{"id":"49484d87-cf6b-49f9-bfcd-4b17be5dff34","arxiv_id":"2507.09899","paper_version":1,"verdict":"CONDITIONAL","confidence":"MODERATE","novelty_score":6.0,"correctness_risk":"medium","formal_verification":"none","parameter_count":0,"one_line_summary":"For distinct integers m < n, if rad(m+i) = rad(n+i) for all i ≤ k, then k ≪ (log n)^{3/2}/(log log n)^{9/2}, improving a 1989 bound.","lead":"Two runs of consecutive integers can have matching products of distinct prime factors, called radicals. This paper proves a much tighter limit on how long such runs can be, and gives the first counts for related matching-radical pairs.","discovery_kind":"extension","skeptic_critique":{"model":"deepseek-v4-flash","headline":"Theorem 4 proof uses the wrong exponent in the contradiction threshold: substituting the printed k=(log n)^3/(log log n)^{9/2} satisfies the derived inequality, so the central bound is unproven as written.","rationale":"The central claim is Theorem 4, and the paper's own proof is the least secure part. The manuscript states that k ∼ C3 (log n)^3/(log log n)^{9/2} leads to a contradiction, but substituting this value into the derived inequality gives an exponent of size (log n)^2, making the right side far larger than n; there is no contradiction. The theorem's stated bound requires the exponent 3/2, which produces an exponent of size C log n and, with a suitably small C3, a real contradiction. This is a one-character typographical error rather than a fundamental mathematical flaw, and the surrounding argument (lcm congruence, adjacent-pair product estimate, application of Stewart-Yu) is otherwise coherent. The reader flagged this typo in their rationale but named the external Stewart-Yu theorem as the weakest assumption; I agree Stewart-Yu is necessary, but because it is a published theorem, the internal exponent error is more immediately load-bearing for the paper as submitted. The same class of typographical issues appears in Theorem 10, where Q is defined with k instead of ℓ, and in Theorem 9's statement, where k√x should be k x^{1/k}. Since the reader already recommended conditional acceptance, this stress-test does not move the verdict: the paper should be accepted only after these corrections are verified.","tokens_in":5076,"tokens_out":23754,"duration_ms":240058,"concrete_test":"Recompute the last display of Section 2 with k = (log n)^3/(log log n)^{9/2}: the right-hand exponent is of order (log n)^2, so the inequality n ≪ exp(...) holds and no contradiction occurs. Then repeat with k = (log n)^{3/2}/(log log n)^{9/2}: the exponent becomes (27/8) C2 C3^{2/3} log n; with C3 chosen so that this coefficient is below 1, the right side is n^{<1}, contradicting n < exp(...) and confirming the required typo correction.","verdict_should_be":"UNCHANGED","load_bearing_attack":"The proof of Theorem 4 derives n ≪ exp(C2 k^{2/3} n^{2/(3k)} ((log k)^3 + ((log n)/k)^3)) and then claims that k ∼ C3 (log n)^3/(log log n)^{9/2} makes this inequality fail. This is arithmetically incorrect: with k = C3 (log n)^3 / (log log n)^{9/2}, the dominant part of the exponent is k^{2/3} (log k)^3 ≈ C3^{2/3} (log n)^2 / (log log n)^3 · 27 (log log n)^3 = 27 C3^{2/3} (log n)^2, so the right side is exp(Θ((log n)^2)) ≫ n and the inequality is satisfied. The contradiction only arises for the theorem-stated exponent 3/2: then the exponent is (27/8) C2 C3^{2/3} log n, and choosing C3 small makes it strictly smaller than log n, flipping the inequality to n < n^{<1}. Thus the submitted proof does not establish Theorem 4 without correcting this threshold. This is an internal gap, not a disagreement with any external result; the Stewart-Yu theorem itself is a valid published input, and the proof's sensitivity to the 1/3 exponent is acknowledged in the paper's own note.","agreement_with_reader":"partial"},"referee_report":{"model":"deepseek-v4-flash","summary":"The paper studies tuples (m, n, k) of positive integers for which rad(m+i) = rad(n+i) for all i ≤ k. Its main result (Theorem 4) claims the bound k ≪ (log n)^{3/2}/(log log n)^{9/2}, improving the earlier exp(O(sqrt(log m log log m))) bound of Balasubramanian, Shorey, and Waldschmidt. The paper also proves an upper bound for F_{k,ℓ}(x), the number of pairs m < n ≤ x with rad(m(m+1)...