# A System of Two Diophantine Inequalities with Primes

Authors: Xue Han; Huafeng Liu; Deyu Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

Let1<d<c<128/119, 1<α<β<61−d/c. In this paper, we prove that there exist positive real numbers N10 and N20 depending on c,d,α,β such that for all real numbers N1>N10, N2>N20 and α≤N2/N1d/c≤β, the system of two Diophantine inequalities p1c+⋯+p6c−N1<N1−1/c128/119−clog109N1,p1d+⋯+p6d−N2<N2−1/d128/119−dlog109N2 is solvable in prime variables p1,…,p6.


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## Body

## 1. Introduction

Assume thatc>1 is not an integer and ε is a sufficiently small positive number. Let Hc denote the least integer s such that the Diophantine inequality(1)p1c+p2c+⋯+psc−N<εis solvable in primes p1,p2,…,ps for sufficiently large N. In 1952, Piatetski-Shapiro [1] proved that(2)limsupc⟶∞Hcclogc≤4.Piatetski-Shapiro also provedHc≤5 for 1<c<3/2. Tolev [2] first improved this result for c close to one. More precisely, Tolev proved that if 1<c<15/14`, the Diophantine inequality(3)p1c+p2c+p3c−N<εNhas prime solutions p1,p2,p3 for large N, where εN=N−1/c15/14−clog9N. Later, this result was improved by many authors (see [3–9]), and many analogous problems of this type were studied (for example, see [10–17]).In 1995, Tolev [18] first considered the system of two Diophantine inequalities with primes(4)p1c+⋯+p5c−N1<ε1∗N1,p1d+⋯+p5d−N2<ε2∗N2.He established that for all real numbersN1,N2 satisfying N1>N10,N2>N20 and α≤N2/N1d/c≤β, system (4) with(5)ε1∗N1=N1−1/c35/34−clog12N1,ε2∗N2=N2−1/d35/34−dlog12N2has solutions in primes p1,…,p5, where c,d,α,β are real numbers satisfying the conditions(6)1<d<c<35/34,1<α<β<51−d/c,and N10,N20 depending on c,d,α,β are sufficiently large numbers. Subsequently, Tolev’s result was improved by Zhai [19] and Zhai and Cao [20]. Now the best result is due to Zhai and Cao [20] who proved that system (4) with(7)ε1∗N1=N1−1/c27/26−clog100N1,ε2∗N2=N2−1/d27/26−dlog100N2is solvable for 1<d<c<27/26.In this paper, we consider the following system of two Diophantine inequalities over primesp1,…,p6:(8)p1c+⋯+p6c−N1<ε1N1,p1d+⋯+p6d−N2<ε2N2,where c>1, d>1 are different numbers but close to 1 and ε1N1,ε2N2 satisfy(9)ε1N1⟶0,asN1⟶∞,ε2N2⟶0,asN2⟶∞.We have to impose a condition on the orders ofN1 and N2 due to the inequality(10)x1c+⋯+x6cd/c≤x1d+⋯+x6d≤61−d/cx1c+⋯+x6cd/c,which holds for every positive x1,…,x6 provided 1<d<c. Our result is as follows.Theorem 1.
Suppose thatc,d,α,β are real numbers and satisfy conditions(11)1<d<c<128/119,(12)1<α<β<61−d/c.
Then, there exist positive real numbersN10, N20 depending on c,d,α,β such that for all real numbers N1>N10,N2>N20 and(13)α≤N2/N1d/c≤β,system (8) with(14)ε1N1=N1−1/c128/119−clog109N1,ε2N2=N2−1/d128/119−dlog109N2has solutions in primes p1,…,p6.Notation. Throughout this paper, the letter p, with or without a subscript, always represents a prime, c and d are real numbers satisfying (11), and α and β are real numbers satisfying (12). η denotes a sufficiently small positive number depending on c and d. χt denotes the characteristic function over the interval −1,1. ρ=β′+iγ′ is the nontrivial zero of the Riemann zeta function ζs. As usual, Λn and τn are the von Mangoldt function and the divisor function, respectively. N1 and N2 are sufficiently large numbers. We set(15)X=N11/c,τ1=X3/4−c−η,τ2=X3/4−d−η,ε1=X−128/119−clog107X,ε2=X−128/119−dlog107X,K1=X128/119−clog−106X,K2=X128/119−dlog−106X,et=e2πit,φt=e−πt2,φδt=δφδt.


