On a binary Diophantine inequality involving prime numbers

Jing Huang; Qian Wang; Rui Zhang; Jing Huang; Qian Wang; Rui Zhang

doi:10.3934/math.2024407

AIMS Mathematics

2024, Volume 9, Issue 4: 8371-8385. doi: 10.3934/math.2024407

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On a binary Diophantine inequality involving prime numbers

School of Mathematics and Statistics, Shandong Normal University, Jinan 250014, China

Received: 15 January 2024 Revised: 21 February 2024 Accepted: 23 February 2024 Published: 28 February 2024
MSC : 11J25

Let $N$ denote a sufficiently large real number. In this paper, we prove that for $1 < c < \frac{104349}{77419}$ , $c\neq\frac{4}{3}$ , for almost all real numbers $T\in(N, 2N]$ (in the sense of Lebesgue measure), the Diophantine inequality $|p_1^c+p_2^c-T| < T^{-\frac{9}{10c}\left(\frac{104349}{77419}-c\right)}$ is solvable in primes $p_1, p_2$ . In addition, it is proved that the Diophantine inequality $|p_1^c+p_2^c+p_3^c+p_4^c-N| < N^{-\frac{9}{10c}\left(\frac{104349}{77419}-c\right)}$ is solvable in primes $p_1, p_2, p_3, p_4$ . This result constitutes a refinement upon that of Li and Cai.

Keywords:

Citation: Jing Huang, Qian Wang, Rui Zhang. On a binary Diophantine inequality involving prime numbers[J]. AIMS Mathematics, 2024, 9(4): 8371-8385. doi: 10.3934/math.2024407

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Abstract

1. Introduction

The Waring-Goldbach problem is that every natural number $n$ can be represented as the type

$n = p^k_1 +\cdots+ p^k_s,$

where $s$ and $k$ are given positive integers and $p_1, \cdots, p_s$ are prime variables. This well-known problem has spawned many analogues, which has attracted a large number of scholars to investigate and obtain many celebrated results. For instance, given $c > 1$ is not an integer and $\varepsilon > 0$ . For every sufficiently large real $N$ , suppose that $h(c)$ is the smallest natural number $s$ satisfying the inequality

$\begin{equation} |p_1^c+p_2^c+\cdots+p_s^c-N| < \varepsilon \end{equation}$

(1.1)

with solutions in primes $p_1, \cdots, p_s$ . Piatetski-Shapiro showed in ^[13] that

$\limsup\limits_{c\rightarrow \infty}\frac{h(c)}{c\log c}\leq4,$

and $h(c) \leq 5$ holds for $1 < c < \frac{3}{2}$ . Based on this result and Vinogradov's three prime theorem, we can conjecture that $h(c)\leq3$ for $c$ near one. Tolev ^[14] first proved this conjecture. More precisely, Tolev ^[15] showed that, if $1 < c < \frac{15}{14}$ , the inequality has solutions in primes $p_1, p_2, p_3$ :

$|p_1^c+p_2^c+p_3^c-N| < \varepsilon,$

where $\varepsilon = N^{-\frac{1}{c}(\frac{15}{14}-c)}\log^9N$ . Subsequently, this result was constantly improved by several authors (see ^{[1,2,3,4,7,8]}).

In 1999, Laporta ^[10] considered the corresponding binary problem. If $1 < c < \frac{15}{14}$ fixed and $\varepsilon = T^{1-\frac{15}{14c}}\log^8T$ , the following inequality

$\begin{equation} |p_1^c+p_2^c-T| < \varepsilon \end{equation}$

(1.2)

has a solution for all real numbers $T\in(N, 2N]\setminus\mathfrak{T}$ with $|\mathfrak{T}| = O\left(N\exp\left(-\frac{1}{3}\left(\frac{\log N}{c}\right)^{\frac{1}{5}}\right)\right)$ . Later, the exponent of $c$ was improved to

$\frac{43}{36} = 1.944444\cdots, \ \ \ \frac{6}{5} = 1.2, \ \ \ \frac{59}{44} = 1.340909\cdots$

by Zhai and Cao ^[16], Kumchev and Laporta ^[9], and Li and Cai ^[11] successively.

In 2003, Zhai and Cao ^[17] first proved that the inequality

$\begin{equation} |p_1^c+p_2^c+\cdots+p_4^c-N| < \varepsilon \end{equation}$

(1.3)

is solvable, where $1 < c < \frac{81}{68}$ . Afterwards, the exponent of $c$ was improved to

$\frac{97}{81} = 1.197530\cdots, \ \ \ \frac{6}{5} = 1.2, \ \ \ \frac{1193}{889} = 1.341957\cdots$

by Mu ^[12], Zhang and Li ^[18], and Li and Zhang ^[19] successively.

Here, we consider the cases $s = 2$ and $s = 4$ in inequality (1.1), and enlarge the exponent of $c$ . Our results are as follows:

Theorem 1. Assume $1 < c < \frac{104349}{77419}$ , $c\neq\frac{4}{3}$ , and $N$ is a sufficiently large real number. Suppose $\varepsilon = T^{-\frac{9}{10c}\left(\frac{104349}{77419}-c\right)}$ , and $B_0(T)$ is the number of solutions of inequality (1.2). Then, for all real numbers $T\in(N, 2N]\setminus\mathfrak{T}$ with $|\mathfrak{T}| = O\left(N\exp\left(-\frac{1}{3}\left(\frac{\log N}{c}\right)^{\frac{1}{5}}\right)\right)$ , we obtain

$B_0(T)\gg\frac{\varepsilon T^{\frac{2}{c}-1}}{\log^2T}.$

Theorem 2. Assume $1 < c < \frac{104349}{77419}$ , $c\neq\frac{4}{3}$ , and $N$ is a sufficiently large real number. Suppose $\varepsilon = T^{-\frac{9}{10c}\left(\frac{104349}{77419}-c\right)}$ , and $\mathcal{B}_0(N)$ is the number of solutions of inequality (1.3). Then, we obtain

$\mathcal{B}_0(N)\gg\frac{\varepsilon N^{\frac{4}{c}-1}}{\log^4N}.$

In this paper, our improvement mainly comes from the estimates of exponential sums. We transform the exponential sum into Type Ⅰ and Type Ⅱ sums by using Heath-Brown's identity. As a result, we improve the previous results by enlarging the upper bound of $c$ . In addition, by using Lemma 2.8, the value of $c = \frac{4}{3}$ can be excluded.

