A question about algebra or number theory? Suppose $r\in \mathbb{Q}$, and $\cos^2(r \pi)\in \mathbb Q$, can we proof $$4\cos^2(r \pi)\in \mathbb Z ?$$
Let $\alpha=\cos(\frac mn\pi)$ with $\gcd(n,m)=1$, $\beta=\sqrt{\alpha^2-1}=i\sqrt{1-\alpha^2}=i\sin(\frac mn\pi)$, $\zeta=\alpha+\beta$. Then by assumption, $$\Bbb \alpha^2\in \Bbb Q,\quad \beta^2\in\Bbb Q,\quad \zeta^n=1.$$
However, instead of $X^n-1$, the minimal polynomial of $\zeta$ is a cyclotomic polynomial of degree $\varphi(n)$. Nevertheless, all roots of this polynomial are roots of unity and hence the constant term is $\pm1$.
As
$$ \zeta^2=(\alpha+\beta)^2=\underbrace{\alpha^2+\beta^2}_{\in\Bbb Q}+2\alpha\beta$$
we have
$$ (\zeta^2-(\alpha^2+\beta^2))^2=4\alpha^2\beta^2\in\Bbb Q$$
i.e., $\zeta$ is also a root of
$$\tag1X^4-2(\alpha^2+\beta^2)X^2+(\alpha^2-\beta^2)^2 $$
and this must be a multiple of the minimal polynomial of $\zeta$. In particular, $\varphi(n)\le 4$.
This leaves us with the following possibilities:


$\varphi(n)=1$: Then $n=1$ or $n=2$ and we deal with $\cos 0$, $\cos \frac\pi2$, $\cos\pi$, $\cos\frac{3\pi}2$, i.e., $\alpha\in \{-1,0,1\}$
$\varphi(n)=2$: Then $n=3$ or $n=4$ or $n=6$ and we deal with $\cos \frac\pi6$, $\cos \frac\pi4$, $\cos \frac\pi3$, $\cos\frac{2\pi}3$, $\cos\frac{3\pi}4$, $\cos\frac{5\pi}6$, i.e., $\alpha\in\{\pm\frac12,\pm\frac{\sqrt2}2,\pm\frac{\sqrt 3}2\}$
$\varphi(n)=3$: This is not possible
$\varphi(n)=4$: Then $n=5$ or $n=8$ or $n=12$. The corresponding cyclotomic polynomials are $X^4+X^3+X^2+X+1$, $X^4+1$, $X^4-X^2+1$. The first does not match $(1)$ and from the others we get $\alpha^2+\beta^2\in\frac12\Bbb Z$ and $\alpha^2-\beta^2=\pm1$, hence indeed 
$$\alpha^2=\frac{(\alpha^2+\beta^2)+(\alpha^2 -\beta^2)}{2}\in\frac14\Bbb Z $$
For $r=k\in\mathbb{Z}$ we have $\cos(r\pi)=\pm1\in\mathbb{Z}$. For $r=\frac13$ we get $\cos(r\pi)=\frac12$. So $4\cos^2(r\pi)=1\in\mathbb{Z}$.

Answer to modified question:  https://en.m.wikipedia.org/wiki/Niven%27s_theorem