## 2. Outline of the Proof

Letλ denote a sufficiently small positive number, whose value depends on c,d,α,β and will be determined more precisely in Lemma 1. Let(16)B=∑λX<p1,…,p6≤Xlogp1⋯logp6χp1c+⋯+p6c−N1ε1logXχp1d+⋯+p6d−N2ε2logX,(17)Sx,y=∑λX<p≤Xlogpexpc+ypd,(18)D=∫−∞∞∫−∞∞S6x,ye−N1x−N2yφε1xφε2ydxdy.We divide the plane into three regions: the neighbourhood of originΩ1, the intermediate region Ω2, and the trivial region Ω3, which are defined as(19)Ω1=x,y:maxx/τ1,y/τ2<1,Ω2=x,y:maxx/τ1,y/τ2≥1,maxx/K1y/K2≤1,Ω3=x,y:maxx/K1y/K2>1.Thus, the integralD can be represented as(20)D=D1+D2+D3,where(21)Di=∫∫ΩiS6x,ye−N1x−N2yφε1xφε2ydxdyi=1,2,3.If we show thatB tends to infinity as X tends to infinity, then Theorem 1 holds. From Lemma 4, it is sufficient to prove that D tends to infinity as X tends to infinity. To do this, noting (20), we shall prove the following:(22)D1≫ε1ε2X6−c−d,(23)D2≪ε1ε2X6−c−dlogX,(24)D3≪1.In Section3, we first give some auxiliary lemmas. Inequality (22) is proved in Section 4. Inequality (23), from which we can get the range of c and d, is proved in Section 5. In Section 6, we complete the proof of Theorem 1.


