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REVIEW 3 major objections 4 minor 16 references

On pairs of consecutive sequences with the same radicals

T0 review · 3 major / 4 minor · reviewed 2026-08-06 · deepseek-v4-flash

Pith's one-line read 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.

desk verdict Real improvement to a 1989 bound, but the proof of Theorem 4 as written does not produce the claimed contradiction; a repairable exponent error. read the letter →

arxiv 2507.09899 v1 pith:IVZC374P submitted 2025-07-14 math.NT

classification math.NT MSC 11A0511N2511J86
keywords radicalsquarefreekernelconsecutiveintegersabcconjectureeffectiveStørmer'stheoremErdősproblemcountingpairs
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

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.

What carries the argument

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 $m \equiv n \pmod{\operatorname{rad}(n(n+1)\cdots(n+k))}$.

What would settle it

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.

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Extended reading notes

Core claim

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.

Load-bearing premise

The whole bound rests on the effective abc-type inequality of Stewart and Yu: for coprime a+b=c, c < exp(C rad(abc)^{1/3}(log rad(abc))^3); if this inequality is false or has a larger exponent on rad(abc), the proof of Theorem 4 collapses.

Editorial extensions

If this is right

  • For any two distinct starting points $m<n$, the run length $k$ with equal radicals satisfies $k < C(\log n)^{3/2}/(\log\log n)^{9/2}$ for an absolute constant $C$.
  • This replaces the previous bound $k=\exp(O(\sqrt{\log m\,\log\log m}))$ with a bound that is polylogarithmic in $n$, so matching-radical runs become provably very short.
  • The counting result $F_{k,\ell}(x) \le x\exp((\ell\log 2+o(1))\log x/\log\log x)$ is the first record for Erdős's question on pairs with equal product radicals.
  • The number of pairs with $k$ consecutive equal radicals is at most $x^{1/k}\exp((C_k+o(1))\log x/\log\log x)$, so the exponent decreases as the block length grows.
  • A strengthening of the Stewart–Yu bound to an exponent below $1/3$ on $\operatorname{rad}(abc)$ would, by the same proof, immediately produce a stronger upper bound on $k$.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • 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.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

3 major / 4 minor

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.

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 (3)
  1. [Section 2, proof of Theorem 4] 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.
  2. [Section 4, proof of Theorem 10] 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.
  3. [Section 3, Theorem 9 and its proof] 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.
minor comments (4)
  1. [Abstract] There is a missing space in 'ifi ≤ k' and an extra closing parenthesis in 'm(m + 1) · · ·(m + k − 1))'.
  2. [Section 2, final paragraph] 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)})'.
  3. [Section 3, Theorem 9] The display for C_k ends with a double period '2/(k − 1), if k is odd..'.
  4. [Throughout] 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.

Circularity Check

0 steps flagged · score 0.0 of 10

No circularity: all load-bearing inputs are external published theorems; no fitted parameter is renamed as a prediction and no self-citation carries the argument.

full rationale

The paper's results are reductions to external, independently established results. Theorem 4 reduces the matching-radical condition to a congruence modulo rad(n(n+1)...(n+k)) and then invokes Stewart and Yu's effective abc-type inequality (Theorem 7, [SY01]); no constant in that theorem is chosen using the paper's own conclusions, and the theorem's assumptions do not include the target result. Theorems 5 and 9 depend on Lehmer's theorem on Pell equations ([Leh64]) and the Robert-Tenenbaum estimate [Ten15], both external. There are no fitted parameters renamed as predictions and no author self-citations carrying the argument. The note at the end of Section 2 explicitly states that the method is sensitive to the 1/3 exponent in the external bound, which makes the dependency transparent rather than circular. I do not treat the apparent exponent mismatch (the Theorem 4 statement says (log n)^{3/2} while the proof's contradiction line says (log n)^3) as circularity; it is a correctness or typographical issue that is out of scope for this pass.

Assumptions & free parameters 0 free parameters · 4 assumptions · 0 invented entities

The central claims rest entirely on established theorems and standard analytic number theory estimates; no free parameters are fitted to data, and no new entities are postulated.

assumptions (4)
  • standard math Stewart-Yu abc-type inequality: for pairwise coprime a+b=c, c < exp(C rad(abc)^{1/3} (log rad(abc))^3)
    Invoked in Section 2 (Theorem 7) and used to bound n ≤ exp(C2 k^{2/3} n^{2/(3k)} (...)). It is a published theorem [SY01], not derived in this paper.
  • standard math Robert-Tenenbaum estimate for N(x,y), the count of n ≤ x with radical ≤ y, in the range log y ≥ (1+o(1)) 2^{-3/2} (log x)^{1/2} (log log x)^{3/2}
    Used in Section 3 (proof of Theorem 9) to estimate N(x, C_k x^{1/k}). Cited to Tenenbaum [Ten15, p.208].
  • standard math Lehmer's theorem on Størmer's problem: for fixed Q, numbers n with rad(n(n+1)) | Q arise from solutions to at most 2^{ω(Q)} Pell equations, each with O(log x) solutions up to x
    Used in Lemma 8 and in the proof of Theorem 10 to bound the number of m for a fixed n.
  • standard math Standard Chebyshev-type estimate: ∏_{p≤k} p^{k/p+1} ≤ e^{k log k + O(k)}
    Used in Section 2 to bound the product of radicals rad(n)...rad(n+k) relative to rad(n...(n+k)).

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Cite this review

Pith. "Pith review of On pairs of consecutive sequences with the same radicals." pith.science (2026). https://pith.science/paper/IVZC374P

@misc{pith2026250709899,
  author       = {Pith},
  title        = {Pith review of: On pairs of consecutive sequences with the same radicals},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/IVZC374P}},
  note         = {Machine review of arXiv:2507.09899}
}
abstract

Let $(m, n, k)$ be a tuple of integers with the property that if $i \leq k$, then $m + i$ and $n + i$ have the same radical. Using a result on the abc Conjecture, we bound $k$ from above, improving a result of Balasubramanian, Shorey, and Waldschmidt. We also bound the number of pairs $(m, n)$ for which $m < n \leq x$ and $m(m + 1) \cdots (m + k - 1))$ and $n(n + 1) \cdots (n + \ell - 1)$ have the same radical and the number of pairs for which $m + i$ and $n + i$ have the same radical for all $i < k$.

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Reference graph

Works this paper leans on

16 extracted references · 16 canonical work pages

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