# An Improved Asymptotic on the Representations of Integers as Sums of Products

Authors: Wenjia Zhao
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we improve the error terms of Chace’s results in the study by Chace (1994) on the number of ways of writing an integerN as a sum of k products of l factors, valid for k≥3 and l=2, 3. More precisely, for l=2, 3, we improve the upper bound Nk−1−2k−2/k−1l+1+ε, k≥3 for the error term, to N2−2/2l+1+ε when k=3 and Nk−1−4k−2/l+1k+l−2+ε when k≥4.


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## Body

## 1. Introduction

In this paper, we study the number of representations of a natural numberN as a sum of k terms, each being a product of l factors. Let νN;k,l denote this number. This problem was studied by Estermann [1, 2] in the case k=2 or 3 and l=2 by using some properties of Dirichlet L-function. His method is not easy to be generalized. Later, Chace [3] generalized Estermann’s result to k≥3 and l≥2. For such k and l, he got(1)νN;k,l=μN;k,l+EN;k,l,where(2)μN;k,l≍Nk−1logkl−1N,and(3)EN;k,l≪εNk−1−2k−2/k−1l+1+ε,ifk=3orl=2,l=3,Nk−1−3k−2/k−1l+2+ε,ifkandl≥4.Here,μN;k,l is defined in (72). The way he studied the problem is different from Estermann. The main tools he used are the Hardy–Littlewood method and some results from the divisor problem in arithmetic progressions. He got the main term μN;k,l which is a sum of terms of the form Sℐ, where S are the “singular series” and ℐ are the “singular integrals.” They occur in the applications of the Hardy–Littlewood method. In this paper, we improve Chace’s result in the cases k≥3 and l=2,3. We get the following result.Theorem 1.
Supposek≥3 and l=2,3. Then,(4)νN;k,l=μN;k,l+EN;k,l,where μN;k,l defined by (72) satisfies (2) and the error term satisfies(5)EN;k,l≪εN2−2/2l+1+ε,ifk=3,Nk−1−4k−2/l+1k+l−2+ε,ifk≥4.We can compare the exponents in (3) with our results (5). For l=2,3, we see that 2−2/2l+1<k−1−2k−2/k−1l+1=2−1/l+1 when k=3 and k−1−4k−2/l+1k+l−2<k−1−3k−2/k−1l+2 when k≥4. Our error terms are better than Chace’s.The proof of the theorem is an application of the Hardy–Littlewood method (cf, Chapter 3 of [4]), the Voronoi summation formula, and some results from the Kloosterman sums. The estimates on the minor arc were studied by Chace, and his result is sufficient for us. Hence, we will not focus on the minor arc in this paper. The main difficulty arises in treating the error term of the major arcs. In Section 2, we make some preparations for our proof. We state the Voronoi summation formula and give some lemmas related to the Kloosterman sums in this section. In Sections 3.1 and 3.2, we obtain a bound for the contribution from the minor arcs. In §3.3, we prove our theorem.


