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Math-5820-A1 · revision r1 · on Math-5820

Reused writeup on Math-5820 (gpt pro 5.2)

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Erdős Problem 687 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. Erdős Problem 687
  5. 1) FORMAL RESTATEMENT.
  6. 2) QUICK LITERATURE/CONTEXT CHECK.
  7. 3) ATTACK PLAN.
  8. 4) WORK.
  9. 5) VERIFICATION.
  10. FINAL.
  11. COMPLETION ESTIMATE

Erdős Problem 687 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/687

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_5.2/687.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/687.tex.

Erdős Problem 687

1) FORMAL RESTATEMENT.
Let $x\ge 2$ be real, and let $\mathcal P(x):=\{p:\ p\text{ prime},\ p\le x\}$. For each $p\in\mathcal P(x)$ choose a residue class $a_p\pmod p$. Say that an integer $n$ is covered if $n\equiv a_p\pmod p$ for at least one prime $p\le x$. Define $Y(x)$ to be the maximal integer $y\ge 0$ for which there exist choices $\{a_p\}_{p\le x}$ such that every integer in $[1,y]$ is covered.

The problem asks for good estimates for $Y(x)$ as $x\to\infty$, in particular whether one can prove $Y(x)=o(x^2)$ or even $Y(x)\ll x^{1+o(1)}$.

2) QUICK LITERATURE/CONTEXT CHECK.
The problem statement already records:
  • an upper bound due to Iwaniec: $Y(x)\ll x^2$,
  • a lower bound due to Ford--Green--Konyagin--Maynard--Tao: $Y(x)\gg x\frac{\log x\log\log\log x}{\log\log x}$,
  • a conjecture of Maier--Pomerance: $Y(x)\ll x(\log x)^{2+o(1)}$.
No additional literature is used or claimed here.
3) ATTACK PLAN.
  • Main reformulation: encode the residue choices by a single shift $t$ modulo the primorial $P(x):=\prod_{p\le x}p$ (Lemma 687.1), so that the problem becomes: maximize the length of an interval of consecutive integers containing no number coprime to $P(x)$.
  • Proof-track: try to upper-bound the longest run of consecutive integers all having a prime factor $\le x$ (i.e. all non-coprime to $P(x)$), using sieve bounds on gaps between $x$-rough numbers.
  • Disproof-track (for very strong conjectures): attempt to build explicit residue choices/shift producing intervals of length close to $x^2$ or larger, guided by constructions for large prime gaps.
4) WORK.

Lemma 687.1 (CRT shift reformulation). Let $P(x):=\prod_{p\le x}p$. Given a choice of residue classes $\{a_p\pmod p\}_{p\le x}$, there exists a unique residue class $t\pmod{P(x)}$ such that \[ t\equiv -a_p \pmod p\quad\text{for every prime }p\le x. \] For this $t$, an integer $n$ satisfies $n\equiv a_p\pmod p$ for some $p\le x$ if and only if $n+t$ is divisible by some prime $p\le x$, i.e. \[ \exists\,p\le x:\ n\equiv a_p\pmod p\quad\Longleftrightarrow\quad \gcd(n+t,P(x))>1. \] Consequently, $Y(x)$ equals the maximum length of a run of consecutive integers all not coprime to $P(x)$: \[ Y(x)=\max_{t\in\mathbb Z}\ \max\{\,y\ge 0:\ \gcd(t+1,P(x))>1,\dots,\gcd(t+y,P(x))>1\,\}. \]

Proof. For distinct primes $p\le x$, the moduli are coprime, so by the Chinese remainder theorem there exists a unique $t\pmod{P(x)}$ solving the simultaneous congruences $t\equiv -a_p\pmod p$.

For a fixed prime $p\le x$ and integer $n$, we have \[ n\equiv a_p\pmod p\quad\Longleftrightarrow\quad n-a_p\equiv 0\pmod p \quad\Longleftrightarrow\quad n+t\equiv 0\pmod p \] because $t\equiv -a_p\pmod p$. Thus $n$ is covered by at least one congruence class $a_p\pmod p$ if and only if $n+t$ is divisible by at least one prime $p\le x$. Since $P(x)$ is the product of all primes $\le x$, this is equivalent to $\gcd(n+t,P(x))>1$.

The final displayed formula for $Y(x)$ is exactly the definition of $Y(x)$ rewritten in terms of this equivalence (covering $[1,y]$ corresponds to choosing $t$ such that $t+1,\dots,t+y$ are all divisible by some prime $\le x$). $\square$

Lemma 687.2 (elementary lower bound $Y(x)\ge x-1$). For every real $x\ge 2$, one has $Y(x)\ge \lfloor x\rfloor-1$.

