Q: Prove this problem with area 
Find all $n$ for which there exists an $n$-regular-polygon with all vertices on a lattice.

$4$ is the only answer (square) and can possibly prove by area.
By Pick's theorem the area of any polygon with all vertices on a lattice $\in\mathbb Q$. Now we want to show that such an $n$-regular-polygon has an area $\not\in\mathbb Q$.
Let half of its side length be $x$ then its area $S=\dfrac{x^2n}{\tan\frac\pi n}$. This can be obtained by dividing the polygon into $n$ isosceles. Let one of its sides be $(x_1,y_1)(x_2,y_2)$ so $x^2=(x_1-x_2)^2+(y_1-y_2)^2\in\mathbb Q$. Therefore, $S\not\in\mathbb Q\Longleftrightarrow \tan\dfrac\pi n\not\in\mathbb Q$. Can we prove this?
