REVIEW 4 major objections 5 minor 1 references
Proof of Existence of Integers Excluding Two Residue Values in a Specific Range
T0 review · 4 major / 5 minor · reviewed 2026-08-10 · deepseek-v4-flash
Pith's one-line read The paper claims that no matter which two residue classes are chosen modulo each prime up to p_k, the interval [p_k^2, p_{k+1}^2] still contains an integer outside all of them.
desk verdict Claims a worst-case sieve existence result but never proves it: the central quantity is a density difference, not a survivor count, and the constant is off. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The carrying object is the Mertens-type product P(x) = ∏_{2<p<x}(1 − 2/p), interpreted as the survival density for an integer to avoid two fixed residue classes modulo each odd prime up to x. The paper proves P(x) = C_2 (ln x)^{-2} (1 + O(1/ln x)) with C_2 ≈ 1.07, and then forms the telescoping difference Δ_k = p_{k+1}^2 P(p_{k+1}^2) − $p_k^{2}$ P($p_k^{2}$). This difference is meant to count the survivors in [$p_k^{2}$, p_{k+1}^2]. The fixed-gap assumption allows the difference to be evaluated by a discrete derivative of the smooth function $p^{2}$/(ln p)^2, giving a positive main term proportional to h p_k/(ln p_k)^2, and that positivity is what yields the existence conclusion.
What would settle it
Check the smallest case stated in the claim: for k=1, the primes up to p_1=2 offer only the two residue classes 0 and 1 modulo 2, which together cover every integer, so the interval [4,9] contains no integer outside them, contradicting the claim.
Extended reading notes
Core claim
The central claim is that for every k, after choosing any two residue values modulo each prime up to p_k, at least one integer in [$p_k^{2}$, p_{k+1}^2] lies outside all the chosen classes. The proof rests on the asymptotic P(x) = C_2 / (ln x)^2 with C_2 ≈ 1.07 for the product over odd primes of (1 − 2/p). Substituting x = $p_k^{2}$ and x = p_{k+1}^2 and subtracting produces Δ_k ~ (C_2 h / 2) p_k / (ln p_k)^2 under the fixed-gap assumption p_{k+1} − p_k = h ≥ 2. Because the main term is positive and unbounded, the author concludes that the covered set cannot exhaust the interval, so an uncovered integer must exist. The manuscript also notes that restricting to odd candidates halves the raw estimate and that the same scheme extends to products ∏(1 − c/p).
Load-bearing premise
The proof depends on treating the telescoping difference Δ_k as the number of integers in [$p_k^{2}$, p_{k+1}^2] that avoid the chosen residue classes, a step that is asserted rather than demonstrated.
Editorial extensions
If this is right
- No choice of two residue classes modulo primes up to p_k can fully cover the interval [p_k^2, p_{k+1}^2]; at least one integer always escapes.
- For intervals with a fixed prime gap h, the guaranteed number of escaping integers grows like h p_k/(ln p_k)^2, so the residual set is nonempty and grows.
- If only odd integers matter, the raw survivor estimate is halved, because roughly half of all integers in large intervals are even.
- The same telescoping product argument, with (1 − 2/p) replaced by (1 − c/p), would give survivor estimates of order h p_k/(ln p_k)^c for any real c > 0.
- Under the average gap p_{k+1} − p_k ≈ ln p_k, the survivor estimate would be of larger order, p_k/ln p_k, than in the fixed-gap case.
Reading between the lines
- The argument as written identifies a difference of density estimates with an integer count, a step the manuscript does not justify; if that identification fails, the existence conclusion is open.
- If the residue-class system includes the modulus 2, the claim as worded fails at k=1, since the two residue classes modulo 2 cover every integer; a corrected statement would need to exclude p=2 explicitly or restrict to odd primes.
- A genuine sieve treatment of the same interval would keep the modulus cutoff at p_k and use P(p_k) rather than P(p_k^2) or P(p_{k+1}^2); doing so would change the estimate and could either rescue or overturn the existence claim.
