A System of Two Diophantine Inequalities with Primes

Abstract

Let , . In this paper, we prove that there exist positive real numbers and depending on such that for all real numbers , and , the system of two Diophantine inequalities is solvable in prime variables .

1. Introduction

Assume that is not an integer and is a sufficiently small positive number. Let denote the least integer such that the Diophantine inequalityis solvable in primes for sufficiently large . In 1952, Piatetski-Shapiro [1] proved that

Piatetski-Shapiro also proved for . Tolev [2] first improved this result for close to one. More precisely, Tolev proved that if , the Diophantine inequalityhas prime solutions for large , where . Later, this result was improved by many authors (see [3–9]), and many analogous problems of this type were studied (for example, see [10–17]).

In 1995, Tolev [18] first considered the system of two Diophantine inequalities with primes

He established that for all real numbers satisfying and , system (4) withhas solutions in primes , where are real numbers satisfying the conditionsand depending on are sufficiently large numbers. Subsequently, Tolev’s result was improved by Zhai [19] and Zhai and Cao [20]. Now the best result is due to Zhai and Cao [20] who proved that system (4) withis solvable for .

In this paper, we consider the following system of two Diophantine inequalities over primes :where , are different numbers but close to 1 and satisfy

We have to impose a condition on the orders of and due to the inequalitywhich holds for every positive provided . Our result is as follows.

Theorem 1. Suppose that are real numbers and satisfy conditionsThen, there exist positive real numbers , depending on such that for all real numbers andsystem (8) withhas solutions in primes .

Notation. Throughout this paper, the letter , with or without a subscript, always represents a prime, and are real numbers satisfying (11), and and are real numbers satisfying (12). denotes a sufficiently small positive number depending on and . denotes the characteristic function over the interval . is the nontrivial zero of the Riemann zeta function . As usual, and are the von Mangoldt function and the divisor function, respectively. and are sufficiently large numbers. We set

2. Outline of the Proof

Let denote a sufficiently small positive number, whose value depends on and will be determined more precisely in Lemma 1. Let

We divide the plane into three regions: the neighbourhood of origin , the intermediate region , and the trivial region , which are defined as

Thus, the integral can be represented aswhere

If we show that tends to infinity as tends to infinity, then Theorem 1 holds. From Lemma 4, it is sufficient to prove that tends to infinity as X tends to infinity. To do this, noting (20), we shall prove the following:

In Section 3, we first give some auxiliary lemmas. Inequality (22) is proved in Section 4. Inequality (23), from which we can get the range of and , is proved in Section 5. In Section 6, we complete the proof of Theorem 1.

3. Auxiliary Lemmas

Lemma 1. Let . There exists depending on such that for the volume of the domain in six-dimensional space defined bywe haveprovided are sufficiently small.

Proof. This lemma is similar to Lemma 1 in Tolev [18]. We can write the volume of asThen, we can fix to get the range of from last two inequalities and getSince and are sufficiently small, we can adjust the value of to ensure that there are intersections between (28) and (29). We may also get this lemma by adjusting the value of and using circle method to estimate

Lemma 2 (see [18], Lemma 2). The function has the following properties:

Lemma 3. Let denote a real number and be a smooth real function on . Suppose that there exists a positive constant such thatwhere the implied absolute constant depends only on . Then, there exists an exponent pair withsuch that

Proof. We can find this lemma in Ivić ([21], pp. 72–79).

Lemma 4. The quantities and satisfy

Proof. From (31) and (32) in Lemma 2, we havewhere we substitute for . Similarly, we haveThen,

Lemma 5. There are two real functions defined in , for and is a monotonous function. WriteIf or for all , thenIf for all , then

Proof. This lemma can be found in Ivić ([21], pp. 56-57).

Lemma 6. Let and ; then,where is the nontrivial zero of .

Proof. This is a well-known explicit formula, which can be found in Karatsuba ([22], p. 80).

4. The Estimate of the Integral

In this section, we give the estimate of the integral and set .