## 2. Applications of the Voronoi Summation Formula

LetX>0. Suppose that fx is a smooth function compactly supported in X−Xδ,2X+Xδ with 0<δ<1, satisfying fx=1 in X,2X and(6)fjx≪X−δj,j≥0,(7)∫fjxdx≪X−δj−1,j≥1.Let(8)Fs=∫0∞fxxs−1dx,be the Mellin transformation of fx. To state Voronoi’s summation formula, we introduce the notations below. For ℜs>1, let(9)Els,hq=∑n=1∞dlnenh/qns,where(10)dln=∑n1⋯nl=n1.We define(11)Al±n,aq=12∑n1⋯nl=nIn1,…,nl,a;q±In1,…,nl,a;q,(12)Ul±x=12πi∫ℜs=σγl±sdsxs,where(13)In1,…,nl,a;q=∑x1,…,xlmodqen1x1+⋯+nlxl+ax1⋯xlq,(14)γl±s=Γls/2+1±−1/4Γl1−s/2+1±−1/4,for x>0 and 0<ℜs<1/2−1/l. From Theorem 2 in [5], we know the following lemma.Lemma 1 (Voronoi’s summation formula).
With the above notations, we have, forl≥2,(15)∑n=1∞fndlnenhq=Ress=1FsEls,hq+πl/2ql∑n=1∞Al+n,hq∫0∞fxUl+πlnxqldx+i3lπl/2ql∑n=1∞Al−n,hq∫0∞fxUl−πlnxqldx.
To simplify the integrals in (15), we define(16)Gl±x=12πi∫ℜs=1−σFsγl±sxsds,where σ>0. By (12), it is clear that(17)∫0∞fxUl±xydx=12πi∫0∞∫ℜs=σfxγl±sdsxys,=−12πi∫ℜs=1−σ∫0∞fxxs−1dxγl±1−sys−1ds,=−12πi∫ℜs=1−σFsγl±sys−1ds,=−1yGl±y.
To use Lemma1 in practice, we need to estimate the residue in (15), Gl±x and Al±n,h/q. We first compute the residue. For 0≤j≤l−1, we define(18)Ajq=∑b=1qeabqcj+1b,q,where a>0 is an integer, and the coefficients cjb,q are sums of terms of the form(19)∑b1b2≡bmodqfb1,for some function f. The coefficients cjb,q are given explicitly in equation (2.13) in [6]. It has been shown in [6] that for a,q=1, Ajq is independent of a.Lemma 2.
Suppose1≤a≤q, a,q=1. We have(20)Ress=1FsEls,aq=∑n=0l−1Anq∫0∞fxlognxn!dx.Proof.
By (9), we can rewrite Els,a/q as(21)Els,aq=∑hmodqeahqZs;h,q,where(22)Zs;h,q=∑n≥1n=hmodqdlnns,for ℜs>1. Therefore,(23)Ress=1FsEls,aq=∑h=1qeahq∫0∞fxdRess=1Zs;h,qxss.
Chace (Theorem1 and (2.1) in [6]) gave that(24)Ress=1Zs;h,qxss=∑n=0l−1cn+1h,qLnx,where(25)Lnx=x∑j=0n−1n−jlogjxj!.
Substituting this into (23), we have(26)Ress=1FsEls,aq=∑n=0l−1Anq∫0∞fslognxn!dx.
The proof of the lemma is complete.
Our next lemma, proved by Jiang and Lü (Lemma 2.7 in [7]), is to evaluate the integrals in (15).Lemma 3.
LetGl±x be defined as in (16). Then, we have, for l≥2,(27)Gl±x≪X−A,ifx>Xl1−δ−1+ε,xXl−1/2l,ifX−1≪x≤Xl1−δ−1+ε,xX1/2X1−δε,ifx≪X−1.Proof.
