How to proove that $\arctan(3/4)$ is not commensurable with $\pi$?

Question

How to prove that $\arctan(3/4)$ is not commensurable with $\pi = 3.14\ldots$?

This is an exercise from E.B.Vinberg "Course of Algebra" (exercise 4.38)

Answer 1

I am not sure what machinery it is expected to be used to solve this problem. Still...

If $\alpha=\arctan\frac{3}{4}$ and $\pi$ were commensurable, then it would be true that $p\alpha=q\pi$ for some integers $p,q\ne 0$. Take $z=\frac{4}{5}+\frac{3}{5}i$: then $z=\cos\alpha+i\sin\alpha$. Thus $z^p=\cos(p\alpha)+i\sin(p\alpha)=\cos(q\pi)+i\sin(q\pi)=\pm 1$.

Now this means that $(5z)^p=(4+3i)^p=\pm 5^q$. Note that $4+3i=i(2-i)^2$, so:

$$i^p(2-i)^{2p}=\pm(2+i)^p(2-i)^p$$

Cancelling $(2-i)^p$, we get:

$$i^p(2-i)^p=\pm(2+i)^p$$

which is impossible because, in the ring $\mathbb Z[i]$ of Gaussian integers, $2-i$ and $2+i$ are distinct prime elements, so the unique factorisation would be violated.