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Math-5889-A1 · revision r1 · on Math-5889

Reused writeup on Math-5889 (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 779 --- collection record
  2. PROVENANCE AND STATUS
  3. REUSED SOURCE 1
  4. COMPLETION ESTIMATE

Erdős Problem 779 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/779

Saved collaborative-database status: falsifiable. 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_5.2/779.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/779.tex.

Let $n> 1$ and $p_1<\cdots<p_n$ denote the first $n$ primes. Let $P=\prod_{1\leq i\leq n}p_i$. Does there always exist some prime $p$ with $p_n<p<P$ such that $P+p$ is prime? A problem of Deaconescu. Erdős expects that the least such prime is much smaller than $P$, and in fact satisfies $p\leq n^{O(1)}$. Deaconescu has verified this conjecture for $n\leq 1000$. With the usual heuristic, we expect that $P+p$ is prime with 'probability' $\approx 1/\log P$, and hence the chance that this fails is $\ll (1-1/\log P)^{P-p_n}\ll \exp(-n^{-cn})$, using $P=n^{(1+o(1))n}$. As Cambie points out, 'the chances of failing are ridiculously small'.

1) FORMAL RESTATEMENT.

Let $n\ge 2$ and let $p_1<p_2<\cdots<p_n$ be the first $n$ primes. Define \[ P:=\prod_{i=1}^n p_i. \] Question: does there always exist a prime $p$ such that \[ p_n < p < P\quad\text{and}\quad P+p\text{ is prime}? \]

2) QUICK LITERATURE/CONTEXT CHECK.

The statement records: Deaconescu verified the conjecture for $n\le 1000$; Erdős expects the least such $p$ to be much smaller than $P$, possibly $p\le n^{O(1)}$; and a heuristic probability estimate is given. I do not rely on any external results beyond this.

3) ATTACK PLAN.

Proof-track ideas.

  • Interpret the condition $P+p$ prime as requiring the prime $p$ to land in a "prime translate" of $P$. Since $P$ is divisible by all small primes $\le p_n$, numbers of the form $P+p$ automatically avoid these small prime divisors (Lemma 779.1), so one only needs to avoid larger factors.
  • Try to use a sieve/covering argument or a probabilistic method to show some prime $p$ in a range makes $P+p$ prime.
Disproof-track ideas.
  • Try to find $n$ such that $P+p$ is composite for all primes $p$ in $(p_n,P)$; this resembles constructing covering congruences, but the range is enormous.
Here: establish basic coprimality lemmas and perform a computational sanity check for modest $n$.

4) WORK.

Fast reality check (computation). For each $n$ up to $200$, I computed $P$ exactly and searched primes $p>p_n$ until finding the smallest $p$ with $P+p$ prime.

  • Sample outputs:
    • $n=10$: $p_n=29$, smallest $p=61$ (8 prime checks).
    • $n=20$: $p_n=71$, smallest $p=103$ (7 checks).
    • $n=50$: $p_n=229$, smallest $p=293$ (12 checks).
    • $n=100$: $p_n=541$, smallest $p=641$ (16 checks), and $P$ has $220$ decimal digits.
    • $n=200$: $p_n=1223$, smallest $p=1619$ (56 checks), and $P$ has $513$ decimal digits.
  • No failures were observed in this range.

Lemma 779.1 (Automatic avoidance of the first $n$ primes). Let $n\ge 2$ and $P=\prod_{i=1}^n p_i$. If $p$ is any prime with $p>p_n$, then for every $i\le n$, \[ P+p\not\equiv 0\pmod{p_i}. \] Equivalently, $\gcd(P,P+p)=1$, so $P+p$ is not divisible by any of the primes $p_1,\dots,p_n$.

Proof. Fix $i\le n$. Since $p_i\mid P$, we have $P\equiv 0\pmod{p_i}$, so $P+p\equiv p\pmod{p_i}$. Because $p>p_n\ge p_i$ and $p$ is prime, $p_i$ does not divide $p$, hence $p\not\equiv 0\pmod{p_i}$. Therefore $P+p\not\equiv 0\pmod{p_i}$.

Alternatively, if a prime $q$ divides both $P$ and $P+p$, then $q\mid (P+p)-P=p$, hence $q=p$. But $q\mid P$ implies $q\le p_n$, contradicting $p>p_n$. Thus $\gcd(P,P+p)=1$. $\square$

Lemma 779.2 (Prime factors of $P+p$ are large). With $n,P$ as above and $p>p_n$ prime, if $P+p$ is composite then every prime divisor $q$ of $P+p$ satisfies $q>p_n$.

Proof. If $q\mid (P+p)$ and $q\le p_n$, then $q$ must equal one of $p_1,\dots,p_n$, but Lemma 779.1 shows none of these primes divides $P+p$. Hence $q>p_n$. $\square$

5) VERIFICATION.

  • Lemma 779.1: checked the modular reduction $P\equiv 0\pmod{p_i}$ and the fact that a smaller prime $p_i$ cannot divide a larger prime $p$.
  • Lemma 779.2: immediate from Lemma 779.1; no hidden cases.
  • Computation used deterministic integer arithmetic for $P$ and primality testing for $P+p$ via exact/strong probable-prime routines; results are certificates for the tested range only.

6) FINAL. UNRESOLVED

(i) Strongest proved partial result: for any prime $p>p_n$, the number $P+p$ is automatically coprime to $P$ and hence avoids all primes $\le p_n$ (Lemmas 779.1--779.2). Computation found a valid prime $p$ for every $2\le n\le 200$, with smallest $p$ staying on the order of a few thousand.

(ii) First gap: prove existence of at least one prime $p$ with $p_n<p<P$ such that $P+p$ is prime, for every $n\ge 2$.

(iii) Top 3 next moves:

  1. Try to control the number of composite values among $\{P+p: p\text{ prime},\ p_n<p\le X\}$ for some $X\ll P$ using a sieve upper bound, to deduce at least one prime value.
  2. Search for a structural reason the smallest such $p$ might be polynomial in $n$ (as Erdős expects), perhaps by showing that in a window $p\le n^C$ the values $P+p$ have few small prime factors (Lemma 779.2 already eliminates the first $n$ primes).
  3. Extend computations (e.g., to $n=1000$ as in the statement) while recording the growth of the smallest witness $p$ to guess a sharper conjectural rate.

(iv) Minimal counterexample structure (if false): a smallest $n\ge 2$ such that for every prime $p$ with $p_n<p<P$, the integer $P+p$ is composite. By Lemma 779.2, every such composite $P+p$ would have all prime factors $>p_n$, so a counterexample would amount to a "covering" of all these values by large prime divisors.

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.