#### Abstract

Let denote the number of even integers , with , such that cannot be written as . We prove that if , then .

#### 1. Introduction

The Waring–Goldbach problem is to study the representation of positive integers as sums of powers of prime numbers. In this paper, we shall focus on the cubic Waring–Goldbach problem. This topic can be traced back to the work of Hua . He proved that almost all integers satisfying certain congruence conditions can be written as cubes of primes, where , and the abovementioned congruence conditions are

Let denote the number of positive numbers not exceeding and cannot be represented as cubes of primes; Hua showed that, for any ,

Throughout, we assume that is a large natural number, and . In this paper, we consider the exceptional set of even integers in the short interval , which cannot be represented as sum of eight cubes of primes. Precisely, let denote the set of natural numbers with , such that cannot be written as the following expression:where are primes. Moreover, we set

The purpose of this paper is to obtain with as small as possible. Zhao  proved that , which implies if . The main result in the paper is as follows.

Theorem 1. Let and be defined as above. For , we have

We shall prove Theorem 1 by means of the Hardy–Littlewood method. The treatment of the integrals on the major arcs is standard, and we will focus on the treatment of the integrals on the minor arcs.

In this paper, we make use of the method of Vaughan  to deal with the equation (see Lemma 3). We also make use of the result of Zhao  on the 10th moment of the Weyl sums over cubes of primes restricted on minor arcs.

Notation. Throughout the paper, denotes a sufficiently small positive number, and let denote a positive constant. We need to point out that , and are allowed to change at different occurrences. With or without subscript, denotes a prime number. Denote by the number of divisors of , and as usual, we write for .

#### 2. Preliminaries

Before we prove Theorem 1, we introduce the following theorem.

Theorem 2. Let be defined in Section 1. Then, we have

We can get Theorems 1 from 2 immediately. Therefore, our task is to prove Theorem 2.

Let

Write

Let be the weighted number of solutions of with and . By orthogonality, one has

We write

We define the set of major arcs as the union of the intervalswith and . We then denote the corresponding set of minor arcs by . So, we can get the lemma.

Lemma 1. For , we have

We will prove Lemma 1 in Section 4.

Lemma 2. For , we have

We can find the proof in (, Section 9).

Lemma 3. Let be the number of solutions ofwith , , and . We have

We will prove Lemma 3 in Section 3.

Proof of Theorem 2. On recalling the set defined in Section 1 and by means of an argument of Wooley (see ), we haveLetwhereSo, one hasWith the help of Lemma 1, we can getWith an application of the Cauchy–Schwarz inequality, one hasBy Lemma 2, we haveBy considering the underlying Diophantine equations and Lemma 3, we getFrom (20) to (23), we haveTherefore, we conclude thatThis completes the proof of Theorem 2.

#### 3. Proof of Lemma 3

We write as the number of solutions of (14) with , and let denote the number of solutions of (14) with . It is easy to find that

Lemma 4. We denote by the number of solutions of the Diophantine equationwith and . Then, we have

The proof of Lemma 1 can follow from an argument of Hooley  (see the proof of Lemma 1 of Parsell  for a sketch of the necessary adjustments to Hooley’s argument).

To calculate , by symmetry, we can assume that . Write . Then, (14) becomes

Since , , and , it follows that . Let

Then,where is defined in (18), and

Let denote the interval and . We may suppose that . Then, with , contained in . Let denote the union of with , , and let . Then, we have

By the same method of the lemma of Vaughan , we can get the following lemma.

Lemma 5. Suppose that and , then

Moreover, we can get the following lemma.

Lemma 6. Let be denoted as above. Then, we haveThe proof can be also found in Vaughan .

Proof of Lemma 3. When , the number of solutions of it satisfiesGiven any one of the possible choices for and with , it follows from Lemma 4 that the number of permissible choices for is , and thus the contribution arising from this class of solutions is . When , on the other hand, the variables satisfy the equation ; the number of the solutions is . Thus, we conclude that the number of solutions of this type is . So, we can getThen, we haveWe consider . By Dirichlet’s theorem on Diophantine approximation, given , we may choose , with and . It is easily verified that . Moreover, since , we have . Therefore, by Lemma 5, we haveHence, we haveWhen , by Lemma 6, we can getBy (40) and (41),In view of (38) and (42), we complete the proof of Lemma 3.

#### 4. Proof of Lemma 1

Let for . We write

We define the set of major arcs as the union of the intervalswith and . It is easy to find that . We then denote the corresponding set of minor arcs by .

Lemma 7. Suppose that the integer satisfies and . Then, one has

The proof can be found in (, Lemma 3.1).

We define the multiplicative function by taking

We have

Therefore, we have the following result.

Lemma 8. Let be a constant. When , one hasfor some constant.

This is due to Zhao (, Lemma 1). Before we deal with the integrals on the minor arcs, we should give an upper bound of . We also quote the following estimate proved by Ren .

Lemma 9. Suppose that is a real number and that with . Let . Then, one can havewhere and and is a constant.

Proof of Lemma 1. When , we haveMoreover, one has when , and when , we have , and by Lemma 7, we haveBy same measure, when and , we can getWhen , by Lemma 9, we haveTherefore, we haveLet . We haveSince , we haveTherefore, by Lemma 8, we obtainfor some absolute constant .
In view of (52) and (57), we complete the proof of Lemma 1.

#### Data Availability

Except references, no data were used to support this study.

#### Conflicts of Interest

The author declares no conflicts of interest.