gpt pro 5.2 attempt on Math-987
mathematicsErdős problemscontributionManuscript
1) FORMAL RESTATEMENT
Let $p_r$ denote the $r$th prime, and write $g_r:=p_{r+1}-p_r$ for the prime gap. For a positive integer $n$, let $P^+(n)$ be the largest prime factor of $n$. The property in the problem is: \[ \exists\ \text{at least two integers }n\ \text{with }p_r<n<p_{r+1}\ \text{and }P^+(n)<g_r. \] The question is whether this happens for infinitely many $r$.
Edge cases: if $g_r=2$ then there are no integers strictly between $p_r$ and $p_{r+1}$; if $g_r=3$ there is exactly one integer in between, so the property cannot hold. Thus any $r$ witnessing the property must have $g_r\ge 4$.
2) QUICK LITERATURE/CONTEXT CHECK
From the provided problem statement: Erdős believed the answer is "yes" but that such $r$ are sparse, and he could show that the density of $r$ for which at least one such $n$ exists is $0$. The statement was also formalised in Lean in the Google DeepMind Formal Conjectures project (as noted in the source file).
3) ATTACK PLAN
- Try to build explicit $r$: a sufficient condition is to find a prime gap in which there are two numbers composed only of primes $<g_r$.
- Local sufficient conditions: if an integer $n$ in the gap is small compared to $g_r$, then having a prime factor $\ge g_r$ is impossible.
- Computation: scan prime gaps up to some bound and count how many $n$ in each gap satisfy $P^+(n)<g_r$.
4) WORK
Lemma 932.1 (a size condition that forces $g_r$-smoothness). Fix $r$ and set $g=g_r$. If $n$ is a composite integer satisfying $p_r<n<p_{r+1}$ and $n<2g$, then every prime factor of $n$ is $<g$. (Equivalently, $P^+(n)<g$.)
Proof. Assume $n$ is composite and $n<2g$. Let $q$ be any prime factor of $n$. If $q\ge g$, then $n=q\cdot m$ for some integer $m\ge 2$ (since $n$ is composite). Hence $n\ge 2q\ge 2g$, contradicting $n<2g$. Therefore every prime factor $q$ of $n$ satisfies $q<g$, as claimed. $\square$
Proposition 932.2 (explicit small example). For $r=4$ we have $(p_r,p_{r+1})=(7,11)$ and $g_4=4$. The integers $8$ and $9$ satisfy $7<8<9<11$ and all their prime factors are $<4$. Hence $r=4$ satisfies the property.
Proof. Compute: $8=2^3$ has prime factor set $\{2\}$, and $9=3^2$ has prime factor set $\{3\}$. Both $2$ and $3$ are strictly less than $g_4=4$. $\square$
FAST REALITY CHECK (local computation up to $200{,}000$). I computed all prime gaps with $p_{r+1}\le 200{,}000$ and counted, for each $r$, how many integers $n$ with $p_r<n<p_{r+1}$ satisfy $P^+(n)<g_r$. In this range I found $20$ values of $r$ for which there are at least two such integers. The first few are:
| $r$ | $p_r$ | $p_{r+1}$ | $g_r$ | first two "good" $n$ |
| 4 | 7 | 11 | 4 | 8,9 |
| 9 | 23 | 29 | 6 | 24,25 |
| 11 | 31 | 37 | 6 | 32,36 |
| 15 | 47 | 53 | 6 | 48,50 |
| 24 | 89 | 97 | 8 | 90,96 |
5) VERIFICATION
- Lemma 932.1: the only use of compositeness is to ensure $m\ge 2$ when writing $n=q m$. If $n$ were prime, it could have $P^+(n)=n\ge g$.
- Proposition 932.2: verified by direct factorization.
6) FINAL
UNRESOLVED
(i) Strongest proved partial result here. A sufficient condition for an integer $n$ in the prime gap to satisfy $P^+(n)<g_r$ is $n<2g_r$ together with compositeness (Lemma 932.1). There are explicit small $r$ with at least two such $n$ (e.g. $r=4$) and computationally many more in small ranges.
(ii) First gap (crisp). Prove that there are infinitely many $r$ such that at least two integers $n$ with $p_r<n<p_{r+1}$ satisfy $P^+(n)<p_{r+1}-p_r$.
(iii) Top 3 next moves.
- Try to explicitly construct prime gaps that contain two numbers whose prime factors are all below the gap size (e.g. by combining known "large gap" constructions with smooth numbers).
- Compute further to much larger bounds to guess the growth rate (frequency) of such $r$, and to look for repeating structural patterns in the good integers.
- Seek a theoretical sufficient condition on the residue class structure of $p_r$ modulo small primes that guarantees two $g_r$-smooth integers in $(p_r,p_{r+1})$.
(iv) What a minimal counterexample would look like. A counterexample to the infinitude statement would mean that beyond some $r_0$, every prime gap $(p_r,p_{r+1})$ contains at most one integer whose prime factors are all $<g_r$. In particular, for large $r$ the interval would force most composites to have a prime factor at least the gap size.