Problem 1229

The answer is yes and we present the following construction: the idea is considering points in the unit circle of the form $P_{n}=(\cos (2 n \theta), \sin (2 n \theta))$ for an appropriate $\theta$. Then the distance $P_{m} P_{n}$ is the length of the chord with central angle $(2 m-2 n) \theta \bmod \pi$, that is, $2|\sin ((m-n) \theta)|$. Our task is then finding $\theta$ such that (i) $\sin (k \theta)$ is rational for all $k \in \mathbb{Z}$; (ii) points $P_{n}$ are all distinct. We claim that $\theta \in(0, \pi / 2)$ such that $\cos \theta=\frac{3}{5}$ and therefore $\sin \theta=\frac{4}{5}$ does the job.

Proof of (i): We know that $\sin ((n+1) \theta)+\sin ((n-1) \theta)=2 \sin (n \theta) \cos \theta$, so if $\sin ((n-1) \theta$ and $\sin (n \theta)$ are both rational then $\sin ((n+1) \theta)$ also is. Since $\sin (0 \theta)=0$ and $\sin \theta$ are rational, an induction shows that $\sin (n \theta)$ is rational for $n \in \mathbb{Z}_{>0}$; the result is also true if $n$ is negative because sin is an odd function.

Proof of (ii): $P_{m}=P_{n} \Longleftrightarrow 2 n \theta=2 m \theta+2 k \pi$ for some $k \in \mathbb{Z}$, which implies $\sin ((n-m) \theta)=$ $\sin (k \pi)=0$. We show that $\sin (k \theta) \neq 0$ for all $k \neq 0$.

We prove a stronger result: let $\sin (k \theta)=\frac{a_{k}}{5^{k}}$. Then

$$\begin{aligned} \sin ((k+1) \theta)+\sin ((k-1) \theta)=2 \sin (k \theta) \cos \theta & \Longleftrightarrow \frac{a_{k+1}}{5^{k+1}}+\frac{a_{k-1}}{5^{k-1}}=2 \cdot \frac{a_{k}}{5^{k}} \cdot \frac{3}{5} \\ & \Longleftrightarrow a_{k+1}=6 a_{k}-25 a_{k-1} . \end{aligned}$$

Since $a_{0}=0$ and $a_{1}=4, a_{k}$ is an integer for $k \geq 0$, and $a_{k+1} \equiv a_{k}(\bmod 5)$ for $k \geq 1($ note that $a_{-1}=-\frac{4}{25}$ is not an integer!). Thus $a_{k} \equiv 4(\bmod 5)$ for all $k \geq 1$, and $\sin (k \theta)=\frac{a_{k}}{5^{k}}$ is an irreducible fraction with $5^{k}$ as denominator and $a_{k} \equiv 4(\bmod 5)$. This proves (ii) and we are done.

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