2. Notation and preliminaries

Now, we will give some notations which are required throughout the paper.

Throughout this paper, suppose that $N$ is a sufficiently large integer, $\Lambda(n)$ stands for the von Mangoldt function, and $|\mathfrak{T}|$ is the cardinality of the set $\mathfrak{T}$ . $a\sim A$ means $A < a\leq 2A$ . As usual, the constants in the $\ll$ -symbols and $O$ -terms are absolute or depend only on $c$ .

In addition, we write

$\begin{align*} &1 < c < \frac{104349}{77419}, \ c\neq\frac{4}{3};\ \ \ \ X = \frac{ N^{\frac{1}{c}}}{4}, \ \ \ \ \iota = X^{\frac{50489}{154838}-\frac{c}{2}}, \\ &\mathcal{\mathcal{K}} = X^{\frac{104349}{77419}-c}, \ \ \ \ e(t) = e^{2\pi it}, \ \ \ \ \varepsilon = N^{-\frac{9}{10c}\left(\frac{104349}{77419}-c\right)}, \\ &\delta = \frac{1}{1000}\left(\frac{104349}{77419}-c\right), \ \ \ \ E = \exp\left(-\left(\frac{\log N}{c}\right)^{\frac{1}{5}}\right), \\ & P = \left(\frac{2}{E^2}\right)^{\frac{1}{3}}\log N, \ \ \ \ \mathfrak{S}(t) = \sum\limits_{p\sim X}e(p^ct)\log p, \\ &\mathcal{I}(t) = \int_X^{2X}e\left(u^c t\right)du. \end{align*}$

Next, we shall recall some preliminary lemmas that are necessary in this paper.

Lemma 2.1 ([15, Lemma 1]). Assume that $\xi(y)$ is a function which is $\omega = [\log X]$ times continuously differentiable and satisfies

$\begin{equation*} \left\{ \begin{aligned} &\xi(y) = 1, \ |y|\leq \frac{4}{5}\varepsilon, \\ &0 < \xi(y) < 1, \ \frac{4}{5}\varepsilon < |y|\leq \varepsilon, \\ &\xi(y) = 0, \ |y|\geq \varepsilon. \end{aligned} \right. \end{equation*}$

For its Fourier transformation

$\Xi(t) = \int_{-\infty}^\infty e(-ty)\xi(y)dy,$

then

$|\Xi(t)|\leq\min\left(\frac{9}{5}\varepsilon, \frac{1}{\pi|t|}, \frac{1}{\pi|t|}\left(\frac{5\omega}{\pi|t|\varepsilon}\right)^\omega\right).$

Lemma 2.2. ([15, Lemma 14]). If $|t|\leq\iota$ , then

$\mathfrak{S}(t) = \mathcal{I}(t)+O\left(X\exp\left(-\log^{\frac{1}{5}}X\right)\right).$

Lemma 2.3 ([5, Lemma 3.1]). There is a trivial bound that

$\begin{equation} \mathcal{I}(t)\ll X^{1-c}|t|^{-1}. \end{equation}$

(2.1)

Lemma 2.4 ([15, Lemma 7]). There exist the following inequalities

$\begin{align} &(i)\int_{|t| < \iota}|\mathfrak{S}(t)|^2dt\ll X^{2-c}\log^3X; \end{align}$

(2.2)

$\begin{align} &(ii)\int_{|t| < \iota}|\mathcal{I}(t)|^2dt\ll X^{2-c}\log^3X; \end{align}$

(2.3)

$\begin{align} &(iii)\int_{n}^{n+1}|\mathfrak{S}(t)|^2dt\ll X\log^3X\ \mathit{\text{uniformly in}}\ n. \end{align}$

(2.4)

Lemma 2.5. We have

$\begin{equation} (i)\int_{-\infty}^\infty \mathcal{I}^2(t)\Xi(t)e(-T t)dt\gg\varepsilon T^{\frac{2}{c}-1}, \end{equation}$

(2.5)

$\begin{equation} (ii)\int_{-\infty}^\infty \mathcal{I}^4(t)\Xi(t)e(-N t)dt\gg\varepsilon X^{4-c}. \end{equation}$

(2.6)

Proof. The above two inequalities (2.5) and (2.6) can be found in ^[9] and ^[17] independently. □

Lemma 2.6 ([6, Lemma 5]). For any complex number $\alpha_n$ , we have the inequality

$\begin{align*} \left|\sum\limits_{H\leq n\leq2H}\alpha_n\right|^2\leq\left(1+\frac{H}{\mathcal{U}}\right)\sum\limits_{0\leq|u|\leq \mathcal{U}}\left(1-\frac{|u|}{\mathcal{U}}\right)\left(\sum\limits_{H\leq n\leq2H-u}\alpha_{n+u}\overline{\alpha_n}\right), \end{align*}$

where $H\geqslant1, \ \mathcal{U}\leq H$ , and $\overline{\alpha_n}$ stands for the conjugate of $\alpha_n$ .

