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Claim. Let $M(R)$ be the supremum of the measures of measurable sets $A$ in the disc of radius $R$ about the origin such that $|a-b|$ is never a positive integer for distinct $a,b\in A$, the quantity Problem 953 asks about. Theorem 1.1 of Przemek Chojecki's note A Poisson–Bessel kernel bound for planar sets avoiding integer distances states that $M(R)\ll R^{1/2}$ for all $R\geq1$, and consequently $M(R)=R^{1/2+o(1)}$ as $R\to\infty$; Theorem 1 of the shorter note The order of growth of planar sets avoiding integer distances states the same. The proof reduces the measurable problem to the uniform estimate $N(X,\delta)\leq C\delta^{-2}X^{1/2}$ for $X\geq1$ and $0<\delta<1/10$ (Theorem 1.2), where $N(X,\delta)$ counts points in a disc of radius $X$ whose pairwise distances stay at least $\delta$ from the integers, and proves that estimate with the positive-definite kernel $K_s(t)=\sum_{k\geq1}(k+2sk^2)e^{-sk}J_0(2\pi kt)$, whose Poisson expansion has non-positive terms away from $s$-neighborhoods of the integers. The matching lower bound $M(R)\gg_\varepsilon R^{1/2-\varepsilon}$ comes from Sárközy's theorem on robust point sets. Both notes read the question as asking for the order of growth of $M(R)$ and take $M(R)=R^{1/2+o(1)}$ as its answer; the problem page's Formulation also asks whether the factor $R^{o(1)}$ can be removed. The notes are carded at chojecki_2026_poisson_bessel_kernel_bound_planar_sets and chojecki_2026_order_growth_planar_sets_avoiding_integer.

Submission note. Posted to the site's forum by Przemek Chojecki on 27 April 2026:

Let $M(R)$ denote the supremum of the measures of measurable sets $A\subset B_R(0)\subset\mathbb R^2$ such that $|a-b|$ is never a positive integer for distinct $a,b\in A$. With GPT-5.5 Pro I've got

$$M(R)\ll R^{1/2} \qquad (R\geq > 1).$$

by using a Poisson-Bessel kernel. Consequently,

$$> M(R)=R^{1/2+o(1)}\qquad (R\to\infty).$$

Here's the note with a proof.

EDIT: streamlined and cleaned version is here following comments by Vjeko.

Covers. The exponent of growth of $M(R)$: jointly with the lower bound $M(R)\gg_\varepsilon R^{1/2-\varepsilon}$ that the site credits, Sárközy's theorem as Koizumi and Kovač adapted it on the site's thread, the theorem settles the exponent $1/2$ in $M(R)=R^{1/2+o(1)}$. The claim's value is answered because the result determines that exponent. Not covered: the constant-factor question, whether $M(R)$ has order exactly $R^{1/2}$, that is, whether the factor $R^{o(1)}$ between the two bounds can be removed; the problem stays open on it.

Claimant. Chojecki posted the bound on the site's discussion thread on 27 April 2026, crediting GPT-5.5 Pro, with the long note; the short note, which Chojecki attributed to GPT-5.5, followed on 28 April 2026. Nat Sothanaphan posted a streamlined write-up of the same argument on 28 April 2026, prepared with GPT-5.5 Thinking (Theorem 2.1 of that write-up is the same bound); it is linked above and carded at sothanaphan_2026_compact_poissonbessel_proof_integer_distance_free.

Depends on. Sárközy's power lower bound, which supplies the lower half of $M(R)=R^{1/2+o(1)}$.

Acceptance. Reviewed: on 8 May 2026 Vjekoslav Kovač, who is independent of the claimant, wrote on the site's thread that the proof Chojecki obtained is correct and that Sothanaphan's simplification presents it well. Not refereed. The site's curator has not credited the result: the site labels the problem OPEN. Allen Hart's Lean development, linked above at a pinned commit, declares Chojecki's long note its primary informal source and proves the upper bound alone (erdos953_upper, stated for closed discs); this corpus has not built or audited it, so formalized is not listed.