## 3. Auxiliary Lemmas

Lemma 1.
Letδ∈α,β. There exists λ>0 depending on c,d,α,β such that for the volume V of the domain in six-dimensional space defined by(25)t1,…,t6>λ,t1c+⋯+t6c−1<μ1,t1d+⋯+t6d−δ<μ2,we have(26)V≫μ1μ2,provided μ1,μ2 are sufficiently small.Proof.
This lemma is similar to Lemma1 in Tolev [18]. We can write the volume of V as(27)V=∫⋯∫t1,…,t6>λt1c+⋯+t6c−1<μ1t1d+⋯+t6d−δ<μ21dt1⋯dt6.
Then, we can fixt1,…,t5 to get the range of t6 from last two inequalities and get(28)1−t1c+⋯+t5c−μ11/c<t6<1−t1c+⋯+t5c+μ11/c,(29)δ−t1d+⋯+t5d−μ21/d<t6<δ−t1d+⋯+t5d+μ21/d.
Sinceμ1 and μ2 are sufficiently small, we can adjust the value of λ to ensure that there are intersections between (28) and (29). We may also get this lemma by adjusting the value of λ and using circle method to estimate(30)t1c+⋯+t6c−1<μ1,t1d+⋯+t6d−δ<μ2.Lemma 2 (see [18], Lemma 2).
The functionφt=e−πt2 has the following properties:(31)φx=∫−∞∞φte−xtdt,(32)χt/ϱ≥φt−e−πϱ2forϱ>0,(33)φt≥e−πfort≤1.Lemma 3.
LetB>1 denote a real number and f be a smooth real function on B,2B. Suppose that there exists a positive constant A=Af such that(34)AB1−j≪jfjx≪jAB1−j,forx∼Bandj∈ℕ,where the implied absolute constant depends only on j. Then, there exists an exponent pair κ,ι with(35)0≤κ≤1/2≤ι<1,such that(36)∑B<n≤B+hefn≪AκBι,for1<h≤B.Proof.
We can find this lemma in Ivić ([21], pp. 72–79).Lemma 4.
The quantitiesB and D satisfy(37)B≥D+O1.Proof.
From (31) and (32) in Lemma 2, we have(38)χp1c+⋯+p6c−N1ε1logX≥φp1c+⋯+p6c−N1ε1−e−πlogX2φp1c+⋯+p6c−N1ε1=φN1−p1c+⋯+p6cε1=∫−∞∞φxe−xN1−p1c+⋯+p6cε1dx=∫−∞∞exp1c+⋯+p6ce−xN1φε1xdx,where we substitute ε1x for x. Similarly, we have(39)χp1d+⋯+p6d−N2ε2logX≥φp1d+⋯+p6d−N2ε2−e−πlogX2φp1d+⋯+p6d−N2ε2=∫−∞∞eyp1d+⋯+p6de−yN2φε2ydy.
Then,(40)B≥∑λX<p1,…,p6≤Xlogp1⋯logp6∫−∞∞∫−∞∞exp1c+⋯+p6c+yp1d+⋯+p6de−xN1−yN2φε1xφε2ydxdy+O1≥D+O1.Lemma 5.
There are two real functionsGx,Fx defined in a,b, Gx≤H for a≤x≤b and Gx/Fx is a monotonous function. Write(41)I=∫abGxeFxdx.
IfF′x≥h>0 or F′x≤−h<0 for all x∈a,b, then(42)I≪H/h.
IfF″x≥h>0 for all x∈a,b, then(43)I≪H/h.Proof.
This lemma can be found in Ivić ([21], pp. 56-57).Lemma 6.
LetΨu=∑n≤uΛn and 2≤t≤u; then,(44)Ψu=u−∑γ′≤tuρρ+Oulog2ut,where ρ=β′+iγ′ is the nontrivial zero of ζs.Proof.
This is a well-known explicit formula, which can be found in Karatsuba ([22], p. 80).


