# Waring–Goldbach Problem of Even Powers in Short Intervals

Authors: Liqun Hu; Tanhui Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we study the average behaviour of the representations ofn=p12+p24+p34+p4k over short intervals for k≥4, where p1,p2,p3,p4 are prime numbers. This improves the previous results.


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## Body

## 1. Introduction

Letn and ki,i=1,…,r be integers; the classical Waring–Goldbach problem considers the representation of n in the form(1)n=p1k1+p2k2+⋯+prkr.This problem has attracted the attention of many authors [1–11]. For sufficiently large integer N, let Ek,N be the exceptional set of expressing positive even integers n up to N in the form(2)n=p12+p24+p34+p4k.Recently, Feng and Ma [2] established that Ek,N≪N1−θk+ε, here(3)θk=132,k=4,148,k=5,164,6≤k≤8,148x,k≥9,where(4)x=k4+1−k42k4−1,9≤k≤19,k4−12k4k4+1,k≥20.Let(5)rkn=∑n=p12+p24+p34+p4klogp1logp2logp3logp4.In our paper, we use the Hardy–Littlewood circle method and some techniques in the works of Languasco and Zaccagnini [3–8] to improve the results of Feng and Ma by studying the average behaviour of rkn over short intervals N,N+P, P=oN.Theorem 1.
LetNbe a sufficiently large integer and1≤P≤N,k≥4be integers. Then, there existsC=Cε≥0for everyε>0, such that(6)∑n=N+1N+Prkn=Γ1/224PN1/k+OkPN1/kexp−CLlogL1/3,as N⟶∞, uniformly for N1−5/6k+ε≤P≤N1−ε.Theorem1 implies that for a sufficiently large integer N, every interval N,N+P contains the integers which are sums of one prime square, two prime quartics, and one prime power in short intervals, where N1−5/6k+ε≤P≤N1−ε for k≥4. This improves Feng and Ma’s results obviously. For example, when k=4, the exceptional set E4,N≪N1−5/24+ε in Theorem 1 is smaller than N1−1/32+ε in Feng and Ma’s results. Comparing the methods in Feng and Ma’s proof process, we replace finite sums with infinite series over primes and get a better result.


## 2. Preliminaries and Lemmas

LetN be a sufficiently large integer, l≥1 be an integer, z=1/N−2πiα, αε−1/2,1/2, L=logN, eα=e2πiα,(7)Slα=∑n=1∞e−nl/NenlαΛn,and(8)Vlα=∑p=2∞e−pl/Neplαlogp.We have(9)z−1≪minN,α−1.We also set(10)Uα,P=∑m=1Pemα.From Montgomery ([11], p. 39), we can find that(11)Uα,P≤minP;α−1.Now, we need some lemmas as follows.Lemma 1 (see [7], Lemma 3).
Letμ>0,z=1/N−2πiα. Then, we have(12)∫−1/21/2z−μe−nαdα=1Γμe−n/Nnμ−1+Oμ1n,uniformly for n≥1.Lemma 2 (see [7], Lemma 1).
Slα−Vlα≪lN1/2l.Lemma 3 (see [7], Lemma 4).
Letε>0be an arbitrarily small positive constant. Then, we have(13)∫−ξξSlα−Γ1/llz1/l2dα≪lN2/l−1exp−c1LlogL1/3,uniformly for 0≤ξ≤N−1+5/6l−ε. Here, c1=c1ε is a positive constant, which does not depend on l.Lemma 4.
Letε>0be an arbitrarily small positive constant andτ>0be a positive constant. Then,(14)∫−ττSlα2dα≪lτN1/l+N2/l−1L3,∫−ττSlα4dα≪lτN2/l+N4/l−1Nε,∫−ττVlα4dα≪lτN2/l+N4/l−1Nε.Proof.
This is Lemma 3.4 in [1] and Lemma 6 in [7].Lemma 5.
Letε>0be a arbitrarily small positive constant,c≥1be a fixed positive constant andN−c≤ω≤N2/l−1. LetIω≔−1/2,−ω∪ω,1/2. Then,(15)∫IwSlα2dαα≪lN2/l−1+εωL3+N1/lL4,∫IwSlα4dαα≪lN4/l−1+εω,∫IwVlα4dαα≪lN4/l−1+εω.Proof.
This is Lemma 3.5 in [1] and Lemma 7 in [7].Lemma 6.
Letc2be a fixed positive constant; we have(16)∫−1/21/2S2αS4α22dα≪NLc2.Proof.
See the proof of Lemma 5 in [7] and Lemma 4.3 in [12] for k=4.