Lemma 7. If , thenand

Proof. We can get this lemma from Lemma 11 and Lemma 12 of Tolev [2].

Lemma 8. If , thenwhere

Proof. Without loss of generality, we consider the case . From , we haveLetWe define three sets of nontrivial zeroes of asand the set may be empty.
We first consider or . In this situation, we consider the following three cases. Case 1. When . In this case, we have Applying Lemma 5, we have Therefore, by (46), we can obtain Case 2. When . In this case, we have Hence, we use (45) and get If the set is empty, then the upper bound is trivial. Case 3. When . In this case, we haveFrom (46), we haveTherefore, from (54)–(58), we obtainNow we consider the remaining case and . We use the trivial estimate of and getwhere . and are estimated analogically as in the previous case. Then, the estimate for is established.

Lemma 9. If , thenwhere

Proof. Notingwe havewhere denotes an integer number. Applying Abel’s transformation, we getThen, Lemma 6 impliesUsing integration by parts, we haveFrom Lemma 8, this lemma follows.

Lemma 10 (see [18], Lemma 8). For defined by (17), we have

Lemma 11 (see [18], Lemma 9). For defined in Lemma 9, we have

Writeand

Lemma 12. For integrals and defined by (70) and (71), respectively, we have

Proof. We haveApplying Lemma 5 with and , from , we can getThen, by Lemma 11, we can get this lemma.

Lemma 13. For the integral defined by (70), we have

Proof. We havewhereFrom (31), we getIn (76), the integral over is convergent, and hence . Applying Lemma 2, we havewhereFrom (13) and Lemma 1, we have

Lemma 14. For the integral in (20), we have

Proof. We haveFrom Lemmas 9–11, we can obtainBy Lemma 12, we haveThen, from Lemma 13, we can get this lemma.

5. The Estimate for the Integral

Lemma 15. Let be arbitrary complex numbers andWriteThen,where

Proof. Without loss of generality, we can consider the case . We define :where will be determined later and . Thus, can be rewritten asBy Cauchy’s inequality, we haveWe rearrange the sums as follows:whereNext, we handle the exponential sum of (93). From the derivatives of , we getThen, by Lemma 3 with the exponent pair (see also [21], p. 77)we haveInserting this upper bound into (93), we havewhereBy Cauchy’s inequality, we obtainFrom (98) and , we haveWe take , i.e.,and then . Inserting this value of into (101), we getwhich yields this lemma.

Lemma 16 (see [2], Lemma 9). Let be arbitrary complex numbers and . WriteThen, there exist satisfying such that

Lemma 17. Assume that ; then,

Proof. Without loss of generality, we can consider the case . Clearly,whereHence, it is sufficient to prove thatThe estimates of and are similar, and thus we focus on the estimate of . Using Vaughan’s identity (see [23]), we havewherewhere and .
For , notingwe havewhereFor , we haveThe second summation over in (115) iswhere , , and . Noting , Lemma 3 with exponent pair givesRecalling (115) and (116), we haveArguing similarly, we can get the following same bound of :For , we divide it into three parts:whereFor , using dyadic subdivision and Lemma 16, we havewherewhere , and . Lemma 15 implieswhere we used the mean value estimatesTherefore, we haveWe estimate and similar to and getFor , we follow a similar argument to and getwhereUsing Lemmas 15 and 16 and the mean value estimatewe getFrom (110) and (118)–(131), we can obtain (109).

Lemma 18 (see [18], Lemma 14). For defined by (17), we have

Lemma 19. For the integral in (20), we have

Proof. By Lemmas 17 and 18, we havewhich yields this lemma.

6. Proof of Theorem 1

In this section, we first give the estimate of in (20) and then complete the proof of Theorem 1.

For the integral in (20), by Lemma 2, we can easily get

From Lemmas 14 and 19 and (135), we know that (22)–(24) follows, respectively. Therefore, recalling Lemma 4 and (20), we can getfrom which we complete the proof of Theorem 1.

Data Availability

The data supporting the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This study was supported by the National Natural Science Foundation of China (grant nos. 11771256 and 11801328).