Lemma 2.7 ([11, Lemma 2.3]). Suppose $|D| > 0$ and $A\leq A'\leq10A$ , then the inequality

$\begin{align*} \sum\limits_{A\leq a\leq A'}e(Da^c)\ll(|D|A^c)^\kappa A^{\lambda-\kappa}+\frac{A}{|D|A^c} \end{align*}$

holds for any exponent pair $(\kappa, \lambda)$ with $0\leq \kappa\leq\frac{1}{2}\leq \lambda\leq1$ .

Lemma 2.8 ([1, Therorem 2]). Suppose $\tau$ , $\upsilon$ are real numbers such that

$\begin{align*} \tau\upsilon (\tau-1)(\upsilon -1)(\tau-2)(\tau+\upsilon -2)(\tau+\upsilon -3) (\tau+2\upsilon -3)(2\tau+\upsilon -4)\neq0. \end{align*}$

Let

${\sum}_I = \sum\limits_{a\leq A}\alpha(a)\sum\limits_{b\in I_a}e\left(Da^\tau b^\upsilon \right),$

where $D > 0$ , $A\geq1$ , $B\geq1$ , $|\alpha(a)|\leq1$ , and $I_a$ is a subinterval of $[B, 2B]$ . Assume $G = DA^\tau B^\upsilon$ . For any $\eta > 0$ , then

$\begin{align*} {\sum}_I\ll&\left(G^{\frac{3}{14}}A^{\frac{41}{56}}B^{\frac{29}{56}}+G^{\frac{1}{5}}A^{\frac{3}{4}}B^{\frac{11}{20}}+G^{\frac{1}{8}}A^{\frac{13}{16}}B^{\frac{11}{16}}\right. \left.+A^{\frac{3}{4}}B+AB^{\frac{3}{4}}+G^{-1}AB\right)(AB)^\eta. \end{align*}$

Lemma 2.9 ([10, Lemma 1]).

$\mathbb{V} = \max\limits_{N < v_2\leq 2N}\int_N^{2N}\left|\int_{\iota < |t|\leq \mathcal{K}}e((v_1-v_2)t)dt\right|dv_1\ll \log N.$

Lemma 2.10 ([10, Lemma 2]). Let the letters $\mathbb{C}$ and $\mathbb{R}$ be the sets of complex and real numbers, respectively. Suppose that $\digamma_1$ and $\digamma_2$ are measurable subsets of $\mathbb{R}^n$ , and

$\|g\|_i = \left(\int_{\digamma_i}|g(t)|^2dt\right)^{\frac{1}{2}}, \ \langle g, f\rangle_i = \int_{\digamma_i}g(t)\overline{f}(t)dt$

stand for the usual norm and inner product in $L^2(\digamma_i, \mathbb{C})$ $(i = 1, 2)$ , respectively.

Assume $\varpi$ is a measurable complex-valued function defined on $\digamma_1\times\digamma_2$ , then

$\sup\limits_{y\in\digamma_1}\int_{\digamma_2}|\varpi(y, t)|dt < +\infty, \ \sup\limits_{t\in\digamma_2}\int_{\digamma_1}|\varpi(y, t)|dt < +\infty.$

Hence,

$\left|\int_{\digamma_1}\phi(y)\langle\psi, \varpi(y, \cdot)\rangle_2dy\right|\ll\|\psi\|_2\|\phi\|_1\left(\sup\limits_{y'\in\digamma_1}\int_{\digamma_1} \left|\langle\varpi(y, \cdot), \varpi(y', \cdot)\rangle_2\right|dy\right)^{\frac{1}{2}},$

where $\phi\in L^2(\digamma_1, \mathbb{C})$ , $\psi\in L^2(\digamma_2, \mathbb{C})$ .

3. The estimation of $\mathfrak{S}(t)$

In this section, we transform the exponential sum into Type Ⅰ and Type Ⅱ sums by taking advantage of Lemma 3.1.

Lemma 3.1 ([6, Lemma 3]). For $3 < L < M < Q < X$ , $\{Q\} = \frac{1}{2}, \ X\geq64Q^2L, \ Q\geq 4L^2, and\ M^3\geq32X$ , if $G(n)$ is a complex valued function satisfying $|G(n)|\leq1$ , then the sum

$\begin{align*} \sum\limits_{X\leq n\leq2X}\Lambda(n)G(n) \end{align*}$

may be decomposed into $O(\log^{10}X)$ sums, each either of type I:

$\sum\limits_{A\leq a\leq2A}\alpha(a)\sum\limits_{B\leq b\leq2B}G(ab)$

with $B > Q$ , $AB\asymp X, \ |\alpha(a)|\ll a^{\eta}$ , or of type II:

$\sum\limits_{A\leq a\leq2A}\alpha(a)\sum\limits_{B\leq b\leq2B}\beta(b)G(ab)$

with $L\ll B\ll M$ , $AB\asymp X, \ |\alpha(a)| \ll a^{\eta}, |\beta(b)| \ll b^{\eta}$ .

Lemma 3.2. For any $\eta > 0$ , $\alpha(a)$ is a sequence of complex numbers satisfying $|\alpha(a)|\ll a^\eta$ . If $\iota\leq|t|\leq \mathcal{K}$ and $A\ll X^{\frac{3685827905}{6017314356}}$ , then

$\mathfrak{S}_I = \sum\limits_{a\leq A}\alpha(a)\sum\limits_{X\leq ab\leq2X}e\left(t(ab)^c\right)\ll X^{\frac{1417193203}{1504328589}+2\eta},$

where $c \in (1, \frac{104349}{77419}]$ and $c\neq\frac{4}{3}$ .

