Can't Place An Equilateral Triangle On A Square Grid

Let the grid points be the integer coordinates in the plane.  Place one triangle corner at the origin (x0=y0=0).
We can choose the triangle side length L and the angle of its baseline a at will.

Assume we have successfully placed the triangle with all three corners on grid points, using some side
length L and angle a.  Then, for the second corner,

x1 = L*cos(a) = I1  (integer)
y1 = L*sin(a) = I2  (integer)

and I1 and I2 must be integers.  Then

tan(a) = I2/I1 = R1  (rational)

Similarly, the third corner is on some other grid  point.  That point lies at angle a+60, and we have

x2 = L*cos(a+60) = I3  (integer)
y2 = L*sin(a+60) = I4  (integer)

and

tan(a+60) = I4/I3 = R2  (rational)

By trig identity for tangent, sum of angles:

tan(a+60) = [tan(a)+tan(60)]/[1-tan(a)*tan(60)]

Now substitute from above:

R2 = [R1 + tan(60)]/[1-R1*tan(60)]

Isolate tan(60):

R2*[1-R1*tan(60)] = R1 + tan(60)

R2 - R1*R2*tan(60) = R1 + tan(60)

[1+R1*R2]*tan(60) = R2-R1

tan(60) = [R2-R1]/[1+R1*R2]  (rational)

But tan(60)=sqrt(3), which is irrational.  Conflict - the third triangle corner cannot lie on a grid point.

Q.E.D.