REVIEW
Primes in short intervals assuming the Riemann hypothesis
Number Theory · Submitted 20 Oct 2026 · 6 references
Pith's verdict: Clean RH-conditional extension of BHP with genuinely new explicit thresholds, but Theorem 2's finite-range step is asserted rather than proved, and Theorem 3 inherits that gap.
T0 · AI review
Why Pith says this
Highlight any text to comment on it.
Referee report
The paper
The paper claims
Under RH, a prime lies in $[x, x+x^{1/2+1/m}]$ for every $x \geq 1$ whenever $m \leq 15.831$ — so consecutive $M$-th powers are always separated by a prime for $M \geq 2.29$.
The reading
What carries the argument
The carrying mechanism is a comparison of two interval lengths. Under RH, a 2019 bound on prime gaps guarantees a prime in the interval $[x, x + \frac{22}{25}\sqrt{x}\,\ln x]$, while the target Baker–Harman–Pintz interval has length $x^{1/2+1/m}$. The two lengths are equal when $x^{1/m} = \frac{22}{25}\ln x$, an equation the note solves two ways: for fixed $m$, $x = \exp(-m\,W_{-1}(-25/(22m)))$, where $W_{-1}$ is the $-1$ branch of the Lambert $W$ function (the inverse of $w e^w$ that tends to $-\infty$); for fixed $x$, $m = \ln x/(\ln\ln x + \ln(22/25))$. Everything above the verified threshold follows from this one identity: once the target interval is the longer one, the guaranteed-prime interval acts as a witness inside it. Below the threshold the witness is the finite computation, and for exponents $m \leq 2$ the Bertrand–Chebyshev theorem covers $[x, 2x]$. The step from Theorem 2 to Theorem 3 is the substitution $x = y^M$ with $M = 2m/(m-2)$, whose point is the identity $M-1 = M(\tfrac12 + \tfrac1m)$, so that $[x, x+x^{1/2+1/m}]$ becomes $[y^M, y^M + y^{M-1}]$, and this interval sits inside $[y^M, (y+1)^M]$ since $(y+1)^M \geq y^M + M y^{M-1}$.
Extended reading notes
Core claim
On the assumption that the Riemann hypothesis is true, the note claims three explicit statements about where primes must lie. Theorem 1: for every $x \geq 3.5\times 10^{90}$ the interval $[x, x+x^{1/2+1/40}]$ contains a prime — the 2001 Baker–Harman–Pintz theorem with its 'sufficiently large' threshold filled in. Theorem 2: for every $x \geq 1$ and every $0 < m \leq 15.831$, the interval $[x, x+x^{1/2+1/m}]$ contains a prime. Theorem 3: for every $y \geq 1$ and every $M \geq 2.29$ there is a prime between $y^M$ and $(y+1)^M$. The small-$x$ half of Theorem 2 is delegated to a computer verification of the upper Oppermann conjecture (a prime between $n^2$ and $n^2+n$) through $n = 7\times 10^{13}$, i.e. $x \leq 4.9\times 10^{27}$; the large-$x$ half is a direct comparison of interval sizes, with crossover at $x_* = \exp(-40\,W_{-1}(-5/176)) \approx 3.4828\times 10^{90}$ for $m=40$ and the exponent bound $m < \ln x_{**}/(\ln\ln x_{**} + \ln(22/25)) = 15.83179525$ at $x_{**} = 4.9\times 10^{27}$. Theorem 3 follows from Theorem 2 through the substitution $x = y^M$ with $M = 2m/(m-2)$, which rewrites the target interval as $[y^M, y^M + y^{M-1}]$, contained in $[y^M, (y+1)^M]$.
Load-bearing premise
The openly assumed Riemann hypothesis is the big price, but the step most exposed to error is the transfer from integers to reals: the note quotes a computer verification (a prime in $(n^2, n^2+n]$ for every integer $n \le 7\times 10^{13}$) and asserts without derivation that $(x, x+\sqrt{x}]$ then contains a prime for every real $x \le 4.9\times 10^{27}$; Theorems 2 and 3 stand or fall on that implication and on the computation that feeds it.
Editorial extensions
If this is right
- The Baker–Harman–Pintz theorem becomes effective under RH: for every $x \geq 3.5\times 10^{90}$ the interval $[x, x+x^{1/2+1/40}]$ provably contains a prime, the first explicit threshold of its kind.
- A single cap on the exponent works uniformly: for every $x \geq 1$ and every $0 < m \leq 15.831$, the interval $[x, x+x^{1/2+1/m}]$ provably contains a prime.
- There is always a prime between consecutive $M$-th powers for every $M \geq 2.29$ and $y \geq 1$, improving the known prime-between-cubes range down toward the prime-between-squares exponent $M = 2$.
- Extending the Oppermann verification to larger $n$ mechanically raises $m$ and lowers $M$ through the explicit crossover formulas; complete verification of the upper Oppermann conjecture would let $m$ grow without bound and push $M$ down to $2$.
