we-solve (beta)

Math-5923-A1 · revision r1 · on Math-5923

Reused writeup on Math-5923 (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 826 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. FORMAL RESTATEMENT
  5. QUICK LITERATURE/CONTEXT CHECK
  6. ATTACK PLAN
  7. WORK
  8. FAST REALITY CHECK (computation for fixed constants).
  9. VERIFICATION
  10. FINAL
  11. COMPLETION ESTIMATE

Erdős Problem 826 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/826

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_5.2/826.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/826.tex.

FORMAL RESTATEMENT

Let $\tau(n)$ denote the number of positive divisors of $n$. The question is:

Problem. Do there exist infinitely many integers $n\ge 1$ such that for all integers $k\ge 1$, \[ \tau(n+k) \ll k, \] meaning: there exists an absolute constant $C>0$ (independent of $n$ and $k$) such that for the given $n$, \[ \tau(n+k)\le Ck\qquad\text{for all }k\ge 1? \]

QUICK LITERATURE/CONTEXT CHECK

The statement says this is a stronger form of another Erdős problem $\#248$. No further results are stated in the file.

ATTACK PLAN

  • Proof track: Try to construct $n$ so that $n+1,n+2,\dots$ systematically avoid having too many divisors relative to their offset $k$. For small $k$ this resembles finding prime/rough patterns in short intervals, which is difficult.
  • Disproof track: Try to show that for every $n$, some $k$ forces $\tau(n+k)$ to be larger than $Ck$ for any fixed $C$. This would require producing very highly composite numbers in every translate, at controlled distance, which also seems difficult.
  • What I will do: Prove elementary lemmas that reduce the quantifier "for all $k\ge 1$" to finitely many checks for a fixed constant $C\ge 2$, and do an exact search for small $n$ for the case $C=2$ and summary counts for $C=3,4,5$ up to $50{,}000$.

WORK

FAST REALITY CHECK (computation for fixed constants).
Define a property $P_C(n)$: $\tau(n+k)\le Ck$ for all $k\ge 1$.
  • For $C=2$, an exact search up to $n\le 1{,}000{,}000$ found precisely \[ \{n\le 10^6: P_2(n)\} = \{1,2,4,6,12,36,60,72,420\}. \] In particular there are only 9 such $n$ up to $10^6$.
  • For $N=50{,}000$, the counts of $n\le N$ with $P_C(n)$ were: \[ \#\{n\le 50{,}000: P_2(n)\}=9, \quad \#\{n\le 50{,}000: P_3(n)\}=82, \quad \#\{n\le 50{,}000: P_4(n)\}=1905, \quad \#\{n\le 50{,}000: P_5(n)\}=2720. \]
These computations do not prove infinitude for any $C$, but they indicate the constraint is very strong for small $C$.

Lemma (Trivial divisor bound). For every integer $m\ge 1$, one has $\tau(m)\le m$.

Proof. Every positive divisor of $m$ is at least $1$ and at most $m$, so there are at most $m$ distinct positive divisors. $\square$

Lemma (For $C\ge 2$, only finitely many $k$ need checking). Fix $C\ge 2$ and $n\ge 1$. If $\tau(n+k)\le Ck$ holds for every integer $k$ with $1\le k\le n$, then it holds for all integers $k\ge 1$.

Proof. Assume $\tau(n+k)\le Ck$ for all $1\le k\le n$.

Let $k>n$. Then $n+k < 2k$ and by the trivial bound $\tau(m)\le m$, \[ \tau(n+k)\le n+k < 2k \le Ck \] because $C\ge 2$. Thus the inequality holds automatically for all $k>n$. $\square$

Lemma (Divisor bound $\tau(m)\le 2\sqrt m$ and an automatic $k$-range). For every $m\ge 1$, one has $\tau(m)\le 2\sqrt m$. Consequently, for any $n\ge 1$, any constant $C>0$, and any integer $k$ such that \[ k \ge \frac{2}{C^2}\left(1+\sqrt{1+C^2 n}\right), \] one has \[ \tau(n+k)\le Ck. \]

Proof. For the bound $\tau(m)\le 2\sqrt m$: pair each divisor $d\le \sqrt m$ with the (distinct) divisor $m/d\ge \sqrt m$. There are at most $\lfloor \sqrt m\rfloor$ divisors $d\le \sqrt m$, hence \[ \tau(m)\le 2\lfloor \sqrt m\rfloor \le 2\sqrt m. \]

Now fix $n\ge 1$ and $C>0$ and assume $k\ge \frac{2}{C^2}(1+\sqrt{1+C^2 n})$. A direct rearrangement shows this is exactly the condition that the quadratic inequality \[ C^2k^2-4k-4n\ge 0 \] holds, i.e. that $C^2k^2\ge 4(n+k)$. Taking square roots (all terms are nonnegative) gives \[ Ck\ge 2\sqrt{n+k}. \] Therefore \[ \tau(n+k)\le 2\sqrt{n+k}\le Ck. \] $\square$

VERIFICATION

  • Lemma (For $C\ge 2$, only finitely many $k$ need checking): checked the strict inequality $n+k<2k$ for $k>n$; the argument needs $C\ge 2$ to conclude $2k\le Ck$.
  • Lemma (Divisor bound $\tau(m)\le 2\sqrt m$ and an automatic $k$-range): checked that the displayed condition on $k$ is equivalent to the quadratic inequality $C^2k^2-4k-4n\ge 0$ and hence to $2\sqrt{n+k}\le Ck$.
  • Computation: For $C=2$, after the finite-check lemma it suffices to check $1\le k\le n$; the search implemented exactly this and cross-checked up to $10^6$.

FINAL

UNRESOLVED.
  1. (i) Strongest proved partial result. For any fixed $C\ge 2$, to verify $\tau(n+k)\le Ck$ for all $k\ge 1$ it suffices to check the finite range $1\le k\le n$ (by the finite-check lemma). Computations show only 9 values of $n\le 10^6$ satisfy the inequality with $C=2$.
  2. (ii) First gap. Prove or disprove that there exists an absolute constant $C$ such that $P_C(n)$ holds for infinitely many $n$.
  3. (iii) Top 3 next moves.
    1. For a given candidate $C$ (say $C=2$ or $3$), search for structural necessary conditions on $n$ coming from the small-$k$ constraints (e.g. $k=1$ forces $n+1$ to have at most $C$ divisors), and try to prove that these conditions can or cannot hold infinitely often.
    2. Use the $\tau(m)\le 2\sqrt m$ lemma to reduce the needed verification range further (from $k\le n$ down to $k\le O(\sqrt n)$ with better constants depending on $C$) and then combine with distributional results about $\tau$ in short intervals (if available in the allowed corpus).
    3. Extend computations to larger $N$ with optimized divisor-count sieves to see whether the set of $n$ satisfying $P_2(n)$ stabilizes (suggesting finiteness) or new examples appear.
  4. (iv) Minimal counterexample structure (if the statement were false). For each fixed $C$, one would need that $P_C(n)$ fails for all sufficiently large $n$. A minimal obstruction would be a mechanism guaranteeing, for every large $n$, the existence of some small $k$ (say $k\le K(C)$) with $\tau(n+k)>Ck$; i.e. a uniform supply of unusually highly composite values among the first $K(C)$ successors of every $n$.

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.