Prove that a regular polygon whose vertices lie on lattice points in $\mathbb{R}^2$ is a square.
Here is my progress: Say $n > 9$ is the smallest positive integer such that a regular $n$-gon has its vertices lying on lattice points (the other base cases are easy to handle. Then note that $n$ must be prime or else if $n = pq$ where $p$ is a prime and $q > 1$, then there exists a $q$-gon whose vertices lie on lattice points. Now note that the center of the regular polygon must satisfy $O \equiv (p, q)$ where $\{ p \}$ and $\{ q \}$ are multiples of $\frac{1}{4}$ (since $n$ is prime, let $n = 2k+1$ and polygon be $A_1A_2 \dots A_{2k+1}$. Then $O \equiv$ midpoint of $A_1B_j$ where $B_j$ is midpoint of $A_d$ and $A_{d+1}$, $d = \frac{k+1}{2}$. Therefore, the result follows) and since each point is a lattice point, it follows that if $r$ is the circumradius of the polygon, then $r^2$ is rational. Thus, $A_1A_2 = 2r\sin \left ( \frac{\pi}{n} \right ) \implies (A_1A_2)^2=4r^2\sin \left ( \frac{\pi}{n} \right )^2$. But $(A_1A_2)^2$ is an integer, so $4r^2\sin \left ( \frac{\pi}{n} \right )^2$ is an integer which means $\sin \left ( \frac{\pi}{n} \right )^2 = \frac{1-\cos\left ( \frac{2\pi}{n} \right )}{2}$ is rational. Thus, from the counterpart of Niven's Theorem for cosine angles, we conclude that $\cos \left ( \frac{2\pi}{n} \right )$ is rational implies that $\frac{2\pi}{n} \in \left \{ 0, \frac{\pi}{2}, \frac{\pi}{3} \right \}$ none of which is possible. So the only value of $n$ which satisfies the propositions of the problems is $\boxed{n=4}$
Is this correct?
