gpt pro 5.2 attempt on Math-6003

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.