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On Sums of Two Coprime k-th Powers

机译:关于两个素数第k个幂的和

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For fixed k≥3, let $rho_k(n)=sum_{n=vert mvert^k+vert lvert^k, {rm g.c.d}(m,l)=1}1.$ It is known that the asymptotic formula $R_k(x)=sum_{nle x}rho_k(n)=c_k x^{2/k}+O(x^{1/k})$ holds for some constant c k . Let E k (x)=R k (x)−c k x 2/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 E k (x)=O(x 4/15+ɛ), which improves the earlier exponent 5/18 due to Nowak. A mean square estimate of E k (x) for k≥6 is also obtained, which implies that E k (x)=Ω(x 1/k−1/k2) for k≥6 under RH.
机译:对于固定k≥3,令$ rho_k(n)= sum_ {n = vert mvert ^ k + vert lvert ^ k,{rm gcd}(m,l)= 1} 1. $已知渐近公式$ R_k(x)= sum_ {nle x} rho_k(n)= c_k x ^ {2 / k} + O(x ^ {1 / k})$保持某个常数ck 。令E k (x)= R k (x)-c k x 2 / k 。如果我们对黎曼Zeta函数ζ(s)的零点分布没有进一步的了解,则目前无法提高指数1 / k。在本文中,我们将证明,如果黎曼假说(RH)是正确的,则E k (x)= O(x 4/15 +ɛ)可以提高早期指数5 / 18由于诺瓦克。还获得了k≥6的E k (x)的均方估计,这意味着E k (x)=Ω(x 1 / k−1 / k2 )在RH下k≥6时。

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