Reused writeup on Math-6020 (gpt pro 5.2)

Erdős Problem 962 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/962

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_5.2/962.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/962.tex.

1) FORMAL RESTATEMENT

For an integer $n\ge 1$, define $k(n)$ to be the largest integer $k\ge 1$ such that there exists an integer $m\le n$ with the property that each of \[ m+1,\;m+2,\;\dots,\;m+k \] is divisible by at least one prime $p>k$.

Equivalently, for each $i\in\{1,\dots,k\}$, the integer $m+i$ has a prime factor exceeding $k$.

Edge case: $k(1)=1$ because taking $m=1$ gives $m+1=2$ divisible by a prime $2>1$, and $k\ge 2$ is impossible since $m+1=2$ is not divisible by a prime $>2$.

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement records (not reproved here) a claimed Erdős lower bound $k(n)\gg_\varepsilon \exp((\log n)^{1/2-\varepsilon})$ and a heuristic upper bound $k(n)=o(n^\varepsilon)$, and mentions an argument (attributed in the problem text to Tao) giving $k(n)\le (1+o(1))n^{1/2}$, plus a stated lower bound due to Tang.

In this solution attempt I do not invoke any results beyond what is explicitly written in the problem text.

3) ATTACK PLAN

Proof-track ideas: - Translate the condition into constraints on prime divisors $>k$ across the block, especially using the fact that a prime $>k$ cannot divide two different numbers among $k$ consecutive integers. - Use those constraints to get elementary necessary conditions (e.g. on $m$ relative to $k$), and attempt to turn them into upper bounds on $k(n)$.

Construction-track ideas: - For small $n$, brute force search for the maximizing $(m,k)$ to get a feel for growth.

4) WORK

FAST REALITY CHECK (computation; exact for the searched ranges):

I brute-forced $k(n)$ for selected $n$ by searching all $m\le n$ and all $k$ up to $3n$ (this upper cutoff did not change the optimum for the reported $n$).

(n, k(n), an m achieving it)
10   2   4
20   4   18
50   4   18
100  6   64
200  9   150
500  12  405
1000 14  735

First n where k(n) increases:
1 -> k=1 (m=1)
4 -> k=2 (m=4)
12 -> k=3 (m=12)
18 -> k=4 (m=18)
54 -> k=5 (m=54)
64 -> k=6 (m=64)
112 -> k=7 (m=112)
150 -> k=9 (m=150)

Lemma 962.1 (Large primes cannot repeat within the block). Assume $m,k$ satisfy Lemma 962's defining property. If a prime $p>k$ divides $m+i$ and $m+j$ for some $1\le i<j\le k$, then this is impossible. Consequently, each prime $p>k$ can divide at most one of the integers $m+1,\dots,m+k$.

Proof. If $p\mid (m+i)$ and $p\mid (m+j)$ then $p\mid (m+j)-(m+i)=j-i$. But $1\le j-i\le k-1<k<p$, so the only multiple of $p$ in this range is $0$, contradiction. Therefore no such pair $i<j$ exists, and a prime $>k$ divides at most one member of the block. $\square$

Lemma 962.2 (A necessary condition: $m\ge k$). If there exist integers $m,k$ with $m+1,\dots,m+k$ each divisible by some prime $>k$, then necessarily $m\ge k$.

Proof. For each $i\in\{1,\dots,k\}$ pick a prime $p_i>k$ dividing $m+i$. By Lemma 962.1 the primes $p_1,\dots,p_k$ are pairwise distinct. Thus $\{p_1,\dots,p_k\}$ is a set of $k$ distinct integers each $>k$, so its minimum possible value (as a multiset of integers) is at least $\{k+1,k+2,\dots,2k\}$ after ordering. In particular, for each $i$ we have $p_i\ge k+1$ and hence $m+i\ge p_i\ge k+1$. Taking $i=1$ gives $m+1\ge k+1$, i.e. $m\ge k$.

(Equivalently, since the interval $(k,m+k]$ has length $m$ and would have to contain $k$ distinct primes $>k$, one must have $m\ge k$.) $\square$

Corollary 962.3 (Trivial upper bound $k(n)\le n$). For every $n\ge 1$, $k(n)\le n$.

Proof. If $k(n)\ge k$, there exists $m\le n$ witnessing the property for this $k$. Lemma 962.2 forces $k\le m\le n$, so $k\le n$. Taking the largest such $k$ gives $k(n)\le n$. $\square$

5) VERIFICATION

- Lemma 962.1: the key inequality is $j-i<k<p$, which holds since $1\le j-i\le k-1$ and $p>k$. - Lemma 962.2: the only substantive step is that $k$ distinct integers each $>k$ must include one $\ge k+1$; using $i=1$ suffices to deduce $m\ge k$. - Computations: the brute force checked all $m\le n$ and verified the condition for each $k$ tested; expanding the search cap from $n$ to $3n$ did not change the reported $k(n)$ for the displayed values.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: Lemma 962.1 (a prime $>k$ cannot divide two numbers in the block) and Lemma 962.2 (any witness must have $m\ge k$), yielding the unconditional bound $k(n)\le n$, plus exact small-$n$ computations listed above.

(ii) First gap (crisp): prove any asymptotically nontrivial upper bound on $k(n)$ (e.g. $k(n)=o(n)$ or $k(n)\ll n^{1/2}$) without invoking results not already stated in the problem.

(iii) Top 3 next moves: 1. Strengthen Lemma 962.2 by exploiting additional structure of the congruences $m\equiv -i\pmod{p_i}$ with many distinct moduli $p_i>k$. 2. Use computation to search for optimizing pairs $(m,k)$ for larger $n$ and look for patterns (e.g. $m$ highly composite), suggesting a constructive lower bound. 3. Try to convert the “distinct primes” constraint into a quantitative limitation by bounding how many primes $>k$ can divide numbers in a length-$k$ interval near $m\le n$.

(iv) Minimal counterexample structure (to the heuristic “$k(n)=o(n^\varepsilon)$ for all $\varepsilon>0$”): a smallest $n$ for which there exists a witness $(m,k)$ with $m\le n$ and $k\ge n^\varepsilon$, meaning a length-$k$ block where every integer has a prime factor exceeding $k$; Lemma 962.1 forces these large primes to be distinct across the block.

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.