- A computational check for small k (e.g., k=2) can be performed within minutes: choose residues, enumerate the interval, and compare the survivor count with Δ_k; the result would show whether the telescoping difference is the right counting tool.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper claims that for every k and every choice of two residue classes modulo each prime p ≤ p_k, there is at least one integer in [p_k^2, p_{k+1}^2] that avoids all chosen classes. To support this, it proves an asymptotic for P(x) = ∏_{2<p<x}(1 − 2/p) with an explicit constant ~C2 ≈ 1.07, defines a telescoping difference Δ_k = p_{k+1}^2 P(p_{k+1}^2) − p_k^2 P(p_k^2), and, under the assumption of a fixed gap p_{k+1} − p_k = h, derives Δ_k ~ (~C2 h / 2) p_k/(ln p_k)^2. The paper then interprets Δ_k as the number of uncovered integers in the interval and concludes that such integers always exist. The central logical step is the identification of a smooth asymptotic density difference with a count of survivors for a specific covering system, and this step is not justified.
Significance. If the existence claim were true, it would be a notable statement about covering systems and sieves; however, the manuscript does not provide a valid proof of that claim. The positive aspects are the transparent derivation of a Mertens-type asymptotic and the elementary telescoping computation, which are checkable and would be useful in a corrected note. But the main theorem is not established: the constant in Theorem 1 is incorrect, the central counting interpretation of Δ_k is invalid, and Theorem 2 depends on an unproved fixed-gap assumption. The result, as it stands, does not rise to a publishable mathematical contribution.
major comments (4)
- [§2, definition of Δ_k and Conclusion] The proof never connects Δ_k to the number of integers in [p_k^2, p_{k+1}^2] that avoid two residue classes modulo primes up to p_k. P(x) is a product over primes < x, so p_k^2 P(p_k^2) is a smooth asymptotic expression evaluated at a point, not a count of survivors modulo the fixed modulus Q = ∏_{p≤p_k} p. The telescoping difference is a derivative-like quantity of order h p_k/(ln p_k)^2, while the interval has length ~2 h p_k and the survivor set is governed by a modulus Q that is far larger than the interval. No inclusion-exclusion, Chinese remainder argument, or covering-system bound is supplied, so the central existence claim does not follow.
- [§2, Theorem 2] Theorem 2 is conditional on the unproven assumption that p_{k+1} − p_k = h for infinitely many k; for h = 2 this is the twin-prime conjecture. The Introduction and Conclusion assert the result for every k, but the only quantitative result proved is conditional on this assumption. Thus the universal existence claim is not established by the theorem.
- [§1, Theorem 1] The constant ~C2 ≈ 1.07 is incorrect. The termwise expansion ln(1 − 2/p) = −2/p − 2/p^2 + O(1/p^3) is valid, but summing the O(1/p^3) tail produces an O(1) constant that contributes to the leading multiplicative constant; it cannot be absorbed into the O(1/ln x) error without changing the constant. The standard asymptotic is ∏_{2<p≤x}(1 − 2/p) ~ (4 e^{−2γ} C_2)/(ln x)^2, where C_2 is the twin-prime constant, giving 4 e^{−2γ} C_2 ≈ 0.832, not 1.07. This invalidates the numerical value used in Δ_k.
- [Introduction and §1 Remark] The claimed existence statement concerns residue classes modulo primes up to p_k, which includes p = 2, but the product P(x) explicitly excludes p = 2. As written, choosing the two residue values 0 and 1 modulo 2 covers every integer, so no integer in the interval survives. The statement is false unless p = 2 is explicitly excluded or the intended domain is restricted to odd primes, which is not stated.
minor comments (5)
- [Abstract and Introduction] The phrase 'residence values' should be 'residue values', and there are several typographical errors such as 'specific' and 'modulo prime numbers is'.