## 4. The Estimate of the IntegralD1

In this section, we give the estimate of the integralD1 and set T=X3/4+119c/512−η.Lemma 7.
If1≤T1≤T, then(45)1T1∑0<γ′≤T1Xβ′≪Xe−logX1/4,and(46)1T1∑0<γ′≤T1Xβ′≪Xe−logX1/4.Proof.
We can get this lemma from Lemma11 and Lemma 12 of Tolev [2].Lemma 8.
Ifmaxx/τ1,y/τ2<1, then(47)Jx,y≪Xe−logX1/5,where(48)Jx,y=∑γ′≤TIρx,y,Iρx,y=∫λXXetcx+tdytρ−1dt.Proof.
Without loss of generality, we consider the case0≤x<τ1,0≤y<τ2. From ρ=β′+iγ′, we have(49)Iρx,y=∫λXXtβ′−1etcx+tdy+γ′2πlogtdt.
Let(50)Ft=tcx+tdy+γ′2πlogt.
We define three sets of nontrivial zeroes ofζs as(51)M1=ρ:γ′≤T,−γ′/2π>3cXcx+dXdy/2,M2=ρ:γ′≤T,cλXcx+dλXdy/2≤−γ′/2π≤3cXcx+dXdy/2,M3=ρ:γ′≤T,−γ′/2π<cλXcx+dλXdy/2,and the set M2 may be empty.
We first considerX−c≤x≤τ1 or X−d≤y≤τ2. In this situation, we consider the following three cases.Case 1. When ρ∈M1. In this case, we have(52)F′t=ctc−1x+dtd−1y+γ′2πt≤1λXcλXcx+dλXdy−−γ′2π<0.Applying Lemma5, we have(53)Iρx,y≪Xβ′−cXcx−dXdy−γ′/2π≪Xβ′γ′.Therefore, by (46), we can obtain(54)∑ρ∈M1Iρx,y≪∑0<γ′≤TXβ′γ′≪logXmax1≤T1≤T1T1∑0<γ′≤T1Xβ′≪Xe−logX1/5.Case 2. When ρ∈M2. In this case, we have(55)F″t=cc−1tc−2x+dd−1td−2y−γ′2πt2≍Xcx+XdyX−2,Iρx,y≪Xβ′Xcx+Xdy.Hence, we use (45) and get(56)∑ρ∈M2Iρx,y≪1Xcx+Xdy∑0<γ′≤Xcx+XdyXβ′≪Xe−logX1/4.If the setM2 is empty, then the upper bound is trivial.Case 3. When ρ∈M3. In this case, we have(57)F′t=ctc−1x+dtd−1y+γ′2πt≫1XcλXcx+dλXdy−−γ′2π>0,Iρx,y≪Xβ′cλXcx+dλXdy+γ′/2π.
From (46), we have(58)∑ρ∈M3Iρx,y≪∑−πcλXcx−πdλXdy<γ′≤TXβ′cλXcx+dλXdy+γ′/2π≪∑−πcλXcx−πdλXdy<γ′≤Xcx+XdyXβ′Xcx+Xdy+∑Xcx+Xdy<γ′≤TXβ′γ′≪logXmax1≤T1≤T1T1∑0<γ′≤T1Xβ′≪Xe−logX1/5.
Therefore, from (54)–(58), we obtain(59)Jx,y≪Xe−logX1/5.
Now we consider the remaining case0≤x≤X−c and 0≤y≤X−d. We use the trivial estimate of Iρx,y and get(60)∑ρ∈M2Iρx,y≪∑ρ∈M2Xβ′≪Xβ0,where β0=maxρ∈M2β′<1. ∑ρ∈M1Iρx,y and ∑ρ∈M3Iρx,y are estimated analogically as in the previous case. Then, the estimate for Jx,y is established.Lemma 9.
Ifmaxx/τ1,y/τ2<1, then(61)Sx,y=Ix,y+OXe−logX1/5,where(62)Ix,y=∫λXXextc+ytddt.Proof.