Proof. Suppose $A\ll X^{\frac{3685827905}{10530300123}}$ , and we use Lemma 2.7 by choosing the exponent pair $\left(\frac{2}{7}, \frac{4}{7}\right)$ , then

$\begin{align*} \mathfrak{S}_I\ll& A^\eta\sum\limits_{a\leq A}\left|\sum\limits_{X\leq ab\leq2X}e\left(t(ab)^c\right)\right|\\ \ll& A^\eta\sum\limits_{a\leq A}\left((|t|X^c)^{\frac{2}{7}}X^{\frac{2}{7}}+\frac{1}{|t|X^{c-1}a}\right)\\ \ll&X^{\frac{1417193203}{1504328589}+2\eta}. \end{align*}$

In addition, suppose $X^{\frac{3685827905}{10530300123}}\ll A\ll X^{\frac{3685827905}{6017314356}}$ , then from Lemma 2.8 we derive that

$\begin{align*} \mathfrak{S}_I = &A^\eta\sum\limits_{a\leq A}\frac{a(m)}{A^\eta}\sum\limits_{X\leq ab\leq2X}e\left(t(ab)^c\right)\\ \ll&\left((\mathcal{K}X^c)^{\frac{3}{14}}A^{\frac{3}{14}}X^{\frac{29}{56}}+(\mathcal{K}X^c)^{\frac{1}{5}}A^{\frac{1}{5}}X^{\frac{11}{20}}+(\mathcal{K}X^c)^{\frac{1}{8}} A^{\frac{1}{8}}X^{\frac{11}{16}}\right.\\ &\left.+A^{-\frac{1}{4}}X+A^{\frac{1}{4}}X^{\frac{3}{4}}+(\iota X^c)^{-1}X\right)X^{2\eta}\\ \ll& X^{\frac{1417193203}{1504328589}+2\eta}. \end{align*}$

Combining the above two cases, we complete the proof of this lemma. □

Lemma 3.3. For any $\eta > 0$ , $\alpha(a)$ and $\beta(b)$ are sequences of complex numbers satisfying $|\alpha(a)|\ll a^\eta$ , $|\beta(b)|\ll b^\eta$ . If $\iota\leq|t|\leq \mathcal{K}$ and $X^{\frac{174270772}{1504328589}}\ll A\ll X^{\frac{427164751875025}{1190813782635796}}$ , then

$\mathfrak{S}_{II} = \sum\limits_{A < a\leq2A}\alpha(a)\sum\limits_{X < ab\leq2X}\beta(b)e\left(t(ab)^c\right)\ll X^{\frac{1417193203}{1504328589}+3\eta}.$

Proof. Let $\mathcal{U} = X^{\frac{174270772}{1504328589}-\eta}$ . Cauchy's inequality and Lemma 2.6 gives us

$\begin{align} |\mathfrak{S}_{II}|&\ll\left(\sum\limits_{b\in I_a}\left|\beta(b)\right|^2\right)^{1/2}\left(\sum\limits_{b\in I_a}\left|\sum\limits_{A < a\leq2A}\alpha(a)e(t(ab)^c)\right|^2\right)^{1/2} \\ &\ll X^{2\eta}\left(\frac{X^{2}}{\mathcal{U}}+\frac{X}{\mathcal{U}}\sum\limits_{1\leq u\leq \mathcal{U}}\sum\limits_{A\leq a\leq 2A-u}\left|E_u\right|\right)^{1/2}, \end{align}$

(3.1)

where $I_a$ stands for a subinterval of $\left[\frac{X}{2A}, \frac{2X}{A}\right]$ , and $E_u = \sum_{b\in I_a}e\left(tb^c((a+u)^c-a^c)\right).$

Following from Lemma 2.7 and choosing the exponent pair $\left(\frac{13}{31}, \frac{16}{31}\right)$ , we obtain the estimate of $E_u$ :

$E_u\ll \left(|t|X^{c-1}u\right)^\frac{13}{31}X^\frac{16}{31}A^{-\frac{16}{31}}+\frac{1}{|t|X^{c-1}u}.$

Then, taking the estimate of $E_u$ into (3.1), we get

$\begin{align*} |\mathfrak{S}_{II}|\ll& X^{2\eta}\left(\frac{X^2}{\mathcal{U}}+\frac{X}{\mathcal{U}}\left(\mathcal{U}^\frac{44}{31}|\mathcal{K}|^\frac{13}{31}X^{\frac{13}{31}c+\frac{3}{31}}A^{\frac{15}{31}}+ X^{1-c}|\iota |^{-1}A\log \mathcal{U}\right)\right)^{1/2}\\ \ll&X^{\frac{1417193203}{1504328589}+3\eta}. \end{align*}$

□

Lemma 3.4. Assume that $\eta$ is any arbitrarily small positive number and $\iota\leq|t|\leq \mathcal{K}$ , then

$\mathfrak{S}(t)\ll X^{\frac{1417193203}{1504328589}+5\eta},$

where $c \in (1, \frac{104349}{77419}]$ and $c\neq\frac{4}{3}$ .

Proof. First of all, we define

$\mathfrak{M}(t) = \sum\limits_{n\sim X}\Lambda(n)e(n^c t).$

Obviously, we can deduce that

$\begin{align} \mathfrak{S}(t) = \mathfrak{M}(t)+O\left(X^\frac{1}{2}\right). \end{align}$

(3.2)

Suppose

$L = X^{\frac{174270772}{1504328589}}, \ M = X^{\frac{427164751875025}{1190813782635796}}, \ Q = \left[X^{\frac{2331486451}{6017314356}}\right]+\frac{1}{2}.$

Following from Lemma 3.1 with $G(n) = e(tn^c)$ , we reduce the sum $\mathfrak{M}(t)$ of type Ⅰ:

$\sum\limits_{A\leq a\leq2A}\alpha(a)\sum\limits_{B\leq b\leq2B}G(ab), \ B > Q$

or of type Ⅱ:

$\sum\limits_{A\leq a\leq2A}\alpha(a)\sum\limits_{B\leq b\leq2B}\beta(b)G(ab), L\ll B\ll M.$

From this combined with Lemmas 3.2 and 3.3, we deduce

$\mathfrak{M}(t)\ll X^{\frac{1417193203}{1504328589}+5\eta}.$

Inserting the bound of $\mathfrak{M}(t)$ into (3.2), we finish the proof of Lemma 3.4. □

4. Proof of Theorem 1

In this section, we shall give the details of the proof of Theorem 1.

