I want to show that for $n\in\mathbb{N}$ and $3\nmid n$ the angle $\alpha = \frac{2\pi}{n}$ is trisectable.
What I use is the following theorem:
An angle $\alpha$ can be trisected using compass and ruler if and only if the polynomial $4X^3-3X-\cos\alpha$ is reducible over $\mathbb{Q}(\cos\alpha)$.
My attempt:
The angle $2\pi$ is trisectable, for the polynomial $4X^3-3X-1$ has root $-1/2$ in $\mathbb{Q}$, so I know that $2\pi/3$ is constructible from compass and ruler. Now if the angle $2\pi/n$ would be constructible, too, then I could reach the conclusion by using that $gcd(3,n)=1$ somehow, although I have no idea how right now.
Yu are right. Given $gcd ~(3, n) = 1$ there are $a, b$ integers such that $3\cdot a + b \cdot n = 1$. So $\frac{1}{3\cdot n} = \frac{a}{3} + \frac{b}{n}$. That is $\frac{1}{3n}$ can be written as an integer linear combination of $\frac{1}{3}$ and $\frac{1}{n}$. Done.