- [§2, heading] The heading '[p^2_k, p^2_{k+1})nd Parity Considerations' should read 'and Parity Considerations'; the 'nd' is a typo.
- [Introduction] The statement 'for any prime p ≤ x' uses x without definition; the later notation uses p_k. Please align the notation so the residue-class condition is stated consistently.
- [§2, discrete difference calculation] In the expansion of 1/(ln(p+h))^2, the displayed intermediate error O(1/(p(ln p)^2)) is not sufficient to justify the final O(1/(p(ln p)^3)); the calculation should explicitly retain the 2h ln p/p term or state the final error directly.
- [References] The manuscript cites only [1] for Mertens' theorem; adding a standard reference for the twin-prime constant and the corrected product asymptotic would help the reader verify the constant issue.
Circularity Check
No circularity found; the proof's defect is an unsupported logical leap, not an input-output reduction.
full rationale
The paper's derivation chain is not circular. Theorem 1 derives an asymptotic for P(x)=∏_{2<p<x}(1−2/p) from Mertens' prime-sum expansions, citing an external standard proof (Villarino), and the constant ~C2 is computed, not fitted to the target claim. Theorem 2 then forms the telescoping difference Δ_k = p_{k+1}^2 P(p_{k+1}^2) − p_k^2 P(p_k^2) and expands it under the additional hypothesis p_{k+1}−p_k = h ≥ 2; this algebra depends only on Theorem 1 and standard asymptotic manipulation. The Conclusion, however, asserts existence of integers in [p_k^2, p_{k+1}^2] avoiding two residue classes modulo each prime up to p_k. That existence statement is not derived from Δ_k: Δ_k is an asymptotic difference of Mertens-product densities, never defined as a survivor count and never connected to arbitrary residue-class selections by inclusion-exclusion or CRT. This is an invalid inference, not circular reasoning: the paper does not assume its target conclusion, does not fit a parameter to the claimed survivor, and contains no load-bearing self-citation. The unproven fixed-gap assumption p_{k+1}−p_k=h≥2 and the literal p=2 counterexample are serious correctness concerns but they are not circularity concerns.
Assumptions & free parameters
assumptions (5)
- standard math Mertens' theorem: Σ_{p≤x} 1/p = ln ln x + M1 + O(1/ln x)
- standard math Σ_p 1/p^2 converges to S ≈ 0.4522474
- domain assumption There are infinitely many k with pk+1 - pk = h ≥ 2
- ad hoc to paper The difference Δ_k equals or bounds the number of integers in [p_k^2, p_{k+1}^2] avoiding two residue classes modulo primes up to p_k
- ad hoc to paper The O(1/p^3) terms in the expansion of ln(1 - 2/p) can be absorbed into the error without affecting the constant
Cite this review
Pith. "Pith review of Proof of Existence of Integers Excluding Two Residue Values in a Specific Range." pith.science (2026). https://pith.science/paper/F73ZAJYO
@misc{pith2026250115707,
author = {Pith},
title = {Pith review of: Proof of Existence of Integers Excluding Two Residue Values in a Specific Range},
year = {2026},
howpublished = {\url{https://pith.science/paper/F73ZAJYO}},
note = {Machine review of arXiv:2501.15707}
}
abstract
This paper investigates the existence of integers that exclude two specific residence values modulo primes up to $p_k$ within the interval $[p_k^2, p_{k+1}^2]$. Using asymptotic results from analytic number theory, we establish bounds on the proportion of integers excluded by the union of residue classes. The findings highlight the density of residue class coverage in large intervals, contributing to the understanding of modular systems and their implications in number theory and related fields.
Reference graph
Works this paper leans on
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[1]
[1] M. B. Villarino, Mertens’ Proof of Mertens’ Theorem , arXiv:math/0504289v3, Universidad de Costa Rica, 2005. 7
work page Pith review arXiv 2005
Reviewed August 10, 2026 · model on record in the stance chip above.
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