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Prime powers dividing products of consecutive integer values of $x^{2^n}+1$

Number Theory 2019-12-10 v3 Combinatorics

Abstract

Let nn be a positive integer and f(x):=x2n+1f(x) := x^{2^n}+1. In this paper, we study orders of primes dividing products of the form Pm,n:=f(1)f(2)f(m)P_{m,n}:=f(1)f(2)\cdots f(m). We prove that if m>max{1012,4n+1}m > \max\{10^{12},4^{n+1}\}, then there exists a prime divisor pp of Pm,nP_{m,n} such that ordp(Pm,n)n2n1_{p}(P_{m,n} )\leq n\cdot 2^{n-1}. For n=2n=2, we establish that for every positive integer mm, there exists a prime divisor pp of Pm,2P_{m,2} such that ordp(Pm,2)4_{p} (P_{m,2}) \leq 4. Consequently, Pm,2P_{m,2} is never a fifth or higher power. This extends work of Cilleruelo who studied the case n=1n=1.

Keywords

Cite

@article{arxiv.1905.13003,
  title  = {Prime powers dividing products of consecutive integer values of $x^{2^n}+1$},
  author = {Stephan Baier and Pallab Kanti Dey},
  journal= {arXiv preprint arXiv:1905.13003},
  year   = {2019}
}

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12 pages