On a binary Diophantine inequality involving prime numbers

1.   Introduction

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

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

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

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 $p_1, p_2, p_3$:

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

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

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

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

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

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

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

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

For its Fourier transformation

then

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

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

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

Lemma 2.5. We have

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

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

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

Let

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

Lemma 2.9 ([10, Lemma 1]).

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

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

Hence,

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

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

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

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

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

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

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

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

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 $E_u$:

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

□

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

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

Proof. First of all, we define

Obviously, we can deduce that

Suppose

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

or of type Ⅱ:

From this combined with Lemmas 3.2 and 3.3, we deduce

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

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

In addition, we write

Lemma 4.1. We have

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

□

Lemma 4.2. We have

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

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

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

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

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

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

From (4.2) and (4.4), we have

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

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

□

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

Lemma 4.4. We have

Proof. Suppose

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

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

Hence, we finish the proof of this lemma. □

Lemma 4.5. We have

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

□

Lemma 4.6. We have

Proof. We obtain

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

From (4.9) and (4.11), we have

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

Lemma 4.6 implies

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

Thus, we finish the proof of Theorem 1.

5.   Proof of Theorem 2

Suppose

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

where

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

then we find that

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

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

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

From Lemma 4.2, we have

From Lemma 2.1, we obtain

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

From (5.1) and (5.9), we have

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.