Prove that an Equilateral cannot have natural number points

Question

Let $ OAB $ be an equilateral triangle with $O(0, 0),\ A(m, n),\ B(x, y)$, where $m, n \in \mathbb{N}^{\ast}$ and $x, y \in \mathbb{R}_{+}$.

Prove that $B$'s coordinates can't be both natural numbers.

So far, using the distance formula, I got the following relations: $$x^2 + y^2 = m^2 + n^2 = 2mx + 2ny$$

Now, I have no idea how to continue, so I would be really grateful if someone could help me. Thanks in advance!

Answer 1

HINT

Note that from your equations, $x^2 + y^2 = m^2 + n^2 = 2mx + 2ny=k$

Then, from $$(x^2+y^2)(m^2+n^2)=(mx+ny)^2+(my-nx)^2$$

We have $$\frac{1}{2}|my-nx|=\frac{\sqrt{3}}{4}(m^2+n^2)$$

But a rational number cannot be an irrational number can it?

NOTE

A slightly simpler proof would be to use the area of $\triangle OAB$, which directly gives $$\frac{1}{2}|my-nx|=\frac{\sqrt{3}}{4}(m^2+n^2)$$

Answer 2

Suppose $x,y$ are integers. From your formula, $x^2+y^2=m^2+n^2=2mx+2ny$, you deduce that $x,y,m,n$ are even. write $x=2x_1, y=2y_1, m=2m_1, n=2n_1.$ You have

${x_1}^2+{y_1}^2={m_1}^2+{n_1}^2=2m_1x_1+2n_1y_1$. Deduce that $x_1,y_1,m_1,n_1$ are even. Recursively you must be able to define $x_i=2x_{i-1},y_i=2y_{i-1},m_i=2m_{i-1},n_i=2n_{i-1}, i\in N$ which are even. This is impossible.