Almost no finite subset of integers containsa $q^{\text{th}}$ power modulo almost every prime

Bhawesh Mishra
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Abstract

Let $q$ be a prime. We give an elementary proof of the fact that for any $k\in\mathbb{N}$, the proportion of $k$-element subsets of $\mathbb{Z}$ that contain a $q^{\text{th}}$ power modulo almost every prime, is zero. This result holds regardless of whether the proportion is measured additively or multiplicatively. More specifically, the number of $k$-element subsets of $[-N, N]\cap\mathbb{Z}$ that contain a $q^{\text{th}}$ power modulo almost every prime is no larger than $a_{q,k} N^{k-(1-\frac{1}{q})}$, for some positive constant $a_{q,k}$. Furthermore, the number of $k$-element subsets of $\{\pm p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_N^{e_N} : 0 \leq e_{1}, e_{2}, \ldots, e_N\leq N\}$ that contain a $q^{\text{th}}$ power modulo almost every prime is no larger than $m_{q,k} \frac{N^{Nk}}{q^N}$ for some positive constant $m_{q,k}$.
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几乎没有有限的整数子集包含$q^{\text{th}}$幂模几乎所有素数
让 $q$ 做一个素数。我们给出了这个事实的初等证明 $k\in\mathbb{N}$的比例 $k$的元素子集 $\mathbb{Z}$ 它包含了 $q^{\text{th}}$ 几乎所有质数的幂模都是零。无论比例是用加法还是乘法来测量,这个结果都成立。更具体地说,是 $k$的元素子集 $[-N, N]\cap\mathbb{Z}$ 它包含了 $q^{\text{th}}$ 几乎所有素数的幂模都不大于 $a_{q,k} N^{k-(1-\frac{1}{q})}$对于某个正常数 $a_{q,k}$. 此外,的数量 $k$的元素子集 $\{\pm p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_N^{e_N} : 0 \leq e_{1}, e_{2}, \ldots, e_N\leq N\}$ 它包含了 $q^{\text{th}}$ 几乎所有素数的幂模都不大于 $m_{q,k} \frac{N^{Nk}}{q^N}$ 对于某个正常数 $m_{q,k}$.
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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来源期刊
CiteScore
0.80
自引率
20.00%
发文量
14
期刊最新文献
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