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Source. Problem 29, p. 14 (Section 7, "Distinct distances with local properties", pp. 13--15), with the definition of $\phi(n,k,\ell)$ and Table 3 on p. 13, of Adam Sheffer, Distinct Distances: Open Problems and Current Bounds, arXiv:1406.1949v3 (2 July 2018), the edition read for the source card.

Statement

Notation (p. 13). For positive integers $k,\ell$, $\phi(n,k,\ell)$ is the least number of distinct distances spanned by a set of $n$ points in the plane in which every $k$ points determine at least $\ell$ distinct distances. The survey credits Erdős with first suggesting its study.

Table 3 (p. 13), row $\phi(n,4,3)$: lower bound $\Omega(n/\log n)$, credited to Guth and Katz; upper bound $O(n/\sqrt{\log n})$, with no citation.

The survey's argument (p. 14). It reads $\phi(n,4,3)$ as the least number of distinct distances among $n$ points that span no square. The $\sqrt n\times\sqrt n$ section of the triangular lattice determines $\Theta(n/\sqrt{\log n})$ distinct distances and contains no square, which the survey takes to give $\phi(n,4,3)=O(n/\sqrt{\log n})$. The lower bound is $\phi(n,4,3)\ge D(n)=\Omega(n/\log n)$.

Problem 29 (p. 14). "Find the asymptotic value of $\phi(n,4,3)$." (quoted)

On the upper bound. A square is not the only four-point set with at most two distinct distances. The triangular lattice contains others: the rhombus made of two equilateral triangles of side 1 has its four sides and its short diagonal of length 1 and its long diagonal of length $\sqrt3$. So the triangular lattice does not have the property that every four points determine at least three distances, and the argument on p. 14 does not prove the upper bound listed in Table 3. Terence Tao pointed this out on the erdosproblems.com discussion thread for Problem 659 on 13 January 2026. This page records the upper bound only as printed.

Read depth

Claims checked: the definition, Table 3, the argument on p. 14 and Problem 29 were read clause by clause on the print. The cited lower bound and lattice count were not checked against their sources here.

Bears on

• Problem 659: the problem asks whether some $n$ planar points, every four of which determine at least three distances, determine $\ll n/\sqrt{\log n}$ distances in total, that is, whether $\phi(n,4,3)=O(n/\sqrt{\log n})$. Table 3 lists this bound, the affirmative answer, but the survey's argument for it fails as stated above, and the survey also poses the asymptotic value as open (Problem 29).