# Distribution ofαp2 Modulo One with Prime Variable p of a Special Form

Authors: Fei Xue; Jinjiang Li; Min Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

LetPr denote an almost-prime with at most r prime factors, counted according to multiplicity. In this paper, it is proved that, for α∈ℝ/ℚ,β∈ℝ, and 0<θ<10/1561, there exist infinitely many primes p, such that αp2+β<p−θ and p+2=P4, which constitutes an improvement upon the previous result.


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## Body

## 1. Introduction and Main Result

LetPr denote an almost-prime with at most r prime factors, counted according to multiplicity. The famous prime twins conjecture states that there exist infinitely many primes p such that p+2 is a prime too. Up to now, this conjecture is still open, but many approximations about this conjecture were established. One of the most interesting results is due to Chen [1], who showed, in 1973, that there exist infinitely many primes p such that p+2=P2.In 1981, Heath-Brown [2] showed that there exist infinitely many arithmetic progressions of four different terms, three of which are primes and the fourth is P2. In 2006, Green and Tao [3] established that there exist infinitely many arithmetic progressions consisting of three different primes p1<p2<p3 such that pj+2=P2 for each j=1,2,3. Later, in 2008, Green and Tao [4] showed that, for any k⩾3, there exist infinitely many arithmetic progressions consisting of k different primes p1<p2<⋯<pk such that pj+2=P2 for each j=1,2,…,k.Suppose that there is a problem including primes and letr⩾2 be an integer. Having in mind Chen’s result, one may consider the problem with primes p, such that p+2=Pr. Many authors investigated several kinds of problems of this type, such as Peneva and Tolev [5], Peneva [6], and Tolev [7–9].Letα be an irrational real number and x denote the distance from x to the nearest integer. Earlier work about the distribution of the fractional parts of the sequence αp was first considered by Vinogradov [10], who showed that, for any real number β, there are infinitely many primes p such that for θ=1/5−ε; then,(1)αp+β<p−θ,where ε denotes arbitrarily small positive number. After that, the first improvement on (1) was due to Vaughan [11], who obtained θ=1/4 in (1) and who also required an additional factor logp8 on the right-hand side of (1). Since then, many authors improved the upper bound of the exponent θ, such as Harman [12, 13], Jia [14, 15], and Heath-Brown and Jia [16]. So far, the best result is given by Matomäki [17] with θ=1/3−ε. Moreover, it seems very natural to consider the sequence αpk for k⩾2, where p denotes a prime variable. Also, many authors studied the fractional parts of the sequence αpk for k⩾2, such as Baker and Harman [18], Harman [19], and Wong [20].In 2010, Todorova and Tolev [21] considered the distribution of αp modulo one with primes of the form specified above and showed that, for θ=1/100, there are infinitely many solutions in primes p to (1) such that p+2=P4. Later, Matomäki [22] showed that this result actually holds with p+2=P2 and θ=1/1000. After that, Shi [23] continued to improve the result of Matomäki [22] and showed that there are infinitely many solutions in primes p to (1) such that p+2=P2 and θ=3/200.Moreover, for the casek=2, Shi and Wu [24] established the result that there exist infinitely many primes p, which satisfy αp2+β<p−θ, such that p+2=P4 and θ=2/375−ε.In this paper, we shall continue to improve the result of Shi and Wu [24] and establish the following theorem.Theorem 1.
Suppose thatα∈ℝ/ℚ,β∈ℝ, and 0<θ<10/1561. Then, there exist infinitely many primes p, which satisfy p+2=P4, such that(2)αp2+β<p−θ.Remark 1.
According to the work of Shi and Wu [24], our improvement comes from using the methods developed by Tolev [9] with more delicate iterative techniques and various bounds for exponential sums, combining with a version of Lemma 2.2 of [25], while the previous method, in dealing exponential sum, e.g., [24], is based on the traditional pattern of exponential sum estimates.


## 2. Notation

LetX be a sufficiently large real number. Set(3)δ=0.307708,ρ=0.23077,η=0.076928,κ=1.4999676,0<θ<101561.Also, we put(4)z=Xη,y=Xρ,D=Xδ,Δ=ΔX=X−θ,H=Δ−1log2X.Throughout this paper, we always denote primes byp and q. ε always denotes an arbitrary small positive constant, which may not be the same at different occurrences. As usual, we use Ωn,φn,μn,andΛn to denote the number of prime factors of n counted according to multiplicity, Euler’s function, Möbius’ function, and Mangold’s function, respectively. We denote by τkn the number of solutions of the equation m1m2…mk=n in natural variables m1,…,mk. Especially, we write τ2n=τn. Let m1,m2,…,mk and m1,m2,…,mk be the greatest common divisor and the least common multiple of m1,m2,…,mk, respectively. Also, we use x and x, respectively, to denote the integer part of x and the distance from x to the nearest integer. fx≪gx means that fx=Ogx; fx≍gx means that fx≪gx≪fx; ex=e2πix; ℒ=logX. Pr always denotes an almost-prime with at most r prime factors, counted according to multiplicity.


## 3. Preliminary Lemmas

Lemma 1.
LetM⩽N<N1⩽M1 and an be any complex numbers. Then, we have(5)∑N<n⩽N1an⩽∫−∞+∞Kθ∑M<m⩽M1ameθmdθ,where(6)Kθ=minM1−M+1,1πθ,1π2θ2,which satisfies(7)∫−∞+∞Kθdθ⩽3log2+M1−M.Proof.
See Lemma 2.2 of [25].Lemma 2.
Let3⩽u<v<w<X, and suppose that w−1/2∈ℕ and that w⩾4u2,X⩾64w2u,andv3⩾32X. Assume further that fn is a complex-valued function. Then, the sum(8)∑X/2<n⩽XΛnfn,can be decomposed into Olog10X sums, each of which is either of Type I:(9)∑M<m⩽M1am∑L<ℓ⩽L1fmℓ,with M<M1⩽2M,L<L1⩽2L,L⩾w, am≪mε,ML≍X, or of Type II:(10)∑M<m⩽M1am∑L<ℓ⩽L1bℓfmℓ,with M<M1⩽2M,L<L1⩽2L,u⩽L⩽v, am≪mε,bℓ≪ℓε,andML≍X.Proof.
See Lemma3 of [26].Lemma 3.
ForP⩾1, we have(11)∑1⩽n⩽Peαn⩽minP,12α.Proof.
See Lemma4 of Chapter VI of [27].Lemma 4.
Suppose thatY1,Y2,andα are real numbers with Y1⩾1andY2⩾1 and that α−a/q⩽q−2 with a,q=1. Then, we have(12)∑n⩽Y1minY1Y2n,1αn≪Y1Y21q+1Y2+qY1Y2log2Y1q.Proof.
See Lemma 2.2 of [28].