(m+k-1)) = rad(n(n+1)...(n+ℓ-1)), namely x exp((ℓ log 2 + o(1)) log x / log log x), and a bound for pairs with rad(m+i) = rad(n+i) for all i < k of the shape x^{1/k} exp(O(log x/log log x)). The arguments rely on the effective abc-type theorem of Stewart and Yu, a theorem of Robert and Tenenbaum on integers with prescribed radical size, and Lehmer's theorem on Størmer's problem.","tokens_in":5192,"tokens_out":14432,"duration_ms":135151,"significance":"If Theorem 4 is correct, it is a substantial improvement over the previous subexponential bound and is the first polylogarithmic upper bound for the length of two consecutive sequences with matching radicals; this would be the paper's main contribution. The counting results for F_{k,ℓ}(x) and for equal-radical consecutive strings appear to be new and are obtained by clean reductions to established external theorems. The paper is concise and transparent: no constants are fitted to the conclusions, and the author explicitly notes that the argument for Theorem 4 depends on the exponent 1/3 in the Stewart–Yu inequality. However, as submitted, the proof of Theorem 4 contains a load-bearing arithmetic error in the final threshold, and the proof of Theorem 10 has a variable mix-up that invalidates the stated exponent. These issues are local and repairable, but they must be corrected before the main claims are established.","major_comments":[{"comment":"The proof derives the inequality n ≪ exp(C2 k^{2/3} n^{2/(3k)} ((log k)^3 + ((log n)/k)^3)) and then claims that taking k ∼ C3 (log n)^3/(log log n)^{9/2} makes this inequality fail. This is arithmetically incorrect: for that k, the dominant term k^{2/3}(log k)^3 is approximately 27 C3^{2/3} (log n)^2, so the right-hand side is exp(Θ((log n)^2)) ≫ n and the inequality is satisfied. The claimed contradiction does occur for the theorem's stated exponent, namely k ∼ C3 (log n)^{3/2}/(log log n)^{9/2}, for which the exponent is approximately (27/8) C2 C3^{2/3} log n and can be made smaller than log n by choosing C3 small. The proof must therefore replace the exponent 3 by 3/2 in the displayed k and adjust the constant threshold; as written, the proof of Theorem 4 does not establish the stated bound.","section":"Section 2, proof of Theorem 4"},{"comment":"In the proof of Theorem 10, Q is defined as n(n+1)...(n+k−1), but the assumption is rad(m(m+1)...(m+k−1)) = rad(n(n+1)...(n+ℓ−1)). To apply Lemma 8 to the condition rad(m(m+1)) | Q, the definition must be Q = n(n+1)...(n+ℓ−1). Correspondingly, the subsequent bound ω(Q) ≲ (k+o(1)) log x / log log x should use ℓ, not k; otherwise the argument yields an exponent proportional to k log 2 rather than the stated ℓ log 2. This is a direct typographical error in a load-bearing quantity, but it must be corrected for the proof to be valid.","section":"Section 4, proof of Theorem 10"},{"comment":"The statement of Theorem 9 gives the bound as k√x exp((C_k+o(1)) log x/log log x), while the proof computes with y = C_k x^{1/k} and obtains N(x, C_k x^{1/k}) = x^{1/k} exp(...), matching the introduction's Theorem 6 statement x^{1/k}. In the proof the line 'Plugging in y = k√x' cannot lead to v = (1 − 1/k) log x; it should read y = C_k x^{1/k}. Additionally, the sentence 'Because rad(n(n+1)) ≤ x^{1/k}' is not justified by the earlier display, which only yields the existence of a pair (n+i, n+i+1) with radical at most k^2 x^{2/k} (for even k) or x^{2/(k−1)} (for odd k); the proof needs to explain how the bound on the number of m follows after shifting to that pair. These