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.

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.

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).