ArcTan(2) a rational multiple of $\pi$? Consider a $2 \times 1$ rectangle split by a diagonal.  Then the two angles
at a corner are ArcTan(2) and ArcTan(1/2), which are about $63.4^\circ$ and $26.6^\circ$.
Of course the sum of these angles is $90^\circ = \pi/2$.
I would like to know if these angles are rational multiples of $\pi$.
It doesn't appear that they are, e.g., $(\tan^{-1} 2 )/\pi$ is computed as

0.35241638234956672582459892377525947404886547611308210540007768713728\
  85232139736632682857010522101960

to 100 decimal places by Mathematica.  But is there a theorem that could be applied here to
prove that these angles are irrational multiples of $\pi$?  Thanks for ideas and/or pointers!
(This question arose thinking about Dehn invariants.)

$\arctan(x)$ is a rational multiple of $\pi$ if and only if the complex number $1+xi$ has the property that $(1+xi)^n$ is a real number for some positive integer $n$.  It is fairly easy to show this isn't possible if $x$ is an integer with $|x|>1$.
This result essentially falls out of the fact that $\mathbb Z[i]$ is a UFD, and the fact that the only specific primes in $\mathbb Z[i]$ are divisors of their conjugates.
You can actually generalize this for all rationals, $|x|\neq 1$, by noting that $(q+pi)^n$ cannot be real for any $n$ if $(q,p)=1$ and $|qp|> 1$.  So $\arctan(\frac{p}q)$ cannot be a rational multiple of $\pi$.
Fuller proof:
If $q+pi=z\in \mathbb Z[i]$, and $z^n$ is real, with $(p,q)=1$, then if $z=u\pi_1^{\alpha_1} ... \pi_n^{\alpha_n}$ is the Gaussian integer prime factorization of $z$ (with $u$ some unit,) $z^n = u^n \pi_1^{n\alpha_1}...\pi_n^{n\alpha_n}$.  But if a Gaussian prime $\pi_i$ is a factor of a rational integer, $z^n$, then the complement, $\bar{\pi}_i$ must also be a factor of $z^n$, and hence must be a factor of $z$.
But if $\pi_i$ and $\bar{\pi}_i$ are relatively prime, that means $\pi_i\bar{\pi}_i=N(\pi_i)$ must divide $z$, which means that $N(\pi_i)$ must divide $p$ and $q$, so $p$ and $q$ would not be relatively prime.
So the only primes which can divide $q+pi$ can be the primes which are multiples of their complements.  But the only such primes are the rational integers $\equiv 3\pmod 4$, and $\pm1\pm i$.  The rational integers are not allowed, since, again, that would mean that $(p,q)\neq 1$, so the only prime factors of $z$ can be $1+i$ (or its unit multiples.)  Since $(1+i)^2 = 2i$, $z$ can have at most one factor of $1+i$, so that means, finally, that $z\in\{\pm 1 \pm i, \pm 1, \pm i\}$.
But then $|pq|=0$ or $|pq|=1$.
