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In this question the author conjectured that a certain angle $\theta$ could be a rational multiple of $2\pi$ only in the case it was an integer multiple of $\pi/2$. In my answer I found an explicit representation for $\theta$: $$ \cos\theta=1+2\cos\left(2r\pi\right), $$ where $r$ is any rational number such that the right hand side of the equation is comprised between $-1$ and $1$.

This equation is certainly satisfied if $\theta=k\pi/2$, with $k$ integer. My question: is it possible to prove that no other solution $\theta$ can be expressed as a rational multiple of $2\pi$?

Intelligenti pauca
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The question seems to be: solve the equation $\cos(A) = 1 + 2 \cos(B)$ where $A$ and $B$ are rational multiples of $\pi$. Some solutions include $(\cos(A),\cos(B)) = (-1,-1)$, $(1,0)$, and $(0,-1/2)$. I claim these are the only solutions.

Another way to view this equation is as follows: solve the equation

$$\zeta^a + \zeta^{-a} = 2 + 2 \zeta^{b} + 2 \zeta^{-b}$$

where $\zeta$ is a root of unity. We may write this as:

$$\zeta^a + \zeta^{-a} - 1 - 1 - \zeta^{b} - \zeta^{b} - \zeta^{-b} - \zeta^{-b} = 0.$$

This is a vanishing sum of $8$ roots of unity (minus a root of unity is a root of unity). A vanishing sum of roots of unity is called primitive if no sub-sum is equal to zero. Any vanishing sum can obviously be written as a sum of primitive vanishing sums. However, for any integer $n$, there are only finitely many vanishing sums of length $n$ up to the obvious operation of multiplying by a root of unity. For references as well as a classification for small $n$, see the following paper of Poonen and Rubinstein:

https://math.mit.edu/~poonen/papers/ngon.pdf

For example, the only vanishing sum of lengths $8$ are:

$$\zeta_7 + \zeta^2_7 + \ldots + \zeta^6_7 + \zeta_6 + \zeta^5_6 = 0,$$ $$\zeta_6 + \zeta^5_6 + \zeta^{4}_{30} + \zeta^{10}_{30} + \zeta^{11}_{30} + \zeta^{17}_{30} + \zeta^{23}_{30} + \zeta^{24}_{30} = 0,$$ $$\zeta_6 + \zeta^5_6 + \zeta^{}_{30} + \zeta^{2}_{30} + \zeta^{12}_{30} + \zeta^{13}_{30} + \zeta^{19}_{30} + \zeta^{20}_{30} = 0.$$

Your sum is not a multiple of either of these types, since it contains equal terms ($-1$ occurs twice) and all the numbers above are distinct. Hence it must contain a primitive sub-sum of length at most $4$. The only vanishing sums of length at most $5$ are

$$1 - 1 = 0, \quad 1 + \zeta_3 + \zeta^2_3 = 0, \quad 1 + \zeta_5 + \ldots \zeta^4_5 = 0.$$

If there is a sum of type $1-1$, that means that two terms are negatives of each other. This forces either:

  1. $\zeta^{a}$ or $\zeta^{-a} = 1$, and then $\cos(A) = 1$ and $\cos(B) = 0$.
  2. $\zeta^{a}$ or $\zeta^{-a}$ is $\zeta^{b}$ or $\zeta^{-b}$, and then $\cos(A) = \cos(B)$, or $\cos(A) = -1$ and $\cos(B) = -1$.
  3. $\zeta^{a} + \zeta^{-a} = 0$, and then $\cos(A) = 0$ and $\cos(B) = -1/2$,
  4. $\zeta^{b} + \zeta^{-b} = 0$, and then $\cos(B) = 0$ and $\cos(A) =1$,
  5. $\zeta^{b}$ or $\zeta^{-b}$ is $-1$, and then $\cos(B) = -1$, and $\cos(A) = -1$.

If there are no primitive sub-sums of length $2$, then the sum must be a multiple of $1 + \zeta_3 + \zeta^2_3$ plus a multiple of $1 + \zeta_5 + \ldots \zeta^4_5$. However, in any such combination, at most one term can occur twice. Since $-1$, $-\zeta^b$, and $-\zeta^{-b}$ each occur twice, this cannot happen. This completes the proof.

Gordon
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    Thank you for this nice answer! Looking at the papers cited by Poonen and Rubinstein I found this one by Conway and Jones which explicitly refers to rational sums of cosines of rational angles: their Theorem 7 can be immediately applied to my case, showing that the only solutions have $A$ and $B$ multiples of $\pi/2$. – Intelligenti pauca Jun 14 '17 at 09:18