inconsistencies affect the statement and proof of a theorem and should be reconciled.","section":"Section 3, Theorem 9 and its proof"}],"minor_comments":[{"comment":"There is a missing space in 'ifi ≤ k' and an extra closing parenthesis in 'm(m + 1) · · ·(m + k − 1))'.","section":"Abstract"},{"comment":"The parenthetical remark contains malformed formulas: 'c < exp((rad(abc)^{1/2+o(1)})' appears to be missing a superscript and closing parenthesis, and 'c < exp(rad(abc))(2/3)+o(1))' should presumably read 'c < exp(rad(abc)^{2/3+o(1)})'.","section":"Section 2, final paragraph"},{"comment":"The display for C_k ends with a double period '2/(k − 1), if k is odd..'.","section":"Section 3, Theorem 9"},{"comment":"The rendering of 'Erdős' and accented characters in the bibliography (e.g., 'Probl`eme') appears as corrupted glyphs; the author should ensure the source compiles to clean text.","section":"Throughout"}],"recommendation":"major_revision","confidential_remarks":"The main theorem is not proven as submitted because of the wrong exponent in the final substitution in the proof of Theorem 4, but the error is clearly local: replacing the exponent 3 by 3/2 restores the argument. The same holds for the k/ℓ mix-up in Theorem 10's proof. The paper's reliance on the Stewart–Yu effective abc-type theorem is explicit and acknowledged. I recommend major revision rather than rejection because the central claims are defensible and the technical gaps are repairable within the manuscript's scope."},"author_rebuttal":null,"desk_editor":{"model":"deepseek-v4-flash","letter":"Main take: the Theorem 4 bound is a real improvement and the proof strategy is sound, but the written proof has a load-bearing exponent error. The submitted k threshold doesn't contradict the derived inequality; the theorem's stated exponent does. Fix that line and the paper works.\n\nThe genuinely new content is the polylog bound (from exp(O(sqrt(log m log log m)))) and the two counting estimates, which appear to be the first recorded for those Erdős questions. The reductions to Stewart–Yu, Robert–Tenenbaum, and Lehmer are transparent and there's no circularity or parameter-fitting. The author even flags the sensitivity to the 1/3 exponent in the abc-type input, which is the kind of honesty you want.\n\nThe soft spots: (1) In the proof of Theorem 4, k ~ C3 (log n)^3/(log log n)^{9/2} is used to force a contradiction, but with that k the right side of the inequality is exp(Theta((log n)^2)), which is much bigger than n, so the inequality holds. The contradiction only appears with the (log n)^{3/2} exponent. This is a mistake in the written argument, not just a typo—the proof as printed does not establish the theorem. It is repairable by changing the exponent, and the author's intended argument is clear. (2) In Section 4, Q is defined as n...(n+k-1) when the equality involves l consecutive integers; that should be l. With k in place, the bound with l log 2 doesn't follow. This one looks like an ordinary typo.\n\nOverall, the paper deserves a serious referee. The main result is likely true and the counting results are worth having. I would send it to review but request corrections to both points before acceptance. Number theorists with an interest in abc-method bounds and Erdős problems are the audience.","headline":"Real improvement to a 1989 bound, but the proof of Theorem 4 as written does not produce the claimed contradiction; a repairable exponent error.","tokens_in":5907,"tokens_out":3377,"would_cite":true,"duration_ms":36435,"reading_group":"yes","serious_thinker":"yes","would_accept_peer_review":true},"rs_alignment":null,"lean_confirmation":null,"pith_extraction":{"msc":["11A05","11N25","11J86"],"pacs":[],"model":"deepseek-v4-flash","headline":"Two strings of consecutive integers can agree in radical for at most a polylogarithmic number of positions, under an effective abc-type inequality, improving the