GPT 6 Astra Ultra attempt on Math-5782
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Erdős Problem #635: independent continuation
Date: 2026-09-23. Research attempt; no claim of a new published result.1) TARGET AND SOURCE AUDIT
For fixed $t\ge1$, maximize $|A|$ for $A\subseteq[1,N]$ subject to $(b-a)\nmid b$ whenever $a,b\in A$ and $b-a\ge t$.All supplied versions considered: 635.tex. The source identifies the correct numbering and proves the already known asymptotic half-density statement using an Elliott inequality. I give an elementary second-moment proof with an explicit error in terms of a prime harmonic sum. The exact extremal function remains open.
2) WORK
Put $\alpha=|A|/N$, $X=\lfloor N^{1/4}\rfloor$, let $\mathcal P=\{p\text{ prime}:t\le p\le X\}$, $P=|\mathcal P|$ and $S=\sum_{p\in\mathcal P}1/p$. Suppose $S>0$. Define \[ U(n)=\sum_{p\in\mathcal P}(1_{p\mid n}-1/p). \] Among multiples of a fixed $p\ge t$, the selected indices cannot be consecutive. Hence at most $N/(2p)+1$ are in $A$, and \[ \sum_{n\in A}U(n)\le N(1/2-\alpha)S+P. \tag{1} \] For uniform $n\in[1,N]$, direct floor-function expansion gives $\mathbb E(1_{p\mid n}-1/p)^2\le1/p$ and, for distinct primes, \[ \left|\mathbb E(1_{p\mid n}-1/p)(1_{q\mid n}-1/q)\right|\le {2\over N}. \] Indeed the three rounding errors have coefficients $1,1/p,1/q$; their sum is at most two. Consequently \[ \mathbb E U^2\le S+3P^2/N. \] Cauchy--Schwarz bounds the absolute value of the left side of $1$ by $N\sqrt{\alpha(S+3P^2/N)}$. Combining and using $\alpha\le1$ gives the explicit bound \[ \alpha\le {1\over2}+{\sqrt{S+3P^2/N}\over S}+{P\over NS}. \] For fixed $t$, the divergence of the sum of reciprocal primes gives $S\to\infty$, while $P^2/N\le N^{-1/2}$. Thus $\alpha\le1/2+o_t(1)$ follows without the quoted sampling theorem. Odd integers supply the matching asymptotic lower density: if $a,b$ are odd and $b-a>0$, their even difference cannot divide the odd number $b$.3) ADVERSARIAL VERIFICATION
The covariance uses distinct primes so that the common multiples have density $1/(pq)$ up to one rounding unit. The prime cutoff keeps the accumulated rounding error small. The proof treats fixed $t$; it is not uniform for arbitrary $t=t(N)$. It re-proves a known part of the problem, rather than solving the exact question.4) FINAL
UNRESOLVED. The quantitative second-moment estimate supplies no matching description of the error $M_t(N)-N/2$, nor exact maximizers for $t\ge2$.5) SOURCES CHECKED
Entire supplied file read. The official indexed pagehttps://www.erdosproblems.com/635 checked on 2026-09-23 confirms that the asymptotic half-density part was already resolved. The proof above is elementary; it does not depend on an unchecked version of Elliott's theorem.