GPT 6 Astra Ultra attempt on Math-5622
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Contents
Erdős Problem 393: factorial factors in a short interval
Statement and scope
For $n\ge2$, let $f(n)$ be the least positive span $m$ for which $n!$ is a product of distinct positive integers in $[A,A+m]$, with both endpoints $A$ and $A+m$ among the factors. How does $f(n)$ grow? We give exact structural reductions and a complete conditional proof that $f(n)\to\infty$ from the $abc$ conjecture. The unconditional growth question remains unresolved in this attempt. Source: https://www.erdosproblems.com/393. The conditional conclusion is known; Luca's more general primary theorem is at https://web.math.pmf.unizg.hr/glasnik/vol_37/no2_04.html. No novelty is claimed.Finite patterns and elementary constraints
For $n\ge3$, the factors $2,3,\ldots,n$ give $f(n)\le n-2$. In any admissible factorization the largest factor $A+m$ is at least $n$, since the product of distinct integers not exceeding $A+m$ is at most $(A+m)!$. There are at most $m+1$ factors, so also \[ A+m\ge(n!)^{1/(m+1)}. \tag{1} \] For fixed $m$, choose a subset $D\subseteq\{0,\ldots,m\}$ containing $0,m$. Every factorization of span $m$ has the exact form \[ n!=P_D(A),\qquad P_D(X)=\prod_{d\in D}(X+d). \tag{2} \] There are only $2^{m-1}$ patterns. This does not bound how large $A$ may be; it reduces the question to finitely many polynomial-factorial equations.In particular $f(n)=1$ exactly when $n!=A(A+1)$, equivalently $4n!+1$ is an odd square. For $m=2$ the only possible patterns are $A(A+2)$ and $A(A+1)(A+2)$. Thus $f(n)\le2$ has a direct exact test using these two equations and the span-one equation. Such tests are finite certificates at a fixed $n$, not proofs that a fixed span occurs only finitely often.
An elementary radical bound
Write $\operatorname{rad}(r)$ for the product of its distinct prime factors, and $\vartheta(y)=\sum_{p\le y}\log p$. We need only \[ \vartheta(n)\le4n\log2. \tag{3} \] For each $j\ge1$, every prime in $(2^{j-1},2^j]$ divides the integer $\binom{2^j}{2^{j-1}}\le2^{2^j}$. Summing the resulting logarithmic bounds through $j=r$ gives $\vartheta(2^r)<2^{r+1}\log2$. Choose $r$ with $2^{r-1}<n\le2^r$ to obtain $3$. Hence $\operatorname{rad}(n!)\le e^{4n\log2}$, with no prime number theorem needed.A conditional escape from every bounded span
Assume the $abc$ conjecture in its following explicit form: for every $\epsilon>0$ there is a constant $C_\epsilon$ such that coprime positive integers $u,v$ with $u+v=w$ satisfy \[ w\le C_\epsilon\operatorname{rad}(uvw)^{1+\epsilon}. \tag{ABC} \] This is a conjectural assumption, not an established theorem or an unconditional input.Fix a span $m$ and suppose $2$ holds. Put $g=\gcd(A,m)$, $u=A/g$, $v=m/g$, $w=(A+m)/g$. These are positive, $u+v=w$, and $\gcd(u,v)=1$. Both endpoint factors divide $n!$, so every prime of $uw$ is at most $n$, while $v$ has only primes dividing $m$. Thus \[ \operatorname{rad}(uvw)\le \operatorname{rad}(m)\operatorname{rad}(n!). \] Apply (ABC) with $\epsilon=1$, and use $g\le m$ and $3$: \[ A+m\le m C_1\operatorname{rad}(m)^2 e^{8n\log2}. \] On the other hand $1$, and the elementary integral estimate $\log(n!)\ge\int_1^n\log t\,dt=n\log n-n+1$, give \[ \frac{n\log n-n+1}{m+1}\le \log(A+m) \le8n\log2+O_m(1). \] This is impossible as $n\to\infty$ with $m$ fixed. For each $M$, the finitely many spans $1\le m\le M$ therefore all eventually disappear, proving $f(n)\to\infty$ under (ABC).