Noting(63)∑λX<pv≤Xv>1logpepcvx+pdvy≪∑p2≤Xlogp∑logλXlogp<v<logXlogp1≪X1/2,we have(64)Sx,y=∑λX<n≤XΛnencx+ndy−∑λX<pv≤Xv>1logpepcvx+pdvy=∑λX<n≤XΛnencx+ndy+OX1/2≔Ux,y+OX1/2,where v denotes an integer number. Applying Abel’s transformation, we get(65)Ux,y=−∫λXXΨt−ΨλXddtetcx+tdydt+ΨX−ΨλXeXcx+Xdy.
Then, Lemma6 implies(66)Ux,y=−∫λXXt−λX−∑γ′≤Ttρ−λXρρ+OXlog2XTddtetcx+tdydt+X−λX−∑γ′≤TXρ−λXρρ+OXlog2XTeXcx+Xdy=−∫λXXt−λX−∑γ′≤Ttρ−λXρρddtetcx+tdydt+X−λX−∑γ′≤TXρ−λXρρeXcx+Xdy+OXlog2XXcτ1+Xdτ2T.
Using integration by parts, we have(67)Ux,y=∫λXXetcx+tdy1−∑γ′≤Ttρ−1dt+OXlog2XXcτ1+Xdτ2T=Ix,y−Jx,y+OXlog2XXcτ1+Xdτ2T.
From Lemma8, this lemma follows.Lemma 10 (see [18], Lemma 8).
ForSx,y defined by (17), we have(68)∫∫Ω1Sx,y4φε1xφε2ydxdy≪ε1ε2X4−c−dlog8X.Lemma 11 (see [18], Lemma 9).
ForIx,y defined in Lemma 9, we have(69)∫−∞∞∫−∞∞Ix,y4φε1xφε2ydxdy≪ε1ε2X4−c−dlog4X.Write(70)H=∫−∞∞∫−∞∞I6x,ye−N1x−N2yφε1xφε2ydxdy,and(71)H1=∫∫Ω1I6x,ye−N1x−N2yφε1xφε2ydxdy.Lemma 12.
For integralsH and H1 defined by (70) and (71), respectively, we have(72)H−H1≪ε1ε2X6−c−dlogX.Proof.
We have(73)H−H1≪∫∫ℝ2/Ω1Ix,y6φε1xφε2ydxdy≪maxℝ2\Ω1Ix,y2∫∫ℝ2Ix,y4φε1xφε2ydxdy.
Applying Lemma5 with Gt=1 and Ft=xtc+ytd, from F″t=cc−1xtc−2+dd−1ytd−2, we can get(74)maxℝ2\Ω1Ix,y≪X5/8+η/2.
Then, by Lemma11, we can get this lemma.Lemma 13.
For the integralH defined by (70), we have(75)H≫ε1ε2X6−c−d.Proof.
We have(76)H=∫⋯∫λX<t1,…,t6≤X∫−∞∞∫−∞∞ext1c+⋯+t6c−N1eyt1d+⋯+t6d−N2×φε1xφε2ydxdydt1⋯dt6=∫⋯∫C1Δdt1⋯dt6+∫⋯∫C2Δdt1⋯dt6,where(77)C1=λX<t1,…,t6≤X:t1c+⋯+t6c−N1ε1<1andt1d+⋯+t6d−N2ε2<1,C2=λX<t1,…,t6≤X:t1c+⋯+t6c−N1ε1≥1ort1d+⋯+t6d−N2ε2≥1,Δ=∫−∞∞∫−∞∞ext1c+⋯+t6c−N1eyt1d+⋯+t6d−N2φε1xφε2ydxdy.
From (31), we get(78)∫−∞∞ext1c+⋯+t6c−N1φε1xdx=φt1c+⋯+t6c−N1ε1,∫−∞∞ext1d+⋯+t6d−N2φε2xdx=φt1d+⋯+t6d−N2ε2.
In (76), the integral over C2 is convergent, and hence ∫⋯∫C2Δdt1⋯dt6=O1. Applying Lemma 2, we have(79)H≫∫⋯∫C11dt1⋯dt6≫X6∫⋯∫C1′1dt1/X⋯dt6/X,where(80)C1′=λ<t1/X,…,t6/X:t1/Xc+⋯+t6/Xc−1<ε1X−candt1/Xd+⋯+t6/Xd−N2/Xd<ε2X−d.
From (13) and Lemma 1, we have(81)H≫ε1ε2X6−c−d.Lemma 14.
For the integralD1 in (20), we have(82)D1≫ε1ε2X6−c−d.Proof.
We have(83)D1−H1≪∫∫Ω1S6x,y−I6x,yφε1xφε2ydxdy≪maxΩ1Sx,y−Ix,y2∫∫Ω1Sx,y4+Ix,y4φε1xφε2ydxdy.
From Lemmas9–11, we can obtain(84)D1−H1≪ε1ε2X6−c−dlogX.
By Lemma12, we have(85)D1=H+Oε1ε2X6−c−dlogX.
Then, from Lemma13, we can get this lemma.


