On Sums of Two Coprime k-th Powers

被引:0
作者
Wenguang Zhai
机构
[1] Shandong Normal University,
来源
Monatshefte für Mathematik | 2005年 / 144卷
关键词
2000 Mathematics Subject Classifications: 11P21; Key words: Riemann Hypothesis, exponential sum;
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摘要
For fixed k≥3, let \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\rho_k(n)=\sum_{n=\vert m\vert^k+\vert l\vert^k, {\rm g.c.d}(m,l)=1}1.$$\end{document} It is known that the asymptotic formula \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$R_k(x)=\sum_{n\le x}\rho_k(n)=c_k x^{2/k}+O(x^{1/k})$$\end{document} holds for some constant ck. Let Ek(x)=Rk(x)−ckx2/k. We cannot improve the exponent 1/k at present if we do not have further knowledge about the distribution of the zeros of the Riemann Zeta function ζ(s). In this paper, we shall prove that if the Riemann Hypothesis (RH) is true, then Ek(x)=O(x4/15+ɛ), which improves the earlier exponent 5/18 due to Nowak. A mean square estimate of Ek(x) for k≥6 is also obtained, which implies that Ek(x)=Ω(x1/k−1/k2) for k≥6 under RH.
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页码:233 / 250
页数:17
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