{"id":"ec074ad0-58c4-414c-aaa5-c5bdd38e71ea","arxiv_id":"2501.15707","paper_version":1,"verdict":"REJECT","confidence":"HIGH","novelty_score":3.0,"correctness_risk":"high","formal_verification":"none","parameter_count":0,"one_line_summary":"The preprint asserts an existence result about integers avoiding two residues modulo primes, but the provided derivations only compute product asymptotics and do not prove the asserted existence.","lead":"This paper claims to prove that every interval between consecutive prime squares contains an integer that avoids two chosen residue classes modulo each small prime, but the proof only derives an asymptotic for a product and never counts such integers. The core existence claim is not established by the mathematics in the preprint.","discovery_kind":"incremental","skeptic_critique":{"model":"deepseek-v4-flash","headline":"The proof's Δ_k is a density difference, not a survivor count, so the claimed worst-case existence is never established; as literally stated the claim also fails for p=2.","rationale":"Read in good faith, the paper aims to prove a worst-case sieve-like covering statement. The technical core (Theorems 1–2) computes an asymptotic for ∏_{2<p<x}(1−2/p) and a telescoping endpoint difference. For that objective, the load-bearing step is the unstated identification of Δ_k with the number of survivors in the interval. The reader flagged this, and I agree it is fatal. The product P(x) uses primes < x, so P(p_{k+1}^2) incorporates all primes up to p_{k+1}^2, not just those ≤ p_k; the density also varies across the interval. Telescoping endpoint values of x^2 P(x) gives no count of integers in a specific short interval for a fixed residue selection. Without a lower-bound sieve or CRT argument uniform over all residue choices, the conclusion does not follow. The fixed-gap assumption in Theorem 2 further restricts the result to hypothetical infinitely many equal gaps. Theorem 1's constant has a separate error: the expansion's 'O(1)' is actually a convergent constant contribution, so exp(1.5 − 2(M+S)) ≈ 1.07 is not the Mertens constant; the exact constant including higher-order terms is about 0.832. The p = 2 issue is even more basic: as stated, the theorem includes modulo 2, where forbidding both residues covers every integer; the paper excludes p = 2 only inside the product. These are internal inconsistencies and missing arguments, not mere disagreement with consensus. The reader's REJECT verdict is appropriate; no change is needed.","tokens_in":4861,"tokens_out":11387,"duration_ms":108714,"concrete_test":"Evaluate k = 1 as literally stated: p_1 = 2, choose r_1 = 0 and r_2 = 1 modulo 2; if the interval [4, 9] contains no integer outside {0, 1} mod 2, the theorem as written is false. For the intended odd-prime version, run an exhaustive CRT/backtracking search for k = 4 over all 3 · 10 · 21 choices of two forbidden residue classes modulo 3, 5, and 7, and check whether [49, 121] always contains a survivor; this directly tests whether a positive-density heuristic can fail for an interval shorter than the modulus period.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The central claim is a worst-case existence statement, but the proof only computes an average-density quantity. Δ_k = p_{k+1}^2 P(p_{k+1}^2) − p_k^2 P(p_k^2) is a telescoping difference of Mertens-product asymptotics; P(x) is the product over primes < x, not over primes ≤ p_k, and P varies on the scale of log x. Thus Δ_k is not the number, nor an asymptotic lower bound for the number, of integers in [p_k^2, p_{k+1}^2] avoiding two residue classes modulo each prime ≤ p_k. No inclusion-exclusion or CRT argument controls the worst-case placement of the allowed residue classes over an interval of length far smaller than ∏_{p≤p_k} p; a positive asymptotic density does not force a survivor for every selection. Theorem 2 also assumes p_{k+1} − p_k = h ≥ 2, an unproven condition for all k (for h = 2 this is the twin-prime conjecture), so it cannot support a universal claim. Finally, as literally stated the claim includes p = 2: choosing the two residue values 0 and 1 modulo 2 covers every integer, so no integer in [p_k^2, p_{k+1}^2] survives. This is inconsistent with the restriction to 2 < p < x in the product. The numerical constant ~C2 ≈ 1.07 is also incorrect because the discarded O(1) tail of the log expansion contributes a nonzero constant; including the higher-order terms gives ~C2 ≈ 0.832, though the sign of the constant error is not the decisive issue.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","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.","tokens_in":5265,"tokens_out":5328,"duration_ms":45762,"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":[{"comment":"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.","section":"§2, definition of Δ_k and Conclusion"},{"comment":"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.","section":"§2, Theorem 2"},{"comment":"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.","section":"§1, Theorem 1"},{"comment":"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.","section":"Introduction and §1 Remark"}],"minor_comments":[{"comment":"The phrase 'residence values' should be 'residue values', and there are several typographical errors such as 'specific' and 'modulo prime numbers is'.","section":"Abstract and Introduction"},{"comment":"The heading '[p^2_k, p^2_{k+1})nd Parity Considerations' should read 'and Parity Considerations'; the 'nd' is a typo.","section":"§2, heading"},{"comment":"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.","section":"Introduction"},{"comment":"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.","section":"§2, discrete difference calculation"},{"comment":"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.","section":"References"}],"recommendation":"reject","confidential_remarks":"The central result is a worst-case existence claim, but the proof only derives a smooth density asymptotic and then identifies it with a survivor count without any justification. This gap, together with the incorrect constant and the reliance on an unproved fixed-gap assumption, cannot be repaired by local revision. I would not invite resubmission unless the author provides a genuine counting or covering argument for the claimed existence statement."