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Math-5619-A2 · revision r1 · on Math-5619

gpt pro 5.2 attempt on Math-5619

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem 388. Equal products of two disjoint blocks of consecutive integers
  2. Formal restatement
  3. Quick literature/context check (browsing available)
  4. Attack plan
  5. Work
  6. Lemma 1 (the longer block must be the earlier one).
  7. Lemma 2 ("smoothness" obstruction).
  8. A computational check (small search).
  9. Where a full solution would likely come from.
  10. Verification
  11. Final
  12. Completion estimate

Problem 388. Equal products of two disjoint blocks of consecutive integers

Formal restatement

For integers $m\ge 0$ and $k\ge 1$ define the rising factorial block \[ P_k(m) \;:=\; \prod_{i=1}^{k}(m+i) \;=\; \frac{(m+k)!}{m!}. \] The problem asks about integer solutions to \begin{equation} \label{wsref-0-65-71-3a-50-33-38-38-2d-6d-61-69-6e} P_{k_1}(m_1)=P_{k_2}(m_2) \end{equation} under the constraints \[ k_1,k_2>3,\qquad m_1,m_2\in\mathbb Z_{\ge 0},\qquad m_1+k_1\le m_2. \] The wording "Are there only finitely many solutions?" is ambiguous:
  • (A) Fixed lengths: for each fixed pair $(k_1,k_2)$ with $k_1,k_2>3$, are there only finitely many pairs $(m_1,m_2)$ satisfying (eq:P388-main) and $m_1+k_1\le m_2$?
  • (B) All parameters: are there only finitely many quadruples $(m_1,k_1,m_2,k_2)$ with $k_1,k_2>3$ and $m_1+k_1\le m_2$ satisfying (eq:P388-main)?
The statement "classify all solutions" suggests (B), while the appearance of fixed $a,b$ in the generalization suggests (A) as a natural first step.

A more general variant mentioned is \begin{equation} \label{wsref-0-65-71-3a-50-33-38-38-2d-61-62} a\,P_{k_1}(m_1)=b\,P_{k_2}(m_2), \end{equation} with fixed nonzero integers $a,b$ and $k_1>2$.

Quick literature/context check (browsing available)

A classical source of examples and discussion is the MathOverflow question "Equal products of consecutive integers"; several small identities are known, e.g. $2\cdot 3\cdot 4\cdot 5\cdot 6 = 8\cdot 9\cdot 10$ and $19\cdot 20\cdot 21\cdot 22 = 55\cdot 56\cdot 57$. The smallest known example with both block lengths $>3$ and with disjointness $m_1+k_1\le m_2$ is \begin{equation} \label{wsref-0-65-71-3a-50-33-38-38-2d-6b-6e-6f-77-6e} 8\cdot 9\cdot 10\cdot 11\cdot 12\cdot 13\cdot 14 \;=\; 63\cdot 64\cdot 65\cdot 66 \;=\;17297280. \end{equation} This example also appears in OEIS A163263 (numbers with multiple non-overlapping consecutive-product representations). The Erdős Problems site (Problem #388) records these examples and highlights the "smoothness" obstruction below.

Attack plan

  1. Prove structural necessary conditions for any solution (e.g. compare lengths; prime-factor "smoothness" constraints).
  2. Do a bounded brute-force search to verify the smallest solutions and gain confidence about the landscape.
  3. Attempt to leverage general finiteness theorems for polynomial Diophantine equations of the form $f(x)=g(y)$ (e.g. Siegel/Bilu--Tichy) to prove finiteness for fixed $(k_1,k_2)$; identify the precise obstruction to converting this into a full classification across all lengths.