Let $\xi(y)$ and $\Xi(t)$ stand for the functions that appear in Lemma 2.1. For $T\in[N, 2N]$ , we write

$\mathfrak{B}(T) = \sum\limits_{p_1, p_2\sim X\atop|p_1^c+p_2^c-T| < \varepsilon}(\log p_1)(\log p_2).$

It suffices to show that $\mathfrak{B}(T)\geq\mathfrak{B}_1(T)$ , where

$\begin{align*} \mathfrak{B}_1(T)& = \sum\limits_{X < p_1, p_2\leq2X}(\log p_1)(\log p_2)\int_{-\infty}^\infty e((p_1^c+p_2^c-T)t)\Xi(t)dt\\ & = \int_{-\infty}^\infty \mathfrak{S}^2(t)\Xi(t)e(-T t)dt\\ & = \left(\int_{|t|\leq\iota}+\int_{\iota < |t|\leq \mathcal{K}}+\int_{|t| > \mathcal{K}}\right)\mathfrak{S}^2(t)\Xi(t)e(-Tt)dt\\ & = Q_1(T)+Q_2(T)+Q_3(T). \end{align*}$

In addition, we write

$P(T) = \int_{-\infty}^\infty \mathcal{I}^2(t)e(-T t)\Xi(t)dt, \ P_1(T) = \int_{-\iota}^\iota \mathcal{I}^2(t)\Xi(t)e(-Tt)dt.$

Lemma 4.1. We have

$\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^2|\Xi(t)|dt\ll X\log^4X.$

Proof. It follows from Lemma 2.1 and (2.4) that

$\begin{align*} &\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^2|\Xi(t)|dt\\ &\ll \varepsilon\sum\limits_{0\leq n\leq\frac{1}{\varepsilon}}\int_n^{n+1}|\mathfrak{S}(t)|^2dt+\sum\limits_{\frac{1}{\varepsilon}-1\leq n\leq \mathcal{K}}\frac{1}{n}\int_n^{n+1}|\mathfrak{S}(t)|^2dt\\ &\ll X\log^4X. \end{align*}$

□

Lemma 4.2. We have

$\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^4|\Xi(t)|dt\ll\varepsilon^{\frac{1}{2}}X^{\frac{10007332969}{3008657178}-\frac{c}{2}+12\delta}.$

Proof. If $V(t)$ is a continuous function defined for $-\mathcal{K}\leq t\leq\mathcal{K}$ , then we find that

$\begin{align} \left|\int_{\iota < |t|\leq \mathcal{K}}\mathfrak{S}(t)V(t)dt\right| & = \left|\sum\limits_{X < p\leq2 X}(\log p)\int_{\iota < |t|\leq \mathcal{K}}V(t)e(p^c t)dt\right|\\ &\leq\sum\limits_{X < p\leq2 X}(\log p)\left|\int_{\iota < |t|\leq \mathcal{K}}V(t)e(p^c t)dt\right|\\ &\leq(\log X)\sum\limits_{X < m\leq2 X}\left|\int_{\iota < |t|\leq \mathcal{K}}V(t)e(m^c t)dt\right|. \end{align}$

(4.1)

Suppose $H(t) = \sum_{m\sim X}e(m^c t)$ . Using Cauchy's inequality, we deduce that

$\begin{align} \left|\int_{\iota < |t|\leq \mathcal{K}}\mathfrak{S}(t)V(t)dt\right| &\leq X^\frac{1}{2}(\log X)\left|\sum\limits_{X < m\leq2 X}\left|\int_{\iota < |t|\leq \mathcal{K}}V(t)e(m^c t)dt\right|^2\right|^\frac{1}{2}\\ & = X^\frac{1}{2}(\log X)\left|\sum\limits_{X < m\leq2 X}\int_{\iota < |t|\leq \mathcal{K}}V(t)e(m^c t)dt\int_{\iota < |y|\leq \mathcal{K}}\overline{V(y)e(m^cy)}dy\right|^\frac{1}{2}\\ & = X^\frac{1}{2}(\log X)\left|\int_{\iota < |y|\leq \mathcal{K}}\overline{V(y)}dy\int_{\iota < |t|\leq \mathcal{K}}V(t)H(t-y)dt\right|^\frac{1}{2}\\ &\leq X^\frac{1}{2}(\log X)\left|\int_{\iota < |y|\leq \mathcal{K}}|V(y)|dy\int_{\iota < |t|\leq \mathcal{K}}|V(t)||H(t-y)|dt\right|^\frac{1}{2}. \end{align}$

(4.2)

Then, we estimate the inner integral in (4.2).

First, we need to consider $H(t-y)$ . Following from Lemma 2.7 and choosing the exponent pair $\left(\frac{19126}{58293}, \frac{31369}{58293}\right)$ , we obtain that the inequality

$H(t)\ll(|t|X^c)^{\frac{19126}{58293}}X^{\frac{12243}{58293}}+\frac{X}{|t|X^c}$

holds for $X^{-c} < |t|\leq2\mathcal{K}$ .

