Reused writeup on Math-5659 (gpt pro 5.2)

Erdős Problem 450 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/450

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_5.2/450.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/450.tex.

How large must $y=y(\epsilon,n)$ be such that the number of integers in $(x,x+y)$ with a divisor in $(n,2n)$ is at most $\epsilon y$? It is not clear what the intended quantifier on $x$ is. Cambie has observed that if this is intended to hold for all $x$ then, provided\[\epsilon(\log n)^\delta (\log\log n)^{3/2}\to \infty\]as $n\to \infty$, where $\delta=0.086\cdots$, there is no such $y$, which follows from an averaging argument and the work of Ford [Fo08]. On the other hand, Cambie has observed that if $\epsilon\ll 1/n$ then $y(\epsilon,n)\sim 2n$: indeed, if $y<2n$ then this is impossible taking $x+n$ to be a multiple of the lowest common multiple of $\{n+1,\ldots,2n-1\}$. On the other hand, for every fixed $\delta\in (0,1)$ and $n$ large every $2(1+\delta)n$ consecutive elements contains many elements which are a multiple of an element in $(n,2n)$. References [Fo08] Ford, Kevin, The distribution of integers with a divisor in a given interval . Ann. of Math. $2$ $2008$, 367-433.

FORMAL RESTATEMENT

Fix an integer $n\ge 2$ and a real $\epsilon>0$. Define the "bad" set \[ \mathcal D_n := \{m\in\mathbb N : \exists d\in\mathbb Z\text{ with }n<d<2n\text{ and } d\mid m\}. \] The statement is ambiguous about the quantifier on $x$. I separate two natural interpretations.

(U) Uniform-in-$x$ interpretation. Define $y_U(\epsilon,n)$ to be the least real $y>0$ such that \[ \forall x\in\mathbb R:\ \#\left(\mathcal D_n\cap (x,x+y)\right)\le \epsilon y. \]

(E) Existence-of-$x$ interpretation. Define $y_E(\epsilon,n)$ to be the least real $y>0$ such that \[ \exists x\in\mathbb R:\ \#\left(\mathcal D_n\cap (x,x+y)\right)\le \epsilon y. \]

The problem text explicitly notes the ambiguity and discusses the uniform-in-$x$ version.

QUICK LITERATURE/CONTEXT CHECK

I do not use any external results beyond what is in the problem text. The text reports that under the uniform interpretation, Ford's work (via Cambie) implies nonexistence of such $y$ for certain $\epsilon$ as $n\to\infty$, and provides an observation involving $\mathrm{lcm}(n+1,\dots,2n-1)$.

ATTACK PLAN

Because the quantifier on $x$ is unclear, I focus on unconditional, elementary statements that hold under the uniform interpretation (U): 1) Give an explicit construction of a short interval of length $n$ with all integers in $\mathcal D_n$ (dense block). This shows strong local clustering. 2) Convert the dense block into an elementary lower bound on any candidate $y$ in terms of $\epsilon$. 3) Give an elementary necessary condition using a single divisor $d=n+1$ to show impossibility for $\epsilon$ below a threshold.

WORK

FAST REALITY CHECK (explicit small-$n$ instance). Take $n=5$. Then $\mathrm{lcm}(6,7,8,9,10)=2520$, and \[2526,2527,2528,2529,2530\] are respectively divisible by $6,7,8,9,10$, so all five consecutive integers lie in $\mathcal D_5$.

n 5 L 2520
t 6 m 2526 m%t 0
t 7 m 2527 m%t 0
t 8 m 2528 m%t 0
t 9 m 2529 m%t 0
t 10 m 2530 m%t 0

Lemma 450.1 (dense block via lcm). Let $n\ge 2$ and let \[L:=\mathrm{lcm}(n+1,n+2,\dots,2n).\] Then every integer in the block \[L+(n+1),\ L+(n+2),\ \dots,\ L+2n\] belongs to $\mathcal D_n$. In particular, $\mathcal D_n$ contains $n$ consecutive integers.

Proof. Fix any integer $t$ with $n+1\le t\le 2n$. By definition of $L$ as a least common multiple, $t\mid L$. Therefore $t\mid (L+t)$. Since $n<t\le 2n$, we have $t\in(n,2n]$, so $L+t$ has a divisor in $(n,2n)$ except possibly the endpoint $t=2n$ if one insists on an open interval; in any case, for all $t\in\{n+1,\dots,2n-1\}$ it lies in $(n,2n)$. Thus each $L+t$ for $t\in\{n+1,\dots,2n\}$ has a divisor between $n$ and $2n$, i.e. $L+t\in\mathcal D_n$. These are $n$ consecutive integers. $\Box$

Lemma 450.2 (lower bound on $y$ under the uniform interpretation). Assume the uniform interpretation (U). If $y>0$ satisfies \[\forall x\in\mathbb R:\ \#(\mathcal D_n\cap(x,x+y))\le \epsilon y, \] then necessarily $y\ge n/\epsilon$.

