牟全武,吕晓东.一个素变量丢番图问题[J].数学年刊A辑,2017,38(1):031~42 |
一个素变量丢番图问题 |
A Diophantine Problem with Prime Variables |
Received:July 02, 2015 Revised:August 13, 2016 |
DOI:10.16205/j.cnki.cama.2017.0004 |
中文关键词: Diophantine inequalities, Davenport-Heilbronn method, Prime |
英文关键词:Diophantine inequalities, Davenport-Heilbronn method, Prime |
基金项目:本文受到陕西省自然科学基础研究计划(No.2016JM1013),西安工程大学博士科研启动基金(No.BS1508)和西安工程大学基础课程质量提升项目(高等数学)的资助. |
|
Hits: 831 |
Download times: 638 |
中文摘要: |
设$k$和$r$是满足$k\geq 3$及$r\geq \psi(k)+1$的正整数,
这里当$3\leq k\leq 4$时, $\psi(k)=2^{k-1}$; 而当$k\geq 5$时, $\psi(k)=\frac{1}{2}k(k+1)$.
假定$\delta$和$\varepsilon$是给定的足够小的正数,
$\lambda_1, \lambda_2, \cdots, \lambda_{r+1}$
是不全同号且两两之比不全为有理数的非零实数.
对于任意实数$\eta$与$0<\sigma<\frac{2^{1-2k}}{r-1}$,
证明了: 存在一个正数序列$X\rightarrow +\infty$, 使得不等式
$$
|\lambda_1p_1^k+\lambda_2p_2^k+\cdots+\lambda_rp_r^k+\lambda_{r+1}p_{r+1}+\eta|<\big(\max_{1\leq j\leq r+1}p_j\big)^{-\sigma}
$$
有$\gg X^{\frac{r}{k}-\frac{2^{1-2k}}{r-1}+\varepsilon}$组素数解
$(p_1, p_2, \cdots, p_{r+1})$,
这里$(\delta X)^\frac{1}{k}\leq p_j\leq X^\frac{1}{k}\,(1\leq j\leq r)$及$\delta X\leq p_{r+1}\leq X$.
这改进了之前的结果. |
英文摘要: |
Let $k$ and $r$ be positive integers with $k\geq 3$ and $r\geq
\psi(k)+1$, where $\psi(k)=2^{k-1}$ for $3\leq k\leq 4$, and
$\psi(k)=\frac{1}{2}k(k+1)$ for $k\geq 5$. Suppose that $\delta$ and
$\varepsilon$ are fixed and sufficiently small positive numbers,
$\lambda_1, \lambda_2, \cdots, \lambda_{r+1}$ are nonzero real
numbers, not all of the same sign and not all in rational ratios.
Then for any real $\eta$ and $0<\sigma<\frac{2^{1-2k}}{r-1}$, it is proved that there exists a positive sequence $X\rightarrow +\infty$, such that
the inequality
$$|\lambda_1p_1^k+\lambda_2p_2^k+\cdots+\lambda_rp_r^k+\lambda_{r+1}p_{r+1}+\eta|<\big(\max_{1\leq j\leq r+1}p_j\big)^{-\sigma}$$
has $\gg X^{\frac{r}{k}-\frac{2^{1-2k}}{r-1}+\varepsilon}$ prime solutions $(p_1, p_2, \cdots, p_{r+1})$ with $(\delta X)^\frac{1}{k}\leq p_j\leq X^\frac{1}{k}\,(1\leq j\leq r)$ and $\delta X\leq p_{r+1}\leq X$.
This gives an improvement of an earlier result. |
View Full Text View/Add Comment Download reader |
Close |
|
|
|
|
|