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Colleague, this one can be desk-rejected. The paper's advertised result—that for every choice of two residue classes modulo each prime up to p_k, some integer in [p_k^2, p_{k+1}^2] avoids them all—is not established. The proof computes a Mertens-type asymptotic for P(x) = ∏_{2<p<x}(1−2/p) and then looks at the telescoping difference Δ_k = p_{k+1}^2 P(p_{k+1}^2) − p_k^2 P(p_k^2). That difference is a change in a smooth density times a growing square, not a count of integers in the interval. P(p_{k+1}^2) runs over primes up to p_{k+1}^2, not up to p_k, so the product has no direct relation to the interval's constraints. No inclusion-exclusion or CRT argument controls worst-case residue choices; a positive average density doesn't force a survivor for every selection.\n\nWhat's real: the log expansion for the product is standard and the binomial expansion of R(p+h)−R(p) is elementary and done cleanly. But the constant is wrong: the correct asymptotic has 4 e^{−2γ} C_2 ≈ 0.832, not exp(1.5−2(M1+S)) ≈ 1.07, because the discarded O(1) tail of the log expansion contributes to the constant. The fixed-gap assumption in Theorem 2 is an unproven twin-prime-type condition, so the result is conditional at best. And the claim as literally stated fails for p=2: choosing residues 0 and 1 mod 2 covers every integer. The product excludes p=2 by necessity, but the theorem statement doesn't.\n\nThe paper is not a serious candidate for peer review. It shows a bit of honest calculation practice, but the central logical step is missing and the main claim is false as stated. The writing also has slips (\"residence values\"). Reject without sending to referees.","headline":"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.","tokens_in":5692,"tokens_out":2471,"would_cite":false,"duration_ms":20448,"reading_group":"no","serious_thinker":"no","would_accept_peer_review":false},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11N05","11N35","11A07"],"pacs":[],"model":"deepseek-v4-flash","headline":"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.","keywords":["residue classes","prime squares","Mertens-type product","telescoping difference","modular arithmetic","prime gaps","analytic number theory","survivor count"],"falsifier":"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.","tokens_in":4651,"feed_emoji":"🔢","tokens_out":15474,"duration_ms":120612,"temperature":0.7,"pith_summary":"The paper sets out to prove that no assignment of two forbidden residue classes modulo each prime up to p_k can cover every integer in the interval [$p_k^{2}$, p_{k+1}^2]. The author cares about this because it would show that such modular coverings are always incomplete, leaving at least one integer untouched in every interval between consecutive prime squares. To get there, the paper derives an asymptotic for the product P(x) = ∏_{2<p<x}(1 − 2/p), which acts as the survival density for a number to miss two residues per odd prime. It then forms the telescoping difference Δ_k = p_{k+1}^2 P(p_{k+1}^2) − $p_k^{2}$ P($p_k^{2}$) and, assuming a fixed gap p_{k+1} − p_k = h, shows Δ_k is positive and of order p_k/(ln p_k)^2, from which it concludes that a survivor exists.","feed_headline":"Two residues per prime can't cover every integer between prime squares","feed_subtitle":"The paper uses a Mertens-type product bound to claim an uncovered integer always exists in each interval.","key_machinery":"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.","core_discovery":"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).","pith_inferences":["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."],"forward_implications":["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."],"supporting_citations":[{"why":"Supplies the standard Mertens expansions for ∑_{p≤x} 1/p and ∑_{p≤x} 1/p^2 used to derive the asymptotic for P(x) in Theorem 1.","marker":"[1]"}],"fun_headline_variants":["Two residues per prime can't cover all integers between prime squares","Always a residue-free integer exists in each prime-square interval","Prime-square gaps always hide a number outside chosen residue classes","Mertens bound proves an uncovered integer in every [p_k^2, p_{k+1}^2]","No residue choice can fill the gap between consecutive prime squares"],"cache_read_input_tokens":3200,"weakest_assumption_plain":"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.","fun_headline_variants_meta":{"raw":{"variants":["Two residues per prime can't cover all integers between prime squares","Always a residue-free integer exists in each prime-square interval","Prime-square gaps always hide a number outside chosen residue classes","Mertens bound proves an uncovered integer in every [p_k^2, p_{k+1}^2]","No residue choice can fill the gap between consecutive prime squares"]},"model":"deepseek-v4-flash","effort":"low","cost_usd":0.000622,"raw_usage":{"total_tokens":2818,"prompt_tokens":820,"completion_tokens":1998,"prompt_tokens_details":{"cached_tokens":384},"prompt_cache_hit_tokens":384,"prompt_cache_miss_tokens":436,"completion_tokens_details":{"reasoning_tokens":1903}},"tokens_in":436,"tokens_out":1998,"duration_ms":14512,"temperature":1.0,"reasoning_tokens":1903,"cache_read_input_tokens":384,"cache_creation_input_tokens":0},"cache_creation_input_tokens":0},"created_at":"2026-08-10T14:02:40.111880+00:00","model_set":{"reader":"deepseek-v4-flash"},"falsifier":"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.","supporting_citations":[{"cited_title":"Mertens' Proof of Mertens' Theorem","cited_arxiv_id":"math/0504289","evidence_quote":"Supplies the standard Mertens expansions for ∑_{p≤x} 1/p and ∑_{p≤x} 1/p^2 used to derive the asymptotic for P(x) in Theorem 1."}],"review_version":1}