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Statement. The lattice

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

contains no four-point set similar to the isosceles trapezoid formed by four vertices of a regular pentagon.

Source. Grayzel, Solution to a Problem of Erdős Concerning Distances and Points, arXiv:2601.09102v2, Lemma 8 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. Let $s$ and $d$ be respectively the side and diagonal lengths in a regular pentagon. Apply Ptolemy's identity to the cyclic quadrilateral formed by four consecutive vertices. Its three consecutive sides have length $s$, its fourth side has length $d$, and both diagonals have length $d$. Hence

$$d^2=s^2+sd.$$

For $\rho=d/s>0$, this says $\rho^2=1+\rho$, so

$$\rho=\frac{1+\sqrt5}{2} \qquad\text{and}\qquad \rho^2=\frac{3+\sqrt5}{2}. \tag{1}$$

The number in (1) is irrational, and similarity preserves this ratio of squared distances.

On the other hand, the difference of any two lattice points is $(u,\sqrt2\,v)$ with $u,v\in\mathbb Z$, so its squared length is

$$u^2+2v^2\in\mathbb Z.$$

The ratio of any two nonzero squared distances determined by points of $L$ is therefore rational. It cannot equal the irrational regular-pentagon ratio in (1). This excludes every similar copy of the trapezoid from $L$. $\square$

Used by. Theorem 5.

Bears on. Problem 659.