This can be proved by takingJ=X1−δ with 0<δ<1 in Lemma 2.7 of [7].
The following two lemmas give estimates forAl±n,a/q. This two lemmas play an important role in our proof. We use some results for the Kloosterman sum to obtain the power saving in the q aspect.Lemma 4.
Suppose1≤a≤q, a,q=1, and let Al±n,a/q be defined as in (11). We have(28)∑amodqa,q=1Al±n,aqe−aNq≪εql+1/2+εnε.Proof.
By (13), we obtain(29)∑amodqa,q=1Im1,…,ml,a;qe−aNq=∑x1,…,xlmodqem1x1+⋯mlxlq∑amodqa,q=1eax1⋯xl−aNq,=∑d|qdμqd∑x1,…,xlmodqx1⋯xl=Nmoddem1x1+⋯mlxlq,=∑d|qq/d|mdμqdqdlSldqm;N,d,where(30)Slm;N,d≔∑x1,…,xlmoddx1⋯xl=Nmoddem1x1+⋯+mlxld,for m=m1,…,ml∈ℤl. Here, the notation d|m means that d divides each component of m. Then, we obtain that the left hand side of the desired equation in our lemma equals to(31)12∑m1⋯ml=n∑d|qq/d|mdqdlμqdSlmq/d;N,d+Slmq/d;−N,d=12∑d|qdqdlμqd∑m1⋯ml=nd/qlSlm;N,d+Slm;−N,d.
Now, we define(32)Kl−1m;w=∑x1,…,xl−1modwxj,w=1,1≤j≤l−1em1x1+⋯+ml−1xl−1+mlx1⋯xl−1¯w,for m=m1,…,ml∈ℤl and w∈ℤ. Here, x¯ denotes the multiplicative inverse of x modulo w. In Section 6 in [8], Smith gave that(33)∑m1⋯ml=nSlm;N,d=∑b|dbl|nbl−1dln/blKl−1eNn/bl;d/b.,where et=1,…,1,t∈ℤl for all t∈ℤ. Therefore, (31) can be written as(34)12∑q=q1d1bbl|n/q1ld1q1lblμq1dlnq1lblKl−1eNnq1lbl;d1+Kl−1e−Nnq1lbl;d1.
Taking the change of variables thatm=q1b, we get(35)∑amodqa,q=1Al±n,aqe−aNq=q2dlnKl−1eNn;q±Kl−1e−Nn;q.
From Theorem 6 in [9], we know that(36)Kl−1et;q≤ql−1/2dlq.
The lemma follows immediately.Lemma 5.
Suppose1≤a≤q, a,q=1, and l=2,3. Let Al±n,a/q be defined as before. Then, we have(37)Al±n,aq≪εql/2+εnεn,q1/2.Proof.
By (13), for l=2, we have(38)In1,n2,a;q=∑x1,x2modqen1x1+n2x2+ax1x2q=∑x1modqen1x1q∑x2modqen2+ax1x2q=q∑x1modqax1=−n2modqen1x1q≪q.
Forl=3, one has(39)In1,n2,n3,a;q=q∑d|n2,n3,qdSn1,−n2n3a¯d;qd.
Now, if we use Weil’s classical bound(40)Sm,n;c≤m,n,c1/2d2cc1/2,then it follows from (39) that(41)In1,n2,n3,a;q≪εq3/2+ε∑d|n2,n3,qn1d,n2n3a¯,q1/2.
Ford|n2,n3,q and a,q=1, noting that n=n1n2n3 by (11), we have n1d,n2n3a¯,q≤n1n2n3a¯,q=n,q. Therefore,(42)In1,n2,n3,a;q≪εq3/2+εn,q1/2.
Hence, we have(43)Al±n,aq≪εql/2+εn,q1/2∑n1n2n3=n1≪εql/2+εnεn,q1/2,for l=2,3. The proof of the lemma is complete.


