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## Regular polygons and lattice points.

Se say that a point in the plane is a lattice point if both its coordinates are integers. Prove that for $n\neq 4$ we can’t find a regular polygon with $n$ egdes, its vertices being lattice points.

Solution:
Looking at Pick’s Formula, we see that two times the area of a polygon with vertices in lattice points is an integer. But if we take an equilateral triangle with side of length $a$ with $a^2 \in \mathbb{Z}$ (distance formula) then its area is $\frac{a^2 \sqrt{3}}{4}$ which is not a rational number. Contradiction. As suggested in the comments, we need to take care of the case $n=6$ separately, but this is a simple consequence of $n=3$.
Suppose now $n \geq 5,\ n\neq 6$ and denote our polygon $P_1...P_n$. Then the angles of such a polygon are greater than $90^\circ$. Therefore, if we construct $Q_i$ such that $\overrightarrow{P_iQ_i}=\overrightarrow{P_{i+1}P_{i+2}}$. Then $Q_1...Q_n$ is a regular polygon with vertices in lattice points, which is strictly inside the initial polygon. If $P_1...P_n$ exists, then we can repeat this construction as many times we want, but this is a contradiction, since there are only a finite number of lattice points inside $P_1..P_n$.

Yes, you are right. The case $n=6$ should be treated separately, but as you point out, it is a consequence of the case $n=3$.