Average Results on the Order of aa modulo pp

preprint2015arXivOpen access

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 xlog⁡xlog⁡log⁡x=o(y)\frac{x}{\log x\log\log x}=o(y), then 1y∑a≤y∑p≤x1la(p)=log⁡x+Clog⁡log⁡x+O(xylog⁡log⁡x)\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 log⁡2x=o(ψ(x))\log^2 x = o(ψ(x)) and x1−δlog⁡3x=o(y)x^{1-δ}\log^3 x = o(y), then 1y∑a<y∑p<xla(p)>xψ(x)1=π(x)+O(xlog⁡xψ(x))+O(x2−δlog⁡2xy).\frac1y \sum_{a<y} \sum_{\substack{{p<x} \\ {l_a(p)>\frac{x}{ψ(x)}}}} 1 = π(x) + O\left(\frac{x\log x}{ψ(x)}\right) + O\left(\frac{x^{2 - δ}\log^2 x}y\right). Furthermore, if x1−δlog⁡3x=o(y)x^{1-δ}\log^3 x = o(y), then 1y∑a<y∑p<xp∤ala(p)=cLi(x2)+O(x2log⁡Ax)+O(x3−δlog⁡2xy)\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 -δ}\log^2 x}y\right) where c=∏p(1−pp3−1).c = \prod_p \left(1-\frac p{p^3-1}\right).

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