Prove that a triangle with lattice points cannot be equilateral

[JoeAllen]

I assumed three points for a triangle P1 = (a, c), P2 = (c, d), P3 = (b, e)

and of course:
a, b, c, d, e∈Z
Using the distance formula between each of the points and setting them equal:
\sqrt { (b - a)^2 + (e - d)^2 } = \sqrt { (c - a)^2 + (d - d)^2 } = \sqrt { (b - c)^2 + (e - d)^2 }(e+d)² = (c-a)² - (b-a)²
(e+d)² = (c-a)² - (b-c)²

c² - 2ac - b² +2ab = -2ac + a² - b² + 2bc
c² + 2ab = a² + 2bc
c(c - 2b) = a(a - 2b)

Thus, for this to be true, a = c. But in this example, the distance between a and c would be 0. Thus, not a triangle and certainly not an equilateral triangle.

Where did I go wrong here? I'm bored waiting for Calculus II in the Fall and I'm going through Courant's Differential and Integral Calculus on my free time until then (Fall term probably starting in August/September, so I'm not worried if it takes a few months to get comfortable with Courant - Calculus I has been a breeze since I already knew most of the content before taking it).

 

[anuttarasammyak]

As for 2D lattice we can make one of the lattice points is (0,0) without losing generality.
Say other points are $(n_1,n_2),(m_1,m_2)$
n_1^2+n_2^2=A
m_1^2+m_2^2=A
(n_1-m_1)^2+(n_2-m_2)^2=A
where A is square of the side length. You will find contradiction in this set of formla.