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Statement. The lattice

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

contains no nondegenerate square.

Source. Grayzel, Solution to a Problem of Erdős Concerning Distances and Points, arXiv:2601.09102v2, Lemma 6 and proof on p. 3. 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 four points of $L$ formed a nondegenerate square. A vector along one side would have the form

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

The vector along an adjacent side is a rotation of $w$ through either $90^\circ$ or $-90^\circ$. Thus, for some $\epsilon\in\{1,-1\}$, it is

$$w'=(-\epsilon\sqrt2\,v,\epsilon u).$$

Both endpoints of that side are in $L$, so their difference $w'$ also lies in $L$. Its first coordinate must be an integer. Hence $\sqrt2\,v\in\mathbb Z$, which forces $v=0$ because $v$ is integral and $\sqrt2$ is irrational. Its second coordinate must belong to $\sqrt2\mathbb Z$. Hence $u\in\sqrt2\mathbb Z$; since $u$ is also an integer, this forces $u=0$.

We obtain $w=0$, contradicting that a side of a nondegenerate square has positive length. $\square$

Used by. Theorem 5.

Bears on. Problem 659.