## 4. Proof of Theorem1

As shown in [21], we take a periodic function χt with period 1 such that(13)0<χt<1,if−Δ<t<Δ,χt=0,ifΔ⩽t⩽1−Δ,which has a Fourier series,(14)χt=Δ+∑k>0gkekt,with coefficients satisfying(15)g0=Δ,gk≪Δ,for allk,∑k>Hgk≪X−1.The existence of such a function is a consequence of a well-known lemma of Vinogradov. For instance, one can see Chapter I, §2 in [27]. Consider the sum(16)Γ≔ΓX=∑X/2<p⩽Xp+2,Pz=1χαp2+βWplogp,where(17)Pz=∏2<p⩽zp,(18)Wp=1−κ∑z<q⩽yq|p+21−logqlogy.LetΓ1 denote the sum of the terms of ΓX in which Wp>0. Then, we have(19)ΓX⩽Γ1.If we denote byΓ2 the sum of the terms of Γ1 in which μp+2=0, it is easy to see that(20)0⩽Γ2≪∑q⩾z∑n⩽Xn+2≡0modq2logn≪logX∑z⩽q⩽X+2Xq2+1≪X1+εz−1+X1/2+ε≪X1−η+ε.By noting the fact that the contribution of the terms (if such terms exist) inΓ1, for which X−2<p⩽X, is OlogX, we deduce that(21)Γ⩽Γ3+OX1−η+ε,where(22)Γ3=∑X/2<p⩽X−2Wp>0,μ2p+2=1p+2,Pz=1χαp2+βWplogp.On the one hand, if we assume that(23)ΓX≫ΔXlogX,then from (21), we obtain(24)Γ3≫ΔXlogX,and thus Γ3>0. Hence, there exists a prime p, which satisfies(25)X2<p⩽X−2,Wp>0,μ2p+2=1,p+2,Pz=1,and such that(26)χαp2+β>0.Combining (13), (25), and (26), we can see that this prime p satisfies(27)αp2+β<p−θ.On the other hand, by the properties of the weightsWp (for example, one can see Chapter 9 of [29]), it is easy to see that if p satisfies (25), then(28)Ωp+2=∑q>z,q|p+21<1κ+∑q>z,q|p+2logqlogy=1κ+logp+2logy⩽1κ+1ρ<5,which implies p+2=P4. Therefore, in order to prove Theorem 1, it is sufficient to show that there exists a sequence Xjj=1∞, which satisfies(29)limj⟶∞Xj=+∞,ΓXj≫ΔXjXjlogXj,j=1,2,3….By (16) and (18), we can write Γ as follows:(30)Γ=Ψ−κΦ,where(31)Ψ=∑X/2<p⩽Xp+2,Pz=1χαp2+βlogp,(32)Φ=∑X/2<p⩽Xp+2,Pz=1χαp2+βlogp∑z<q⩽yq|p+21−logqlogy.Next, we shall give lower bound estimate ofΨ and upper bound estimate of Φ by using lower bound linear sieve and upper bound linear sieve, respectively. First, we consider Ψ. Let λ−d be the lower bounds for Rosser’s weights of level D. Hence, for any positive integer d, there holds(33)λ−d⩽1,λ−d=0,ifd>Dorμd=0,(34)∑d|nλ−d⩽∑d|nμd=1,ifn=1,0,ifn∈ℕ,n>1.Also, we shall use the fact if2<s<4, then there holds(35)∑d|Pzλ−dφd⩾Πz2eγlogs−1s+OlogX−1/3,where(36)Πz=∏2<p⩽z1−1p−1.Now, we take(37)s=logDlogz=δη=7692719232∈2,4,in (35). By (31) and (34), we obtain(38)Ψ=∑X/2<p⩽Xχαp2+βlogp∑d|p+2,Pzμd⩾∑X/2<p⩽Xχαp2+βlogp∑d|p+2,Pzλ−d=∑d|Pzλ−d∑X/2<p⩽Xp+2≡0moddχαp2+βlogp=Ψ1,say.From (32), we have(39)Ψ1=∑d|Pzλ−d∑X/2<p⩽Xp+2≡0moddΔ+∑k>0gkeαp2k+βklogp=Δ∑d|Pzλ−d∑X/2<p⩽Xp+2≡0moddlogp+∑d|Pzλ−d∑k>0gkeβk∑X/2<p⩽Xp+2≡0moddeαp2klogp=Δ∑d|Pzλ−d∑X/2<p⩽Xp+2≡0moddlogp+∑d|Pzλ−d∑0<k⩽HΔ−1gkeβk∑X/2<p⩽Xp+2≡0moddeαp2klogp+∑d|Pzλ−d∑k>Hgkeβk∑X/2<p⩽Xp+2≡0moddeαp2klogp.By (33) and the fact that λ−d=0 for d>D, we obtain(40)∑d|Pzλ−d∑k>Hgkeβk∑X/2<p⩽Xp+2≡0moddeαp2klogp≪∑d|Pzλ−d∑k>Hgk∑X/2<p⩽Xp+2≡0moddlogp≪∑d⩽D1φd≪logD≪logX.Therefore, we obtain(41)Ψ1=ΔΨ2+Ψ3+OlogX,where(42)Ψ2=∑d|Pzλ−d∑X/2<p⩽Xp+2≡0moddlogp,Ψ3=∑d|Pzλ−d∑0<k⩽Hck∑X/2<p⩽Xp+2≡0moddeαp2klogp,ck=Δ−1gkeβk≪1.ForΨ2, by Bombieri–Vinogradov’s mean value theorem (see Chapter 28 of [30]) and (33), we derive that(43)Ψ2=X2∑d|Pzλ−dφd+OXlog2X.It follows from Mertens’ prime number theorem (see [31]) that(44)Πz≍1logz.Then, from (35), (43), and (44), we obtain(45)Ψ2⩾eγXΠzlogs−1s+OXlog4/3X,where s is defined by (37). For Ψ3, we shall investigate it in Section 5.Now, we study the sumΦ, which is defined by (32). We rewrite Φ in the following form:(46)Φ=∑z<q<y1−logqlogy∑X/2<p⩽Xp+2≡0moddp+2,Pz=1χαp2+βlogp.In order to give upper bound estimate ofΦ, we shall apply an upper bound linear sieve. Let λq+d be the upper bounds for Rosser’s weights of level D/q. Hence, for any positive integer d, we have(47)λq+d⩽1,λq+d=0,ifd>Dqorμd=0,(48)∑d|nλq+d⩾∑d|nμd=1,ifn=1,0,ifn∈ℕ,n>1.Also, we shall use the fact, for1<s1<3, and there holds(49)∑d|Pzλq+dφd⩽Πz2eγs1+OlogX−1/3.For primeq in the sum Φ, we take(50)s1=logD/qlogz.Then, it is easy to check that1<s1<3, and thus, (49) holds. By (46)–(48), we obtain(51)Φ=∑z<q<y1−logqlogy∑X/2<p⩽Xp+2≡0moddχαp2+βlogp∑d|p+2,Pzμd⩽∑z<q<y1−logqlogy∑X/2<p⩽Xp+2≡0moddχαp2+βlogp∑d|p+2,Pzλq+d=∑z<q<y1−logqlogy∑d|Pzλq+d∑X/2<p⩽Xp+2≡0moddχαp2+βlogp=∑m⩽Dνm∑X/2<p⩽Xp+2≡0modmχαp2+βlogp≕Φ1,where(52)νm=∑z<q<yd|Pzm=dq1−logqlogyλq+d.Ifm⩽z, then νm=0. If z<m⩽D, by (17) and (52), we know that the representation m=dq with z<q<y and d|Pz is unique. Thus, it is easy to see that(53)νm⩽1.From (14), we obtain(54)Φ1=∑m⩽Dνm∑X/2<p⩽Xp+2≡0modmΔ+∑k>0gkeαp2k+βklogp=Δ∑m⩽Dνm∑X/2<p⩽Xp+2≡0modmeαp2klogp+∑m⩽Dνm∑k>0gkeβk∑X/2<p⩽Xp+2≡0modmeαp2klogp=Δ∑m⩽Dνm∑X/2<p⩽Xp+2≡0modmlogp+Δ∑m⩽Dνm∑0<|k|⩽HΔ−1gkeβk∑X/2<p⩽Xp+2≡0modmeαp2klogp+∑m⩽Dνm∑k>Hgkeβk∑X/2<p⩽Xp+2≡0modmeαp2klogp.By (15) and (53), we obtain(55)∑m⩽Dνm∑k>Hgkeβk∑X/2<p⩽Xp+2≡0modmeαp2klogp≪∑k>Hgk∑m⩽D∑X/2<p⩽Xp+2≡0modmlogp≪∑m⩽D1φm≪logX.Thus, we derive that(56)Φ1=ΔΦ2+Φ3+OlogX,where(57)Φ2=∑m⩽Dνm∑X/2<p⩽Xp+2≡0modmlogp,Φ3=∑m⩽Dνm∑0<k⩽Hck∑X/2<p⩽Xp+2≡0modmeαp2klogp,ck=Δ−1gkeβk≪1.By Bombieri–Vinogradov’s mean value theorem and (53), we have(58)Φ2=X2∑m⩽Dνmφm+OXlog2X.Using (49) and (52), we obtain(59)∑m⩽Dνmφm=∑z<q<y1−logqlogy∑d|Pzλq+dφqd=∑z<q<y1−logqlogy1q−1∑d|Pzλq+dφd⩽∑z<q<y1−logqlogy1q−1Πz2eγlogD/qlogz−1+OlogX−1/3.Therefore, by (44), (58), and (59), we have(60)Φ2⩽eγXΠz∑z<q<y1−logqlogy1q−1logD/qlogz−1+OXlogX4/3−ε.Now, we find a lower bound for the sumΓ. From (30), (38), (41), (45), (51), (56), and (60), we derive that(61)Γ⩾eγΔXΠzS+OΔXlogX4/3−ε+OΔΨ3−κΦ3,where(62)S=logs−1s−κ∑z<q<y1−logqlogy1q−1logD/qlogz−1,s=logDlogz.Moreover, by partial summation and the prime number theorem, it is easy to show that(63)S=S0+O1logX,where(64)S0=logs−1s−κη∫ηρ1u−1ρ1δ−udu.According to simple numerical calculation, we know that(65)S0⩾0.000032113949.From (44), (61), and (63), we obtain(66)Γ⩾eγΔXΠzS0+OΔXlogX4/3−ε+OΔΨ3−κΦ3.We shall illustrate that ifX runs over a suitable sequence, which tends to infinity, then the second error term in (66) can be absorbed. Hence, we need the following lemma.Lemma 5.
Suppose thatα∈ℝ/ℚ and δ,θ,D,andH are defined in (3) and (4). Let ξdandck be complex numbers defined for d⩽Dand0<k⩽H, respectively, which satisfy(67)ξd≪1,ck≪1.
Then, there exists a sequenceXjj=1∞ satisfying limj⟶∞Xj=+∞, such that the sum SX is defined by(68)SX=∑d⩽Dξd∑1⩽k⩽Hck∑X/2<p⩽Xp+2≡0moddlogpeαp2k,which satisfies(69)SXj≪Xjlog2Xj,j=1,2,3,….The proof of Lemma5 will be given in Section 5. From (42) and (57), we know that Ψ3−κΦ3 can be represented as a sum of type (68) with(70)ξd=λ∗d−κνd,where(71)λ∗d=λ−d,ifd|Pz,0,otherwise.According to Lemma5 and (66), there exists a sequence Xjj=1∞, which tends to infinity, such that(72)ΓXj⩾eγΔXjΠzS0+OΔXjlogXj4/3−ε.From (44) and (72), we know that there exists a positive constant c>0 such that(73)ΓXj⩾cΔXjXjlogXj>0,j=1,2,3,….This completes the proof of Theorem1.


