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The claim. Call $A\subseteq\{1,\ldots,N\}$ square-sum-free if no nonempty subset of $A$ has a square sum, and let $SF(N)$ be the largest size of such a set. There is an absolute constant $C$ such that $SF(N)\le N^{1/3}(\log N)^C$ for all $N\ge2$ (Theorem 1.4). With Erdős's lower bound $SF(N)\gg N^{1/3}$, from the first $k$ multiples of a prime $p$ of order $N^{2/3}$ with $1+\cdots+k<p$ (the paper's Example 1.2), this gives $SF(N)=N^{1/3+o(1)}$, which the paper's abstract presents as the answer to the question of Problem 587. The bound is derived from Theorem 1.5: for large $N$, every $p<N^{2/3}(\log N)^{-C}$ and every $A\subseteq\{1,\ldots,\lfloor N/p\rfloor\}$ of size $N^{1/3}(\log N)^C$, some subset sum of $A$ equals $pz^2$ for an integer $z$. Earlier upper bounds were Alon's $O(N/\log N)$, Lipkin's $O(N^{3/4+o(1)})$, Alon and Freiman's $O(N^{2/3+o(1)})$ and Sárközy's $O(\sqrt{N\log N})$, as the paper's introduction recounts. The source is H. H. Nguyen and V. H. Vu, Squares in sumsets, in An Irregular Mind (Szemerédi is 70), Bolyai Society Mathematical Studies 21, Springer (2010), 491--524, DOI 10.1007/978-3-642-14444-8_14; arXiv:0811.1311, submitted 9 November 2008 (the date this page is named by) and revised 29 October 2009. The paper is carded as [[../library/integer_sequences/nguyen_2010_squares_sumsets/_index|Nguyen and Vu (2010)]]. Question 1.1, Example 1.2 and the statements of Theorems 1.4 and 1.5 are cited from the arXiv version; nothing here is independently reviewed by this project.

What is settled. The size of the largest square-sum-free subset of $\{1,\ldots,N\}$ is $N^{1/3+o(1)}$: the exponent is determined. The exact value of $SF(N)$ and the power of the logarithm between $N^{1/3}$ and $N^{1/3}(\log N)^C$ are not; the site describes the result as solving the problem in essence, and its label SOLVED attaches to the order of magnitude. This page records the claim at that scope, which is how the site reads the question.

Reported gap. A Lean development in Boris Alexeev's repository plby/lean-proofs (August 2026; informal authors Nguyen and Vu, formal authors Codex and GPT-5.6 Sol, as its header names them), linked above as a formalization, gives a checked counterexample ($d=2$, $q=4$, residue $2$) to a step in the proof of Lemma 4.2 (Section 6). The step takes the number of $|s|\le4MN/q$ with $d\mid r_a+sq$ to be $4MN/(qd)+O(1)$, where $r_a\in[0,q-1]$ is the residue with $ar_a\equiv r\pmod q$, and this fails when $\gcd(d,q)>1$: the development's audit shows that no bounded error term holds there without a coprimality condition. The development replaces the lemma with a corrected finite form, a divisor envelope in place of the paper's zero-residue estimate, and proves the bound of Theorem 1.4 by that route; it describes itself as an unconditional formalization of the paper's bounds that does not assume the lemma as published. It was neither built nor audited here, so the published proof carries this reported gap and the repair is not verified in this corpus. The same repository's independent reconstruction of a stronger bound is the pending claim [[problems/integer_sequences/E0587/claims/2026_08_27_alexeev|Alexeev's log-log page]].

Acceptance. Reviewed: the site's curator, Thomas Bloom, independent of the authors, labels the problem SOLVED, and the commentary credits the solution to this paper with the bound $\lvert A\rvert\ll N^{1/3}(\log N)^{O(1)}$, after Erdős's $N^{1/3}$ construction; page last edited 27 October 2025, so the credit predates the gap report of August 2026; the problem's thread carries no comments and no proof claims. Publication: a chapter of a Springer volume in the Bolyai Society Mathematical Studies series; the volume's refereeing is not documented here, so the publication is cited and not counted as refereed. The site also points to Problem 438 as related.

Depends on. Nothing in this wiki: the theorem is proved within the paper, whose card is linked above.