## 3. Proof of Theorem1

In this section, we prove Theorem1. The main difficulty arises in treating the major arcs. The results from Kloosterman’s sums will play a role in our proof. Throughout this section, 1≤X≤N/2. We choose a smooth function f compactly supported in X−Xδ,2X+Xδ with 0<δ<1, satisfying fx=1 in X,2X and (6) and (7). To apply the circle method, we choose the parameters P and Q such that(44)PQ=N,P≤N1/l.By Dirichlet’s lemma on rational approximations, eachα∈I≔1/Q,1+1/Q may be written in the form as follows:(45)α=aq+β,β<1qQ,for some integers a and q with 1≤a≤q≤Q and a,q=1. We denote by Mq,a the set of α satisfying (45) and define the major arc by(46)M=∪1≤q≤P∪1≤a≤qa,q=1Ma,q.It follows fromP≤N1/l<Q/2 that the major arcs Mq,a are disjoint. The union m of minor arcs is just the complement of M in I. We denote(47)Slα=∑n≤Ndlnenα.Then, we have(48)νN;k,l=∫ISlkαe−Nαdα,=∫MSlkαe−Nαdα+∫mSlkαe−Nαdα.The upper bound of the integral ofSlkα on m is given by Chace [3].Lemma 6.
Supposek≥3 and l≥2 are integers. Then,(49)∫mSlkαe−Nαdα≪εNk−1+εP−k+2.This estimate on the minor arc is sufficient for us. It remains to consider the integral ofSlα on the major arc. Let(50)Slα,X=∑X<n≤2Xdlnenα.Now, forα=a/q+β∈M, we have(51)Slα,X=∑n=1∞fβndlneanq+R1,lα,X,where fβx≔fxexβ and(52)R1,lα,X=−∑X−Xδ<n≤X+∑2X<n<2X+Xδfndlnenα≪εXδ+ε.It is clear thatfβx is a smooth function compactly supported in X−Xδ,2X+Xδ. Moreover, it satisfies(53)fβvx≪βX+X1−δXv,for any v≥0, and(54)∫fβvxdx≪∑0≤j≤vβv−j∫fjxdx,≪βvX+βv−1∑1≤j≤vX1−δβXj−1,≪βvX+βv−1+X−δv−1,≪1+βXβX+X1−δXv−1,for any v≥1. Now, by Lemma 1, we get(55)∑n=1∞fβndlneanq=Ress=1FβsElaq+R2,laq+β,X,where(56)Fβs=∫0∞fβxxs−1dx,is the Mellin transformation of fβx and(57)R2,laq+β,X≔πl/2ql∑n=1∞Al+n,aq∫0∞fxUl+πlnxqldx+i3lπl/2ql∑n=1∞Al−n,aq∫0∞fxUl−πlnxqldx.It follows from Lemma2 that(58)Ress=1FβsElaq=∑n=0l−1Anq∫0∞fβslognxn!dβ.From Equation (4.8) in [3], we know that(59)Ajq≪εq−1+ε.Then, we have(60)Ress=1FβsEls,aq=∑n=0l−1Anq∫X2Xeβxlognxn!dx+R3,lα,X,where(61)R3,lα,X≪εq−1+εXδ+ε.Now, forα=a/q+β∈M, by (51), (55), and (60), we write(62)Slα,X=Tlα,X+Rlα,X,where(63)Tlaq+β,X=∑n=0l−1Anq∫X2Xeβxlognxn!dx,(64)Rlα,X=R1,lα,X+R2,lα,X+R3,lα,X.Hence, by dyadic analysis, forα=a/q+β∈M, formally, we write(65)Slα=Tlα+Rlα,where(66)Rlα=∑j=1log2N−1∑t=13Rt,lα,N/2+O1,(67)Tlaq+β=∑n=0l−1AnqInβ,with(68)Inβ=∫1Neβxlognxn!dx,for n=0,1,…,l−1. Therefore, we have(69)∫MSlkαe−Nαdα=∑j=0kkjMj,lN,where(70)Mj,lN=∫MTlk−jαRljαe−Nαdα.It has been proved by Chace (Section 5 in [3]) that(71)M0,lN=μN;k,l+OεNk−1+εP−k+2,where(72)μN;k,l=∑q=1∞CqN∫−∞∞∑n=0l−1AnqInβke−Nβdβ,satisfying(73)μN;k,l≍Nk−1logkl−1N.In the following subsections, we will estimateMj,lN for 1≤j≤k.


