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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. Theorem 1.1 of Hung M. Bui, Kyle Pratt and Alexandru Zaharescu, Power savings for counting solutions to polynomial-factorial equations, Adv. Math. 422 (2023), Paper No. 109021, 32 pp., states that for a fixed polynomial P∈Z[X]P\in\mathbb Z[X] of degree r≥2r\ge2 and a fixed nonzero integer ss there is a constant CC with

#{N≤n<2N: s⋅n!=P(x) for some x∈Z}≤C N33/34.\#\{N\le n<2N:\ s\cdot n!=P(x)\text{ for some }x\in\mathbb Z\}\le C\,N^{33/34}.

Remark 1.4 and Proposition 3.2 say that the method gives the exponent 122−16+ε=0.97056…12\sqrt2-16+\varepsilon=0.97056\ldots (Proposition 3.2 states it as the hypothesis θ≤17−122−ε\theta\le17-12\sqrt2-\varepsilon), which 33/34=0.97058…33/34=0.97058\ldots approximates. The proof splits the solutions into (2r+1)(2r+1)-tuples, finds by pigeonhole three solutions with small gaps in one residue class modulo rr, turns each such triple into a simultaneous rational approximation to values of algebraic functions, in the manner of Berend and Osgood, and bounds the number of such approximations by Diophantine and Padé approximation. The source card is Bui, Pratt and Zaharescu 2023; the arXiv posting is the preprint link.

Consequence for the problem. If f(n)=mf(n)=m in Problem 393, then n!=∏s∈S(a+s)n!=\prod_{s\in S}(a+s) for some a≥1a\ge1 and some S⊆{0,…,m}S\subseteq\{0,\ldots,m\} containing 00 and mm, so n!=PS(a)n!=P_S(a) for one of the finitely many polynomials PS(X)=∏s∈S(X+s)P_S(X)=\prod_{s\in S}(X+s), each of degree ∣S∣≥2|S|\ge2. Summing the theorem over these polynomials and over dyadic ranges gives, with Fm(N)F_m(N) the number of n≤Nn\le N with f(n)=mf(n)=m, Fm(N)≪mN33/34F_m(N)\ll_m N^{33/34}, as the site's remarks state; this sharpens the o(N)o(N) of Berend and Osgood 1992.

Covers. The counting bound only: Fm(N)≪mN33/34F_m(N)\ll_m N^{33/34} for each fixed mm. It does not settle whether f(n)=1f(n)=1 infinitely often, which is open unconditionally, nor the growth of f(n)f(n) along every nn.

Acceptance. Refereed: Advances in Mathematics, volume 422 (June 2023), article 109021; the Crossref record of the DOI gives these data. The site labels the problem OPEN, so its remark crediting the result is commentary on an open problem and not acceptance, and no reviewed evidence is listed. The proof is not checked here.

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