Powers in $\prod\limits_{k=1}^n (ak^{2^l\cdot3^m}+b)$

Type: Article

Publication Date: 2012-03-01

Citations: 2

DOI: https://doi.org/10.7169/facm/2012.46.1.1

Abstract

Let $f(x)=ax^{2^l\cdot3^m}+b\in \mathbb{Z}[x]$ be a polynomial with $l\geq 1, l+m\geq 2, ab\neq 0$ and such that $f(k)\neq 0$ for any $k\geq 1$. We prove, under $ABC$ conjecture, that the product $\prod_{k=1}^n f(k)$ is not a $2^l\cdot3^m$-th power for $n$ large enough.

Locations

  • Functiones et Approximatio Commentarii Mathematici - View - PDF

Similar Works

Action Title Year Authors
+ PDF Chat On the greatest prime factor of $prod^{x}_{k=1}f(k)$ 1990 Péter L. Erdős
Andrzej Schinzel
+ PDF Chat On the greatest prime factor of $(ax^m + by^n)$ 1980 T. N. Shorey
+ The sum $\sum_{k=0}^{q-1}\binom{2k}{k}$ for q a power of 3 2010 Sandro Mattarei
+ Primes of the form $${kM^n+n}$$ k M n + n 2019 Xiaotao Sun
+ PDF Chat Powers and Polynomials in ${\Bbb Z}_m$ 1999 Lorenz Halbeısen
Norbert Hungerbühler
+ On the power of standard polynomial to M_{a,b}(E) 2016 Fernanda G. de Paula
Sérgio Mota Alves
+ PDF Chat The greatest prime factor of $a^n - b^n$ 1975 C. L. Stewart
+ Largest prime factor of $n^2+1$ 2019 Jori Merikoski
+ On the divisibility of $a^n \pm b^n$ by powers of $n$ 2013 Salvatore Tringali
+ The $p$-exponent of the $K(1) 2003 Michael J. Fisher
+ PDF Chat On the greatest prime factor of ab + 1 2009 Kaisa Matomäki
+ On the greatest prime factor of ab + 1 2014 Étienne Fouvry
+ On the powerful numbers in $\prod_{x=g}^{f}(x^k\pm h^k)$ 2023 Qingjie Zhang
Chuanze Niu
+ $$\mathfrak {cP}$$-Baer Polynomial Extensions 2024 Nasibeh Aramideh
A. Moussavi
+ On the Prime Factors of \binom2nn 1975 P. Erdős
R. L. Graham
Imre Z. Ruzsa
E. G. Straus
+ On a problem of Chen and Liu concerning the prime power factorization of $n!$ 2011 Johannes F. Morgenbesser
Thomas Stoll
+ The Primes in k(ϱ) 1951 Balth. van der Pol
Pierre Speziali
+ The Bateman-Horn constant for $x^3+x+1$ 2011 Timothy Foo
+ Primes in Beatty sequence 2019 C. G. Karthick Babu
+ PDF Chat On the congruence $$1^m + 2^m + \cdots + m^m\equiv n \pmod {m}$$ 1 m + 2 m + ⋯ + m m ≡ n ( mod m ) with $$n\mid m$$ n ∣ m 2014 J. M. Grau
Antonio M. Oller‐Marcén
Jonathan Sondow