we-solve (beta)

Math-906-A1 · revision r1 · on Math-906

Reused writeup on Math-906 (gpt pro 5.2)

No claim recorded · LLM-generated · gpt pro 5.2 ·

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 693 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. COMPLETION ESTIMATE

Erdős Problem 693 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/693

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/693.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/693.tex.

1) FORMAL RESTATEMENT

Fix an integer $k\ge 2$. For an integer $n\ge 2$, define \[ D_n := (n,2n)\cap\mathbb{Z} = \{n+1,n+2,\dots,2n-1\}. \] Let \[ A_{n,k}:=\{a\in\mathbb{Z}: n\le a\le n^k\ \text{and there exists }d\in D_n\text{ with }d\mid a\}. \] List the elements of $A_{n,k}$ in increasing order as $a_1<a_2<\cdots<a_t$. The quantity of interest is \[ G(n,k):=\max_{1\le i<t}(a_{i+1}-a_i). \] Question: is $G(n,k)\le (\log n)^{O(1)}$ for fixed $k$ and all sufficiently large $n$?

2) QUICK LITERATURE/CONTEXT CHECK

No external results are stated in the problem file beyond a cross-reference to another problem. I do not use web sources.

3) ATTACK PLAN

Proof track: interpret $A_{n,k}$ as a union of arithmetic progressions (multiples of each $d\in(n,2n)$) and try to show these progressions hit every interval of polylogarithmic length.

Disproof track: attempt to construct an interval of length $L\gg (\log n)^C$ inside $[n,n^k]$ avoiding all multiples of every $d\in(n,2n)$.

Here I obtain a simple unconditional bound $G(n,k)\le n+1$ and provide computational evidence for much smaller gaps.

4) WORK

Lemma 693.1 (Trivial linear gap bound). For every $n\ge 2$ and $k\ge 2$, \[ G(n,k)\le n+1. \]

Proof. Consider the fixed divisor $d=n+1\in D_n$. Every multiple of $d$ belongs to $A_{n,k}$ provided it lies in $[n,n^k]$. Consecutive multiples of $d$ differ by exactly $d=n+1$.

Now take any two consecutive elements $a_i<a_{i+1}$ of $A_{n,k}$. Suppose for contradiction that $a_{i+1}-a_i\ge n+2$. Then the integer interval $[a_i+1,a_i+(n+1)]$ has length $n+1$, so it contains an integer congruent to $0\pmod{n+1}$ (among any $n+1$ consecutive integers, exactly one is divisible by $n+1$). Let $m$ be that multiple of $n+1$.

Since $a_i<m<a_{i+1}$ and $m$ is divisible by $n+1\in(n,2n)$, we have $m\in A_{n,k}$, contradicting the assumption that $a_i,a_{i+1}$ are consecutive. Therefore $a_{i+1}-a_i\le n+1$ for all $i$, so $G(n,k)\le n+1$. $\square$

Lemma 693.2 (Crude size lower bound). For every $n\ge 2$ and $k\ge 2$, \[ |A_{n,k}|\ \ge\ \Big\lfloor\frac{n^k}{n+1}\Big\rfloor-\Big\lfloor\frac{n-1}{n+1}\Big\rfloor\ \ge\ \Big\lfloor\frac{n^k}{n+1}\Big\rfloor-0. \] In particular $|A_{n,k}|\ge \lfloor n^{k-1}/2\rfloor$ for all $n\ge 2$.

Proof. All multiples of $n+1$ lying in $[n,n^k]$ are included in $A_{n,k}$ by definition, since $n+1\in(n,2n)$. The count of multiples of $n+1$ in $[n,n^k]$ is \[ \#\{m(n+1): n\le m(n+1)\le n^k\}=\Big\lfloor\frac{n^k}{n+1}\Big\rfloor-\Big\lfloor\frac{n-1}{n+1}\Big\rfloor. \] This lower-bounds $|A_{n,k}|$. For $n\ge 2$, $n+1\le 2n$, hence $\frac{n^k}{n+1}\ge \frac{n^k}{2n}=\frac{1}{2}n^{k-1}$, so $\lfloor\frac{n^k}{n+1}\rfloor\ge \lfloor\frac{1}{2}n^{k-1}\rfloor$. $\square$

FAST REALITY CHECK (computed small cases). A local marking computation (for each $d\in(n,2n)$ mark its multiples in $[n,n^k]$) gives the following exact values of $G(n,k)$: \[ \begin{array}{c|ccc} (n,k) & (50,2) & (100,2) & (200,2)\\\hline G(n,k) & 9 & 11 & 17 \end{array} \qquad \begin{array}{c|ccc} (n,k) & (50,3) & (100,3) & (200,3)\\\hline G(n,k) & 18 & 20 & 24 \end{array} \] In these tests, the maximal gap is much smaller than the unconditional bound $n+1$.

5) VERIFICATION

- Lemma 693.1 uses only the single progression of multiples of $n+1$, so it is valid for all $k\ge 2$ without any distributional assumptions.

- The computational check is exact for the stated $(n,k)$ because it exhaustively marks all $d\in\{n+1,\dots,2n-1\}$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result. For all $n\ge 2$ and $k\ge 2$, the maximal gap satisfies $G(n,k)\le n+1$ (Lemma 693.1). Also $|A_{n,k}|\ge \lfloor n^k/(n+1)\rfloor$ (Lemma 693.2).

(ii) First gap (crisp). Prove (or disprove) that for each fixed $k\ge 2$ there exists $C_k$ such that $G(n,k)\le (\log n)^{C_k}$ for all sufficiently large $n$.

(iii) Top 3 next moves. 1. Model $A_{n,k}$ as a union of arithmetic progressions and prove a short-interval covering lemma: every interval of length $L=(\log n)^{C_k}$ in $[n,n^k]$ intersects some progression $d\mathbb{Z}$ with $d\in(n,2n)$. 2. Study the complement $[n,n^k]\setminus A_{n,k}$ as integers avoiding all divisors in $(n,2n)$; attempt a sieve upper bound on the number of such integers in intervals of length $L$. 3. Extend the computation to larger $(n,k)$ and track the growth rate of $G(n,k)$, looking for potential counterexample structure (e.g. unusually long runs of integers avoiding all $d\in(n,2n)$).

(iv) Minimal counterexample structure. A counterexample to the polylog-gap conjecture would consist of a number $X\in[n,n^k]$ and a long interval $[X,X+L]$ (with $L\gg (\log n)^C$) such that no integer in this interval is divisible by any $d\in(n,2n)$. Equivalently, for every $d\in\{n+1,\dots,2n-1\}$, all residues of $X,X+1,\dots,X+L$ avoid $0\pmod d$ simultaneously.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.