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.)
 A: There's actually an incredible way of proving it that is quite elementary.
First, we note that $\arctan(2)$ can be represented in the complex number $1 + 2i$ as $\sqrt{5} e^{i \arctan(2)}$
The question of proving whether $\arctan(2)$ is a rational multiple of $\pi$ is now framed as checking if the complex number $(1 + 2i)^n$ is ever real for all $n \in \mathbb{N}$. If it is real, then the argument of $\sqrt{5} e^{i \cdot n *\arctan(2)}$ must be a multiple of pi. In essence, if $(1 + 2i)^n$ is ever real, then $\arctan(2) \cdot n = m \cdot \pi \implies \arctan(2) = \frac{m}{n} \pi$.
Why was it useful to phrase the problem in this manner? Let's try and define a recurrence relation for the terms of $(1 + 2i)^n$. Suppose $a_1 = 1$ and $a_2 = 2$, then define $a_n$ to be the real part of $(1 + 2i)^n$ and $b_n$ to be the imaginary part.
$a_{n+1} + ib_{n+1} = (1+2i) \cdot (1 + 2i)^n = (1+2i) \cdot (a_n + ib_n) \\
a_{n+1} = a_n - 2b_n \\
b_{n+1} = 2a_n + b_n$
Okay, I don't like our recursion having two variables in it. Let's try and mess with it a bit.
$a_n = a_{n-1} - 2b_{n-1} \\
b_n = 2a_{n-1} + b_{n-1}$
Substituting the first equation into the $b_{n+1}$ equation, we get
$b_{n+1} = 2a_{n-1} - 4b_{n-1} + b_n$
Now we want to get rid of $a_{n-1}$, so let's substitute the second equation after solving for $a_{n-1}$
$b_{n+1} = 2b_n - 5b_{n-1}$
Nice, we got it all in one variable.
It may look tempting to solve the recurrence relation, but we can do better.
Let's take both sides mod 5.
$b_{n+1} \equiv 2b_n \pmod 5$
Why did we do this? Well, we know the first term, $b_1$, is 2. Therefore, every term after that will just be a power of 2 (mod 5).
$b_n \equiv 2^n \pmod 5$
If there exists some term $b_k = 0$, then it will be 0 (mod 5), because 5 divides 0. (If $b_k = 0$, then that means there exists a natural number $k$ such that $(1+2i)^k$ is real)
But powers of 2 can never be congruent to 0 (mod 5). Therefore, $(1+2i)^n$ is never real and thus $\arctan(2)$ is not a rational multiple of $\pi$.
A: Lemma: If $x$ is a rational multiple of $\pi$ then $2 \cos(x)$ is an algebraic integer.
Proof
$$\cos(n+1)x+ \cos(n-1)x= 2\cos(nx)\cos(x) \,.$$
Thus
$$2\cos(n+1)x+ 2\cos(n-1)x= 2\cos(nx)2\cos(x) \,.$$
It follows from here that $2 \cos(nx)= P_n (2\cos(x))$, where $P_n$ is a monic polynomial of degree $n$ with integer coefficients.
Actually $P_{n+1}=XP_n-P_{n-1}$ with $P_1(x)=X$ and $P_0(x)=1$.
Then, if $x$ is a rational multiple of $\pi$ we have $nx =2k \pi$ for some $n$ and thus, $P_n(2 \cos(x))=1$.

Now, coming back to the problem. If $\tan(x)=2$ then $\cos(x) =\frac{1}{\sqrt{5}}$. Suppose now by contradiction that $x$ is a rational multiple of $\pi$. Then  $2\cos(x) =\frac{2}{\sqrt{5}}$ is an algebraic integer, and so is its square $\frac{4}{5}$. But this number is algebraic integer and rational, thus integer, contradiction....
P.S. If $\tan(x)$ is rational, and $x$ is a rational multiple of $\pi$, it follows exactly the same way that $\cos^2(x)$ is rational, thus $4 \cos^2(x)$ is  algebraic integer and rational. This shows that $2 \cos(x) \in \{ 0, \pm 1, \pm 2 \}$.....
A: $\arctan(x)$ 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 $|pq|=1$.