## 5. Proof of Lemma5

In this section, we shall prove Lemma5. Since α∈ℝ\ℚ, by Dirichlet’s approximation theorem, there exist infinitely many integers A and natural numbers Q with A,Q=1 such that(74)α−AQ<1Q2.For each suchQ, we choose X in a suitable way, i.e., as in (144). In this way, we construct our sequence Xjj=1∞.First, we have(75)SX=W+OHX1/2+ε,where(76)W=∑X/2<n⩽XΛn∑1⩽k⩽Hckeαn2k∑d⩽Ddn+22dξd.According to Lemma2, by taking u=2−7Xδ/2,v=27X1/3,andw=X1/2−δ/4, it is easy to see that the sum W can be decompose into Olog10X sums, each of which is either of Type I,(77)SI=∑M<m⩽M1am∑L<ℓ⩽L1X/2<mℓ⩽X∑1⩽|k|⩽Hckeαm2ℓ2k∑d⩽Ddmℓ+22dξd,with M1⩽2M,L1⩽2L,L⩾w,am≪mε,andML≍X, or of Type II(78)SII=∑M<m⩽M1am∑L<ℓ⩽L1X/2<mℓ⩽Xbℓ∑1⩽k⩽Hckeαm2ℓ2k∑d⩽Ddmℓ+22dξd,with M1⩽2M,L1⩽2L,u⩽L⩽v,am≪mε,bℓ≪ℓε,andML≍X.Next, we shall deal with the sums of Type I and Type II in the following sections, respectively.


