It’s fairly obvious that a square is a regular lattice polygon, but it’s fair from obvious that it’s the only one. In the past semester, we actually proved that it’s the only regular polygon on the lattice without invoking any complicated mathematics.

Sketch of Proof:

We first show that regular triangles and hexagons can’t exist. The easiest way is to apply the irrationality of sine to either the area or a rotational perspective. Regular hexagons cannot exist because it’s composed of equilateral triangles.

For any other polygons, we apply infinite descent. Let P1P2…PN be the vertices of the polygons, then consider the vector P2P3 applied to P1. Since it’s a lattice vector applied to a lattice point, we will end up with another lattice point. Applying this procedure to all of the lattice points will result in a smaller lattice polygon! This concludes the proof.

Now what are the implications, and some other problems to consider?

Well for one, what other lattices will give us more regular polygons? For example a triangular lattice will give a triangle and a hexagon as the regular polygons. Is there any others? Does there contain a maximal lattice with such numbers?

Furthermore, does this imply the irrationality of sines and cosines of certain angles? If a regular polygon in the lattice exist, then the trig functions are rational. Is the converse true? I’m pretty certain it is, but haven’t worked it out yet.