# Exceptional Sets for Sums of Prime Cubes in Short Interval

Authors: Gongrui Chen
Journal: Journal of Mathematics
 (2021)

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## Abstract

LetEN,X denote the number of even integers n, with N−X≤n≤N, such that n cannot be written as n=p13+⋯+p83. We prove that if X>N1/36+ɛ, then EN,X=oX.


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## Body

## 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 [1]. He proved that almost all integers satisfying certain congruence conditions can be written as s cubes of primes, where s=5,6,7,8, and the abovementioned congruence conditions are(1)N5=n∈ℕ:n≢1mod2,n≢0,±2mod9,n≢0mod7,N6=n∈ℕ:n≢0mod2,n≢±1mod9,N7=n∈ℕ:n≢1mod2,n≢0mod9,N8=n∈ℕ:n≢0mod2.LetEsN denote the number of positive numbers n∈Ns not exceeding N and cannot be represented as s cubes of primes; Hua showed that, for any A>0,(2)EsN≪NlogN−A.Throughout, we assume thatN is a large natural number, and X<N. In this paper, we consider the exceptional set of even integers n in the short interval N−X≤n≤N, which cannot be represented as sum of eight cubes of primes. Precisely, let ξN,X denote the set of natural numbers n∈N8 with N−X≤n≤N, such that n cannot be written as the following expression:(3)n=p13+p23+⋯+p83,where pi are primes. Moreover, we set(4)EN,X=cardξN,X.The purpose of this paper is to obtainEN,X=oX with X as small as possible. Zhao [2] proved that E8N≪N1/6+ɛ, which implies EN,X=oX if X>N1/6+ɛ. The main result in the paper is as follows.Theorem 1.
Letθ>1/36 and EN,X be defined as above. For X>Nθ, we have(5)EN,X=oX.We shall prove Theorem1 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 [3] to deal with the equation x13+y13+y23+z1=x23+y33+y43+z2 (see Lemma 3). We also make use of the result of Zhao [2] on the 10th moment of the Weyl sums over cubes of primes restricted on minor arcs.Notation. Throughout the paper, ɛ denotes a sufficiently small positive number, and let A,c denote a positive constant. We need to point out that A, c and ɛ are allowed to change at different occurrences. With or without subscript, p denotes a prime number. Denote by dn the number of divisors of n, and as usual, we write ex for e2πix.


## 2. Preliminaries

Before we prove Theorem1, we introduce the following theorem.Theorem 2.
LetE=EN,N5/6 be defined in Section 1. Then, we have(6)E≪N1/36+ɛ.We can get Theorems1 from 2 immediately. Therefore, our task is to prove Theorem 2.Let(7)U=N161/3.Write(8)gα=∑U≤p≤2Ulogpep3α,fα=∑U5/6≤p≤2U5/6logpep3α.LetRn be the weighted number of solutions of n=p13+⋯+p63+p73+p83 with U≤p1,…,p6≤2U and U5/6≤p7,p8≤2U5/6. By orthogonality, one has(9)Rn=∫01gα6fα2e−nαdα.We write(10)P=U2/5,Q=U3P−1.We define the set of major arcsM as the union of the intervals(11)Mq⋅a=α∈Q−1,1+Q−1:qα−a≤Q−1,with 1≤a≤q≤P and a,q=1. We then denote the corresponding set of minor arcs by m=Q−1,1+Q−1\M. So, we can get the lemma.Lemma 1.
Forα∈M, we have(12)∫Mgα6fα2e−nαdα≫U14/3.We will prove Lemma1 in Section 4.Lemma 2.
Forα∈m, we have(13)∫mgα10dα≪U7−1/4+ϵ.We can find the proof in ([2], Section 9).Lemma 3.
LetS be the number of solutions of(14)x13+y13+y23+z1=x23+y33+y43+z2,with U≤xi≤2U, U5/6≤yi≤2U5/6, and zi∈ξN,N5/6. We have(15)S≪U1+ɛU5/3E+U55/36E2.We will prove Lemma3 in Section 3.Proof of Theorem 2.
On recalling the setξN,N5/6 defined in Section 1 and by means of an argument of Wooley (see [4]), we have(16)∑n∈ξN,N5/6∫01gα6fα2e−nαdα=0.
Let(17)RM=∑n∈ξN,N5/6∫Mgα6fα2e−nαdα,Rm=∫mgα6fα2h−αdα,where(18)hα=∑n∈ξN,N5/6enα.
So, one has(19)Rm=RM.
With the help of Lemma1, we can get(20)Rm≫U14/3E.
With an application of the Cauchy–Schwarz inequality, one has(21)∫mgα6fα2h−αdα≤∫mgα10dα1/2∫01gαfα2h−α2dα1/2.
By Lemma2, we have(22)∫mgα10dα≪U7−1/4+ϵ.
By considering the underlying Diophantine equations and Lemma3, we get(23)∫01gαfα2h−α2dα≪U1+ɛU5/3E+U55/36E2.
From (20) to (23), we have(24)U14/3E≪U4−1/8+5/6+ɛE1/2+U4−1/8+55/72+ɛE.
Therefore, we conclude that(25)E≪U1/12+ɛ≪N1/36+ɛ.
This completes the proof of Theorem 2.


