Prove (again?!) that a triangle whose vertex points with integer coordinates can't be equilateral? I'm reading Courant's Calculus, there is the following exercise to prove. I've seen other questions with answers, but they (the answers) were somewhat different of mine. I want to know if my "proof" works. 

In an ordinary system of rectangular coordinates, the points for with both coordinates are integers are called lattice points. Prove that a triangle whose vertussies are lattice points cannot be equilateral.

I did this: Taking the following triangle:
$ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad $  
Then we must have:
$$m=2x \\ m=\sqrt{x^2+y^2}$$
Now,
$$m^2=4x^2 \\ m^2=x^2+y^2$$
As $\displaystyle x=\frac{m}{2}$, we have:
$$m^2=4x^2 \\ 3m^2=y^2$$
And:
$$y=\sqrt{3}m$$
Now suppose we take an integer $m_0>0$, $\displaystyle x=\frac{m_0}{2}$ and $y=\sqrt{3}m_0$. As $m_0$ is an integer, it can be uniquely factored in a product of primes $m_0=2^{\alpha_1}\cdot 3^{\alpha_2}\dots p^{\alpha_n}$ with $\alpha_i\in \Bbb{N}$. Now:
$$y=\sqrt{3}(2^{\alpha_1}\cdot 3^{\alpha_2}\dots p^{\alpha_n})=3^{\frac{1}{2}}(2^{\alpha_1}\cdot 3^{\alpha_2}\dots p^{\alpha_n})=2^{\alpha_1}\cdot 3^{\alpha_2+\frac{1}{2}}\dots p^{\alpha_n}$$
Contradiction because $\alpha_2 + \frac{1}{2}\not\in \Bbb{N}$. With this, we actually took integers $\geq 2$, it remains to be proved that $m_0=1$ also won't provide an equilateral triangle, but this is trivial because: 
$$y=\sqrt{3}\cdot 1= \sqrt{3}\not\in \Bbb{N}$$
Is my proof correct?
 A: You were on the right trail, the irrationality of $\sqrt{3}$ is crucial, here. If such a triangle were possible, we'd have two integer vectors $(x_1,y_1)$ and $(x_2,y_2)$ forming an angle of $60°$. That means that their scalar product would be $\cos 60°=1/2$ times the product of their lengths, and the area of the parallelogram formed by them would be $\sin 60°=\sqrt{3}/2$ times the product of their lengths. The squares of those equations would be $$(x_1y_1+x_2y_2)^2=\frac{1}{4}(x^2_1+x^2_2)(y^2_1+y^2_2)$$ and $$(x_1y_2-x_2y_1)^2=\frac{3}{4}(x^2_1+x^2_2)(y^2_1+y^2_2).$$ Dividing them gives $$\left(\frac{x_1y_2-x_2y_1}{x_1y_1+x_2y_2}\right)^2=3,$$ i.e. $3$ would be the square of a rational number.
A: It's not restrictive to assume that one vertex is the origin, because a translation moving a lattice point to a lattice point also moves all lattice points to lattice points.
Let $(x_1,y_1)$ and $(x_2,y_2)$ be the other two vertices; then the area of the triangle is
$$
\frac{1}{2}
\left|
  \det\begin{bmatrix} x_1 & x_2 \\ y_1 & y_2 \end{bmatrix}
\right|
$$
On the other hand, the area is also $\frac{1}{2}l^2\sin\frac{\pi}{3}$, where $l$ is the length of the sides and
$$
l^2=(x_1-x_2)^2+(y_1-y_2)^2
$$
This implies
$$
\frac{\sqrt{3}}{2}=\frac{|x_1y_2-x_2y_1|}{(x_1-x_2)^2+(y_1-y_2)^2}
$$
is a rational number.
