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Math-6113-A1 · revision r1 · on Math-6113

Reused writeup on Math-6113 (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 1059 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. FORMAL RESTATEMENT
  5. QUICK LITERATURE/CONTEXT CHECK
  6. ATTACK PLAN
  7. WORK
  8. VERIFICATION
  9. FINAL
  10. COMPLETION ESTIMATE

Erdős Problem 1059 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1059

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_5.2/1059.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1059.tex.

FORMAL RESTATEMENT

Let $p$ be a prime. Define \[ \mathcal F(p)=\{k! : k\in\mathbb Z_{\ge 1},\ k!<p\}. \] We ask whether there exist infinitely many primes $p$ such that \[ \forall m\in\mathcal F(p),\quad p-m\ \text{is composite}. \] Throughout, "composite" means an integer $>1$ that is not prime, so the condition excludes $p-m=1$.

QUICK LITERATURE/CONTEXT CHECK

The problem statement (from Guy's collection) gives examples $p=101,211$ and suggests an easier related variant about general integers $n$ in a factorial interval. I do not use or claim any results beyond what is explicitly in the problem statement.

ATTACK PLAN

Proof-track idea: try to force each difference $p-k!$ to be composite by congruences $p\equiv k!\pmod{q_k}$ so that $q_k\mid (p-k!)$, and then appeal to infinitude of primes in an arithmetic progression. The obstruction is that the set of factorials to control grows with $p$. Disproof-track idea: search small primes for a counter-pattern (e.g. show only finitely many). Computations suggest the property is not rare among primes up to $10^6$, so disproof by small counterexample seems unlikely.

WORK

FAST REALITY CHECK. For $p\in\{2,3,5,7,11\}$ the property fails:
  • $p=2$: $1!<2$ and $2-1!=1$ (not composite).
  • $p=3$: $2!<3$ and $3-1!=2$ (prime), $3-2!=1$ (not composite).
  • $p=5$: $2!<5$ and $5-2!=3$ (prime).
  • $p=7$: $2!<7$ and $7-2!=5$ (prime).
  • $p=11$: $3!=6<11$ and $11-3!=5$ (prime).
A minimal local search over primes up to $10^6$ found many examples. Exact output: 
primes p <= 1,000,000 with property: 7874
first 30 examples:
[101, 211, 367, 409, 419, 461, 557, 673, 709, 769,
 937, 967, 1009, 1201, 1259, 1709, 1831, 1889, 2141, 2221,
 2309, 2351, 2411, 2437, 2539, 2647, 2837, 2879, 3011, 3019]

distribution by m(p)=max{k: k!<p} among these primes p<=1,000,000:
 m=4: 1
 m=5: 8
 m=6: 48
 m=7: 383
 m=8: 3000
 m=9: 4434
(These numbers were produced by a direct sieve-and-check script.)

Lemma 1059.1 (interval reduction). Let $p$ be a prime satisfying the property, and let \[ m:=m(p):=\max\{k\in\mathbb Z_{\ge 1}:\ k!<p\}. \] Then $m!<p<(m+1)!$, the condition to check is exactly the finite set $k=1,2,\dots,m$, and moreover $p\neq m!+1$ and $p-m!$ is a composite integer in the range $2\le p-m!\le m\cdot m!$.

Proof. By definition of $m$, we have $m!<p$ and $(m+1)!\ge p$, hence $m!<p\le (m+1)!-1$; in particular $p<(m+1)!$. The set of factorials $k!$ less than $p$ is exactly $\{1!,2!,\dots,m!\}$, so the property is precisely that $p-k!$ is composite for all $1\le k\le m$. For $k=m$, the condition says $p-m!$ is composite. Composite integers are $\ge 4$ except for $4,6,8,9,\dots$, but in any case composite means $>1$, so $p-m!\ge 2$. In particular $p\neq m!+1$. Finally, since $p<(m+1)!=(m+1)m!$, we have $p-m!<m\cdot m!$, hence $p-m!\le m\cdot m!$ because it is integral. $\square$

Proposition 1059.2 (finite-factorial forcing). Fix an integer $L\ge 1$. There exist infinitely many primes $p$ such that for every $k\in\{1,2,\dots,L\}$ the integer $p-k!$ is composite.

Proof. Choose pairwise distinct primes $q_1,q_2,\dots,q_L$ such that $q_k>k$ for each $k$. Then $q_k\nmid k!$ because every prime factor of $k!$ is $\le k$. For each $k$, impose the congruence \[ p\equiv k!\pmod{q_k}. \] By the Chinese remainder theorem (CRT), there exists an integer $r$ modulo \[ Q:=\prod_{k=1}^L q_k \] such that $r\equiv k!\pmod{q_k}$ for all $k$. Because $q_k\nmid k!$, we have $r\not\equiv 0\pmod{q_k}$ for each $k$, hence $\gcd(r,Q)=1$. By Dirichlet's theorem on primes in arithmetic progressions, there are infinitely many primes $p$ with \[ p\equiv r\pmod{Q}. \] Fix such a prime $p$. For each $k\le L$, we have $p\equiv k!\pmod{q_k}$, hence $q_k\mid (p-k!)$. Since there are infinitely many such primes, we may further restrict to those with \[ p>\max_{1\le k\le L}(k!+q_k). \] For these primes, $p-k!>q_k\ge 2$, so $p-k!$ has a nontrivial proper divisor $q_k$ and is therefore composite. This holds simultaneously for all $k=1,2,\dots,L$. $\square$

VERIFICATION

I checked the edge cases $p=2,3,5,7,11$ by direct evaluation of the defining condition. In Proposition 1059.2, the only number-theoretic input beyond CRT is Dirichlet's theorem; the coprimality hypothesis $\gcd(r,Q)=1$ is verified by choosing $q_k>k$ so that $q_k\nmid k!$. For the "composite" requirement, I explicitly ensured $p-k!>q_k$ by restricting to primes beyond $\max(k!+q_k)$.

FINAL

UNRESOLVED
  1. (i) Strongest proved partial result. For every fixed $L$ there are infinitely many primes $p$ for which $p-k!$ is composite for all $1\le k\le L$ (Proposition 1059.2). Computations show $7874$ such primes $p\le 10^6$.
  2. (ii) First gap (crisp). Upgrade Proposition 1059.2 to handle the moving target $L=m(p)=\max\{k:k!<p\}$, i.e. find infinitely many primes $p$ such that $p-k!$ is composite for all $k\le m(p)$. Equivalently: for infinitely many $m$ find a prime $p$ with $m!<p<(m+1)!$ such that $p-k!$ is composite for every $1\le k\le m$.
  3. (iii) Top 3 next moves. $1$ Computationally: for each $m$ search primes $p\in(m!, (m+1)!)$ satisfying the full condition for $k\le m$, to see whether examples persist for larger $m$. $2$ Try to combine the CRT forcing with an upper bound for the least prime in an arithmetic progression (a "least prime in AP" theorem) strong enough to guarantee a prime $p\equiv r\pmod Q$ below $(m+1)!$ when $Q$ is built from the needed congruences for $k\le m$. $3$ Attempt a different forcing that reuses moduli across many factorials (e.g. a structured covering by small primes) so that the modulus grows more slowly with $m$.
  4. (iv) Minimal counterexample structure. A disproof would require that beyond some threshold, every prime $p$ admits some $k$ with $1\le k!<p$ for which $p-k!$ is either prime or $1$. Such a counterexample would likely exploit that $k!$ quickly becomes very large, so only $k\ll \log p/\log\log p$ are relevant.

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.