Work

Lemma 1 (the longer block must be the earlier one).
Assume $m_1+k_1\le m_2$ and (eq:P388-main) holds with $k_1,k_2\ge 1$. Then $k_1>k_2$.
Proof: Because $m_1+k_1\le m_2$, every factor on the right is at least $m_2+1\ge m_1+k_1+1$, while every factor on the left is at most $m_1+k_1$. Hence \[ P_{k_2}(m_2)=\prod_{j=1}^{k_2}(m_2+j) \;\ge\; (m_1+k_1+1)^{k_2}. \] Also \[ P_{k_1}(m_1)=\prod_{i=1}^{k_1}(m_1+i) \;\le\; (m_1+k_1)^{k_1}. \] If $k_1\le k_2$ then \[(m_1+k_1+1)^{k_2} \;\ge\; (m_1+k_1+1)^{k_1} \;>\; (m_1+k_1)^{k_1} \;\ge\; P_{k_1}(m_1),\] contradicting $P_{k_1}(m_1)=P_{k_2}(m_2)$. Therefore $k_1>k_2$. ∎
Lemma 2 ("smoothness" obstruction).
Let $A:=m_1+k_1$. If (eq:P388-main) holds, then every prime factor of each integer in the right block $\{m_2+1,\dots,m_2+k_2\}$ is $\le A$. Equivalently, the entire right block consists of $A$-smooth integers.
Proof: All prime factors of $P_{k_1}(m_1)=\prod_{i=1}^{k_1}(m_1+i)$ are $\le m_1+k_1=A$, since each factor $m_1+i\le A$. Equality (eq:P388-main) forces the same set of primes (with the same multiplicities) to divide $P_{k_2}(m_2)=\prod_{j=1}^{k_2}(m_2+j)$, hence every prime dividing any $m_2+j$ must be $\le A$. ∎
A computational check (small search).
A brute-force search over \[ 4\le k_1,k_2\le 10,\quad 0\le m_1\le 500,\quad m_1+k_1\le m_2\le 500 \] finds exactly one solution with both lengths $>3$, namely (eq:P388-known) (i.e. $m_1=7,k_1=7,m_2=62,k_2=4$). This does not prove uniqueness, but it supports the conjecture recorded on the Erdős Problems site that (eq:P388-known) might be the only solution under the stated constraints.
Where a full solution would likely come from.
Lemmas 1--2 reduce any putative classification to understanding long runs of consecutive $A$-smooth integers above $A$ (since $m_2+1>A$). One plausible route is to combine:
  • quantitative bounds on the maximal length of a run of $A$-smooth integers in $[x,x+L]$ with $x$ much larger than $A$,
  • together with the size constraint imposed by (eq:P388-main), which forces the growth rates of $P_{k_1}(m_1)$ and $P_{k_2}(m_2)$ to match.
I did not find a way to complete this program in a gap-free way.

Verification

  • Lemma 1 uses only monotone inequalities and the disjointness condition $m_1+k_1\le m_2$; the strict inequality step $(A+1)^{k_1}>(A)^{k_1}$ is immediate.
  • Lemma 2 is immediate from prime-factor containment.
  • The identity (eq:P388-known) was verified by direct multiplication (in a separate computation).

Final

UNRESOLVED.
  1. (i) Strongest proved partial result: Any solution of (eq:P388-main) with $m_1+k_1\le m_2$ must satisfy $k_1>k_2$ and the entire right block $\{m_2+1,\dots,m_2+k_2\}$ is $(m_1+k_1)$-smooth (Lemmas 1--2). In addition, a bounded brute-force search (specified above) finds only the solution (eq:P388-known) with $k_1,k_2>3$ in that range.
  2. (ii) First gap / obstruction: Turning the smoothness obstruction into a global finiteness or uniqueness theorem requires nontrivial control of long runs of $A$-smooth numbers above $A$ that is uniform in the parameters and compatible with the size constraint from (eq:P388-main).
  3. (iii) Top 3 next moves: $1$ Translate (eq:P388-main) into a statement about smoothness of a short interval and try to apply strongest available bounds on smooth numbers in short intervals. $2$ For fixed $(k_1,k_2)$, treat $P_{k_1}(x)=P_{k_2}(y)$ as an integral-point problem on an algebraic curve and attempt an effective determination of all integral points (in small degrees this reduces to elliptic/hyperelliptic methods). $3$ Expand the computational search using modular sieving informed by Lemma 2 (restricting to candidate intervals whose elements are all $A$-smooth).
  4. (iv) What a minimal counterexample (to "only finitely many solutions") would look like: It would require an infinite family of pairs of disjoint blocks with $k_1>k_2\ge 4$ in which the later block forms unusually long runs of $A$-smooth numbers relative to $A=m_1+k_1$, while simultaneously matching the magnitude constraint $\log P_{k_1}(m_1)\approx \log P_{k_2}(m_2)$.

Completion estimate

What remains is either (a) an unconditional finiteness/uniqueness theorem for (eq:P388-main) under $k_1,k_2>3,\ m_1+k_1\le m_2$, or (b) an explicit infinite family (or even a second distinct example). Any such proof likely needs deep input on smooth numbers or an effective resolution of integral points on families of curves.