## 5. The Estimate for the IntegralD2

Lemma 15.
Letam,bn be arbitrary complex numbers and(86)maxx/τ1,y/τ2≥1,maxx/K1,y/K2≤1,X1/4<R≤X1/2,X1/4<L<L1≤2L,LR≤X.
Write(87)W=∑L<n≤L1∑X1/4<m≤Rambnemncx+mndy.
Then,(88)W≪ℬ1/2X101/238log−13X,where(89)A=∑X1/4<m≤Ram2,ℬ=∑L<n≤L1bn2.Proof.
Without loss of generality, we can consider the caseτ1≤x≤K1,0<y≤K2. We define Ri0≤i≤Q:(90)R0=X1/4,Ri+1=minRi+s,R,RQ=R,where s∈1,R will be determined later and Q≪R/s. Thus, W can be rewritten as(91)W=∑L<n≤L1∑1≤i≤Q∑Ri−1<m≤Riambnemncx+mndy.
By Cauchy’s inequality, we have(92)W2≤ℬQ∑L<n≤L1∑1≤i≤Q∑Ri−1<m1,m2≤Riam1a¯m2em1c−m2cncx+m1d−m2dndy.
We rearrange the sums as follows:(93)W2≪ℬQ∑L<n≤L1∑1≤i≤Q∑Ri−1<m≤Riam2+ℬQ∑1≤i≤Q∑Ri−1<m1,m2≤Rim1≠m2am1am2∑L<n≤L1em1c−m2cncx+m1d−m2dndy≪ℬQAL+∑1≤h≤s∑1≤i≤Q∑Ri−1<m≤Ri−hamam+h∑L<n≤L1efn,where(94)fn=m+hc−mcncx+m+hd−mdndy.
Next, we handle the exponential sum∑L<n≤L1efn of (93). From the derivatives of f, we get(95)fjn≍m+hc−mcLc−1x+m+hd−mdLd−1yL1−j.
Then, by Lemma3 with the exponent pair (see also [21], p. 77)(96)BA2BA2B0,1=13/40,22/40,we have(97)∑L<n≤L1efn≪m+hc−mcLc−1x+m+hd−mdLd−1y13/40L22/40≪m+hc−mcLc−1K1+m+hd−mdLd−1K213/40L22/40≪s13/40R13c−1/40L13c+9/40K113/40+R13d−1/40L13d+9/40K213/40.
Inserting this upper bound into (93), we have(98)W2≪ℬQAL+s13/40R13c−1/40L13c+9/40K113/40+R13d−1/40L13d+9/40K213/40∑0,where(99)∑0∑1≤h≤s∑1≤i≤Q∑Ri−1<m≤Ri−hamam+h.
By Cauchy’s inequality, we obtain(100)∑0≤∑1≤h≤s∑X1/4<m≤R−hamam+h≤∑1≤h≤s∑X1/4<m≤R−h|am|21/2∑X1/4<m≤R−ham+h21/2≪sA.
From (98) and Q≪R/s, we have(101)W2≪AℬLR1/s+s13/40R13c−1/40L13c−31/40K113/40+R13d−1/40L13d−31/40K213/40.
We take1/s=s13/40R13c−1/40L13c−31/40K113/40+R13d−1/40L13d−31/40K213/40, i.e.,(102)s=R131−c/53L31−13c/53K1−13/53+R131−d/53L31−13d/53K2−13/53,and then s∈1,R. Inserting this value of s into (101), we get(103)W2≪ABX101/119log−26X,which yields this lemma.Lemma 16 (see [2], Lemma 9).
Letam,bn be arbitrary complex numbers and L<L1≤2L,L≤X. Write(104)V=∑L<n≤L1∑X1/4<m≤X/nambnemncx+mndy.
Then, there existam′,bn′ satisfying am′≤am,bn′≤bn such that(105)V≪logX∑L<n≤L1∑X1/4<m≤X/Lam′bn′emncx+mndy.Lemma 17.
Assume thatmaxx/τ1,y/τ2≥1,maxx/K1,y/K2≤1; then,(106)Sx,y≪X110/119log−9X.Proof.
Without loss of generality, we can consider the caseτ1≤x≤K1,0<y≤K2. Clearly,(107)Sx,y=V0x,y−V1x,y+OX1/2,where(108)V0x,y=∑X1/4<n≤XΛnencx+ndy,V1x,y=∑X1/4<n≤λXΛnencx+ndy.
Hence, it is sufficient to prove that(109)V0x,y,V1x,y≪X110/119log−9X.
