If the length of the sides is sqrt(n) for an odd integer n , then consecutive vertices have different parity of x+y . If the length of the sides is sqrt(2n), then all vertices have the same parity of x+y , so are contained in a square sublattice of index 2. In terms of these units, the length becomes sqrt(n). So by repeating this process, we can always assume that n is odd. Any equilateral polygon in a lattice must be able to be colored with vertices of alternating color. But a circular alternating pattern is impossible with an odd number of vertices, thus the heptagon is imposible.

Draw the smallest possible octagon on the vertices.

Turn the sides into vectors.

Translate these vectors to the origin.

A new, smaller octagon is produced.

Contradiction! (This works for any regular n-gon larger than
the hexagon)