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(b-a).  If a and b are integral lattice points, then so is b-a (i.e., its real and imaginary parts are integers).  However, the imaginary part of z is irrational; whence z(b-a) 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 2005-03-03 05:08:09