Given that $\cos\left(\dfrac{2\pi m}{n}\right) \in \mathbb{Q}$ prove $\cos\left(\dfrac{2\pi}{n}\right) \in \mathbb{Q}$ Given that $\cos\left(\dfrac{2\pi m}{n}\right) \in \mathbb{Q}$, $\gcd(m,n) = 1$, $m \in \mathbb{Z}, \, n \in \mathbb{N}$ prove that $\cos\left(\dfrac{2\pi}{n}\right) \in \mathbb{Q}.$
I know nothing about how to attack the problem. I believe I need to suppose that $\cos\left(\dfrac{2\pi}{n}\right) \not \in \mathbb{Q}$ and somehow show that $\cos\left(\dfrac{2\pi m}{n}\right) \not \in \mathbb{Q}$ what would have been a contradiction. Could you give me some hints?
 A: HINT:

*

*For all $\theta\in\mathbb R, m\in\mathbb N$, $\cos(m\theta)$ can be expressed as a polynomial of $\cos(\theta)$ (with rational coefficients), so if $\cos(\theta)$ is rational, so is $\cos(m\theta)$.

*If $m,n\in\mathbb Z$ and $\gcd(m,n)=1$, there exists $k\in\mathbb N$ such that $mk\equiv 1 \bmod  n$.

*The cosine function is periodic modulo $2\pi$, that is, $\cos(\theta + 2\pi) = \cos(\theta)$ for all $\theta\in\mathbb R$.

Can you combine these facts to come up with a proof of the desired result?
A: Let $\zeta=\exp(2\pi i/n)$ be a primitive $n$-th root of unity. You know that $\Re(\zeta^m)\in\mathbb Q$ and you want to show that $\Re(\zeta)\in\mathbb Q$. Since $m$ and $n$ are coprime $\zeta^m$ is a primitive $n$-th root of unity as well. This implies that we have $(\zeta^m)^k=\zeta$ for a $k\in\mathbb N$.
So it is enough (even better) to prove that for a number $\zeta \in \mathbb C$ with $|\zeta|=1$ and $\Re(\zeta) \in \mathbb Q$ you have $\Re(\zeta^n) \in \mathbb Q$ for all $n \in \mathbb N$.
We have
$$\zeta^n=(x+iy)^n  = \sum_{k=0}^n  {{n}\choose{k}} x^k (iy)^k $$
The real part of this is given by only giving a shit on the even $k$. But then we have an even power of $y$ and we have $y^2=1-x^2 \in \mathbb Q$, so everything is in $\mathbb Q$.