## 3. Proof of Lemma 3

We writeS1 as the number of solutions of (14) with x13=x23, and let S2 denote the number of solutions of (14) with x13≠x23. It is easy to find that(26)S=S1+S2.Lemma 4.
We denote byTk the number of solutions of the Diophantine equation(27)y13+y23−y33−y43=k,with 1≤yi≤U5/6 and k≠0. Then, we have(28)Tk≪kɛU36/55+ɛ.The proof of Lemma1 can follow from an argument of Hooley [5] (see the proof of Lemma 1 of Parsell [6] for a sketch of the necessary adjustments to Hooley’s argument).To calculateS2, by symmetry, we can assume that x2>x1. Write x2=x1+h. Then, (14) becomes(29)h3x12+3x1h+h2=y13+y23−y33−y43+z1−z2.Sincey13+y23≤16U5/2, z1−z2≤16U5/2, and x12≥U2, it follows that h<11U1/2. Let(30)ghα=∑U≤x≤2Ueαh3x2+3xh+h2,Gα=∑h<11U1/2ghα.Then,(31)S2≤∫01GαFα4hα2dα,where hα is defined in (18), and(32)Fα=∑U5/6≤y≤2U5/6ey3α.LetNq,a denote the interval a/q−q−1U−3/2,a/q+q−1U−3/2 and U=U−3/2,1+U−3/2. We may suppose that U≥4. Then, Nq,a with 1≤a≤q≤U, a,q=1 contained in U. Let N denote the union of Nq,a with 1≤a≤q≤U, a,q=1, and let n=U\N. Then, we have(33)S2≤∫NGαFα4hα2dα+∫nGαFα4hα2dα.By the same method of the lemma of Vaughan [3], we can get the following lemma.Lemma 5.
Suppose thatα−a/q≤q−2 and a,q=1, then(34)Gα≪UɛU3/2q1/2+U+U1/4q1/2.Moreover, we can get the following lemma.Lemma 6.
LetN be denoted as above. Then, we have(35)∫NGαFα4dα≪U5/2+ɛ.
The proof can be also found in Vaughan [3].Proof of Lemma 3.
Wheny13+y23+z1=y33+y43+z2, the number R of solutions of it satisfies(36)R=∫01Fα4hα2dα.
Given any one of theOE2 possible choices for z1 and z2 with z1≠z2, it follows from Lemma 4 that the number of permissible choices for yi is OU55/36+ɛ, and thus the contribution arising from this class of solutions is OU55/36+ɛE2. When z1=z2, on the other hand, the variables yi satisfy the equation y13+y23=y33+y43; the number of the solutions is OU5/3+ɛ. Thus, we conclude that the number of solutions of this type is OU5/3+ɛE. So, we can get(37)R≪UɛU5/3E+U55/36E2.
Then, we have(38)S1≪U1+ɛU5/3E+U55/36E2.
We considern. By Dirichlet’s theorem on Diophantine approximation, given α∈n, we may choose a, q with a,q=1,α−a/q≤q−1U−3/2 and q≤U3/2. It is easily verified that 1≤a≤q. Moreover, since α∉N, we have q>U. Therefore, by Lemma 5, we have(39)Gα≪U1+ɛ.
Hence, we have(40)∫nGαFα4hα2dα≪U1+ɛ∫01Fα4hα2dα≪U1+ɛU5/3E+U55/36E2.
Whenα∈N, by Lemma 6, we can get(41)∫NGαFα4hα2dα≪E2∫NGαFα4dα≪U5/2+ɛE2.
By (40) and (41),(42)S2≪U1+ɛU5/3E+U55/36E2.
In view of (38) and (42), we complete the proof of Lemma 3.


