Is it possible to make a regular 3-polygon by selecting $3$ points in $S$

Question

I would appreciate if somebody could help me with the following problem:

Let $\mathbb{Z}$ be the set of all integers and let $ S = \mathbb{Z} \times \mathbb{Z} $.

Question:

1). Is it possible to make a regular 3-polygon by selecting $3$ points in $S$ ?

2). Is it possible to make a regular 5-polygon by selecting $3$ points in $S$ ?

3). Is it possible to make a regular $n$-polygon by selecting $n$ points in $S$ ?

Answer 1

Hints:

A regular $\,3$-gon is just an equilateral triangle. To simplify things assume one of the vertices of this triangle is $\,(0,0)\,$ and one of the sides is on the positive $\,x$-axis. For a second side we need a straight line through the origin forming an angle of $\,60^\circ\,$ with the positive $\,x$-axis, i.e. the line

$$y=\sqrt 3\,x$$

If an element of $\,(n,m)\in S\,$ is on the above line, then

$$m=\sqrt 3\,n\,\,.\;\text{ Do you think this is possible for}\;\; \,m,n\in \Bbb Z\;?\ldots$$

Do something similar to the above for the regular pentagon: we need then a line with slope equal to $\,\tan 108^\circ\,$ and etc.

Answer 2

If it is possible to make an $n$-gon, then the center of that $n$-gon has coordinates in $\frac1n\mathbb Z$ (it is the average of $n$ integers). You can scale the original $n$ gon with a factor of $n$. Then the center of this scaled $n$-gon has integer coordinates. Thus after translating the $n$-gon to move its center to the origin, we see that the vertices are $$\tag1(x_k,y_k)=\left(r\cos(\frac{2\pi}n\cdot k+\alpha),r\sin(\frac{2\pi}n\cdot k+\alpha)\right)$$ with some (possibly irrational) $r>0$, $\alpha\in[0,2\pi)$. One can even simplify further by considering the $n$-gon with vertices $$\tag2(x_k',y_k') = (x_kx_1+y_ky_1,x_ky_1-y_kx_1),\quad k=1,\ldots,n$$ (this is just a scaled and rotated version of the original). Since $(x_1',y_1')=(x_1^2+y_1^2,0)$, we have in effect made the $r$ in $(1)$ an integer and $\alpha=0$. Therefore, both $\sin\frac{2\pi}n=\frac{y_2}{\sqrt{x^2+y^2}}$ and $\cos\frac{2\pi}n=\frac{x_2}{\sqrt{x^2+y^2}}$ are at most quadratic irrationals. Then consider the possible minimal polynomials of the complex number $z=\frac{x_2+iy_2}{\sqrt{x^2+y^2}}$.

Answer 3

The sides of lattice triangles have rational slopes, and using the addition formula for $\tan$ one deduces that all three $\tan\alpha_i$ are rational. In particular $\alpha=60^\circ$ cannot be realized.