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Source. Theorem 2.2, p. 2, of Jozsef Solymosi and Frank de Zeeuw, On a question of Erdős and Ulam, arXiv:0806.3095v2 (14 January 2009), published in Discrete Comput. Geom. 43 (2010), no. 2, 393-401, the version named on the source card; the proof is Section 4, pp. 6-7.

Read depth. Claims checked: the statement and the remark after it (p. 2) were read clause by clause on the printed page. The proof was read for structure only. Nothing here is independently reviewed.

Statement

Rational sets are as in Theorem 2.1: planar point sets with all pairwise distances rational.

Theorem 2.2 (p. 2, quoted). "If a rational set $S$ has infinitely many points on a line or on a circle, then all but $4$ resp. $3$ points of $S$ are on the line or on the circle."

In the corpus's words: if a rational set has infinitely many points on some line, at most 4 of its points lie off that line; if it has infinitely many points on some circle, at most 3 of its points lie off that circle.

The remark after the theorem (p. 2) notes that both bounds are attained: a construction of Huff gives an infinite rational set with all but 4 points on a line, and an inversion with rational radius centred at one of the 4 points off the line turns it into one with all but 3 points on a circle. The paper says the theorem answers questions of Guy (Problem D20 of Unsolved Problems in Number Theory) and of Pach (Section 5.11 of Brass, Moser and Pach, Research Problems in Discrete Geometry) (p. 1).

Proof pointer

Section 4 (pp. 6-7). The circle case follows from the line case: inverting a rational set with infinitely many points on a circle $C$ and at least 4 off it, centred at a point of the set on $C$ with rational radius, gives a rational set with infinitely many points on a line and, with the centre added, 5 points off it. For the line case, suppose the $x$-axis holds infinitely many points and 5 or more lie off it; three of them may be taken on one side, at $(0,1)$, $(a_1,b_1)$ and $(a_2,b_2)$. Each further point $(x,0)$ of the set then gives a rational point of $y^2=(x^2+1)((x-a_1)^2+b_1^2)((x-a_2)^2+b_2^2)$, whose right side has no repeated roots, so the curve has genus 2 and Faltings' theorem leaves only finitely many such points.

Dependencies

Faltings' theorem (cited, p. 2); Lemma 3.3 (inversion centred at a point of a rational set with rational radius keeps the rest of the set rational, p. 3).

Bears on

• Problem 212: a dense set has infinitely many points off any line or circle, so by this theorem a dense rational set, if one exists, has only finitely many points on each line and each circle; with Theorem 2.1 it meets every real algebraic curve finitely often. The paper does not draw this consequence out and does not settle the problem.