## 3. Proof of Theorem1

Letε>0 and P≥AN1−5/6k+ε for k≥4, where A=N2ε. From equation (5), we can get(17)∑n=N+1N+Pe−n/Nrkn=∫−1/21/2V2αV4α2VkαU−α,Pe−Nαdα.We find it also convenient to set(18)Elα=Slα−Γ1/llz1/l.LetIA,P=−1/2,−A/P∪A/P,1/2; we can get(19)∑n=N+1N+Pe−n/Nrkn=∫−A/PA/PΓ1/kΓ1/224kz1+1/kU−α,Pe−Nαdα+∫−A/PA/PΓ1/kkz1/kS2αS4α2−Γ1/22z1/22U−α,Pe−Nαdα+∫−A/PA/PEkαS2αS4α2U−α,Pe−Nαdα+∫−1/21/2VkαV2αV4α2−S2αS4α2U−α,Pe−Nαdα+∫−1/21/2S2αS4α2Vkα−SkαU−α,Pe−Nαdα+∫IA,PSkαS2αS4α2U−α,Pe−Nαdα≔ℱ1+ℱ2+ℱ3+ℱ4+ℱ5+ℱ6.Now, we begin to estimate these terms. First, we estimateℱ1. By the approximation e−n/N=e−1+OPN−1, Lemma 1 , and equations (9) and (11), we can get(20)ℱ1=Γ1/kΓ1/224kΓ1+1/k∑n=N+1N+Pn1/ke−n/N+OkPN+Ok∫A/P1/2dαα2+1/k=Γ1/224ePN1/k+OkP2N1/k−1+PA1+1/k+N1/k.Next, we estimateℱ2.(21)∫−A/PA/PΓ1/kkz1/kS2αS4α2−Γ1/22z1/22U−α,Pe−Nαdα=∫−A/PA/PΓ1/kkz1/kS2α2−Γ1/224zU−α,Pe−Nαdα+∫−A/PA/PΓ1/kkz1/kS2αS4α2−S2α2U−α,Pe−Nαdα=Ε1+Ε2.FromS2α≪N1/2 and f2−g2=2ff−g−f−g2, we can get(22)Ε1≪kN1/2∫−A/PA/PE2αU−α,Pz1/kdα+∫−A/PA/PE2α2U−α,Pz1/kdα≔Ε11+Ε12.By Lemma3 and equation (11), for every ε>0, then there exists c1=c1ε>0 such that(23)Ε12≪kPN1/k∫−A/PA/PE2α2dα≪kPN1/kexp−c1LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4. By Lemma 3, equation (11), and the Cauchy–Schwarz inequality, we obtain that(24)E11≪kN1/2∫−A/PA/P|E2α|2dα1/2P∫−A/PA/Pdαz2/k1/2≪kPN1/2exp−c12LlogL1/3∫−1/N1/NN2/kdα+∫1/NA/Pdαα2/k1/2≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4. Hence, by equations (23) and (24), for every ε>0, then there exists c1=c1ε>0 such that(25)Ε1≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4.(26)Ε2=∫−A/PA/PΓ1/kkz1/kS2αS4α2−S2α2U−α,Pe−Nαdα=Ε21+Ε22.By Lemma4, equation (11), and Skα≪kN1/k, we have(27)Ε21≪kN1/k+1/2∫−A/PA/PS4α2U−α,Pdα≪kN1/k+1/2∫−A/PA/PS4α4dα1/2∫−1/P1/PP2dα+∫1/PA/Pα−2dα1/2≪kN1/k+1/2APN1/2+ε1/2P+PA1/2≪kA1/2N1/k+3/4+ε,and(28)Ε22≪kN1/k∫−A/PA/PS2α2U−α,Pdα≪kPN1/k∫−A/PA/PS2α4dα1/2∫−A/PA/Pdα1/2≪kPN1/kAPN1+ε1/2AP1/2≪kAN1/k+1/2+ε.Hence, by equations (27) and (28), we have(29)Ε2≪kAN1/k+1/2+ε+A1/2N1/k+3/4+ε≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4. Thus, from equations (25) and (29), for every ε>0, there exists c1=c1ε>0 such that(30)ℱ2≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4Now, we estimateℱ3. By equations (11) and (19), Lemmas 3 and 6, and the Cauchy–Schwarz inequality, we obtain that for every ε>0, there exists c1=c1ε>0 such that(31)ℱ3≪kP∫−1/21/2S2αS4α22dα1/2∫−A/PA/PEkα2dα1/2≪kPN1/2Lc2/2∫−A/PA/PEkα2dα1/2≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4.Next, we estimateℱ4.