### 5.1. The Estimate of Type II Sums

In this section, we shall deal with the estimate of the sums of Type II. First, we have(79)SII=∑1⩽|k|⩽Hck∑M<m⩽M1am∑L<ℓ⩽L1X/2<mℓ⩽Xbℓeαm2ℓ2k∑d⩽Ddmℓ+22dξd≪Xε∑1⩽k⩽H∑M<m⩽M1∑L<ℓ⩽L1X/2<mℓ⩽Xbℓeαm2ℓ2k∑d⩽Ddmℓ+22dξd.By Cauchy’s inequality, we obtain(80)SII2≪XεHM∑1⩽k⩽H∑M<m⩽M1∑L<ℓ⩽L1X/2<mℓ⩽Xbℓeαm2ℓ2k∑d⩽Ddmℓ+22dξd2=XεHM∑1⩽k⩽H∑M<m⩽M1∑L<ℓ⩽L1X/2<mℓ⩽X∑d1,d2⩽Dmℓ1+2≡0modd1mℓ2+2≡0modd2d1d1,2=1bℓ1bℓ2¯ξd1ξd2¯eαm2kℓ12−ℓ22≪XεHM∑1⩽k⩽H∑L<ℓ1,ℓ2⩽L1∑d1,d2⩽Dd1d2,2=1ℓ1,d1=ℓ2,d2=1∑M′<m⩽M1′mℓ1+2≡0modd1mℓ2+2≡0modd2eαm2kℓ12−ℓ22≪XεHM∑1⩽k⩽H∑L<ℓ1,ℓ2⩽L1∑d1,d2⩽Dd1d2,2=1ℓ1,d1=ℓ2,d2=1V,where(81)V=∑M′<m⩽M1′mℓ1+2≡0modd1mℓ2+2≡0modd2eαm2kℓ12−ℓ22,M′=maxM,X2ℓ1,X2ℓ2,M1′=minM1,Xℓ1,Xℓ2.If the system of the congruence,(82)mℓ1+2≡0modd1,mℓ2+2≡0modd2,has no solution, then V=0. Assume that (82) has a solution. Then, there exists an f0=f0ℓ1,ℓ2,d1,d2 such that (82) is equivalent to m≡f0modd1,d2. In this case, we have(83)V=∑M′<m⩽M1′m≡f0modd1,d2eαm2kℓ12−ℓ22=∑M′−f0d1,d2<r⩽M1′−f0d1,d2eαf0+rd1,d22kℓ12−ℓ22=∑M′−f0d1,d2<r⩽M1′−f0d1,d2eαr2d1,d22+2f0rd1,d2kℓ12−ℓ22=∑R<r⩽R1eαr2d1,d22+2f0rd1,d2kℓ12−ℓ22,where(84)R=M′−f0d1,d2,R1=M1′−f0d1,d2.The contribution ofV with ℓ1=ℓ2 to SII2 is(85)≪XεHM2∑1⩽k⩽H∑L<ℓ1,ℓ2⩽L1ℓ1=ℓ2∑d1,d2⩽D1d1,d2≪XεH2M2L∑d1,d2⩽D1d1,d2≪XεH2M2L∑h⩽D2τ2hh≪XεH2M2L≪X1+εH2M.Therefore, we have(86)SII2≪X1+εH2M+XεHM∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V.Moreover, by Cauchy’s inequality again, we obtain(87)SII4≪X2+εH4M2+XεH3M2∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=11d1,d2×∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2L2∑d1,d2⩽Dd1d2,2=11d1,d2×∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2L2∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2L2⋅Σ0,where(88)Σ0=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑R<r1,r2⩽R1eαr12−r22d1,d22+2f0r1−r2d1,d2kℓ12−ℓ22.ForΣ0, we have(89)Σ0=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1,s2∑R<r1,r2⩽R1r1−r2=s1r1+r2=s21eαs1s2d1,d22+2f0s1d1,d2kℓ12−ℓ22=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1,s2:s1≡s2mod22R<s2+s1⩽2R12R<s2−s1⩽2R1eαs1s2d1,d22+2f0s1d1,d2kℓ12−ℓ22=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2∑|s1|⩽2R1−2Re2αf0s1d1,d2kℓ12−ℓ22×∑s2:s2≡s1mod22R<s2+s1⩽2R12R<s2−s1⩽2R1eαs1s2d1,d22kℓ12−ℓ22.Set(90)D0=X50/3θ.Then, we divideΣ0 into two parts(91)Σ0=Σ1+Σ2,where Σ1 denotes the part of Σ0 which satisfies d1,d2⩽D0, while Σ2 denotes the remaining part of Σ0 which satisfies d1,d2>D0. We set s2=s1+2t in Σ1 and Σ2 and derive that(92)Σ1⩽∑1⩽k⩽H∑d1,d2⩽D0d1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1⩽2R1−2R∑R′<t⩽R1′e2αs1td1,d22kℓ12−ℓ22,(93)Σ2⩽∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2>D0d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑|s1|⩽2R1−2R∑R′<t⩽R1′e2αs1td1,d22kℓ12−ℓ22,where(94)R′=maxR−s1,R,R1′=minR1−s1,R1.First, we consider the upper bound forΣ1. Let Σ11 and Σ12 denote the contribution of the right-hand side of (92) for s1≠0 and s1=0, respectively. Trivially, there holds(95)Σ12≪HML2∑d1,d2⩽D0d1d2,2=11≪HML2D02≪D02HXL.ForΣ11, by Lemma 3, we have(96)Σ11≪∑1⩽k⩽H∑d1,d2⩽D0d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2×∑0<s⩽2M/d1,d2minMd1,d2,12αsd1,d22kℓ12−ℓ22≪∑1⩽k⩽H∑h⩽D02h∑d1,d2⩽D0d1,d2=h1∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑0<s⩽2M/hminMh,12αsh2kℓ12−ℓ22≪D02∑1⩽k⩽H∑h⩽D02∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑0<|s|⩽2MminM,12αsh2kℓ12−ℓ22≪D02∑1⩽k⩽H∑h⩽D02∑t1,t2∑L<ℓ1,ℓ2⩽L1ℓ1−ℓ2=t1ℓ1+ℓ2=t2ℓ1≠ℓ21∑0<s⩽2MminM,12αsh2kt1t2≪D02∑1⩽k⩽H∑h⩽D02∑1⩽t1⩽L1⩽t2⩽4L∑0<s⩽2MminM,12αsh2kt1t2≪D02∑1⩽k⩽H∑h⩽D02∑1⩽t1,t2⩽4L∑1⩽|s|⩽2MminM,12αsh2kt1t2≪D02∑1⩽m⩽64D04HML2τ7mminM,1αm.By Lemma4, we have(97)∑1⩽m⩽64D04HML2τ7mminM,1αm≪Xε∑1⩽m⩽64D04HML2min64D04HM2L2m,1αm≪XεD04HM2L21Q+1M+QD04HM2L2≪XεHX2D04Q+HXLD04+Q.Combining (92), (93), (96), and (97) and by noting the fact that ML≍X, we obtain(98)Σ1≪XεHX2D06Q−1+HXLD06+QD02.Now, we consider the estimate ofΣ2. According to (93), by a splitting argument, we have(99)Σ2≪ℒmaxD0≪T≪D2TΣ21,where(100)Σ21=Σ21T=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1T<d1,d2⩽2T∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1⩽2R1−2R∑R′<t⩽R1′e2αs1td1,d22kℓ12−ℓ22.By Lemma1, we have(101)Σ21=Σ21T⩽∑1⩽k⩽H∑d1,d2⩽DT<d1,d2⩽2T∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T×∫−∞+∞Kθ∑M/4T<t⩽4M/Te2αstd1,d22kℓ12−ℓ22+θtdθ≕∫−∞+∞Kθ⋅Σ22θ,Tdθ,where(102)Kθ=min15M4T+1,1πθ,1π2θ2,Σ22θ,T=∑1⩽k⩽H∑d1,d2⩽DT<d1,d2⩽2T∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T×∑M/4T<t⩽4M/Te2αstd1,d22kℓ12−ℓ22+θt.According to (7) and (101), it is easy to see that(103)Σ21≪ℒmax0⩽θ⩽1Σ22θ,T.ForΣ22θ,T, we have(104)Σ22θ,T=∑1⩽k⩽H∑T<h⩽2T∑d1,d2⩽Dd1,d2=h1∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T×∑M/4T<t⩽4M/Te2αsth2kℓ12−ℓ22+θt≪∑1⩽k⩽H∑T<h⩽2Tτ2h∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T∑M/4T<t⩽4M/Te2αsth2kℓ12−ℓ22+θt=∑1⩽k⩽H∑T<h⩽2Tτ2h∑t1,t2∑L<ℓ1,ℓ2⩽L1ℓ1−ℓ2=t1ℓ1+ℓ2=t2ℓ1≠ℓ21×∑s⩽2M/T∑M/4T<t⩽4M/Te2αsth2kt1t2+θt≪∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽t1,t2⩽4L∑s⩽2M/T∑M/4T<t⩽4M/Te2αsth2kt1t2+θt≪ℒ3HML2+∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽t1,t2⩽4L∑1⩽s⩽2MT×∑M/4T<t⩽4M/Te2αsth2kt1t2+θt≪ℒ3HML2+∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽m⩽32ML2/Tτ3m×∑M/4T<t⩽4M/Te2αth2km+θt=ℒ3HML2+Σ23,say.It follows from Cauchy’s inequality that(105)Σ232≪H∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽m⩽32ML2/Tτ3m∑M/4T<t⩽4M/Te2αth2km+θt2≪H∑1⩽k⩽H∑T<h⩽2Tτ4h∑T<h⩽2T∑1⩽m⩽32ML2/Tτ3m×∑M/4T<t⩽4M/Te2αth2km+θt2≪H∑T<h⩽2Tτ4h∑1⩽m⩽32ML2/Tτ32m∑1⩽k⩽H∑T<h⩽2T×∑1⩽m⩽32ML2/T∑M/4T<t⩽4M/Te2αth2km+θt2≪ℒ23HML2∑1⩽k⩽H∑T<h⩽2T∑1⩽m⩽32ML2/T∑M/4T<t⩽4M/T2αth2km+θte2=ℒ23HML2⋅Σ24,say.ForΣ24, we have(106)Σ24=∑1⩽k⩽H∑T<h⩽2T∑1⩽m⩽32ML2/T∑M/4T<t1,t2⩽4M/Te2αh2km+θt1−t2≪∑1⩽k⩽H∑1⩽m⩽32ML2/T∑M/4T<t1,t2⩽4M/T∑T<h⩽2Te2αh2kmt1−t2≪HM2L2T+MT∑1⩽k⩽H∑1⩽m⩽32ML2/T∑1⩽n⩽4M/T∑T<h⩽2Te2αh2kmn≪HM2L2T+MT∑1⩽k⩽H∑1⩽s⩽256M2L2/T2τ3s∑T<h⩽2Teαh2ks=HM2L2T+MT⋅Σ25,say.By Cauchy’s inequality, we deduce that(107)Σ252≪H∑1⩽k⩽H∑1⩽s⩽256M2L2/T2τ3s∑T<h⩽2Teαh2ks2≪H∑1⩽s⩽256M2L2/T2τ32s∑1⩽k⩽H∑1⩽s⩽256M2L2/T2∑T<h⩽2Teαh2ks2≪ℒ8HM2L2T2∑1⩽k⩽H∑1⩽s⩽256M2L2/T2∑T<h1,h2⩽2Teαksh12−h22≪ℒ8H2M4L4T3+ℒ8HM2L2T2∑1⩽k⩽H∑1⩽s⩽256M2L2/T2∑T<h1,h2⩽2Th1≠h2eαksh12−h22=ℒ8H2M4L4T3+ℒ8HM2L2T2⋅Σ26,say.ForΣ26, from Lemma 3, we have(108)Σ26=∑1⩽k⩽H∑t1,t2∑T<h1,h2⩽2Th1−h2=t1h1+h2=t2h1≠h21∑1⩽s⩽256M2L2/T2eαkst1t2≪∑1⩽k⩽H∑1⩽t1,t2⩽4T∑1⩽s⩽256M2L2/T2eαkst1t2≪∑1⩽k⩽H∑1⩽t1,t2⩽4TminM2L2T2,1αkt1t2≪∑1⩽n⩽16HT2τ3nminM2L2T2,1αn.It follows from Lemma4 that(109)∑1⩽n⩽16HT2τ3nminM2L2T2,1αn≪Xε∑1⩽n⩽16HT2min16HM2L2n,1αn≪XεHM2L21Q+T2M2L2+QHM2L2≪XεHX2Q−1+HT2+Q.From (107), (108), and (109), we obtain(110)Σ25≪XεHX2T3/2+HX2TQ1/2+HX+H1/2XQ1/2T.Putting (110) into (106), we obtain(111)Σ24≪XεHX2T+HX2MT5/2+HX2MT2Q1/2+HXMT+H1/2XQ1/2MT2.Combining (105) and (111), one has(112)Σ23≪XεHX3/2L1/2T1/2+HX2T5/4+H1/2X2TQ1/4+H3/4X3/2Q1/4T.Inserting (112) into (104), we derive that(113)Σ22θ,T≪XεHXL+HX3/2L1/2T1/2+HX2T5/4+H1/2X2TQ1/4+H3/4X3/2Q1/4T,which combines (99) and (103) to obtain(114)Σ2≪XεmaxD0≪T≪D2HXLT+HX3/2L1/2T1/2+HX2T−1/4+H1/2X2Q−1/4+H3/4X3/2Q1/4≪XεHXLD2+HX3/2L1/2D+HX2D0−1/4+H1/2X2Q−1/4+H3/4X3/2Q1/4.From (91), (98), and (114), we obtain(115)Σ0≪XεD06HX2Q−1+D06HXL+D02Q+HXLD2+HX3/2L1/2D+HX2D0−1/4+H1/2X2Q−1/4+H3/4X3/2Q1/4,which combines (87) yields(116)SII≪XεHXL−1/2+D03/2HXQ−1/4+D03/2HX3/4L1/4+D01/2Q1/4H3/4X1/2+HX3/4L1/4D1/2+HX7/8L1/8D1/4+HXD0−1/16+H7/8XQ−1/16+H15/16X7/8Q1/16≪XεHXu−1/2+D03/2HXQ−1/4+D03/2HX3/4v1/4+D01/2Q1/4H3/4X1/2+HX3/4v1/4D1/2+HX7/8v1/8D1/4+HXD0−1/16+H7/8XQ−1/16+H15/16X7/8Q1/16.


