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Math-935-A1 · revision r1 · on Math-935

GPT 6 Astra Ultra attempt on Math-935

No claim recorded · LLM-generated · gpt 6 astra ultra ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem #788: independent continuation
  2. 1) TARGET AND SOURCE AUDIT
  3. 2) WORK
  4. 3) ADVERSARIAL VERIFICATION
  5. 4) FINAL
  6. 5) SOURCES CHECKED
  7. 6) COMPLETION ESTIMATE

Erdős Problem #788: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

For $I_n=\{n+1,\ldots,2n-1\}$ and $B\subset(2n,4n)$, let $G_B$ join distinct $x,y\in I_n$ when $x+y\in B$. Estimate the extremal quantity $f(n)=\min_B(|B|+\alpha(G_B))$ in the supplied reformulation.

All supplied versions considered: 788.tex. The source transfers the Alon--Pham random Cayley-sum bound to obtain exponent $3/5$. I verify the primary theorem, make the logarithmic loss explicit, and make the argument robust to whether loops are counted in the Cayley-sum independence convention.

2) WORK

The proof of Alon--Pham Theorem 4 supplies an independence bound \[ C p^{-3/2}(\log q)^{19/4} \] with high probability for a random generating set $S$ in an abelian group of order $q$. Choose an odd prime $4n<q<8n$, so all sums of distinct vertices of $I_n$ have no modular wraparound, and include group elements in $S$ independently with probability $p$.

Put $B=S\cap(2n,4n)$. If the theorem uses the usual loopless convention, its restriction directly bounds $\alpha(G_B)$. If its convention also excludes a vertex when $2x\in S$, take any independent set $A$ in the loopless interval graph and remove those vertices. The remaining set has $(A+A)\cap S=\varnothing$ even on the diagonal. Since doubling is a bijection modulo odd $q$, at most $|S|$ vertices were removed. Thus in either convention the safe bound is \[ \alpha(G_B)\le C p^{-3/2}(\log q)^{19/4}+|S|. \] With high probability $|S|\le2pq$, and the simultaneous favorable events therefore give \[ f(n)\le4pq+C p^{-3/2}(\log q)^{19/4}. \] Choose \[ p=n^{-2/5}(\log n)^{19/10}. \] Then $p\le1/2$ eventually, $pq\to\infty$, and both terms have order \[ \boxed{f(n)\ll n^{3/5}(\log n)^{19/10}.} \] The logarithmic exponent follows from $19/4-(3/2)(19/10)=19/10$. This is an explicit form of the source's $n^{3/5+o(1)}$ transfer, not a better power exponent. The loop correction costs only another $O(pn)$ term.

3) ADVERSARIAL VERIFICATION

The input is a theorem about polynomially sparse random generating sets, and its explicit proof was checked at the chosen $p$, rather than interpreting a fixed-$p$ statement as uniform. The diagonal correction avoids silently identifying two independence conventions. Bertrand's prime choice changes constants only.

4) FINAL

UNRESOLVED. The interval extremal exponent remains between the square-root lower scale and this $3/5$ upper scale. No argument reaches exponent $1/2$ or a sharp asymptotic.

5) SOURCES CHECKED

Entire supplied source read. Checked Alon--Pham, https://arxiv.org/html/2509.02561v1, Theorem 4 and Section 3.1, on 2026-09-23. The displayed $19/4$ factor occurs in its proof. The explicit optimization and the conservative diagonal correction are derived here.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 20%