Combining with the trivial upper bound $H(t-y)\ll X$ , we get

$\begin{align} H(t-y)\ll\min\left((|t-y|X^c)^{\frac{19126}{58293}}X^{\frac{12243}{58293}}+\frac{X}{|t-y|X^c}, X\right). \end{align}$

(4.3)

Next, inserting the estimate of $H(t-y)$ into the inner integral in (4.2), we find that

$\begin{align} &\int_{\iota < |t|\leq \mathcal{K}}|V(t)||H(t-y)|dt\\ \ll&\int_{\iota < |t|\leq \mathcal{K}\atop|t-y|\leq X^{-c}}|V(t)||H(t-y)|dt+\int_{\iota < |t|\leq \mathcal{K}\atop X^{-c} < |t-y|\leq2\mathcal{K}}|V(t)||H(t-y)|dt\\ \ll&X\int_{\iota < |t|\leq \mathcal{K}\atop|t-y|\leq X^{-c}}|V(X)|dt+\int_{\iota < |t|\leq \mathcal{K}\atop X^{-c} < |t-y|\leq2\mathcal{K}}|V(t)|\left((|t-y|X^c)^{\frac{19126}{58293}}X^{\frac{12243}{58293}}+\frac{1}{|t-y|X^{c-1}}\right)dt\\ \ll&X\max\limits_{\iota < |t|\leq \mathcal{K}}|V(t)|\int_{|t-y|\leq X^{-c}}dt+X^{\frac{981206597}{1504328589}}\int_{\iota < |t|\leq \mathcal{K}}|V(t)|dt \end{align}$

(4.4)

$\begin{align} &+X^{1-c}\max\limits_{\iota < |t|\leq \mathcal{K}}|V(t)|\int_{X^{-c} < |t-y|\leq2\mathcal{K}}\frac{1}{|t-y|}dt\\ \ll&X^{\frac{981206597}{1504328589}}\int_{\iota < |t|\leq \mathcal{K}}|V(t)|dt+X^{1-c}(\log X)\max\limits_{\iota < |t|\leq \mathcal{K}}|V(t)|. \end{align}$

(4.5)

From (4.2) and (4.4), we have

$\begin{align} \left|\int_{\iota < |t|\leq \mathcal{K}}\mathfrak{S}(t)V(t)dt\right| \ll &X^{\frac{2485535186}{3008657178}}(\log X)\int_{\iota < |t|\leq \mathcal{K}}|V(t)|dt\\ &+X^{1-\frac{c}{2}}(\log^{\frac{3}{2}}X)\left|\max\limits_{\iota < |t|\leq \mathcal{K}}|V(t)|\int_{\iota < |t|\leq \mathcal{K}}|V(t)|dt\right|^\frac{1}{2}. \end{align}$

(4.6)

Taking $V(t) = |\Xi(t)|\overline{\mathfrak{S}(t)}|\mathfrak{S}(t)|$ , from Lemmas 3.4 and 4.1 and (4.6), we get

$\begin{align} &\int_{\iota < |t|\leq \mathcal{K}}|\Xi(t)||\mathfrak{S}(t)|^3 dt\\ = &\int_{\iota < |t|\leq \mathcal{K}}V(t)\mathfrak{S}(t)dt\\ \ll&X^{\frac{1242767598}{1504328589}}(\log X)\int_{\iota < |t|\leq \mathcal{K}}|\Xi(t)||\mathfrak{S}(t)|^2dt+\varepsilon^{\frac{1}{2}}X^{\frac{2921521792}{1504328589}-\frac{c}{2}+6\delta}\left|\int_{\iota < |t|\leq \mathcal{K}}|\Xi(t)||\mathfrak{S}(t)|^2dt\right|^{\frac{1}{2}}\\ \ll&X^{\frac{2747096182}{1504328589}}\log^5X+\varepsilon^{\frac{1}{2}}X^{\frac{7347372173}{3008657178}-\frac{c}{2}+7\delta}\\ \ll&X^{\frac{2747096182}{1504328589}+7\delta}. \end{align}$

(4.7)

Next, taking $V(t) = |\Xi(t)|\overline{\mathfrak{S}(t)}|\mathfrak{S}(t)|^2$ , by Lemma 3.4, (4.6), and (4.7), we obtain

$\begin{align*} &\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^4|\Xi(t)|dt\\ = &\int_{\iota < |t|\leq \mathcal{K}}V(t)\mathfrak{S}(t)dt\\ \ll&X^{\frac{1242767598}{1504328589}}(\log X)\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^3|\Xi(t)|dt+\varepsilon^{\frac{1}{2}}X^{\frac{2420078929}{1002885726}-\frac{c}{2}+8\delta}\left|\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^3|\Xi(t)|dt\right|^{\frac{1}{2}}\\ \ll&X^{\frac{3989863780}{1504328589}+8\delta}+\varepsilon^{\frac{1}{2}}X^{\frac{10007332969}{3008657178}-\frac{c}{2}+12\delta}\\ \ll&\varepsilon^{\frac{1}{2}}X^{\frac{10007332969}{3008657178}-\frac{c}{2}+12\delta}. \end{align*}$

□

Lemma 4.3 ([11, Lemma 3.3]). We have

$\int_N^{2N}|Q_1(T)-P_1(T)|^2dT\ll\varepsilon^2N^{\frac{4}{c}-1}E^{\frac{1}{3}}.$

Lemma 4.4. We have

$\int_N^{2N}|Q_2(T)|^2dT\ll\varepsilon^2N^{\frac{4}{c}-1}E^{\frac{1}{3}}.$

Proof. Suppose

$\digamma_1 = \{T:T\sim N\}, \ \digamma_2 = \{t:\iota < |t|\leq \mathcal{K}\}.$

Taking advantage of Lemma 2.10 with $\psi(t) = \Xi(t)\mathfrak{S}^2(t), \varpi(t, T) = e(T t), \phi(T) = \overline{Q_2(T)}$ , we obtain

