) + 1 = 2\cos\left( \frac{4\pi}{3}(3a \pm 1) \right) + 1 = 0$).