- The analysis marks $3.5\times 10^{90}$ as the region where an unconditional effective $m=40$ theorem would have to operate, since the RH-conditional mechanism cannot easily do better.
Reading between the lines
- The whole large-$x$ content reduces to one crossover identity, $x^{1/m} = \frac{22}{25}\ln x$; any future sharpening of the RH-conditional prime-gap constant, or any longer verified range, would feed through the same formula and upgrade $m$ and $M$ without any other change to the proof.
- The part of the chain below $x = 4.9\times 10^{27}$ rests on a finite but enormous computation; the transfer from integer squares to every real $x$ is the one step a reader cannot redo by hand, and the most natural first target for independent checking.
- Run under the weaker 7/8 quasi-Riemann hypothesis, the same comparison gives only $m \geq 8/3$ and $M \geq 8$, as the author notes; the gap between those numbers and $m = 15.831$, $M = 2.29$ is a quantitative measure of what the full Riemann hypothesis buys.
- A natural extension is to rerun the identical crossover computation against any future explicit prime-gap bound, conditional or not: nothing in the argument would change beyond substituting the new constants in the Lambert-$W$ identity.
Original abstract
In 2001 Baker, Harman, and Pintz proved, for $x$ sufficiently large, the existence of at least one prime in the sub-linear interval $[x,x+x^{\frac{1}{2} +\frac{1}{40}}]$. But just how large is sufficiently large? In the current note I shall make the Baker--Harman--Pintz theorem effective, but at an extremely high price. Contingent upon assuming validity of the Riemann hypothesis (RH), I shall show that the Baker--Harman--Pintz theorem holds for $x \geq 3.5\times 10^{90}$. Furthermore, again assuming RH, I shall also show that for any $0< m\leq15.831$ and $x \geq 1$ there is at least one prime in the interval $[x,x+x^{\frac{1}{2} +\frac{1}{m}}]$, and that for $M\geq 2.29$ and $y \geq 1$ there is at least one prime in the interval $[y^M, (y+1)^M]$.
The compiled record
The paper's own sentences, arranged, kept beside the review and separate from it. Each label is drawn from a closed vocabulary; the number beside it is the probability the classifier gave that label. Numbers in the tables are the authors', checked by code. No sentence in this record was written by a model.
The abstract, sentence by sentence
In 2001 Baker, Harman, and Pintz proved, for $x$ sufficiently large, the existence of at least one prime in the sub-linear interval $[x,x+x^{\frac{1}{2} +\frac{1}{40}}]$. But just how large is sufficiently large? In the current note I shall make the Baker--Harman--Pintz theorem effective, but at an extremely high price. Contingent upon assuming validity of the Riemann hypothesis (RH), I shall show that the Baker--Harman--Pintz theorem holds for $x \geq 3.5\times 10^{90}$. Furthermore, again assuming RH, I shall also show that for any $0< m\leq15.831$ and $x \geq 1$ there is at least one prime in the interval $[x,x+x^{\frac{1}{2} +\frac{1}{m}}]$, and that for $M\geq 2.29$ and $y \geq 1$ there is at least one prime in the interval $[y^M, (y+1)^M]$.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | In 2001 Baker, Harman, and Pintz proved, for $x$ sufficiently large, the existence of at least one prime in the sub-linear interval $[x,x+x^{\frac{1}{2} +\frac{1}{40}}]$. | background 1.00 | 0.02 | 0.05 | asserted 1.00 |
| s1 | But just how large is sufficiently large? | gap 0.99 | 0.07 | 0.02 | unclear 0.92 |
| s2 | In the current note I shall make the Baker--Harman--Pintz theorem effective, but at an extremely high price. | contribution 0.56 | 0.59 | 0.06 | asserted 0.97 |
| s3 | Contingent upon assuming validity of the Riemann hypothesis (RH), I shall show that the Baker--Harman--Pintz theorem holds for $x \geq 3.5\times 10^{90}$. | result 0.91 | 0.93 | 0.47 | asserted 1.00 |
| s4 | Furthermore, again assuming RH, I shall also show that for any $0< m\leq15.831$ and $x \geq 1$ there is at least one prime in the interval $[x,x+x^{\frac{1}{2} +\frac{1}{m}}]$, and that for $M\geq 2.29$ and $y \geq 1$ there is at least one prime in the interval $[y^M, (y+1)^M]$. | result 0.97 | 0.92 | 0.38 | asserted 1.00 |
What was examined, and what was not
- arXiv listing abstract v1
- full text
- figures and tables
- references and citing sentences
- relations to other papers
- P abstract
- S sections
- T body sentences
- C citations
- R cross-paper relations
References
Discussion
Disagree with the review, or know something it missed? Leave a comment, highlight any passage above to comment on it, or challenge this review formally.