Erdős Problem #1138
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Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
Let $p_n$ denote the increasing primes and $d(x)=\max_{p_n<x}(p_{n+1}-p_n)$. The proposed assertion is that for every fixed $C>1$, uniformly over $x/2<y<x$, \[ \pi(y+Cd(x))-\pi(y)\sim \frac{Cd(x)}{\log y}. \] We disprove this universal assertion. The proof reconstructs the record-gap argument reported at https://www.erdosproblems.com/1138 and in the linked Sunder--Kumrawat--Cheri write-up, which credits GPT-5.5 assistance: https://sourish-kumrawat.github.io/papers/Erdos_1138.pdf.
Restatement from Math-6192-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-6192
Erdős Problem 1138: a prime-gap obstruction
Statement and scope.
Let $p_n$ denote the increasing primes and $d(x)=\max_{p_n<x}(p_{n+1}-p_n)$. The proposed assertion is that for every fixed $C>1$, uniformly over $x/2<y<x$, \[ \pi(y+Cd(x))-\pi(y)\sim \frac{Cd(x)}{\log y}. \] We disprove this universal assertion. The proof reconstructs the record-gap argument reported at https://www.erdosproblems.com/1138 and in the linked Sunder--Kumrawat--Cheri write-up, which credits GPT-5.5 assistance: https://sourish-kumrawat.github.io/papers/Erdos_1138.pdf.Record gaps.
Prime gaps are unbounded, since factorial blocks give arbitrarily long strings of composite integers. Consequently there are infinitely many increasing record gaps $d=q-p$, where $p,q$ are consecutive primes and no earlier prime gap is larger. The prime number theorem implies $d=o(p)$ along these records: for each fixed $\varepsilon>0$ there is a prime in $(p,(1+\varepsilon)p)$ for all sufficiently large $p$.Set $x=p+1$ and $y=p-2d$. Then $d(x)=d$; this includes the gap beginning at $p$, and there is no further prime below $x$. For sufficiently large records, $x/2<y<x$. The two endpoints \[ y+2d=p,\qquad y+\tfrac52d=p+\tfrac12d \] have the same prime counting function, because $(p,q)$ contains no prime. Hence the two corresponding numerators have a common value $M$.
Incompatible asymptotics.
Put $T=d/\log y>0$. The proposed asymptotics for the two fixed constants $C=2$ and $C=5/2$ would require simultaneously \[ M/(2T)\longrightarrow1,\qquad M/((5/2)T)\longrightarrow1, \] which is impossible. More quantitatively, for every $M\geq0$, \[ \max\left\{\left|\frac{M}{2T}-1\right|, \left|\frac{M}{(5/2)T}-1\right|\right\}\geq\frac19. \] Otherwise the two allowed intervals for $M/T$, whose common boundary would be $20/9$, are disjoint. One of these two fixed constants therefore fails along infinitely many of the records, by the infinite pigeonhole principle.This refutes the assertion quantified over every $C>1$. It does not claim that the displayed argument singles out which of $2$ and $5/2$ fails, or that every fixed $C$ fails.