Can an equilateral triangle have vertices at integral lattice points?
Integral lattice points are such points as (101, 254) or (3453, 12), but not points such as (123.4, 1) or (√2, 5)
If you can't find a solution in the 2D Cartesian plane, can you find one in a 3 (or more) dimensional space?
Thinking of the plane as the complex plane, we may regard its points as complex numbers. Let a, b, and c be the vertices of an equilateral triangle in counterclockwise order in the complex plane. Let z be the complex number which rotates, by multiplication, a point 60 degrees counterclockwise about the origin. Then the point c can be written as c=a+z(ba). If a and b are integral lattice points, then so is ba (i.e., its real and imaginary parts are integers). However, the imaginary part of z is irrational; whence z(ba) has at least one irrational coordinate. Adding that to a results in a point with at least one irrational coordinate as well, since a has an integer coordinates. Hence c has at least one irrational coordinate. So, no, an equilateral triangle in the plane cannot have all its vertices at integral lattice points.

Posted by McWorter
on 20050303 05:08:09 