### 5.2. The Estimate of Type I Sums

In this section, we shall deal with the estimate of the sums of Type I. First, we have(117)SI=∑1⩽k⩽Hck∑d⩽Dd,2=1ξd∑M<m⩽M1am∑L′<ℓ⩽L1′mℓ+2≡0moddeαm2ℓ2k,where(118)L′=maxL,X2m,L1′=minL1,Xm.By a splitting argument, there holds(119)SI≪Xε⋅max1≪T≪DΣ3,where(120)Σ3=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑L′<ℓ⩽L1′mℓ+2≡0moddeαm2ℓ2k.Form,d=1, there exists m¯, which satisfies 0⩽m¯⩽d−1, such that mm¯≡1modd. Therefore, the equation mℓ+2≡0modd is equivalent to ℓ≡−2m¯modd, i.e., ℓ=−2m¯+dr for some r∈ℤ. Then, it follows from Cauchy’s inequality that(121)Σ32≪HMT∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑L′<ℓ⩽L1′mℓ+2≡0moddeαm2ℓ2k2≪HMT∑1⩽k⩽H∑T<d⩽2Td,2∑M<m⩽M1m,d∑L′+2m¯/d<r⩽L1′+2m¯/deαm2−2m¯+dr2k2=HMT∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑L′+2m¯/d<r⩽L1′+2m¯/deαm2d2r2−4m¯drk2=HMT∑1⩽k⩽H∑T<d⩽2Td,2=1=1∑M<m⩽M1m,d=1=1×∑L′+2m¯/d<r⩽L1′+2m¯/deαm2d2r12−r22−4m¯dr1−r2k.Set(122)R=L′+2m¯d,R1=L1′+2m¯d.Then, we have(123)Σ32≪XεH2M2LT+HMT∑1⩽k⩽H∑T<d⩽2Td,2∑M<m⩽M1m,d=1×∑R<r1,r2⩽R1r1≠r2eαm2d2r12−r22−4m¯dr1−r2k≪XεH2M2LT+HMT⋅Σ31,where(124)Σ31=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑R<r1,r2⩽R1r1≠r2eαm2d2r12−r22−4m¯dr1−r2k.ForΣ31, we have(125)Σ31=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑s1,s2∑R<r1,r2⩽R1r1−r2=s1r1+r2=s2r1≠r21eαm2d2s1s2−4m¯ds1k=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑s1,s22R<s1+s2⩽2R12R<s2−s1⩽2R1s1≡s2mod2,s1≠0eαm2d2s1s2−4m¯ds1k≪∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑1⩽s1⩽4L/T∑s2:s2≡s1mod22R−s1<s2⩽2R1−s12R+s1<s2⩽2R1+s1eαm2d2s1s2k≪∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑1⩽s1⩽4L/T∑R−s1<t⩽R1−s1R<t⩽R1e2αm2d2s1tk.Next, we will discuss the estimate of the right-hand side of (125) in two cases.Case 1.
Suppose thatMT⩽D0, and under this condition, there holds 1≪M,T≪D0. By Lemma 4, we have(126)Σ31≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s⩽8LminL,12αm2d2sk≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s⩽8Lmin256HM2T2L22m2d2sk,12αm2d2sk≪∑1⩽n⩽256HM2T2Lτ7nminHM2T2L2n,1αn≪XεHX2T21Q+1L+QHX2T2≪XεHX2D02Q+HXMTT+Q≪XεHX2D02Q+HXD02+Q.
From (119), (123), and (126), we derive that, under the condition MT⩽D0, there holds(127)SI≪XεHX1/2D01/2+HXD03/2Q−1/2+HX1/2D03/2+H1/2D01/2Q1/2.Case 2.
Now, we suppose thatMT>D0. Set(128)R′=maxR,R−s1,R1′=minR1,R1−s1.
Applying Lemma1 to (125), we have(129)Σ31≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s1⩽4L/T∑R′<t⩽R1′e2αm2d2s1tk≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s1⩽4L/T∫−∞+∞Kθ1∑L/4T<t⩽4L/Te2αm2d2s1tk+θtdθ=∫−∞+∞K1θ⋅Σ32θ,Tdθ,where(130)K1θ=min15L4T+1,1πθ,1π2θ2,Σ32θ,T=∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s1⩽4L/T∑L/4T<t⩽4L/Te2αm2d2s1tk+θt.
According to (7) and (129), it is easy to see that(131)Σ31≪ℒ⋅max0⩽θ⩽1Σ32θ,T.
ForΣ32θ,T, we have(132)Σ32θ,T≪∑1⩽k⩽H∑MT<h⩽4MTτh∑1⩽s1⩽4L/T∑L/4T<t⩽4L/Te2αh2stk+θt.
It follows from Cauchy’s inequality that(133)Σ32θ,T2≪H∑MT<h⩽4MTτ2h∑1⩽s⩽4L/T1∑1⩽k⩽H∑MT<h⩽4MT×∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/Te2αh2stk+θt2≪XεHML∑1⩽k⩽H∑MT<h⩽4MT∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/Te2αh2sk+θt1−t2≪XεHML∑1⩽k⩽H∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/T∑MT<h⩽4MTe2αh2skt1−t2≪XεH2M2L3T+XεHML⋅Σ33,where(134)Σ33=∑1⩽k⩽H∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/T∑MT<h⩽4MTe2αh2skt1−t2.
ForΣ33, we have(135)Σ33=∑1⩽k⩽H∑1⩽s⩽4L/T∑1⩽r1⩽4L/T1⩽r2⩽8L/T∑4L/T<t1,t2⩽4L/Tt1−t2=r1t1+t2=r21∑MT<h⩽4MTe2αh2skr1≪LT∑1⩽k⩽H∑1⩽s⩽4L/T∑1⩽s⩽4L/T∑MT<h⩽4MTe2αh2skr1≪LT∑1⩽n⩽32HL2/T2τ4n∑MT<h⩽4MTeαh2n.
Therefore, by Cauchy’s inequality, one has(136)Σ332≪L2T2∑1⩽n⩽32HL2/T2τ42n∑1⩽n⩽32HL2/T2∑MT<h⩽4MTeαh2n2≪XεHL4T4∑1⩽n⩽32HL2/T2∑MT<h1,h2⩽4MTeαh12−h22n≪XεH2ML6T5+XεHL4T4⋅Σ34,where(137)Σ34=∑MT<h1,h2⩽4MTh1≠h2∑1⩽n⩽32HL2/T2eαh12−h22n.
ForΣ34, by Lemma 3, we have(138)Σ34=∑1⩽|t1|,|t2|⩽8MT∑MT<h1,h2⩽4MTh1−h2=t1h1+h2=t21∑1⩽n⩽32HL2/T2eαt1t2n≪∑1⩽t1,t2⩽8MT∑1⩽n⩽32HL2/T2eαt1t2n≪∑1⩽t1,t2⩽8MTminHL2T2,1αt1t2≪∑1⩽t⩽64M2T2τtminHL2T2,1αt.
It follows from Lemma4 that(139)∑1⩽t⩽64M2T2τtminHL2T2,1αt≪Xε∑1⩽t⩽64M2T2min64HM2L2t,1αt≪XεHM2L21Q+T2HL2+QHM2L2≪XεHX2Q+M2T2+Q.
From (136), (138), and (139), we derive that(140)Σ33≪XεHX1/2L5/2T5/2+HXL2T2Q1/2+H1/2XLT+H1/2L2Q1/2T2,which combines (133) yields(141)Σ32θ,T≪XεHXL1/2T1/2+HX3/4L5/4T5/4+HXLTQ1/4+H3/4X1/2LQ1/4T.
From (123), (131), and (141), we obtain(142)Σ3≪XεHXL−1/2T1/2+HXL−1/4T1/4+HX7/8L1/8T−1/8+HXQ−1/8+H7/8Q1/8X3/4,from which and (139), we derive that, under the condition MT>D0, and there holds(143)SI≪Xεmax1≪T≪DHXT1/2L1/2+HXT1/4L1/4+HX7/8L1/8T1/8+HXQ1/8+H7/8Q1/8X3/4≪XεHXD1/2w1/2+HXD1/4w1/4+HX7/8L1/8M1/8MT1/8+HXQ1/8+H7/8Q1/8X3/4≪XεHXw−1/2D1/2+HXw−1/4D1/4+HXD0−1/8+HXQ−1/8+H7/8Q1/8X3/4.


