Equilateral triangle whose vertices are lattice points?

Let the vertices of our triangle be $(0,0)$, $(a,b)$, and $(c,d)$, where $a$, $b$, $c$, and $d$ are integers. If all edge lengths are the same, then $$a^2+b^2=c^2+d^2=(a-c)^2+(b-d)^2.$$ Minor manipulation turns this into $$a^2+b^2=c^2+d^2=2ac+2bd.$$

Now we use my favourite identity, which was known more than a millenium ago in India, and even earlier by Diophantus, and so has often been called the Fermat Identity: $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2.\qquad\qquad(\ast)$$ This identity can be easily verified by expanding both sides, or more conceptually by noting that the norm of the product of two complex numbers is the product of the norms.

Let $N=a^2+b^2=c^2+d^2=**2(ac+bd)**$. Then $ac+bd=N/2$. The identity $(\ast)$ now gives $$N^2=\frac{N^2}{4}+(ad-bc)^2$$ or equivalently $$3N^2=4(ad-bc)^2.$$ This is impossible, since $3$ times the perfect square $N^2$ cannot be a square unless $N=0$, which gives a very tiny triangle.

It is possible, but you need three dimensions in order to do it.

Consider $\bigtriangleup v_{1}v_{2}v_{3}$ with:

$v_{1}=(1,0,0)$

$v_{2}=(0,1,0)$

$v_{3}=(0,0,1)$

For $a,b\in{1,2,3}$, $a\neq b$, $d(v_{a},v_{b})=\sqrt{2}$, therefore the triangle is equilateral. It is not possible (as other answers indicate) to have an equilateral triangle with integer coordinates for the vertices in a two dimensional square lattice (a grid is just a 2d lattice).

Solution 1 (by me): Assume WLOG that two of the points are $(0,0), (m,n), m,n \in \mathbb{Q}$. Then the third point is $(m/2 - n \sqrt{3} / 2, n/2 + m \sqrt{3}/2)$, which is not a rational point.

Solution 2 (by a friend): The determinant formula for area is rational, so if the all three points are rational points, then the area of the triangle is also rational, so whereas the area of an equilateral triangle with side length s is $\frac{s^2 \sqrt{3}}{4}$, which is irrational since $s^2$ is an integer.

Note that the above solutions both generalize from integer points to rational points.

You can also use Pick's theorem for integer points.

My favorite proof: Pick's theorem says such a triangle would have rational area, but the area is a half integral multiple of $\sqrt{3}$.

I found the same problem on page 13 in Richard Courant's classic textbook Differential and Integral Calculus:

2.* In an ordinary system of rectangular co-ordinates, the points for which both co-ordinates are integers are called lattice points. Prove that a triangle whose vertices are lattice points cannot be equilateral

My proposed solution is to set the origin at a vertex. This can be done w.l.o.g. For every lattice point $P : (p,q)$ we get an equilateral triangle with vertex $P$ and a third vertex $X$ (we also get another triangle with $P$ and $X'$, see figure) where both vertices lie on the circumference of a circle with radius $r =\sqrt(p^2+q^2)$. The angle of the $OP$ edge is $\alpha$, and $\sin a = \frac q r, \ \cos \alpha = \frac p r$.

Figure: Two equilateral triangles for every lattice point $P$

We get the coordinates of $X$ on the circle (and in a similar way for $X'$):

$(x,y) = (r \cos(\alpha+\frac \pi 3), r \sin(\alpha + \frac \pi 3))$

The addition formulas for sine and cosine give,

$$(x,y) = \left(r \left[ \cos \alpha \cos \frac \pi 3 - \sin \alpha \sin \frac \pi 3 \right], r \left[ \sin \alpha \cos \frac \pi 3 + \cos \alpha \sin \frac \pi 3 \right] \right) = \left( r \left[ \frac p {2r} -\frac {q \sqrt 3} {2r} \right], r \left[ \frac q {2r} + \frac {p \sqrt 3} {2r} \right] \right) = \left( \frac p 2 - \frac{ \sqrt 3 q} 2, \frac q 2 + \frac{\sqrt 3 p} 2 \right)$$

If $ p,q \in \Bbb Z$ then $X$ is not a lattice point.