Proof. Let $X:=\lfloor x\rfloor$ and set $t=1$ in the reformulation of Lemma 687.1. We claim that every integer $n\in[1,X-1]$ is covered by the residue choices $a_p\equiv -1\equiv p-1\pmod p$ for all primes $p\le x$. Indeed, for such $n$ we have $n+1\le X\le x$, hence $n+1$ has some prime divisor $p\le n+1\le x$. Then $p\mid (n+1)$ implies $n\equiv -1\equiv p-1\pmod p$, so $n$ lies in the chosen class for that prime $p$. Thus $[1,X-1]$ is fully covered and $Y(x)\ge X-1$. $\square$

Lemma 687.3 (trivial upper bound from counting totatives). Let $P(x)=\prod_{p\le x}p$ and $\varphi$ be Euler's totient function. Then \[ Y(x)\le P(x)-\varphi(P(x)). \]

Proof. By Lemma 687.1, for any shift $t$ the pattern of coprimality of $t+1,t+2,\dots$ with $P(x)$ is periodic modulo $P(x)$. In a complete residue system modulo $P(x)$, there are exactly $\varphi(P(x))$ residues coprime to $P(x)$ and hence $P(x)-\varphi(P(x))$ residues that are not coprime.

If an interval of consecutive integers contains no number coprime to $P(x)$, then within one period of length $P(x)$ it can cover at most all the non-coprime residues, i.e. it cannot have length exceeding $P(x)-\varphi(P(x))$; otherwise, among the first $P(x)-\varphi(P(x))+1$ distinct residues in $[1,P(x)]$ one would be coprime (pigeonhole, since only $P(x)-\varphi(P(x))$ residues are non-coprime). Thus the maximal such run length is at most $P(x)-\varphi(P(x))$, which equals the stated bound on $Y(x)$. $\square$

FAST REALITY CHECK (local computation). Using the Lemma 687.1 reformulation, I computed $Y(x)$ for small $x$ by scanning residues modulo $P(x)$ and finding the maximal gap between consecutive totatives of $P(x)$. The exact values for integer $x\le 19$ are: 

x= 2  Y(x)= 1
x= 3  Y(x)= 3
x= 4  Y(x)= 3
x= 5  Y(x)= 5
x= 6  Y(x)= 5
x= 7  Y(x)= 9
x= 8  Y(x)= 9
x= 9  Y(x)= 9
x=10  Y(x)= 9
x=11  Y(x)=13
x=12  Y(x)=13
x=13  Y(x)=21
x=14  Y(x)=21
x=15  Y(x)=21
x=16  Y(x)=21
x=17  Y(x)=25
x=18  Y(x)=25
x=19  Y(x)=33
These are exact for the computed $x$ and are consistent with the easy bound $Y(x)\ge x-1$.

5) VERIFICATION.
  • Lemma 687.1 is a direct CRT equivalence; the key point is that choosing one residue class mod each prime is equivalent to choosing a single shift $t$ modulo their product.
  • Lemma 687.2 is checked: for each $n\le x-1$, the integer $n+1$ has some prime factor $\le x$, guaranteeing coverage by $a_p\equiv -1$.
  • Lemma 687.3 is very weak asymptotically because $P(x)$ grows super-exponentially in $x$, but it is logically correct.
FINAL.
UNRESOLVED

(i) Strongest proved partial result. An exact reformulation: $Y(x)$ is the maximum length of a run of consecutive integers all not coprime to the primorial $P(x)=\prod_{p\le x}p$ (Lemma 687.1). In particular, $Y(x)\ge \lfloor x\rfloor-1$ (Lemma 687.2) and $Y(x)\le P(x)-\varphi(P(x))$ (Lemma 687.3).

(ii) First gap (crisp statement). Prove a nontrivial upper bound improving $Y(x)\ll x^2$ (e.g. $Y(x)=o(x^2)$) from only the covering/CRT formulation, without assuming any results not already in the problem statement.

(iii) Top 3 next moves.

  1. Translate $Y(x)$ bounds into bounds on maximal gaps between integers coprime to $P(x)$, and attempt to apply/derive sieve bounds on gaps between $x$-rough numbers (integers with no prime factor $\le x$).
  2. Perform more computations for larger $x$ (as far as feasible) to see whether $Y(x)$ appears closer to $x(\log x)^2$ or to $x^2$.
  3. Try to prove that any run of length $L$ with all numbers divisible by primes $\le x$ forces the existence of a prime gap of comparable length; then import (only) those prime-gap bounds explicitly allowed by the problem statement.

(iv) What a minimal counterexample would likely look like. A counterexample to $Y(x)=o(x^2)$ would require a sequence $x_j\to\infty$ and shifts $t_j$ such that the interval $[t_j+1,t_j+Y(x_j)]$ contains no integer coprime to $P(x_j)$, with $Y(x_j)$ of order $c\,x_j^2$ or larger; equivalently, extremely long gaps between integers coprime to the primorials.

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.