Erdős Problem #215
PROVED LEANdatabase updated 2026-08-24
The statement as posed is true, and a proof exists. A solution has also been checked by machine, in Lean.
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Does there exist \(S\subseteq \mathbb{R}^2\) such that every set congruent to \(S\) (that is, \(S\) after some translation and rotation) contains exactly one point from \(\mathbb{Z}^2\)? An old question of Steinhaus. Erdős was 'almost certain that such a set does not exist'. In fact, such a set does exist, as proved by Jackson and Mauldin [JaMa02]. Their construction depends on the axiom of choice. The plane is identified with \(\mathbb{C}\): the sets congruent to \(S\) are the sets \(uS + t\) with \(\lvert u\rvert = 1\) and \(t \in \mathbb{C}\), and \(\mathbb{Z}^2\) is the set of Gaussian integers.
Statement as formalized in Formal Conjectures (Google DeepMind, Apache-2.0) · their label: research solved. The authoritative wording is on erdosproblems.com/215.Prizenone
Topicsgeometry
Statement in LeanFormal Conjectures
Solution in Leanyes
Original pageerdosproblems.com/215
Attempts
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