(32)ℱ4=∫−1/21/2VkαV2αV4α2−S2αS4α2U−α,Pe−Nαdα=∫−1/21/2VkαV4α2V2α−S2αU−α,Pe−Nαdα+∫−1/21/2VkαS2αV4α2−S4α2U−α,Pe−Nαdα≔U1+U2.By Lemma2 and Vkα≪kN1/k, we obtain that(33)U2=∫−1/21/2VkαV4α−S4αS2αV4α+S4αU−α,Pe−Nαdα≪kN1/k+1/8∫−1/21/2S2αV4α+S2αS4αU−α,Pe−Nαdα≔N1/k+1/8J1+J2.By Lemmas4 and 5, equation (11), and the Cauchy–Schwarz inequality, we have(34)J1=∫−1/21/2S2αV4αU−α,Pe−Nαdα≪P∫−1/P1/PS2αV4αdα+∫I1,PS2αV4ααdα≪P∫−1/P1/PS2α2dα1/2∫−1/P1/PV4α2dα1/2+∫I1,PS2α2αdα1/2∫I1,PV4α2αdα1/2≪P1PN1/2+11/2L3/2∫−1/P1/PV4α4dα1/4∫−1/P1/Pdα1/4+N1/2L4+PL31/2∫I1,PV4α4αdα1/4∫I1,P1αdα1/4≪P3/4Nε/4L3/2+N1/4+ε/4P1/4L9/4≪P3/4Nε/4L3/2,provided that P≥AN1−5/6k+ε for k≥4.Similarly, we also have(35)J2=∫−1/21/2S2αS4αU−α,Pe−Nαdα≪P∫−1/P1/PS2αS4αdα+∫I1,PS2αS4ααdα≪P∫−1/P1/PS2α2dα1/2∫−1/P1/PS4α2dα1/2+∫I1,PS2α2αdα1/2∫I1,PS4α2αdα1/2≪P3/4Nε/4L3/2,provided that P≥AN1−5/6k+ε for k≥4.Hence, by equations (34) and (35), we have(36)U2≪kP3/4N1/k+1/8+ε,provided that P≥AN1−5/6k+ε for k≥4.By Lemma2, equation (11), and Vkα≪kN1/k, we have(37)U1=∫−1/21/2VkαV4α2V2α−S2αU−α,Pe−Nαdα≪kN1/k+1/4∫−1/21/2V4α2U−α,Pdα≪kN1/k+1/4P∫−1/P1/PV4α2dα+∫I1,PV4α2αdα≔N1/k+1/4ℳ1+ℳ2.By Lemmas4 and 5 and the Cauchy–Schwarz inequality, we have(38)ℳ1≪P∫−1/P1/PV4α4dα1/2∫−1/P1/Pdα1/2≪P1/2Nε/2,and(39)ℳ2≪∫I1,PVkα4αdα1/2∫I1,P1αdα1/2≪P1/2Nε,provided that P≥AN1−5/6k+ε for k≥4. Then, we have(40)U1≪kP1/2N1/k+1/4+ε,provided that P≥AN1−5/6k+ε for k≥4.Hence, by equations (36) and (40), we have(41)ℱ4≪kP1/2N1/k+1/4+ε+P3/4N1/k+1/8+ε≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4.Now, we estimateℱ5. By Lemmas 4 and 5, equation (11), and a partial integration, we have(42)ℱ5≪kPN1/2k+1/2∫−1/P1/PS4α2dα+N1/2k+1/2∫I1,PS4α2αdα≪kPN1/2k+1/2N1/4P+N−1/2L3+PNεN1/2k+1/2≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4.Finally, we estimateℱ6. Clearly, by Lemma 5, equation (11), and Cauchy–Schwarz inequality, we have(43)ℱ6≪kN1/k∫IA,PS2α2αdα1/2∫IA,PS4α4αdα1/2≪kN1/kN1/2L4+PAL31/2NεPA1/2≪kN1/kN1/2+εL4PA+PA2NεL31/2≪kPAN1/k+ε/2L2/3≪kPN1/kexp−c12LlogL1/3,provided that P≥AN1−5/6k+ε for k≥4.Completion of the proof. Letk≥4. By equations (19)–(37) and (40)–(43), we obtain that, for every ε>0, there exists C=Cε>0, such that(44)∑n=N+1N+Pe−n/Nrkn=Γ1/424ePN1/k+OkPN1/kexp−CLlogL1/3,provided that P≥AN1−5/6k+ε with A=N2ε. From e−n/N=e−1+OPN−1, for n∈N+1,N+P,1≤P≤N and every ε>0, there exists C=Cε>0, such that(45)e−1∑n=N+1N+Prkn=Γ1/224ePN1/k+OkPN1/kexp−CLlogL1/3+OkPN∑n=N+1N+Prkn,provided that N1−5/6k+ε≤P≤N1−ε for k≥4. Using e≤en/N≤e2 and equation (44), the last error term is ≪kP2N1/k−1. Thus, we have(46)∑n=N+1N+Prkn=Γ1/224PN1/k+OkPN1/kexp−CLlogL1/3,uniformly for N1−5/6k+ε≤P≤N1−ε. Now, Theorem 1 follows.



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*Source: 9693218-2021-12-16.xml*