Equilateral triangle whose vertices are lattice points? Is it possible to construct an equilateral triangle with vertices on lattice points?

I think the answer is no, but how can I prove this?

I started with a triangle with coordinates $(0,0)$ $(a,b)$ and $(c,d)$.
Equating the size of the 3 sides, I get

$a^{2}+b^{2}=c^{2}+d^{2}=2ab+2cd$

How should I continue?



I see there are solutions based on the fact that the angle between two edges can not be 60°. Is it possible to have a solution based on the fact that the length of the edges can not be the same?
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.
You can start like this. Without loss of generality, let the three points be $(0,0), (0,a)$ and $(b,a/2)$. You can do this because you can always rotate and translate the axis to get these points. Now for all the points to be integral, you need $a$ to be even. This is the first constraint. Secondly, from the basic trigonometry, 

$$ \tan \theta = \frac{2b}{a} = \sqrt{3}. $$ From this, you get $b = a\sqrt{3}/2$, which is irrational.