一个素变量丢番图问题
A Diophantine Problem with Prime Variables

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.
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.