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 →
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 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.
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
- 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.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
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)
- [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.
- [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.
- [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)
- [Abstract] There is a missing space in 'ifi ≤ k' and an extra closing parenthesis in 'm(m + 1) · · ·(m + k − 1))'.
- [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)})'.
- [Section 3, Theorem 9] The display for C_k ends with a double period '2/(k − 1), if k is odd..'.
- [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
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
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)
- 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}
- 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
- standard math Standard Chebyshev-type estimate: ∏_{p≤k} p^{k/p+1} ≤ e^{k log k + O(k)}
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 $i < k$.
Reference graph
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