Diophantine equation $\cos(2\pi x)\cos(2\pi y) = \cos(2\pi z)$

Question

While working on finite order elements of $\operatorname{SO}_n$, I meet this question:

Find all identities of the form $\cos(2\pi x)\cos(2\pi y) = \cos(2\pi z)$ with $x, y, z$ rational numbers.

As to background: this is more or less the equation that the product of three finite order elements of $\operatorname{SO}_3(\Bbb R)$ is the identity.

From experimental results, it seems that the only possible solutions are:

• $0 \cdot * = 0$;
• $\pm1 \cdot * = \pm*$;
• $\pm\frac1{\sqrt 2} \cdot \pm\frac1{\sqrt 2} = \pm\frac1 2$.

So, are these the only solutions and how can one prove that?

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I abbreviate $\cos(2\pi x)$ to $c(x)$. Using the identity $c(x)c(y) = \frac{c(x - y) + c(x + y)}2$, we can (up to renaming the variables) get an equivalent form of the equation: $c(x) + c(y) = 2c(z)$.

From here I don't find a proper method from my usual arsenal. I tried to look at ramification of the corresponding cyclotomic fields but got nowhere.

Answer 1

The solution to the Dipohantine equation:

You have, indeed, found all the solutions. Put $\alpha = e^{2 \pi i x}$, $\beta = e^{2 \pi i y}$, $\gamma = e^{2 \pi i z}$, so you want $$(\alpha+\alpha^{-1})(\beta+\beta^{-1}) = 2 (\gamma+\gamma^{-1}).$$ Let $K$ be a cyclotomic field large enough to contain $\alpha$, $\beta$ and $\gamma$, and let $\mathfrak{p}$ be a prime of $\mathcal{O}_K$ lying above $2$. Let $v$ be the $\mathfrak{p}$-adic valuation, normalized so that $v(2)=1$.

If $\zeta$ is a primitive $m$-th root of unity, then $$v(\zeta+\zeta^{-1}) = \begin{cases} 1 & m=1, 2 \\ \infty & m = 4 \\ 1/k & m = 2^{k+1}, k \geq 2 \\ 0& \text{otherwise} \\ \end{cases}.$$ (See this problem for a very similar analysis.) So we want to solve $$a+b=1+c$$ with $a$, $b$, $c$ in $\{ 0,1,1/k, \infty \}$. The solutions are $1+\ast = 1 + \ast$, $1/2 + 1/2 = 1+0$ and $\infty+\ast = 1+\infty$. It's easy to check that these lift to your solution.

This is not the condition for three rotations in $SO_3$ to have product $1$:

However, something has gone wrong earlier in your analysis. If $\theta_1$, $\theta_2$, $\theta_3 \in [0, \pi]$ obey the triangle inequalities (meaning $\theta_1+\theta_2 \geq \theta_3$, $\theta_1+\theta_3 \geq \theta_2$ and $\theta_2+\theta_3 \geq \theta_1$) then there are rotations by $\theta_1$, $\theta_2$ and $\theta_3$ radians with product $1$. So there are tons of additional solutions to this problem.

Proof sketch: I find this computation easiest in quaternions. Rotation by angle $\theta$ corresponds to a quaternion of the form $$\cos \tfrac{\theta}{2} + \sin \tfrac{\theta}{2} (p i+qj + r k)$$ with $p^2+q^2+r^2=1$. The real part of $\left( \cos \tfrac{\theta_1}{2} + \sin \tfrac{\theta_1}{2} (p_1 i+q_1 j + r_1 k) \right)\left( \cos \tfrac{\theta_2}{2} + \sin \tfrac{\theta_2}{2} (p_2 i+q_2 j + r_2 k) \right)$ is $$\cos \tfrac{\theta_1}{2} \cos \tfrac{\theta_2}{2} - \sin \tfrac{\theta_1}{2} \sin \tfrac{\theta_2}{2} (p_1 p_2 + q_1 q_2 + r_1 r_2). \qquad (\ast)$$ Since $p_1^2+q_1^2+r_1^2 = p_2^2+q_2^2+r_2^2 = 1$, the dot product $p_1 p_2 + q_1 q_2 + r_1 r_2$ can be anywhere between $-1$ and $1$, so $(\ast)$ can be anywhere between $\cos \tfrac{\theta_1}{2} \cos \tfrac{\theta_2}{2} - \sin \tfrac{\theta_1}{2} \sin \tfrac{\theta_2}{2} = \cos \tfrac{\theta_1+\theta_2}{2}$ and $\cos \tfrac{\theta_1}{2} \cos \tfrac{\theta_2}{2} + \sin \tfrac{\theta_1}{2} \sin \tfrac{\theta_2}{2} = \cos \tfrac{\theta_1-\theta_2}{2}$. Using that all my angles are in $[0, \pi]$, this corresponds to $\theta_3$ being anywhere between $|\theta_1-\theta_2|$ and $\theta_1+\theta_2$.