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Claim. There is a constant c>0c>0 such that every set of NN integers contains a Sidon subset of size at least cN1/2cN^{1/2} for all sufficiently large NN: ℓ(N)≫N1/2\ell(N)\gg N^{1/2} in the notation of Problem 530. The theorem of J. Komlós, M. Sulyok and E. Szemerédi, Linear problems in combinatorial number theory, Acta Math. Acad. Sci. Hungar. 26 (1975), no. 1--2, 113--121, cited as [KSS75] on the problem page, compares, for a fixed translation-invariant linear relation, the largest relation-free subset guaranteed in every nn-element set of integers, g(n)g(n), with the largest relation-free subset of {1,…,n}\{1,\ldots,n\}, f(n)f(n): for all sufficiently large nn, g(n)≥f(n)/(8α6)g(n)\ge f(n)/(8\alpha^6), where α\alpha is the largest row ℓ1\ell^1-norm of the relation's coefficients. The Sidon condition forbids the translation-invariant relations a+b=c+da+b=c+d and 2a=c+d2a=c+d among distinct elements, each with α≤4\alpha\le4; the paper counts it among its translation-invariant relations (Remark 2, p. 114), and its residue reductions and translation step preserve every solution of both at once, so the bound applies with α=4\alpha=4; and f(n)=(1+o(1))n1/2f(n)=(1+o(1))n^{1/2} for Sidon sets by the Erdős--Turán upper bound and Singer's construction, so ℓ(N)≥(2−15+o(1))N1/2\ell(N)\ge(2^{-15}+o(1))N^{1/2} for sets of integers. Library home komlos_1975_linear_problems_combinatorial_number_theory; result page translation_invariant_theorem. The paper states its theorem for sets of integers, while the problem is posed for finite sets of real numbers; the transfer is the standard one, recorded here as the page's own remark: a finite set of reals is Freiman isomorphic of order 22 to a finite set of integers, and such an isomorphism carries Sidon subsets to Sidon subsets, so the integer bound gives the real one with the same constant. Bailleul and Riblet (2026) perform the same transfer by Dirichlet approximation, and the site's commentary credits the real bound N1/2≪ℓ(N)N^{1/2}\ll\ell(N) to [KSS75] directly.

Covers. The lower bound ℓ(N)≫N1/2\ell(N)\gg N^{1/2} alone. With the upper bound ℓ(N)≤(1+o(1))N1/2\ell(N)\le(1+o(1))N^{1/2} from A={1,…,N}A=\{1,\ldots,N\} it fixes the order of ℓ(N)\ell(N) at N1/2N^{1/2}, the problem's first question, and it supersedes Erdős's earlier ℓ(N)≫N1/3\ell(N)\gg N^{1/3}. It does not determine the constant, so the problem's second question, whether ℓ(N)∼N1/2\ell(N)\sim N^{1/2}, stays open. The later explicit constants, Abbott's c<2/25c<2/25 (1990) and Bailleul and Riblet's 1/(33)+o(1)1/(3\sqrt3)+o(1) (2026), improve the constant only and are recorded on the problem page without claim pages.

Depends on. No page of this wiki: the comparison theorem and its proof are the paper's own, and the Sidon inputs f(n)=(1+o(1))n1/2f(n)=(1+o(1))n^{1/2} are the classical Erdős--Turán and Singer results.

Acceptance. Refereed: the paper is a journal publication in Acta Mathematica Academiae Scientiarum Hungaricae, volume 26, issue 1--2 (1975), received 20 November 1973, the refereed evidence; the citation gives the year without a month or day, so this page is dated to the first day of that year. The site's curator, Thomas Bloom, credits the bound N1/2≪ℓ(N)N^{1/2}\ll\ell(N) to this paper in the problem page's commentary, but the site labels the problem OPEN, so that credit is not reviewed evidence. The library's result pages reconstruct the paper's proof chain and await independent review; they award nothing by this corpus.