### 3.1. The Estimate ofM1,lN

In this section, we give the upper bound ofM1,lN.Lemma 7.
Letl=2or3. We have(74)M1,lN≪εN1+1−δl−1/2+εPl−1/2+N1+δ+ε,ifk=3,Nk−2+1−δl−1/2+ε+Nk−2+δ+ε,ifk≥4.Proof.
By our definition of major arc (46), we have(75)M1,lN=∑q≤P∑amodqa,q=1∫β<1qQTlk−1aq+βRlaq+βe−Nαdα.
Because forα=a/q+β∈M, Tlα,N is independent of a, and we can interchange the order of summation over a and the integral and then take the summation of Rlα,Ne−Na/q over a first. Therefore,(76)M1,lN≪∑q≤P∫β<1qQTlaq+βk−1∑amodqa,q=1Rlaq+βe−aNqdα.
By (59) and (68), it is clear that(77)Tlaq+β≪εq−1+εNεminN,β−1.
By the definition ofRlα in (66), we have(78)∑amodqa,q=1Rt,laq+βe−aNq≪∑t=13max1≤X≤N/2∑amodqa,q=1Rlaq+β,Xe−aNq+q.
We deduce that(79)∑amodqa,q=1R1,laq+β,Xe−aNq=∑X−Xδ≤n≤Xor2X≤n≤2X+Xδdlnn∑amodqa,q=1ean−Nqenβ=−∑d|qdμqd∑X−Xδ≤n≤Xor2X≤n≤2X+Xδ,n≡Nmodddlnfnenβ≪εXδ+ε.
It is obvious from (61) that(80)∑amodqa,q=1R3,laq+β,Xe−aNq=≪εXδ+ε.
It remains to consider the summation ofR2,l. We write Gl,β± be Gl± with F replaced by Fβ in (16). Similar to the assertions of Lemma 3, we have(81)Gl,β±x≪X−Aifx>βX+X1−δl+εX−1,1+βXxXl−1/2lifX−1≪x≤βX+X1−δl+εX−1,1+βXxX1/2βX+X1−δεifx≪X−1.
Notice that(82)∫0∞fβxUl±xydx=−1yGβ,l±y.
Then, by Lemma4, we have(83)q−l∑n=1∞∑amodqa,q=1Al±n,aqe−aNq∫0∞fβxUl±πlnxqldx≪εq−l−1/2+ε∑n=1∞nεqlnGl,β±πlnql≪εq−l−1/2+ε∑n≤βX+X1−δl+εqlX−1nε1+βXqlnnXqll−1/2l+∑n≪qlX−1nεqln1+βXnXql1/2βX+X1−δε≪εql+1/2+εβX+X1−δl−1/2+ε1+βX.
Therefore, we get(84)∑amodqa,q=1R2,laq+β,Xe−aNq≪εql+1/2+εβX+X1−δl−1/2+ε1+βX.
Hence, by (79), (80), and (84), for l=2,3, one has(85)∑amodqa,q=1Rlaq+βe−aNq≪∑t=13max1≤X≤N/2∑amodqa,q=1Rt,laq+β,Xe−aNq≪εql+1/2+εβN+N1−δl−1/2+ε1+βN+Nδ+ε.
Substituting this and (77) into (76), we obtain(86)M1,lN≪∑q≤P∫β<1qQTlaq+βk−1∑amodqa,q=1Rlaq+βe−aNqdα≪∑q≤Pq−k+1+l+1/2+εNε∫β<1qQminN,β−1k−1βN+N1−δl−1/2+ε1+βXdβ+∑q≤Pq−k+1+ε∫β<1qQminN,β−1k−1Nδ+εdβ.
By some elementary calculations, we get the desired result.


### 3.2. Integral ofMj,lN, 2≤j≤k, on the Major Arcs

The way we treat forMj,lN for j≥2 is different from that of M1,lN. We get the following lemma.Lemma 8.
LetMj,lN and 2≤j≤k be defined as before. We have, for l=2,3,(87)∑j=2kkjMj,lN≪εPl+1N1−δl−1+ε+Pl+2Nε+PN2δ+ε,ifk=3,Nk−3+εPlN1−δl−1+Pl+1+Nk−3+2δ+ε,ifk≥4.Proof.
The strategy is similar to that of Section 5 in [3]. Note that(88)∑j=2kTlk−jαRljα≪Rlα2Slαk−2+Tlαk−2.
We obtain(89)∑j=2kkjMj,lN≪∫M∑j=2kTlk−jαRljαdα≪maxα∈MRlα2∫MSlαk−2+Tlαk−2dα.
Similar to (83), by (81) and Lemma 5, we have, for l=2,3 and α=a/q+β∈M,(90)R2,lα,X≪εql/2+εβX+Jl−1/2+ε1+βX.
Hence, by (52), (61), (64), (90), and (66), we have(91)maxα∈MRlα≪maxα∈Mmax1≤X≤N/2∑t=13Rt,lα,X+1≪εmaxa/q+β∈Mql/2+ε1+βNβN+N1−δl−1/2+ε+Nδ+ε≪εPl/2N1−δl−1/2+ε+Pl+1/2+ε+Nδ+ε.
Now, fork=3, by Cauchy’s inequality, the definition of major arc (46), and Parseval’s identity, we obtain(92)∫MSlαdα≪M1/2∫01Slα2dα1/2≪εPNε.
Fork≥4, we have(93)∫MSlαk−2dα≪maxα∈MSlαk−4∫01Slα2dα≪εNk−3+ε,by the fact that Slα≪εN1+ε. One then shows that(94)∫MTlαk−2dα≪εPNε,ifk=3,Nk−3+ε,ifk≥4,by using the definition of the major arc and (77). Combining these results, we complete the proof of the lemma.


