Let S⊂R^2 be a set such that for all x,y∈ S, operatornamedist(x,y)≠1. Must there exist four points p_0,p_1,p_2,p_3∈R^2 setminus S forming a congruent copy of the unit square with vertices (0,0),(1,0),(1,1),(0,1)?

Erdős Problem #214 — Unit squares in the complement of a unit-distance-avoiding set

The problem asks whether every subset of the plane with no unit-distance pair forces a unit square entirely in its complement. No proposer or date is identified in the retrieved sources.

Known results

A 2006 paper states that the vertices of a unit square are Jackson: every subset $S\subseteq\mathbb{R}^2$ has a congruent copy $Y$ with $|Y\cap S|\ne 1$ . This property alone does not imply the desired conclusion.

The claimed deduction is invalid

The retrieved discussion says a unit square has at most one vertex in $S$ , but forbidding unit-distance pairs permits two opposite vertices of a square to lie in $S$ . Thus the Jackson property allows $|Y\cap S|=2$ and does not produce a square disjoint from $S$ .

Current status (as of May 2026): The problem remains open; the Jackson-property result is relevant but does not establish it, and no valid proof or counterexample was found.

Erdős Problem #214 — Unit squares in the complement of a unit-distance-avoiding set

Primary source

Additional references

Progress summary

Solutions 0