### 5.3. Proof of Lemma5

From (116), (127), and (143), by taking(144)Q=X4138/15θ,then we deduce that, under conditions (3) and (4), there holds(145)SI≪X1−ϖandSII≪X1−ϖ,for some ϖ>0. This completes the proof of Lemma 5.


## 5.1. The Estimate of Type II Sums

In this section, we shall deal with the estimate of the sums of Type II. First, we have(79)SII=∑1⩽|k|⩽Hck∑M<m⩽M1am∑L<ℓ⩽L1X/2<mℓ⩽Xbℓeαm2ℓ2k∑d⩽Ddmℓ+22dξd≪Xε∑1⩽k⩽H∑M<m⩽M1∑L<ℓ⩽L1X/2<mℓ⩽Xbℓeαm2ℓ2k∑d⩽Ddmℓ+22dξd.By Cauchy’s inequality, we obtain(80)SII2≪XεHM∑1⩽k⩽H∑M<m⩽M1∑L<ℓ⩽L1X/2<mℓ⩽Xbℓeαm2ℓ2k∑d⩽Ddmℓ+22dξd2=XεHM∑1⩽k⩽H∑M<m⩽M1∑L<ℓ⩽L1X/2<mℓ⩽X∑d1,d2⩽Dmℓ1+2≡0modd1mℓ2+2≡0modd2d1d1,2=1bℓ1bℓ2¯ξd1ξd2¯eαm2kℓ12−ℓ22≪XεHM∑1⩽k⩽H∑L<ℓ1,ℓ2⩽L1∑d1,d2⩽Dd1d2,2=1ℓ1,d1=ℓ2,d2=1∑M′<m⩽M1′mℓ1+2≡0modd1mℓ2+2≡0modd2eαm2kℓ12−ℓ22≪XεHM∑1⩽k⩽H∑L<ℓ1,ℓ2⩽L1∑d1,d2⩽Dd1d2,2=1ℓ1,d1=ℓ2,d2=1V,where(81)V=∑M′<m⩽M1′mℓ1+2≡0modd1mℓ2+2≡0modd2eαm2kℓ12−ℓ22,M′=maxM,X2ℓ1,X2ℓ2,M1′=minM1,Xℓ1,Xℓ2.If the system of the congruence,(82)mℓ1+2≡0modd1,mℓ2+2≡0modd2,has no solution, then V=0. Assume that (82) has a solution. Then, there exists an f0=f0ℓ1,ℓ2,d1,d2 such that (82) is equivalent to m≡f0modd1,d2. In this case, we have(83)V=∑M′<m⩽M1′m≡f0modd1,d2eαm2kℓ12−ℓ22=∑M′−f0d1,d2<r⩽M1′−f0d1,d2eαf0+rd1,d22kℓ12−ℓ22=∑M′−f0d1,d2<r⩽M1′−f0d1,d2eαr2d1,d22+2f0rd1,d2kℓ12−ℓ22=∑R<r⩽R1eαr2d1,d22+2f0rd1,d2kℓ12−ℓ22,where(84)R=M′−f0d1,d2,R1=M1′−f0d1,d2.The contribution ofV with ℓ1=ℓ2 to SII2 is(85)≪XεHM2∑1⩽k⩽H∑L<ℓ1,ℓ2⩽L1ℓ1=ℓ2∑d1,d2⩽D1d1,d2≪XεH2M2L∑d1,d2⩽D1d1,d2≪XεH2M2L∑h⩽D2τ2hh≪XεH2M2L≪X1+εH2M.Therefore, we have(86)SII2≪X1+εH2M+XεHM∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V.Moreover, by Cauchy’s inequality again, we obtain(87)SII4≪X2+εH4M2+XεH3M2∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=11d1,d2×∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2L2∑d1,d2⩽Dd1d2,2=11d1,d2×∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2L2∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2V2≪X2+εH4M2+XεH3M2L2⋅Σ0,where(88)Σ0=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑R<r1,r2⩽R1eαr12−r22d1,d22+2f0r1−r2d1,d2kℓ12−ℓ22.ForΣ0, we have(89)Σ0=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1,s2∑R<r1,r2⩽R1r1−r2=s1r1+r2=s21eαs1s2d1,d22+2f0s1d1,d2kℓ12−ℓ22=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1,s2:s1≡s2mod22R<s2+s1⩽2R12R<s2−s1⩽2R1eαs1s2d1,d22+2f0s1d1,d2kℓ12−ℓ22=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2∑|s1|⩽2R1−2Re2αf0s1d1,d2kℓ12−ℓ22×∑s2:s2≡s1mod22R<s2+s1⩽2R12R<s2−s1⩽2R1eαs1s2d1,d22kℓ12−ℓ22.Set(90)D0=X50/3θ.Then, we divideΣ0 into two parts(91)Σ0=Σ1+Σ2,where Σ1 denotes the part of Σ0 which satisfies d1,d2⩽D0, while Σ2 denotes the remaining part of Σ0 which satisfies d1,d2>D0. We set s2=s1+2t in Σ1 and Σ2 and derive that(92)Σ1⩽∑1⩽k⩽H∑d1,d2⩽D0d1d2,2=1d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1⩽2R1−2R∑R′<t⩽R1′e2αs1td1,d22kℓ12−ℓ22,(93)Σ2⩽∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1d1,d2>D0d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑|s1|⩽2R1−2R∑R′<t⩽R1′e2αs1td1,d22kℓ12−ℓ22,where(94)R′=maxR−s1,R,R1′=minR1−s1,R1.First, we consider the upper bound forΣ1. Let Σ11 and Σ12 denote the contribution of the right-hand side of (92) for s1≠0 and s1=0, respectively. Trivially, there holds(95)Σ12≪HML2∑d1,d2⩽D0d1d2,2=11≪HML2D02≪D02HXL.ForΣ11, by Lemma 3, we have(96)Σ11≪∑1⩽k⩽H∑d1,d2⩽D0d1,d2∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2×∑0<s⩽2M/d1,d2minMd1,d2,12αsd1,d22kℓ12−ℓ22≪∑1⩽k⩽H∑h⩽D02h∑d1,d2⩽D0d1,d2=h1∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑0<s⩽2M/hminMh,12αsh2kℓ12−ℓ22≪D02∑1⩽k⩽H∑h⩽D02∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑0<|s|⩽2MminM,12αsh2kℓ12−ℓ22≪D02∑1⩽k⩽H∑h⩽D02∑t1,t2∑L<ℓ1,ℓ2⩽L1ℓ1−ℓ2=t1ℓ1+ℓ2=t2ℓ1≠ℓ21∑0<s⩽2MminM,12αsh2kt1t2≪D02∑1⩽k⩽H∑h⩽D02∑1⩽t1⩽L1⩽t2⩽4L∑0<s⩽2MminM,12αsh2kt1t2≪D02∑1⩽k⩽H∑h⩽D02∑1⩽t1,t2⩽4L∑1⩽|s|⩽2MminM,12αsh2kt1t2≪D02∑1⩽m⩽64D04HML2τ7mminM,1αm.By