The estimates ofV0x,y and V1x,y are similar, and thus we focus on the estimate of V0x,y. Using Vaughan’s identity (see [23]), we have(110)V0x,y=S1−S2−S3,where(111)S1=∑g≤X1/4μg∑l≤X/gloglelgcx+lgdy,S2=∑k≤X1/2∑r≤X/kckekrcx+krdy,S3=∑X1/4<m≤X3/4∑X1/4<n≤X/mamΛnemncx+mndy,where ck≤logk and am≤τm.
ForS2, noting(112)∑X1/4<k≤X1/2∑r≤X1/4ckekrcx+krdy≪X3/4logX,we have(113)S2=S21+S22+OX3/4logX,where(114)S21=∑k≤X1/4∑r≤X/kckekrcx+krdy,S22=∑X1/4<k≤X1/2∑X1/4<r≤X/kckekrcx+krdy.
ForS1, we have(115)S1≤∑g≤X1/4∑l≤X/gloglelgcx+lgdy.
The second summation overl in (115) is(116)≪log2XmaxL2∈L,L1∑L<l≤L2ehl,where L<L1≤2L, L1≤X/g, and hl=lgcx+lgdy. Noting hjl≍Lc−1gcx+Ld−1gdyL1−j, Lemma 3 with exponent pair 13/40,22/40 gives(117)∑L<l≤L1ehl≪Lc−1gcK1+Ld−1gdK213/40L22/40≪X13c/40L9/40K113/40+X13d/40L9/40K213/40.
Recalling (115) and (116), we have(118)S1≪X110/119log−9X.
Arguing similarly, we can get the following same bound ofS21:(119)S21≪X110/119log−9X.
ForS3, we divide it into three parts:(120)S3=W1+W2+W3,where(121)W1=∑X1/2<n≤X3/4∑X1/4<m≤X/namΛnemncx+mndy,W2=∑X1/2<n≤X3/4∑X1/4<m≤X/nanΛmemncx+mndy,W3=∑X1/4<n≤X1/2∑X1/4<m≤X1/2amΛnemncx+mndy.
ForW1, using dyadic subdivision and Lemma 16, we have(122)W1≪log2XW1′L,where(123)W1′L=∑L<n≤L1∑X1/4<m≤X/Lam′bn′emncx+mndy,where X1/2≤L<L1≤2L≤X3/4,am′≤am≤τm, and bn′≤Λn. Lemma 15 implies(124)W1′L≪XLlog3X⋅LlogX1/2X101/238log−13X=X110/119log−11X,where we used the mean value estimates(125)∑m≤yτ2m≪ylog3y,∑n≤yΛ2n≪ylogy.
Therefore, we have(126)W1≪X110/119log−9X.
We estimateW2 and W3 similar to W1 and get(127)W2,W3≪X110/119log−9X.
ForS22, we follow a similar argument to S3 and get(128)S22=U1+U2,where(129)U1=∑X1/4<r≤X1/2∑X1/4<k≤X1/2ckekrcx+krdy,U2=∑X1/2<r≤X3/4∑X1/4<k≤X/rckekrcx+krdy.
Using Lemmas15 and 16 and the mean value estimate(130)∑k≤ylog2k≪ylog2y,we get(131)U1,U2≪X110/119log−9X.
From (110) and (118)–(131), we can obtain (109).Lemma 18 (see [18], Lemma 14).
ForSx,y defined by (17), we have(132)∫−∞∞∫−∞∞Sx,y4φε1xφε2ydxdy≪X2log6X.Lemma 19.
For the integralD2 in (20), we have(133)D2≪ε1ε2X6−c−dlogX.Proof.
By Lemmas17 and 18, we have(134)D2≪maxΩ2Sx,y2∫−∞∞∫−∞∞Sx,y4φε1xφε2ydxdy≪X110/119log−9X2X2log6X≪X458/119log−12X,which yields this lemma.


## 6. Proof of Theorem1

In this section, we first give the estimate ofD3 in (20) and then complete the proof of Theorem 1.For the integralD3 in (20), by Lemma 2, we can easily get(135)D3≪1.From Lemmas14 and 19 and (135), we know that (22)–(24) follows, respectively. Therefore, recalling Lemma 4 and (20), we can get(136)B≫ε1ε2X6−c−d,from which we complete the proof of Theorem 1.



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*Source: 6613947-2021-02-15.xml*