Triangle In A Haystack

Let $\left(A,B,C\right)$ be the vertices and $\left(a,b,c\right)$ be the sides of any right triangle, where $C$ has coordinates $\left(0,0\right)$ . Then, without any loss in generality (i.e., more than we need), we consider all points $A,B$ with coordinates

$A=\left(\sqrt{3}{A}_{x},\;{A}_{y}\right)$
$B=\left(\sqrt{3}{B}_{x},\;{B}_{y}\right)$

where ${A}_{x},\;{A}_{y},\;{B}_{x},\;{B}_{y}$ are integers, so that we have

$3{{A}_{x}}^{2}+{{A}_{y}}^{2}={a}^{2}$
$3{{B}_{x}}^{2}+{{B}_{y}}^{2}={b}^{2}$
$3{{A}_{x}}^{2}+{{A}_{y}}^{2} + 3{{B}_{x}}^{2}+{{B}_{y}}^{2}={c}^{2}$

Thus, $a,b,c$ have either integer lengths or square root of integers lengths. But no sum of square roots of integers is ever rational, so all $a,b,c$ must be integers, i.e., must be a Pythagorean triple. A property of all Pythagorean triples without a common factor is that one and only one of $a,b,c$ is a multiple of $3$ . Given the expressions for $a,b,c$ above, it is an impossibility for either $a$ or $b$ to be a multiple of $3$ . This leaves $c$ . In order for $c$ to be a multiple of $3$ , then the sum of two integer squares ${{A}_{y}}^{2},\; {{B}_{y}}^{2}$ has to be a multiple of $3$ . But this is an impossibility as well if neither is a multiple of $3$ . Hence no three lattices points on the honeycomb lattice can form a right triangle with a rational perimeter .