Lemma4, we have(97)∑1⩽m⩽64D04HML2τ7mminM,1αm≪Xε∑1⩽m⩽64D04HML2min64D04HM2L2m,1αm≪XεD04HM2L21Q+1M+QD04HM2L2≪XεHX2D04Q+HXLD04+Q.Combining (92), (93), (96), and (97) and by noting the fact that ML≍X, we obtain(98)Σ1≪XεHX2D06Q−1+HXLD06+QD02.Now, we consider the estimate ofΣ2. According to (93), by a splitting argument, we have(99)Σ2≪ℒmaxD0≪T≪D2TΣ21,where(100)Σ21=Σ21T=∑1⩽k⩽H∑d1,d2⩽Dd1d2,2=1T<d1,d2⩽2T∑L<ℓ1,ℓ2⩽L1ℓ1,d1=ℓ2,d2=1ℓ1≠ℓ2×∑s1⩽2R1−2R∑R′<t⩽R1′e2αs1td1,d22kℓ12−ℓ22.By Lemma1, we have(101)Σ21=Σ21T⩽∑1⩽k⩽H∑d1,d2⩽DT<d1,d2⩽2T∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T×∫−∞+∞Kθ∑M/4T<t⩽4M/Te2αstd1,d22kℓ12−ℓ22+θtdθ≕∫−∞+∞Kθ⋅Σ22θ,Tdθ,where(102)Kθ=min15M4T+1,1πθ,1π2θ2,Σ22θ,T=∑1⩽k⩽H∑d1,d2⩽DT<d1,d2⩽2T∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T×∑M/4T<t⩽4M/Te2αstd1,d22kℓ12−ℓ22+θt.According to (7) and (101), it is easy to see that(103)Σ21≪ℒmax0⩽θ⩽1Σ22θ,T.ForΣ22θ,T, we have(104)Σ22θ,T=∑1⩽k⩽H∑T<h⩽2T∑d1,d2⩽Dd1,d2=h1∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T×∑M/4T<t⩽4M/Te2αsth2kℓ12−ℓ22+θt≪∑1⩽k⩽H∑T<h⩽2Tτ2h∑L<ℓ1,ℓ2⩽L1ℓ1≠ℓ2∑s⩽2M/T∑M/4T<t⩽4M/Te2αsth2kℓ12−ℓ22+θt=∑1⩽k⩽H∑T<h⩽2Tτ2h∑t1,t2∑L<ℓ1,ℓ2⩽L1ℓ1−ℓ2=t1ℓ1+ℓ2=t2ℓ1≠ℓ21×∑s⩽2M/T∑M/4T<t⩽4M/Te2αsth2kt1t2+θt≪∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽t1,t2⩽4L∑s⩽2M/T∑M/4T<t⩽4M/Te2αsth2kt1t2+θt≪ℒ3HML2+∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽t1,t2⩽4L∑1⩽s⩽2MT×∑M/4T<t⩽4M/Te2αsth2kt1t2+θt≪ℒ3HML2+∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽m⩽32ML2/Tτ3m×∑M/4T<t⩽4M/Te2αth2km+θt=ℒ3HML2+Σ23,say.It follows from Cauchy’s inequality that(105)Σ232≪H∑1⩽k⩽H∑T<h⩽2Tτ2h∑1⩽m⩽32ML2/Tτ3m∑M/4T<t⩽4M/Te2αth2km+θt2≪H∑1⩽k⩽H∑T<h⩽2Tτ4h∑T<h⩽2T∑1⩽m⩽32ML2/Tτ3m×∑M/4T<t⩽4M/Te2αth2km+θt2≪H∑T<h⩽2Tτ4h∑1⩽m⩽32ML2/Tτ32m∑1⩽k⩽H∑T<h⩽2T×∑1⩽m⩽32ML2/T∑M/4T<t⩽4M/Te2αth2km+θt2≪ℒ23HML2∑1⩽k⩽H∑T<h⩽2T∑1⩽m⩽32ML2/T∑M/4T<t⩽4M/T2αth2km+θte2=ℒ23HML2⋅Σ24,say.ForΣ24, we have(106)Σ24=∑1⩽k⩽H∑T<h⩽2T∑1⩽m⩽32ML2/T∑M/4T<t1,t2⩽4M/Te2αh2km+θt1−t2≪∑1⩽k⩽H∑1⩽m⩽32ML2/T∑M/4T<t1,t2⩽4M/T∑T<h⩽2Te2αh2kmt1−t2≪HM2L2T+MT∑1⩽k⩽H∑1⩽m⩽32ML2/T∑1⩽n⩽4M/T∑T<h⩽2Te2αh2kmn≪HM2L2T+MT∑1⩽k⩽H∑1⩽s⩽256M2L2/T2τ3s∑T<h⩽2Teαh2ks=HM2L2T+MT⋅Σ25,say.By Cauchy’s inequality, we deduce that(107)Σ252≪H∑1⩽k⩽H∑1⩽s⩽256M2L2/T2τ3s∑T<h⩽2Teαh2ks2≪H∑1⩽s⩽256M2L2/T2τ32s∑1⩽k⩽H∑1⩽s⩽256M2L2/T2∑T<h⩽2Teαh2ks2≪ℒ8HM2L2T2∑1⩽k⩽H∑1⩽s⩽256M2L2/T2∑T<h1,h2⩽2Teαksh12−h22≪ℒ8H2M4L4T3+ℒ8HM2L2T2∑1⩽k⩽H∑1⩽s⩽256M2L2/T2∑T<h1,h2⩽2Th1≠h2eαksh12−h22=ℒ8H2M4L4T3+ℒ8HM2L2T2⋅Σ26,say.ForΣ26, from Lemma 3, we have(108)Σ26=∑1⩽k⩽H∑t1,t2∑T<h1,h2⩽2Th1−h2=t1h1+h2=t2h1≠h21∑1⩽s⩽256M2L2/T2eαkst1t2≪∑1⩽k⩽H∑1⩽t1,t2⩽4T∑1⩽s⩽256M2L2/T2eαkst1t2≪∑1⩽k⩽H∑1⩽t1,t2⩽4TminM2L2T2,1αkt1t2≪∑1⩽n⩽16HT2τ3nminM2L2T2,1αn.It follows from Lemma4 that(109)∑1⩽n⩽16HT2τ3nminM2L2T2,1αn≪Xε∑1⩽n⩽16HT2min16HM2L2n,1αn≪XεHM2L21Q+T2M2L2+QHM2L2≪XεHX2Q−1+HT2+Q.From (107), (108), and (109), we obtain(110)Σ25≪XεHX2T3/2+HX2TQ1/2+HX+H1/2XQ1/2T.Putting (110) into (106), we obtain(111)Σ24≪XεHX2T+HX2MT5/2+HX2MT2Q1/2+HXMT+H1/2XQ1/2MT2.Combining (105) and (111), one has(112)Σ23≪XεHX3/2L1/2T1/2+HX2T5/4+H1/2X2TQ1/4+H3/4X3/2Q1/4T.Inserting (112) into (104), we derive that(113)Σ22θ,T≪XεHXL+HX3/2L1/2T1/2+HX2T5/4+H1/2X2TQ1/4+H3/4X3/2Q1/4T,which combines (99) and (103) to obtain(114)Σ2≪XεmaxD0≪T≪D2HXLT+HX3/2L1/2T1/2+HX2T−1/4+H1/2X2Q−1/4+H3/4X3/2Q1/4≪XεHXLD2+HX3/2L1/2D+HX2D0−1/4+H1/2X2Q−1/4+H3/4X3/2Q1/4.From (91), (98), and (114), we obtain(115)Σ0≪XεD06HX2Q−1+D06HXL+D02Q+HXLD2+HX3/2L1/2D+HX2D0−1/4+H1/2X2Q−1/4+H3/4X3/2Q1/4,which combines (87) yields(116)SII≪XεHXL−1/2+D03/2HXQ−1/4+D03/2HX3/4L1/4+D01/2Q1/4H3/4X1/2+HX3/4L1/4D1/2+HX7/8L1/8D1/4+HXD0−1/16+H7/8XQ−1/16+H15/16X7/8Q1/16≪XεHXu−1/2+D03/2HXQ−1/4+D03/2HX3/4v1/4+D01/2Q1/4H3/4X1/2+HX3/4v1/4D1/2+HX7/8v1/8D1/4+HXD0−1/16+H7/8XQ−1/16+H15/16X7/8Q1/16.


