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There are a few nice proofs of the following fact: the only regular polygon that can embedded in the integer lattice is the square.

The proof I am thinking about is as follows: Let $n \in \{5,7,8,9,\dots\}$. Suppose $P_1,P_2,\dots P_n$ are vertices of a regular $n$-gon embedded in the integer lattice. Translate $P_1$ by $\vec{P_2P_3}$ to obtain the point $Q_1$, translate $P_2$ by $\vec{P_3P_4}$ to obtain $Q_2$, etc.

My question is: how does one prove carefully that $Q_1,\dots, Q_n$ forms a regular $n$-gon. I have tried to compute the side lengths of the resulting polygon explicitly but this turns in to a bit of a mess. I was curious if there was a more elegant way to argue here.

Thanks in advance for your help.

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Playing it fast and loose with notation, we have that $$Q_i = P_i + \overrightarrow{P_{i+1}P_{i+2}} = P_i + P_{i+2} - P_{i+1}$$ Then$$\begin{align*}\overrightarrow{Q_i Q_{i+1}} &= Q_{i+1} - Q_i \\ &= \Big(P_{i+1} + P_{i+3} - P_{i+2}\Big) - \Big(P_i + P_{i+2} - P_{i+1}\Big)\\ &= \Big(P_{i+3}-P_{i+2}\Big) -\Big(P_{i+2} - P_{i+1}\Big) + \Big(P_{i+1} - P_i\Big)\\ &=\overrightarrow{P_{i+2}P_{i+3}} - \overrightarrow{P_{i+1}P_{i+2}} + \overrightarrow{P_i P_{i+1}}\end{align*}$$ the magnitude of which is clearly equal for all $i$ by symmetry, whence $Q_1, \ldots, Q_n$ forms a regular $n$-gon.

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  • $\begingroup$ Nice nice nice thank you so much for your help. Ah I’m sorry I’m realizing I was overthinking this a lot. By the way, an interesting fact: the ratio of the new side length to the old side length is $2\sin(\pi / n) < 1$. I’m curious if one can prove this via the formula you found above. $\endgroup$ Commented Jun 9, 2024 at 1:26
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    $\begingroup$ After playing around with the given set-up in Desmos, it appears that the side length of the original $n$-gon is $2\sin(\pi/n)$, while the side length of the new $n$-gon is $2\sin(3\pi/n) - 4\sin(\pi/n)$. Link to the Desmos model here $\endgroup$ Commented Jun 9, 2024 at 17:11

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