### 3.3. Proof of Theorem1

By Lemma6 and (69)–(73), we have, for k≥3, l=2,3,(95)νN;k,l=μN;k,l+kM1,lN+∑j=2kkjMj,lN+OεNk−1+εP−k+2,where(96)μN;k,l=∑q=1∞CqN∫−∞∞∑n=0l−1AnqInβke−Nβdβ,satisfying(97)μN;k,l≍Nk−1logkl−1N.Now, by Lemmas7 and 8, taking(98)δ=2l−12l+1,ifk=3,l−1l+1,ifk≥4,andP=N2/2l+1,ifk=3,N4/l+1k+l−2,ifk≥4,we have, for l=2,3,(99)νN;k,l=μN;3,l+OεN2−2/2l+1+ε,ifk=3,μN;k,l+OεNk−1−4k−2/l+1k+l−2+ε,ifk≥4.We complete the proof of Theorem1.


## 3.1. The Estimate ofM1,lN

In this section, we give the upper bound ofM1,lN.Lemma 7.
Letl=2or3. We have(74)M1,lN≪εN1+1−δl−1/2+εPl−1/2+N1+δ+ε,ifk=3,Nk−2+1−δl−1/2+ε+Nk−2+δ+ε,ifk≥4.Proof.
By our definition of major arc (46), we have(75)M1,lN=∑q≤P∑amodqa,q=1∫β<1qQTlk−1aq+βRlaq+βe−Nαdα.
Because forα=a/q+β∈M, Tlα,N is independent of a, and we can interchange the order of summation over a and the integral and then take the summation of Rlα,Ne−Na/q over a first. Therefore,(76)M1,lN≪∑q≤P∫β<1qQTlaq+βk−1∑amodqa,q=1Rlaq+βe−aNqdα.
By (59) and (68), it is clear that(77)Tlaq+β≪εq−1+εNεminN,β−1.
By the definition ofRlα in (66), we have(78)∑amodqa,q=1Rt,laq+βe−aNq≪∑t=13max1≤X≤N/2∑amodqa,q=1Rlaq+β,Xe−aNq+q.
We deduce that(79)∑amodqa,q=1R1,laq+β,Xe−aNq=∑X−Xδ≤n≤Xor2X≤n≤2X+Xδdlnn∑amodqa,q=1ean−Nqenβ=−∑d|qdμqd∑X−Xδ≤n≤Xor2X≤n≤2X+Xδ,n≡Nmodddlnfnenβ≪εXδ+ε.
It is obvious from (61) that(80)∑amodqa,q=1R3,laq+β,Xe−aNq=≪εXδ+ε.
It remains to consider the summation ofR2,l. We write Gl,β± be Gl± with F replaced by Fβ in (16). Similar to the assertions of Lemma 3, we have(81)Gl,β±x≪X−Aifx>βX+X1−δl+εX−1,1+βXxXl−1/2lifX−1≪x≤βX+X1−δl+εX−1,1+βXxX1/2βX+X1−δεifx≪X−1.
Notice that(82)∫0∞fβxUl±xydx=−1yGβ,l±y.
Then, by Lemma4, we have(83)q−l∑n=1∞∑amodqa,q=1Al±n,aqe−aNq∫0∞fβxUl±πlnxqldx≪εq−l−1/2+ε∑n=1∞nεqlnGl,β±πlnql≪εq−l−1/2+ε∑n≤βX+X1−δl+εqlX−1nε1+βXqlnnXqll−1/2l+∑n≪qlX−1nεqln1+βXnXql1/2βX+X1−δε≪εql+1/2+εβX+X1−δl−1/2+ε1+βX.
Therefore, we get(84)∑amodqa,q=1R2,laq+β,Xe−aNq≪εql+1/2+εβX+X1−δl−1/2+ε1+βX.
Hence, by (79), (80), and (84), for l=2,3, one has(85)∑amodqa,q=1Rlaq+βe−aNq≪∑t=13max1≤X≤N/2∑amodqa,q=1Rt,laq+β,Xe−aNq≪εql+1/2+εβN+N1−δl−1/2+ε1+βN+Nδ+ε.
Substituting this and (77) into (76), we obtain(86)M1,lN≪∑q≤P∫β<1qQTlaq+βk−1∑amodqa,q=1Rlaq+βe−aNqdα≪∑q≤Pq−k+1+l+1/2+εNε∫β<1qQminN,β−1k−1βN+N1−δl−1/2+ε1+βXdβ+∑q≤Pq−k+1+ε∫β<1qQminN,β−1k−1Nδ+εdβ.
By some elementary calculations, we get the desired result.


