Reused writeup on Math-6003 (gpt pro 5.2)

Erdős Problem 935 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/935

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_5.2/935.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/935.tex.

1) FORMAL RESTATEMENT

Fix an integer $\ell\ge 1$. For a positive integer $m$ with prime factorisation \[m=\prod_{p} p^{v_p(m)},\] define its powerful part \[Q_2(m):=\prod_{p:\,v_p(m)\ge 2} p^{v_p(m)}.\] Equivalently, $Q_2(m)$ is the largest powerful (squarefull) divisor of $m$, and $Q_2(m)=m/\prod_{p:\,v_p(m)=1}p$. For $n\in\mathbb N$ set \[P_{n,\ell}:=\prod_{i=0}^{\ell} (n+i).\] The questions are:

(a) For every $\varepsilon>0$ and $\ell\ge 1$, does there exist $N(\varepsilon,\ell)$ such that for all $n\ge N(\varepsilon,\ell)$, \[Q_2(P_{n,\ell})<n^{2+\varepsilon}?\]

(b) For $\ell\ge 2$, is \[\limsup_{n\to\infty} \frac{Q_2(P_{n,\ell})}{n^2}=+\infty?\]

(c) For $\ell\ge 2$, is \[\lim_{n\to\infty} \frac{Q_2(P_{n,\ell})}{n^{\ell+1}}=0?\]

Edge cases: $n$ is intended large; $Q_2(1)=1$. All questions are about the asymptotic regime $n\to\infty$ with $\ell$ fixed.

2) QUICK LITERATURE/CONTEXT CHECK

The problem statement itself notes: (i) Erdős wrote that (a), if true, seems very difficult to prove. (ii) A result of Mahler implies $\limsup_{n\to\infty} Q_2(P_{n,\ell})/n^2\ge 1$ for every fixed $\ell\ge 1$. I do not use any other external results.

3) ATTACK PLAN

Proof track ideas: - Decompose $Q_2(P_{n,\ell})$ prime-by-prime via $v_p(P_{n,\ell})=\sum_{i=0}^{\ell} v_p(n+i)$. - Separate primes $p\le \ell$ (which can divide multiple factors) from primes $p>\ell$ (which can divide at most one factor), and try to bound the contribution of each regime.

Disproof/construction ideas: - Try to make several of $n,n+1,\dots,n+\ell$ very squareful simultaneously (e.g. consecutive powerful numbers), so that $Q_2(P_{n,\ell})$ becomes large. - Use CRT constraints $n\equiv -i\pmod{q_i^2}$ to force large square divisors in each factor.

Given current tools, I can rigorously prove only structural lemmas and provide computational evidence; I do not reach a full proof or counterexample to (a)–(c).

4) WORK

Fast reality check (explicit computation). Using direct prime-factorisation computations (sympy factorisation) of $Q_2(P_{n,\ell})$ for small ranges: - For $\ell=1$: $\max_{1\le n\le 500} Q_2(P_{n,1})/n^2 = 1.125$ attained at $n=8$. - For $\ell=2$: $\max_{1\le n\le 100000} Q_2(P_{n,2})/n^2 = 338.0344897959184$ attained at $n=9800$, where $Q_2(9800\cdot 9801\cdot 9802)=32464832400$. - For $\ell=3$: $\max_{1\le n\le 50000} Q_2(P_{n,3})/n^2 = 432.77173469387753$ attained at $n=1680$. - For $\ell=4$: $\max_{1\le n\le 20000} Q_2(P_{n,4})/n^2 = 48078.584176085664$ attained at $n=6724$. Sample values suggesting $Q_2(P_{n,\ell})/n^{\ell+1}$ is small for these ranges: for $\ell=2$, at $n=100,200,500$ the ratios are approximately $2\cdot 10^{-4},\ 5\cdot 10^{-5},\ 8\cdot 10^{-6}$, respectively.

Lemma 935.1 (Large primes hit at most one factor). Let $\ell\ge 1$ and $p$ be a prime with $p>\ell$. Then among $n,n+1,\dots,n+\ell$ there is at most one integer divisible by $p$.

Proof. Assume for contradiction that $p\mid (n+i)$ and $p\mid (n+j)$ for some $0\le i<j\le \ell$. Then $p\mid (n+j)-(n+i)=j-i$. But $1\le j-i\le \ell<p$, so $j-i$ is a positive integer strictly smaller than $p$, hence cannot be divisible by the prime $p$. Contradiction. QED.

