Does a solution to the equation $\sin(mx)= \frac{x}{n \pi}$ exist for $x, n, m \in \mathbb{N}$?

Question

I attempted to prove that no such solution exists. More generally, I tried to show that the equation $\pi \sin(p) = q$ has no solutions when $p,q \in \mathbb{Q^+}$. I explored approaches involving Niven's theorem and attempted to proceed similarly to this question. However, I was unable to make further progress beyond recognizing that both $\pi$ and $sin(mx)$ are both transcendental.

That is a valid point. I have revised my question. Thank you:) — sunghoon kim, Feb 12 '25 at 14:58
The question is a more specific version of , "Can $ \sin(k),\ k\in\mathbb{N}$ be a rational multiple of $\frac{1}{\pi}?"$ But then, why not first ask, "Can $ \sin(k),\ k\in\mathbb{N}$ be a rational multiple of $\pi?"$ — Adam Rubinson, Feb 12 '25 at 15:15

Qiaochu Yuan · Accepted Answer · 2025-02-12T17:10:29.473

The answer is no conditional on Schanuel's conjecture, which implies that $\pi$ and $e^i = \cos 1 + i \sin 1$ are algebraically independent (take $z_1 = i, z_2 = \pi i$). Unconditionally I believe it's an open question whether $\pi$ and, say, $e$ are algebraically independent, so I expect nearly every variant of this question to also be an open question.

Generally speaking I think the "standard conjecture" is that

if an irrational number doesn't have a "good reason" to be algebraic then it's transcendental, and
if a set of irrational numbers doesn't have a "good reason" to be algebraically dependent then it's algebraically independent

although I don't know if anyone's formally written down or named a conjecture along these lines, and this is way, way beyond what is provable.

(Also, it's notationally confusing to write $x$ for an integer; less confusing to reserve $x$ for a formal or real variable.)

Does a solution to the equation $\sin(mx)= \frac{x}{n \pi}$ exist for $x, n, m \in \mathbb{N}$?

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