Proof that lattice points can't form an equilateral triangle

[Figaro]

From Courant's Differential and Integral Calculus p.13,
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.

Proof: Let $A=(0,0), B=(\frac{a}{2},b), C=(a,0)$ be points in a rectangular coordinate system where $a$ and $b$ are integers. Equilateral triangles have angles 60° at every corner, and we know that $tan(60)=\sqrt3$. So, the distance from the point $B$ to the line $\bar {AC}$ is $\frac{a}{2}tan(60) = \frac{\sqrt3}{2}a$ which is an irrational number and contradicts the fact that the distance from the point $B$ to the line $\bar {AC}$ is an integer.

But according to other post and resources this proof is flawed. Can anyone help me with this?

 

[jedishrfu]

Have you tried to prove using the Pythagorean theorem?

Construct the triangle with sides 2a in length and height b?

The A vertex is at (0,0) and the B vertex is at (a,b) and the C vertex is at (2a,0)

This preserves a and b as integers and using the Pythagorean theorem removes the need for the tangent.

 

[Figaro]

Have you tried to prove using the Pythagorean theorem?

Construct the triangle with sides 2a in length and height b?

The A vertice is at (0,0) and the B vertice is at (a,b) and the C vertice is at (2a,0)

This preserves a and b as integers and using the Pythagorean theorem removes the need for the tangent.

Actually, I want to use the hint that Courant gave "Hint: Use the irrationality of $sin(60)=\frac{\sqrt3}{2}$". I used tan(60) but the idea is the same, what does he want me to do?

 

[jedishrfu]

I think I see the flaw now. Basically you want to use lattice points as the vertices that doesn't mean the the sides of the constructed equilateral triangle will have integer measure though only that the sides are all equal in measure.

You could select two vertices as A at (0,0) and B at (1,1) so that the side of the proposed equilateral triangle is $\sqrt{2}$ and you have to show that the third vertex cannot be a lattice point.

For your proof then choose A to be (0,0) and B to be (a,b) where a and b are integer values and then determine the third lattice point in terms of these two.

 

[Figaro]

I think I see the flaw now. Basically you want to use lattice points as the vertices that doesn't mean the the sides of the constructed equilateral triangle will have integer measure though only that the sides are all equal in measure.

You could select two vertices as A at (0,0) and B at (1,1) so that the side of the proposed equilateral triangle is $\sqrt{2}$ and you have to show that the third vertex cannot be a lattice point.

For your proof then choose A to be (0,0) and B to be (a,b) where a and b are integer values and then determine the third lattice point in terms of these two.

How come it cannot be a lattice point? Suppose the side $\bar {AB}$ is $\sqrt2$, then the second side should be $\sqrt2$ also but then the third would not be $\sqrt2$ which shows that it is not an equilateral triangle.

 

[jedishrfu]

You've started with the premise that you have an equilateral triangle and you fix a side via (0,0) and (a,b) vertices and then construct the third vertex from them. The third vertex of the equilateral triangle coordinates won't be integers and hence it can't be a lattice point.