$\begin{align} \int_N^{2N}|Q_2(T)|^2dT = &\int_{\digamma_1}\overline{Q_2(T)}\langle \Xi(t)\mathfrak{S}^2(t), e(tT)\rangle_2dT\\ \ll&\left(\int_{\digamma_2}|\Xi(t)\mathfrak{S}^2(t)|^2dt\right)^{\frac{1}{2}}\left(\int_{\digamma_1}|\overline{Q_2(T)}|^2dT\right)^{\frac{1}{2}} \left(\sup\limits_{t'\in\digamma_1}\int_{\digamma_1}\left|\langle e(tT), e(t'T)\rangle_2\right|dt\right)^{\frac{1}{2}}. \end{align}$

(4.8)

According to Lemmas 2.1, 2.9, and 4.2, we get

$\begin{align*} \int_N^{2N}|Q_2(T)|^2dT\ll&\mathbb{V}\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^4|\Xi(t)|^2dt\\ \ll&\varepsilon(\log N)\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^4|\Xi(t)|dt\\ \ll&\varepsilon^{\frac{3}{2}}X^{\frac{10007332969}{3008657178}-\frac{c}{2}+12\delta}\log N\\ \ll&\varepsilon^2N^{\frac{4}{c}-1}E^{\frac{1}{3}}. \end{align*}$

Hence, we finish the proof of this lemma. □

Lemma 4.5. We have

$\int_N^{2N}|Q_3(T)|^2dT\ll N.$

Proof. It follows from the estimate of $\Xi(t)$ in Lemma 2.1 that

$\begin{align*} \int_N^{2N}|Q_3(T)|^2dT\ll&N\left|\int_{t\geq\mathcal{K}}|\mathfrak{S}^2(t)||\Xi(t)|dt\right|^2\\ \ll&NX^4\left|\int_{t\geq\mathcal{K}}\left(\frac{5\omega}{\pi t\varepsilon}\right)^\omega\frac{dt} {t}\right|^2\\ \ll&NX^4\left(\frac{5\omega}{\pi \mathcal{K}\varepsilon}\right)^{2\omega}\\ \ll&N. \end{align*}$

□

Lemma 4.6. We have

$\int_N^{2N}|\mathfrak{B}_1(T)-P(T)|^2dT\ll\varepsilon^2N^{\frac{4}{c}-1}E^{\frac{1}{3}}.$

Proof. We obtain

$\begin{align} \int_N^{2N}|\mathfrak{B}_1(T)-P(T)|^2 = &\int_N^{2N}|Q_1(t)-P_1(T)+Q_2(t)+Q_3(t)+P_1(T)-P(T)|^2 \end{align}$

(4.9)

$\begin{align} \leq&\int_N^{2N}|Q_1(T)-P_1(T)|^2dT+\int_N^{2N}|Q_2(T)|^2dT\\ &+\int_N^{2N}|Q_3(T)|^2dT+\int_N^{2N}|P(T)-P_1(T)|^2dT. \end{align}$

(4.10)

For the last integral, we will use Lemmas 2.1 and 2.3 to estimate and obtain

$\begin{align} \int_N^{2N}|P(T)-P_1(T)|^2dT\ll&\int_N^{2N}\int_{|t| > \iota}|\mathcal{I}(t)|^4|\Xi(t)|^2dtdT\\ \ll&NX^{4-4c}\int_{|t| > \iota}|\Xi(t)|^2|t|^{-4}dt\\ \ll&\varepsilon^2NX^{4-4c}\iota^{-3}. \end{align}$

(4.11)

From (4.9) and (4.11), we have

$\begin{align} \int_N^{2N}|\mathfrak{B}_1(T)-P(T)|^2dT\ll&\int_N^{2N}|Q_1(T)-P_1(T)|^2dT+\int_N^{2N}|Q_2(T)|^2dT\\ &+\int_N^{2N}|Q_3(T)|^2dT+\varepsilon^2NX^{4-4c}\iota^{-3}\\ \ll&\varepsilon^2N^{\frac{4}{c}-1}E^{\frac{1}{3}}, \end{align}$

(4.12)

where Lemmas 4.3, 4.4, and 4.5 are employed. □

Lemma 4.6 implies

$\begin{align} \mathfrak{B}_1(T) = P(T)+O\left(\varepsilon N^{\frac{2}{c}-1}E^\frac{1}{9}\right), \end{align}$

(4.13)

where $T\in(N, 2N]\backslash \mathfrak{T}$ with $|\mathfrak{T}| = O\left(NE^{1/3}\right)$ .

Then, it follows from (4.13) and (2.5) that

$\begin{align} B_0(T)\geq\frac{\mathfrak{B}(T)}{\log^22X}\geq\frac{\mathfrak{B}_1(T)}{\log^22X}\geq\frac{P(T)}{\log^22X}\gg\frac{\varepsilon T^{\frac{2}{c}-1}}{\log^2T}. \end{align}$

(4.14)

Thus, we finish the proof of Theorem 1.