Lemma 935.2 (Characterisation of the $p>\ell$ contribution to $Q_2$). Let $\ell\ge 1$, $p$ prime with $p>\ell$, and $P_{n,\ell}=\prod_{i=0}^{\ell}(n+i)$. Then either $v_p(P_{n,\ell})=0$, or there is a unique $i\in\{0,\dots,\ell\}$ such that \[v_p(P_{n,\ell})=v_p(n+i).\] In particular, for $p>\ell$, the prime $p$ appears in $Q_2(P_{n,\ell})$ (i.e. $v_p(P_{n,\ell})\ge 2$) if and only if $p^2\mid (n+i)$ for that unique index $i$.

Proof. By Lemma 935.1, there is at most one index $i$ with $p\mid (n+i)$. If none exists, then $v_p(P_{n,\ell})=0$. If such an $i$ exists, then $p\nmid (n+j)$ for all $j\ne i$, hence $v_p(n+j)=0$ for $j\ne i$. By additivity of $p$-adic valuation on products, \[v_p(P_{n,\ell})=\sum_{j=0}^{\ell} v_p(n+j)=v_p(n+i).\] The final statement is then immediate from the definition of $Q_2$: for any prime $p$, it contributes to $Q_2$ exactly when its valuation is at least $2$. QED.

Lemma 935.3 (Crude bound for small primes). Fix $\ell\ge 1$. For any prime $p\le \ell$ and any $n\ge 1$, \[v_p(P_{n,\ell}) \le (\ell+1)\,\Big\lfloor \log_p(n+\ell)\Big\rfloor. \] Consequently, \[\prod_{p\le \ell} p^{v_p(P_{n,\ell})} \le (n+\ell)^{(\ell+1)\pi(\ell)}. \]

Proof. For each $i\in\{0,\dots,\ell\}$ we have $n+i\le n+\ell$. If $p^{k}\mid (n+i)$ then $p^{k}\le n+\ell$, hence $k\le \lfloor \log_p(n+\ell)\rfloor$. Thus $v_p(n+i)\le \lfloor \log_p(n+\ell)\rfloor$ for every $i$, and summing over $i$ gives the valuation bound. Exponentiating and multiplying over the finitely many primes $p\le \ell$ yields \[\prod_{p\le \ell} p^{v_p(P_{n,\ell})} \le \prod_{p\le \ell} p^{(\ell+1)\log_p(n+\ell)} = \prod_{p\le \ell} (n+\ell)^{\ell+1}=(n+\ell)^{(\ell+1)\pi(\ell)}. \] (Here we used that $p^{\log_p(n+\ell)}=n+\ell$.) QED.

5) VERIFICATION

- Lemma 935.1: the only place where $p>\ell$ is used is to ensure $j-i < p$. This is correct. - Lemma 935.2: uses additivity $v_p(\prod m_i)=\sum v_p(m_i)$, valid for integers. - Lemma 935.3: checks that $v_p(m)\le \lfloor\log_p m\rfloor$ for $m\ge1$, since $p^{v_p(m)}\le m$. This is correct. - Computations: the reported maxima depend on the search ranges stated. They are not claims about the true global maxima.

6) FINAL

UNRESOLVED (i) Strongest proved partial result: For fixed $\ell$, primes $p>\ell$ contribute to $Q_2(P_{n,\ell})$ only via square divisibility inside a single factor (Lemma 935.2), and primes $p\le \ell$ have total contribution at most $(n+\ell)^{(\ell+1)\pi(\ell)}$ (Lemma 935.3). (ii) First gap: Prove or disprove that for each fixed $\ell\ge1$ and every $\varepsilon>0$ one has $Q_2(P_{n,\ell})=O(n^{2+\varepsilon})$ as $n\to\infty$. (iii) Top 3 next moves: 1. Prove a lower bound on the product of primes that divide $P_{n,\ell}$ to exponent exactly $1$ (equivalently, a lower bound on $\prod_{p: v_p(P_{n,\ell})=1} p$) to upper-bound $Q_2(P_{n,\ell})=P_{n,\ell}/\prod_{v_p=1}p$. 2. Control how often several of $n,n+1,\dots,n+\ell$ can be simultaneously very squareful; even a quantitative bound on the count of $i$ with a large square divisor of $n+i$ would feed into (a) and (c). 3. Extend the brute-force search for $\ell=2,3$ to locate families with unusually large $Q_2(P_{n,\ell})/n^2$ and look for a CRT-driven parametrisation. (iv) Minimal counterexample structure: A counterexample sequence to (a) would require infinitely many $n$ for which many primes (including some larger than $\ell$) occur with valuation $\ge2$ in $P_{n,\ell}$, i.e. several of the factors $n+i$ must contain large square divisors and the small primes $p\le \ell$ must contribute unusually large valuations 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.