## 4. Proof of Lemma 1

LetL=logUA for A>0. We write(43)P′=L,Q′=U3L−1.We define the set of major arcsM′ as the union of the intervals(44)M′q⋅a=α∈Q′−1,1+Q′−1:qα−a≤Q′−1,with 1≤a≤q≤P and a⋅q=1. It is easy to find that M′⊆M. We then denote the corresponding set of minor arcs by m′=Q′−1,1+Q′−1\M′.Lemma 7.
Suppose that the integern satisfies N/2≤n≤N and n∈N6. Then, one has(45)∫M′gα6e−nαdα≫U3.The proof can be found in ([3], Lemma 3.1).We define the multiplicative functionwq by taking(46)wp3u+v=3p−u−1/2,whenu≥0andv=1,p−u−1,whenu≥0andv=2,3.We have(47)q−1/2≤wq≪q−1/3.Therefore, we have the following result.Lemma 8.
Letc be a constant. When m≥2, one has(48)∑q≤mwq2dqc≪logmA,for some A constant.This is due to Zhao ([2], Lemma 1). Before we deal with the integrals on the minor arcs, we should give an upper bound of gα. We also quote the following estimate proved by Ren [7].Lemma 9.
Suppose thatα is a real number and that 1≤a≤q with a,q=1. Let β=α−a/q. Then, one can have(49)gα≪dqc0logUcU1/2V+U4/5+UV−1,where c0=1/2+log3/log2 and V=q1+U3α−a/q and c is a constant.Proof of Lemma 1.
Whenα∈M′, we have(50)∫M′gα6fαe−nαdα=∑U5/6≤p≤2U5/6logp∫M′gα6e−n−p3αdα.
Moreover, one hasn−p3∈N6 when n∈N7, and when N−2N5/6≤n≤N, we have N/2≤n−p3≤N, and by Lemma 7, we have(51)∫M′gα6fαe−nαdα≫U3∑U5/6≤p≤2U5/6logp≫U23/6.
By same measure, whenN−N5/6≤n≤N and n∈N8, we can get(52)∫M′gα6fα2e−nαdα≫U14/3.
Whenα∈m′∩M, by Lemma 9, we have(53)gα≪U4/5+Uq1+U3α−a/qdcqlogUc≪dqclogUcUq1+U3α−a/q.
Therefore, we have(54)∫m′∩Mgα6fα2dα≪U4L−2∫m′∩MdqclogUcU2fα2q1+U3α−a/qdα.
Letα=a/q+β. We have(55)∑1≤a≤qfaq+β2=q∑p13≡p23modqU5/6≤p1,p2≤2U5/6ep13−p23β.
Sinceq<U5/6, we have(56)∑1≤a≤qfaq+β2≪U5/3q−1∑b13≡b23modq1≤b1,b2<qb1b2,q=11≪U5/3∑b3≡1modq1≤b<q1≪U5/3dqc′.
Therefore, by Lemma8, we obtain(57)∫m′∩Mgα6fα2dα≪U6+5/3L−2∑q≤U2/5wq2dqc+c′∫β<U3−2/511+βU3dβ≪U14/3L−2+ɛlogUA,for some absolute constant A.
In view of (52) and (57), we complete the proof of Lemma 1.



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