# If $\cos p\alpha$ and $\cos q\alpha$ are rational with $p,q$ relatively prime, then $\cos \alpha$ is rational, or $\alpha$ is a multiple of $\pi / 6$.

Would you please help me solve Exercise 2, which I repeat here:

Suppose that $p$ and $q$ are relatively-prime positive integers. Show that if $\cos p \alpha$ and $\cos q \alpha$ are rational, then $\cos \alpha$ is rational or $\alpha$ is a multiple of $\pi / 6$.

The exercise is a supplement to An Introduction to the Theory of Numbers by Niven, Zuckerman, and Montgomery.

There are two approaches to solve the problem. The first is to assume that $\cos \alpha$ is irrational and prove that $\alpha$ must then be a multiple of $\pi / 6$. In this case, I know from Exercise 1 that $\cos (p + q) \alpha$ is irrational, which is equivalent to the fact that $\sin p \alpha \sin q \alpha$ is irrational, deduced from my solution to Exercise 1. I can construct examples that satisfy this approach but cannot devise a general proof.

The other approach is to assume that $\alpha$ is not a multiple of $\pi / 6$ and prove that $\cos \alpha$ must then be rational. Once again, I can construct examples but cannot figure out how to prove it in general.

Since $\cos(p\alpha)=T_p(\cos \alpha)$, with $T_p\in\mathbb{Z}[x]$ being the $p$-th Chebyshev polynomial of the first kind, $\cos(p\alpha)\in\mathbb{Q}$ implies that $\cos(mp\alpha)\in\mathbb{Q}$ for any $m\in\mathbb{N}$. Similarly, $\cos(nq\alpha)\in\mathbb{Q}$ for any $n\in\mathbb{N}$. Since $p$ and $q$ are coprime, for some choice of $(m,n)\in\mathbb{N}^2$ we have $mp-nq=1$ (Bézout Lemma). Now $$\cos(\alpha) = \cos((mp-nq)\alpha) = \underbrace{\cos(mp\alpha)}_{\in\mathbb{Q}}\underbrace{\cos(nq\alpha)}_{\in\mathbb{Q}}+\sin(mp\alpha)\sin(nq\alpha)$$ implies $$\big[\cos(\alpha)-\underbrace{\cos(nq\alpha)\cos(mp\alpha)}_{\in\mathbb{Q}}\big]^2 \in \mathbb{Q}$$ by the Pythagoran theorem. In particular, $\cos(\alpha)$ is an algebraic number over $\mathbb{Q}$ with degree $\leq 2$. Assuming that $\cos(\alpha)\not\in\mathbb{Q}$, we have that $\alpha$ is a rational multiple of $\pi$ (otherwise $\cos(\alpha)$ would not be integral) of the form $\alpha = \frac{2\pi u}{v}$, with $\varphi(v)=4$ (the degree of $\cos\frac{2\pi u}{v}$ is $\varphi(v)/2$ since the degree of a primitive $v$-th root of unity is $\varphi(v)$). So if $\cos\alpha\not\in\mathbb{Q}$ we have $v=5$ or $v=12$, but in the former case we do not have two coprime integers $p,q$ such that $\cos(p\alpha)$ and $\cos(q\alpha)$ are simultaneously rational.
• Would you please explain why "$\alpha$ is a rational multiple of $\pi$." You write that "otherwise $\cos \alpha$ would not be integral," i.e., $\cos \alpha$ has been assumed to be integral, but at this point in the proof, we have assumed that $\cos \alpha$ is of the form $r + \sqrt s$ where $r$ and $s$ are rational numbers and $s$ is not zero nor square. I know that if $\alpha$ is a rational multiple of $\pi$, then $\cos \alpha$ is an algebraic number, but you are somehow using that statement's converse, which I am not sure is true. – Maurice P Sep 29 '17 at 16:50
• @MauriceP: I think a concrete example would clarify this point. Let us consider $\alpha=\arctan(1/2)$, for instance. $\cos(\alpha)$ is an algebraic number of degree $2$, but $\cos(n\alpha)$ is never rational (unless $n=0$), which is required by the assumptions. – Jack D'Aurizio Sep 29 '17 at 16:51