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Statement. The lattice

$$L=\{(x,\sqrt2\,y):x,y\in\mathbb Z\}$$

contains no nondegenerate equilateral triangle.

Source. Grayzel, Solution to a Problem of Erdős Concerning Distances and Points, arXiv:2601.09102v2, Lemma 7 and proof on p. 4. See the arXiv v2 PDF.

Verification scope. Author-recorded; this component belongs to the proof chain recorded on the single living [[distance_problems/grayzel_2026_solution_problem_erdos_concerning_distances_points/theorem_1|Current verification]] record on Theorem 1, where an independent review is reported but its report is not retained in this repository.

Proof. Suppose that $p,q,r\in L$ formed a nondegenerate equilateral triangle. Put

$$w=q-p=(u,\sqrt2\,v), \qquad u,v\in\mathbb Z, \qquad (u,v)\ne(0,0).$$

The vector $r-p$ must be obtained by rotating $w$ through either $60^\circ$ or $-60^\circ$. If $\epsilon\in\{1,-1\}$ records the sign of this rotation, then

$$R_{\epsilon60}w =\left( \frac{u-\epsilon\sqrt6\,v}{2}, \frac{\epsilon\sqrt3\,u+\sqrt2\,v}{2} \right). \tag{1}$$

Because $r-p\in L$, its second coordinate equals $\sqrt2\,k$ for some $k\in\mathbb Z$. Equation (1) therefore gives

$$\epsilon\sqrt3\,u=\sqrt2\,(2k-v)\in\mathbb Q(\sqrt2). \tag{2}$$

We have $\sqrt3\notin\mathbb Q(\sqrt2)$. Indeed, if $\sqrt3=a+b\sqrt2$ for rational $a,b$, then squaring and comparing the rational and $\sqrt2$ parts gives $ab=0$ and $a^2+2b^2=3$. The cases $a=0$ and $b=0$ would say respectively that $3/2$ or $3$ is a square in $\mathbb Q$, both impossible by unique factorization.

It follows from (2) that $u=0$. The first coordinate in (1) then becomes $-\epsilon\sqrt6\,v/2$. It must be an integer because $r-p\in L$, and the irrationality of $\sqrt6$ forces $v=0$. This contradicts $w\ne0$. The same calculation covered both rotation signs, so no nondegenerate equilateral triangle lies in $L$. $\square$

Used by. Theorem 5.

Bears on. Problem 659.