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Problem 1088

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claims/: The 4 claim pages of Problem 1088, one per claimant's result; the problem's standing derives from them.


Statement. Let fd(n)f_d(n) be the minimal mm such that any set of mm points in Rd\mathbb{R}^d contains a set of nn points such that any two determined distances are distinct. Estimate fd(n)f_d(n). In particular, is it true that, for fixed n≥3n\geq 3,

fd(n)=2o(d)?f_d(n)=2^{o(d)}?

Status. Open, in the site's label (OPEN; page last edited 8 April 2026). The site's remarks credit exact values and orders of growth for small cases, recorded as partial claims in claims/; the question whether fd(n)=2o(d)f_d(n)=2^{o(d)} is open for every n≥4n\ge4, so the standing in the frontmatter, derived from the claim pages, is open.

Source. erdosproblems.com/1088, accessed 2026-09-04. Cite as: T. F. Bloom, Erdős Problem #1088, https://www.erdosproblems.com/1088.

References.

  • [Cr62] Croft, H. T., 99-point and 77-point configurations in 33-space. Proc. London Math. Soc. (3) (1962), 400-424.
  • [Er75f] Erdős, Paul, On some problems of elementary and combinatorial geometry. Ann. Mat. Pura Appl. (4) (1975), 99-108.

Formalization. Statement in formal-conjectures.

Current assessment

The site's formulation (page last edited 8 April 2026) asks for an estimate of fd(n)f_d(n), the least mm such that every mm points of Rd\mathbb{R}^d contain nn points whose pairwise distances are all distinct, and in particular whether fd(n)=2o(d)f_d(n)=2^{o(d)} for each fixed n≥3n\ge3. The general estimate is open; four results settle instances.

The case n=3n=3. Three points have three distinct distances exactly when they do not form an isosceles triangle, so fd(3)−1f_d(3)-1 is the largest size of an isosceles set in Rd\mathbb{R}^d, the subject of Problem 503. The solution of Monthly problem E735 gives f2(3)=7f_2(3)=7 ([[problems/discrete_geometry/E1088/claims/1947_04_01_erdos_kelly|claim page]]), and Croft [Cr62] proved f3(3)=9f_3(3)=9 (claim page), both accepted on their journal publication. Blokhuis's bound for isosceles sets gives fd(3)≤(d+1)(d+2)/2+1f_d(3)\le(d+1)(d+2)/2+1, and two-distance sets give a lower bound of the same order, so fd(3)=d2/2+O(d)f_d(3)=d^2/2+O(d) and the answer for n=3n=3 is yes; Erdős [Er75f] had written that he and Straus could not prove this even for n=3n=3. That claim page (Blokhuis) is pending, since the source is a CWI Tract and not a journal publication.

The case d=1d=1. Points of the line have distinct distances exactly when they form a Sidon set, the subject of Problem 530, and f1(n)≍n2f_1(n)\asymp n^2: the upper bound is the theorem of Komlós, Sulyok and Szemerédi ([[problems/discrete_geometry/E1088/claims/1975_01_01_komlos_sulyok_szemeredi|claim page]], accepted on its journal publication) and the lower bound the Erdős–Turán bound for Sidon subsets of {1,…,N}\{1,\ldots,N\}. The constant is open.

General bounds. The site's remarks call fd(n)≤nOd(1)f_d(n)\le n^{O_d(1)} easy, and Erdős [Er75f] reports an unpublished bound fd(n)≤cndf_d(n)\le c_n^d of Erdős and Straus. Neither settles an instance, so neither has a claim page. The behavior of fd(n)f_d(n) for fixed dd as n→∞n\to\infty is Problem 1208. The question whether fd(n)=2o(d)f_d(n)=2^{o(d)} is open for every n≥4n\ge4.

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