GPT 6 Astra Ultra attempt on Math-783
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Contents
Erdős Problem #367: full parts of consecutive integers
1. The two proposed bounds
Define \[ B_r(n)=\prod_{\substack{p^a\Vert n\\a\ge r}}p^a. \] For each fixed $k$, the main question is whether $\prod_{i=0}^{k-1}B_2(n+i)\le n^{2+o(1)}$. The stronger proposal is an $O_k(n^2)$ upper bound. We reconstruct van Doorn's counterexample to the stronger bound, using the explicit algebraic-integer computation given in Tao's discussion. It gives $\gg n^2\log n$ infinitely often for $k=3$. The main $n^{2+o(1)}$ question remains unresolved.2. Consecutive powerful numbers from a Pell recurrence
Let $\alpha=3+\sqrt8$, so $\alpha^{-1}=3-\sqrt8$. Write \[ \alpha^j=x_j+y_j\sqrt8,\qquad j\ge1. \] Multiplication by $\alpha$ proves inductively that $x_j,y_j$ are positive integers; conjugation gives $x_j^2-8y_j^2=1$. Set $n_j=8y_j^2$. Then \[ n_j+1=x_j^2,\quad B_2(n_j)=n_j,\quad B_2(n_j+1)=n_j+1, \] and direct expansion gives \[ 4(n_j+2)=\alpha^{2j}+6+\alpha^{-2j}. \tag{1} \]3. Force a growing full divisor of the third integer
In the ring $\mathbb Z[\sqrt8]$, \[ \alpha^{\pm3}=99\pm35\sqrt8 =-1+5(20\pm7\sqrt8). \] If $u=-1+5^t v$ with $t\ge1$, the binomial expansion of $u^5$ shows $u^5\equiv-1\pmod{5^{t+1}}$: the linear term has factor $5^{t+1}$, and every term of degree at least two has at least this divisibility as well. Induction therefore gives \[ \alpha^{\pm3\cdot5^{t-1}}\equiv-1\pmod{5^t}. \] Choose the positive integer \[ j_t=\frac{3\cdot5^{t-1}-1}{2}. \] Substituting $2j_t=3\cdot5^{t-1}-1$ in $1$ yields \[ 4(n_{j_t}+2) \equiv-\alpha^{-1}+6-\alpha=0\pmod{5^t}. \] The congruence is in the quadratic-integer ring, and its left side is an ordinary integer. Divisibility of its integer coefficient, and invertibility of $4$ modulo $5^t$, give $5^t\mid n_{j_t}+2$ in $\mathbb Z$. For $t\ge2$, the prime $5$ therefore occurs in the full part and $B_2(n_{j_t}+2)\ge5^t$.Also $n_j<\alpha^{2j}$, so \[ \log n_{j_t}<3\cdot5^{t-1}\log\alpha,\qquad 5^t>\frac5{3\log\alpha}\log n_{j_t}. \] As $t\to\infty$ the $n_{j_t}$ are distinct and tend to infinity. Consequently \[ \prod_{i=0}^2B_2(n_{j_t}+i) \ge n_{j_t}(n_{j_t}+1)5^t \gg n_{j_t}^2\log n_{j_t}. \tag{2} \] The first case $t=2$ has $j_t=7$ and $n_{j_t}=13051463048$. The infinite sequence, rather than one large numerical ratio, is what disproves an upper bound with an unspecified constant.
4. Consequences and the higher-fullness wording
For $k=1,2$ one has trivially $\prod_{i<k}B_2(n+i)\le n(n+1)\le2n^2$. For every fixed $k\ge3$, the extra factors are at least $1$, so $2$ disproves the stronger $O_k(n^2)$ proposal. Thus that proposal holds precisely for $k\le2$.The question also discusses $B_r$ for $r\ge3$. Its displayed wording with every $\varepsilon>0$ cannot literally give an unbounded ratio to $n^{1+\varepsilon}$ for fixed $k$: always \[ \prod_{i<k}B_r(n+i)\le\prod_{i<k}(n+i)=O_k(n^k), \] so the ratio is bounded when $\varepsilon\ge k-1$. The nontrivial interpretation, also identified in the source discussion, asks for some positive $\varepsilon=\varepsilon(r,k)$. No proof of that interpretation for general $r,k$ is claimed here.
5. Assessment and references
The logarithm in $2$ is $n^{o(1)}$, so it is fully compatible with the main proposed upper bound. That bound is unresolved in this attempt, even though the stronger variant has been completely classified.Sources: https://www.erdosproblems.com/367 and https://www.erdosproblems.com/forum/thread/367, checked 2026-09-25. The latter contains van Doorn's construction and Tao's explicit congruence derivation of 20 November 2025. The previous GPT Pro 5.2 and 5.4 reconstructions were independently checked. The full argument is reproduced above with its scope separated from the main question; no novelty or formal Lean verification is claimed.