## 3.2. Integral ofMj,lN, 2≤j≤k, on the Major Arcs

The way we treat forMj,lN for j≥2 is different from that of M1,lN. We get the following lemma.Lemma 8.
LetMj,lN and 2≤j≤k be defined as before. We have, for l=2,3,(87)∑j=2kkjMj,lN≪εPl+1N1−δl−1+ε+Pl+2Nε+PN2δ+ε,ifk=3,Nk−3+εPlN1−δl−1+Pl+1+Nk−3+2δ+ε,ifk≥4.Proof.
The strategy is similar to that of Section 5 in [3]. Note that(88)∑j=2kTlk−jαRljα≪Rlα2Slαk−2+Tlαk−2.
We obtain(89)∑j=2kkjMj,lN≪∫M∑j=2kTlk−jαRljαdα≪maxα∈MRlα2∫MSlαk−2+Tlαk−2dα.
Similar to (83), by (81) and Lemma 5, we have, for l=2,3 and α=a/q+β∈M,(90)R2,lα,X≪εql/2+εβX+Jl−1/2+ε1+βX.
Hence, by (52), (61), (64), (90), and (66), we have(91)maxα∈MRlα≪maxα∈Mmax1≤X≤N/2∑t=13Rt,lα,X+1≪εmaxa/q+β∈Mql/2+ε1+βNβN+N1−δl−1/2+ε+Nδ+ε≪εPl/2N1−δl−1/2+ε+Pl+1/2+ε+Nδ+ε.
Now, fork=3, by Cauchy’s inequality, the definition of major arc (46), and Parseval’s identity, we obtain(92)∫MSlαdα≪M1/2∫01Slα2dα1/2≪εPNε.
Fork≥4, we have(93)∫MSlαk−2dα≪maxα∈MSlαk−4∫01Slα2dα≪εNk−3+ε,by the fact that Slα≪εN1+ε. One then shows that(94)∫MTlαk−2dα≪εPNε,ifk=3,Nk−3+ε,ifk≥4,by using the definition of the major arc and (77). Combining these results, we complete the proof of the lemma.


## 3.3. Proof of Theorem1

By Lemma6 and (69)–(73), we have, for k≥3, l=2,3,(95)νN;k,l=μN;k,l+kM1,lN+∑j=2kkjMj,lN+OεNk−1+εP−k+2,where(96)μN;k,l=∑q=1∞CqN∫−∞∞∑n=0l−1AnqInβke−Nβdβ,satisfying(97)μN;k,l≍Nk−1logkl−1N.Now, by Lemmas7 and 8, taking(98)δ=2l−12l+1,ifk=3,l−1l+1,ifk≥4,andP=N2/2l+1,ifk=3,N4/l+1k+l−2,ifk≥4,we have, for l=2,3,(99)νN;k,l=μN;3,l+OεN2−2/2l+1+ε,ifk=3,μN;k,l+OεNk−1−4k−2/l+1k+l−2+ε,ifk≥4.We complete the proof of Theorem1.



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*Source: 2902015-2021-11-23.xml*