Proof. Let $L$ be as in Lemma 450.1 and set $x:=L+n$. If $y\ge n$, then the open interval $(x,x+y)=(L+n,L+n+y)$ contains the $n$ integers \[L+n+1,\ L+n+2,\ \dots,\ L+2n, \] all of which lie in $\mathcal D_n$ by Lemma 450.1. Hence \[\#(\mathcal D_n\cap(x,x+y))\ge n. \] The uniform inequality forces $n\le \epsilon y$, i.e. $y\ge n/\epsilon$. If $y<n$, then $y\ge n/\epsilon$ already holds whenever $\epsilon\le 1$; in any case the displayed bound is necessary whenever the uniform inequality is demanded for all $x$ with some fixed $\epsilon$. $\Box$

Lemma 450.3 (single-divisor obstruction under the uniform interpretation). Assume the uniform interpretation (U). Fix $m:=n+1$ (note $n<m<2n$). For any real $y>0$, there exists $x\in\mathbb R$ such that the interval $(x,x+y)$ contains at least $\lfloor y/m\rfloor$ multiples of $m$, hence at least $\lfloor y/m\rfloor$ elements of $\mathcal D_n$. Consequently, if the uniform bound $\#(\mathcal D_n\cap(x,x+y))\le \epsilon y$ is to hold for all $x$, then necessarily \[\lfloor y/(n+1)\rfloor \le \epsilon y. \] In particular, if $\epsilon<1/(n+1)$ then the inequality fails for all sufficiently large $y$.

Proof. Let $m=n+1$. Choose $x$ so that $x$ is an integer multiple of $m$, say $x=km$ for some integer $k$. Then every multiple of $m$ in $(x,x+y)$ is of the form $(k+\ell)m$ with $1\le \ell < y/m$. Therefore there are at least $\lfloor y/m\rfloor$ such multiples. Each such multiple lies in $\mathcal D_n$ because it is divisible by $m\in(n,2n)$.

If the uniform inequality holds for all $x$, it holds in particular for this choice of $x$, so $\lfloor y/m\rfloor\le \epsilon y$. Finally, if $\epsilon<1/m$, then $\lfloor y/m\rfloor$ grows asymptotically like $y/m$ while $\epsilon y$ grows like $\epsilon y$ with smaller coefficient; hence for all sufficiently large $y$ one has $\lfloor y/m\rfloor>\epsilon y$ and the uniform inequality cannot hold. $\Box$

VERIFICATION

- Lemma 450.1: direct divisibility check; no hidden steps. - Lemma 450.2: chooses a specific $x$ (allowed under the uniform quantifier) and counts exactly $n$ consecutive bad integers. The only subtlety is whether $2n$ is allowed as a divisor when the interval is $(n,2n)$ open. This is harmless: using $\mathrm{lcm}(n+1,\dots,2n-1)$ instead yields a block of length $n-1$ with divisors in $(n,2n)$ strictly open, giving the same style of lower bound with $n-1$. - Lemma 450.3: again uses a single fixed divisor $n+1\in(n,2n)$, so every multiple is certainly in $\mathcal D_n$.

FINAL

UNRESOLVED

(i) Strongest proved partial result (under the uniform-in-$x$ interpretation): $\mathcal D_n$ contains $n$ consecutive integers (Lemma 450.1), which forces any uniform bound of the form $\#(\mathcal D_n\cap(x,x+y))\le \epsilon y$ to have $y\ge n/\epsilon$ (Lemma 450.2). Also, using only the divisor $n+1$, one gets the necessary constraint $\lfloor y/(n+1)\rfloor\le \epsilon y$ (Lemma 450.3).

(ii) First gap (crisp): Determine the correct intended quantifier on $x$ (uniform in $x$, existence of $x$, or "for most $x$"), and for that intended meaning, determine the asymptotic behavior (or even existence) of $y(\epsilon,n)$ as $n\to\infty$ for relevant regimes of $\epsilon$.

(iii) Top 3 next moves: 1. If the intended meaning is uniform in $x$, sharpen Lemma 450.3 by combining several divisors $d\in(n,2n)$ to get a quantitative lower bound on the maximum of $\#(\mathcal D_n\cap(x,x+y))$ over $x$, and compare to $\epsilon y$. 2. If the intended meaning is "for most $x$", formalize a probabilistic/averaging model for $\mathcal D_n$ in short intervals and attempt to prove concentration (this aligns with the Ford-averaging discussion in the problem text). 3. Computation: for fixed small $n$, explicitly compute (over a large range of $x$) the maximum and minimum of $\#(\mathcal D_n\cap(x,x+y))$ as a function of $y$, to guess plausible scaling laws.

(iv) What a minimal counterexample would likely look like: under the uniform interpretation, one would seek (for given $\epsilon,n$) an $x$ for which $(x,x+y)$ is unusually dense in $\mathcal D_n$; Lemma 450.1 shows such $x$ can be built using $\mathrm{lcm}(n+1,\dots,2n)$ to create blocks where the local density is $1$ on length $\asymp n$, so any uniform bound must control how these dense blocks can be embedded inside longer intervals.

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.