English

Average Results on the Order of $a$ modulo $p$

Number Theory 2021-02-10 v3

Abstract

Let a>1a>1 be an integer. Denote by la(p)l_a(p) the multiplicative order of aa modulo primes pp. We prove that if xlogxloglogx=o(y)\frac{x}{\log x\log\log x}=o(y), then 1yaypx1la(p)=logx+Cloglogx+O(xyloglogx)\frac 1 y \sum_{a\leq y}\sum_{p\leq x}\frac{1}{l_a(p)}=\log x + C\log\log x+O\left(\frac x {y \log\log x}\right) which is an improvement over a theorem by Felix ~\cite{Fe}. Additionally, we also prove two other average results If log2x=o(ψ(x))\log^2 x = o(\psi(x)) and x1δlog3x=o(y)x^{1-\delta}\log^3 x = o(y), then 1ya<yp<xla(p)>xψ(x)1=π(x)+O(xlogxψ(x))+O(x2δlog2xy).\frac1y \sum_{a<y} \sum_{\substack{{p<x} \\ {l_a(p)>\frac{x}{\psi(x)}}}} 1 = \pi(x) + O\left(\frac{x\log x}{\psi(x)}\right) + O\left(\frac{x^{2 - \delta}\log^2 x}y\right). Furthermore, if x1δlog3x=o(y)x^{1-\delta}\log^3 x = o(y), then 1ya<yp<xpala(p)=cLi(x2)+O(x2logAx)+O(x3δlog2xy)\frac1y\sum_{a<y} \sum_{\substack{{p<x} \\ {p\nmid a}}}l_a(p) = c\textrm{Li}(x^2) + O\left( \frac{x^2}{\log^A x} \right) + O\left(\frac{x^{3 -\delta}\log^2 x}y\right) where c=p(1pp31).c = \prod_p \left(1-\frac p{p^3-1}\right).

Keywords

Cite

@article{arxiv.1509.01752,
  title  = {Average Results on the Order of $a$ modulo $p$},
  author = {Sungjin Kim},
  journal= {arXiv preprint arXiv:1509.01752},
  year   = {2021}
}
R2 v1 2026-06-22T10:50:01.066Z