## 5.2. The Estimate of Type I Sums

In this section, we shall deal with the estimate of the sums of Type I. First, we have(117)SI=∑1⩽k⩽Hck∑d⩽Dd,2=1ξd∑M<m⩽M1am∑L′<ℓ⩽L1′mℓ+2≡0moddeαm2ℓ2k,where(118)L′=maxL,X2m,L1′=minL1,Xm.By a splitting argument, there holds(119)SI≪Xε⋅max1≪T≪DΣ3,where(120)Σ3=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑L′<ℓ⩽L1′mℓ+2≡0moddeαm2ℓ2k.Form,d=1, there exists m¯, which satisfies 0⩽m¯⩽d−1, such that mm¯≡1modd. Therefore, the equation mℓ+2≡0modd is equivalent to ℓ≡−2m¯modd, i.e., ℓ=−2m¯+dr for some r∈ℤ. Then, it follows from Cauchy’s inequality that(121)Σ32≪HMT∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑L′<ℓ⩽L1′mℓ+2≡0moddeαm2ℓ2k2≪HMT∑1⩽k⩽H∑T<d⩽2Td,2∑M<m⩽M1m,d∑L′+2m¯/d<r⩽L1′+2m¯/deαm2−2m¯+dr2k2=HMT∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑L′+2m¯/d<r⩽L1′+2m¯/deαm2d2r2−4m¯drk2=HMT∑1⩽k⩽H∑T<d⩽2Td,2=1=1∑M<m⩽M1m,d=1=1×∑L′+2m¯/d<r⩽L1′+2m¯/deαm2d2r12−r22−4m¯dr1−r2k.Set(122)R=L′+2m¯d,R1=L1′+2m¯d.Then, we have(123)Σ32≪XεH2M2LT+HMT∑1⩽k⩽H∑T<d⩽2Td,2∑M<m⩽M1m,d=1×∑R<r1,r2⩽R1r1≠r2eαm2d2r12−r22−4m¯dr1−r2k≪XεH2M2LT+HMT⋅Σ31,where(124)Σ31=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑R<r1,r2⩽R1r1≠r2eαm2d2r12−r22−4m¯dr1−r2k.ForΣ31, we have(125)Σ31=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑s1,s2∑R<r1,r2⩽R1r1−r2=s1r1+r2=s2r1≠r21eαm2d2s1s2−4m¯ds1k=∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑s1,s22R<s1+s2⩽2R12R<s2−s1⩽2R1s1≡s2mod2,s1≠0eαm2d2s1s2−4m¯ds1k≪∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑1⩽s1⩽4L/T∑s2:s2≡s1mod22R−s1<s2⩽2R1−s12R+s1<s2⩽2R1+s1eαm2d2s1s2k≪∑1⩽k⩽H∑T<d⩽2Td,2=1∑M<m⩽M1m,d=1∑1⩽s1⩽4L/T∑R−s1<t⩽R1−s1R<t⩽R1e2αm2d2s1tk.Next, we will discuss the estimate of the right-hand side of (125) in two cases.Case 1.
Suppose thatMT⩽D0, and under this condition, there holds 1≪M,T≪D0. By Lemma 4, we have(126)Σ31≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s⩽8LminL,12αm2d2sk≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s⩽8Lmin256HM2T2L22m2d2sk,12αm2d2sk≪∑1⩽n⩽256HM2T2Lτ7nminHM2T2L2n,1αn≪XεHX2T21Q+1L+QHX2T2≪XεHX2D02Q+HXMTT+Q≪XεHX2D02Q+HXD02+Q.
From (119), (123), and (126), we derive that, under the condition MT⩽D0, there holds(127)SI≪XεHX1/2D01/2+HXD03/2Q−1/2+HX1/2D03/2+H1/2D01/2Q1/2.Case 2.
Now, we suppose thatMT>D0. Set(128)R′=maxR,R−s1,R1′=minR1,R1−s1.
Applying Lemma1 to (125), we have(129)Σ31≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s1⩽4L/T∑R′<t⩽R1′e2αm2d2s1tk≪∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s1⩽4L/T∫−∞+∞Kθ1∑L/4T<t⩽4L/Te2αm2d2s1tk+θtdθ=∫−∞+∞K1θ⋅Σ32θ,Tdθ,where(130)K1θ=min15L4T+1,1πθ,1π2θ2,Σ32θ,T=∑1⩽k⩽H∑T<d⩽2T∑M<m⩽M1∑1⩽s1⩽4L/T∑L/4T<t⩽4L/Te2αm2d2s1tk+θt.
According to (7) and (129), it is easy to see that(131)Σ31≪ℒ⋅max0⩽θ⩽1Σ32θ,T.
ForΣ32θ,T, we have(132)Σ32θ,T≪∑1⩽k⩽H∑MT<h⩽4MTτh∑1⩽s1⩽4L/T∑L/4T<t⩽4L/Te2αh2stk+θt.
It follows from Cauchy’s inequality that(133)Σ32θ,T2≪H∑MT<h⩽4MTτ2h∑1⩽s⩽4L/T1∑1⩽k⩽H∑MT<h⩽4MT×∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/Te2αh2stk+θt2≪XεHML∑1⩽k⩽H∑MT<h⩽4MT∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/Te2αh2sk+θt1−t2≪XεHML∑1⩽k⩽H∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/T∑MT<h⩽4MTe2αh2skt1−t2≪XεH2M2L3T+XεHML⋅Σ33,where(134)Σ33=∑1⩽k⩽H∑1⩽s⩽4L/T∑L/4T<t1,t2⩽4L/T∑MT<h⩽4MTe2αh2skt1−t2.
ForΣ33, we have(135)Σ33=∑1⩽k⩽H∑1⩽s⩽4L/T∑1⩽r1⩽4L/T1⩽r2⩽8L/T∑4L/T<t1,t2⩽4L/Tt1−t2=r1t1+t2=r21∑MT<h⩽4MTe2αh2skr1≪LT∑1⩽k⩽H∑1⩽s⩽4L/T∑1⩽s⩽4L/T∑MT<h⩽4MTe2αh2skr1≪LT∑1⩽n⩽32HL2/T2τ4n∑MT<h⩽4MTeαh2n.
Therefore, by Cauchy’s inequality, one has(136)Σ332≪L2T2∑1⩽n⩽32HL2/T2τ42n∑1⩽n⩽32HL2/T2∑MT<h⩽4MTeαh2n2≪XεHL4T4∑1⩽n⩽32HL2/T2∑MT<h1,h2⩽4MTeαh12−h22n≪XεH2ML6T5+XεHL4T4⋅Σ34,where(137)Σ34=∑MT<h1,h2⩽4MTh1≠h2∑1⩽n⩽32HL2/T2eαh12−h22n.
ForΣ34, by Lemma 3, we have(138)Σ34=∑1⩽|t1|,|t2|⩽8MT∑MT<h1,h2⩽4MTh1−h2=t1h1+h2=t21∑1⩽n⩽32HL2/T2eαt1t2n≪∑1⩽t1,t2⩽8MT∑1⩽n⩽32HL2/T2eαt1t2n≪∑1⩽t1,t2⩽8MTminHL2T2,1αt1t2≪∑1⩽t⩽64M2T2τtminHL2T2,1αt.
It follows from Lemma4 that(139)∑1⩽t⩽64M2T2τtminHL2T2,1αt≪Xε∑1⩽t⩽64M2T2min64HM2L2t,1αt≪XεHM2L21Q+T2HL2+QHM2L2≪XεHX2Q+M2T2+Q.
From (136), (138), and (139), we derive that(140)Σ33≪XεHX1/2L5/2T5/2+HXL2T2Q1/2+H1/2XLT+H1/2L2Q1/2T2,which combines (133) yields(141)Σ32θ,T≪XεHXL1/2T1/2+HX3/4L5/4T5/4+HXLTQ1/4+H3/4X1/2LQ1/4T.
From (123), (131), and (141), we obtain(142)Σ3≪XεHXL−1/2T1/2+HXL−1/4T1/4+HX7/8L1/8T−1/8+HXQ−1/8+H7/8Q1/8X3/4,from which and (139), we derive that, under the condition MT>D0, and there holds(143)SI≪Xεmax1≪T≪DHXT1/2L1/2+HXT1/4L1/4+HX7/8L1/8T1/8+HXQ1/8+H7/8Q1/8X3/4≪XεHXD1/2w1/2+HXD1/4w1/4+HX7/8L1/8M1/8MT1/8+HXQ1/8+H7/8Q1/8X3/4≪XεHXw−1/2D1/2+HXw−1/4D1/4+HXD0−1/8+HXQ−1/8+H7/8Q1/8X3/4.


## 5.3. Proof of Lemma5

From (116), (127), and (143), by taking(144)Q=X4138/15θ,then we deduce that, under conditions (3) and (4), there holds(145)SI≪X1−ϖandSII≪X1−ϖ,for some ϖ>0. This completes the proof of Lemma 5.



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