5. Proof of Theorem 2

Suppose

$\mathcal{B}(N) = \sum\limits_{X < p_1, \cdots, p_4\leq2X\atop|p_1^c+\cdots+p_4^c-N| < \varepsilon}(\log p_1)\cdots(\log p_4).$

It follows from the definition of $\xi(y)$ and $\Xi(t)$ in Lemma 2.1 that

$\begin{equation} \mathcal{B}(N)\geq\mathcal{B}_1(N), \end{equation}$

(5.1)

where

$\begin{align} \mathcal{B}_1(N)& = \int_{-\infty}^\infty \mathfrak{S}^4(t)\Xi(t)e(-N t)dt\\ & = \left(\int_{-\iota}^\iota+ \int_{\iota < |t|\leq \mathcal{K}}+\int_{|t| > \mathcal{K}}\right)\mathfrak{S}^4(t)\Xi(t)e(-N t)dt\\ & = \mathcal{Q}_1(N)+\mathcal{Q}_2(N)+\mathcal{Q}_3(N). \end{align}$

(5.2)

First of all, we need to estimate $\mathcal{Q}_1(N)$ . Let

$\begin{align*} \mathcal{P}_1(N) = &\int_{-\iota}^\iota \mathcal{I}^4(t)\Xi(t)e(-tN)dt, \\ \mathcal{P}(N) = &\int_{-\infty}^\infty \mathcal{I}^4(t)\Xi(t)e(-tN)dt, \end{align*}$

then we find that

$\begin{align} \mathcal{Q}_1(N)& = \int_{-\iota}^\iota\mathfrak{S}^4(t)\Xi(t)e(-N t)dt\\ & = \mathcal{P}(N)+(\mathcal{P}_1(N)-\mathcal{P}(N))+(\mathcal{Q}_1(N)-\mathcal{P}_1(N)). \end{align}$

(5.3)

For the second integral in (5.3), we use Lemmas 2.1 and 2.3 to estimate and obtain

$\begin{align} |\mathcal{P}_1(N)-\mathcal{P}(N)|\ll&\int_{-\iota}^\iota|\mathcal{I}(t)|^4|\Xi(t)|dt\\ \ll&X^{4-4c}\int_{|t| > \iota}|\Xi(t)||t|^{-4}dt\\ \ll&\varepsilon X^{4-4c}\iota^{-3}\\ \ll&\frac{\varepsilon X^{4-c}}{\log X}. \end{align}$

(5.4)

For the third integral in (5.3), we use Lemmas 2.1 and 2.2 to estimate and get

$\begin{align} |\mathcal{Q}_1(N)-\mathcal{P}_1(N)| \ll&\int_{|t|\leq\iota}|\mathfrak{S}^4(t)-\mathcal{I}^4(t)||\Xi(t)|dt\\ \ll&\varepsilon\max\limits_{-\iota\leq t\leq\iota}|\mathfrak{S}(t)-\mathcal{I}(t)|\int_{|t|\leq\iota}(|\mathfrak{S}(t)|+|\mathcal{I}(t)|)(|\mathfrak{S}(t)|^2+|\mathcal{I}(t)|^2)dt\\ \ll&\varepsilon X\exp\left(-\log^{\frac{1}{5}}X\right)X \int_{|t|\leq\iota}(|\mathfrak{S}(t)|^2+|\mathcal{I}(t)|^2)dt\\ \ll&\varepsilon X^2\exp\left(-\log^{\frac{1}{5}}X\right)X^{2-c}(\log^3X)\\ \ll&\varepsilon X^{4-c}\exp\left(-\log^{\frac{1}{5}}X\right), \end{align}$

(5.5)

where (2.2) and (2.3) in Lemma 2.4 are utilized. Combining (2.6) and (5.3)–(5.5), we derive that

$\begin{align} \mathcal{Q}_1(N)\gg\varepsilon X^{4-c}. \end{align}$

(5.6)

From Lemma 4.2, we have

$\begin{align} \mathcal{Q}_2(N)\ll&\int_{\iota < |t|\leq \mathcal{K}}|\mathfrak{S}(t)|^4|\Xi(t) |dt\\ \ll&\varepsilon^{\frac{1}{2}}X^{\frac{10007332969}{3008657178}-\frac{c}{2}+12\delta}\\ \ll&\frac{\varepsilon X^{4-c}}{\log X}. \end{align}$

(5.7)

From Lemma 2.1, we obtain

$\begin{align} \mathcal{Q}_3(N)\ll&\int_{-\mathcal{K}}^\mathcal{K}|\mathfrak{S}^4(t)||\Xi(t)|dt\\ \ll&X^4\int_{-\mathcal{K}}^\mathcal{K}\left(\frac{5\omega}{\pi|t|\varepsilon}\right)^\omega\frac{dt }{|t|}\\ \ll&X^4\left(\frac{5\omega}{\pi\mathcal{K}\varepsilon}\right)^\omega\\ \ll&1. \end{align}$

(5.8)

It follows from (5.2), (5.6)–(5.8) that

$\begin{align} \mathcal{B}_1(N)\gg\varepsilon X^{4-c}. \end{align}$

(5.9)

From (5.1) and (5.9), we have

$\begin{align} \mathcal{B}_0(N)\geq\frac{\mathcal{B}(N)}{\log^42X}\geq\frac{\mathcal{B}_1(N)}{\log^42X}\gg\frac{\varepsilon N^{\frac{4}{c}-1}}{\log^4N}. \end{align}$

(5.10)

Hence, we finish the proof of Theorem 2.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The paper is supported by National Natural Science Foundation of China (Grant No. 12171286).

Conflict of interest

The authors declare that they have no conflict of interest.

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AIMS Mathematics

On a binary Diophantine inequality involving prime numbers

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Abstract

1. Introduction

2. Notation and preliminaries

3. The estimation of $\mathfrak{S}(t)$

4. Proof of Theorem 1

5. Proof of Theorem 2

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On a binary Diophantine inequality involving prime numbers

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1. Introduction

2. Notation and preliminaries

3. The estimation of S(t) \mathfrak{S}(t)

4. Proof of Theorem 1

5. Proof of Theorem 2

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3. The estimation of $\mathfrak{S}(t)$