previous subexponential bound.","keywords":["radical","squarefree kernel","consecutive integers","abc conjecture","effective abc","Størmer's theorem","Erdős problem","counting pairs"],"falsifier":"Exhibit distinct positive integers m < n and an index k with k exceeding, say, $10^{3}$ (log n)^{3/2}/(log log n)^{9/2} and rad(m+i)=rad(n+i) for all i ≤ k; no such example is known. Computationally, checking all n up to $10^{6}$ by sieving radicals and comparing aligned blocks would certify the bound's practical range.","tokens_in":4703,"feed_emoji":"🔢","tokens_out":6512,"duration_ms":64964,"temperature":0.7,"pith_summary":"The paper studies how long two disjoint strings of consecutive integers can be when each aligned pair has the same radical (the largest squarefree divisor). Its main result is that, under an effective form of the abc bound proved by Stewart and Yu, the number $k$ of aligned terms must satisfy $k \\ll (\\log n)^{3/2}/(\\log\\log n)^{9/2}$, where $n$ is the larger starting point. This is a polylogarithmic bound, improving the earlier $\\exp(O(\\sqrt{\\log m\\,\\log\\log m}))$ bound of Balasubramanian, Shorey, and Waldschmidt. The paper also counts pairs of starting points whose block products have equal radicals, giving the first recorded upper bounds for one of Erdős's questions.","feed_headline":"Matching-radical streaks: length capped by a power of log n","feed_subtitle":"New proof uses an effective abc bound to beat the old subexponential limit and counts Erdős's pairs.","key_machinery":"The load-bearing mechanism is the effective abc-type inequality of Stewart and Yu (Theorem 7): for pairwise coprime $a+b=c$, $c < \\exp(C\\operatorname{rad}(abc)^{1/3}(\\log\\operatorname{rad}(abc))^3)$. It is applied to a pair $(n+m,n+m+1)$ found by a pigeonhole step, turning a congruence-divisibility structure into an exponential upper bound on $n$ in terms of $k$. The argument also uses the elementary fact that $\\operatorname{rad}(n)\\cdots\\operatorname{rad}(n+k) \\le e^{k\\log k+O(k)}\\operatorname{rad}(n(n+1)\\cdots(n+k))$ and the derived congruence $m \\equiv n \\pmod{\\operatorname{rad}(n(n+1)\\cdots(n+k))}$.","core_discovery":"The central claim is Theorem 4: for distinct positive integers $m < n$ with $\\operatorname{rad}(m+i)=\\operatorname{rad}(n+i)$ for every $i \\le k$, one has $k \\ll (\\log n)^{3/2}/(\\log\\log n)^{9/2}$. The proof starts from the observation that $m \\equiv n \\pmod{\\operatorname{rad}(n(n+1)\\cdots(n+k))}$ and that this modulus is less than $n$, so the product of the radicals of the $n$-block is at most $e^{k\\log k+O(k)}n$. Pigeonholing forces two adjacent entries with small radical product, and applying the effective abc inequality $c < \\exp(C\\operatorname{rad}(abc)^{1/3}(\\log\\operatorname{rad}(abc))^3)$ to the triple $(1, n+m, n+m+1)$ yields a contradiction for larger $k$. The paper states explicitly that the argument needs the exponent $1/3$ in the abc bound; with the earlier exponent $2/3$ the method gives no bound at all.","pith_inferences":["The proof suggests a general recipe: any improved effective abc inequality translates directly into a shorter allowed run of equal radicals; testing this would mean re-running the same pigeonhole argument with a smaller exponent.","The author's Conjecture 11 implies the Theorem 5 bound is far from sharp; computing $F_{2,2}(x)$ for larger $x$ would show how far the $x^{1+o(1)}$ upper bound sits above reality.","A similar congruence-elimination argument could apply to other multiplicative kernels, such as the largest prime factor or the squarefree kernel of shifted products, as long as an effective abc-type bound is available.","If one could exhibit infinitely many pairs with $k$ on the order of $(\\log n)^{3/2}/(\\log\\log n)^{9/2}$, then Theorem 4 would be essentially optimal; no such construction is known."],"forward_implications":["For any two distinct starting points $m