Let [α] denote the integer part of the real number α, N be a sufficiently large integer and (κ,λ) be the exponent pair. In this paper, we prove that for 1<c<3+3κ−λ3κ+2, the Diophantine equation [pc1]+[pc2]+[pc3]=N is solvable in prime variables p1,p2,p3. If we take (κ,λ)=(81242,132242), we can get the range 1<c<837727, which improves the previous result of Cai.
Citation: Jing Huang, Ao Han, Huafeng Liu. On a Diophantine equation with prime variables[J]. AIMS Mathematics, 2021, 6(9): 9602-9618. doi: 10.3934/math.2021559
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Let [α] denote the integer part of the real number α, N be a sufficiently large integer and (κ,λ) be the exponent pair. In this paper, we prove that for 1<c<3+3κ−λ3κ+2, the Diophantine equation [pc1]+[pc2]+[pc3]=N is solvable in prime variables p1,p2,p3. If we take (κ,λ)=(81242,132242), we can get the range 1<c<837727, which improves the previous result of Cai.
Diophantine equation is a classical problem in number theory. Let [α] denote the integer part of the real number α and N be a sufficiently large integer. In 1933, Segal [27,28] firstly studied additive problems with non-integer degrees, and proved that there exists a k0(c)>0 such that the Diophantine equation
[xc1]+[xc2]+⋯+[xck]=N | (1.1) |
is solvable for k>k0(c), where c>1 is not an integer. Later, Deshouillers [5] improved Segal's bound of k0(c) to 6c3(logc+14) with c>12. Further Arkhilov and Zhitkov [1] refined Deshouillers's result to 22c2(logc+4) with c>12. Afterwards, many results of various Diophantine equations were established (e.g., see [7,10,14,16,17,18,19,21,25,41,42]). In particular, Laporta [17] in 1999 showed that the equation
[pc1]+[pc2]=N | (1.2) |
is solvable in primes p1, p2 provided that 1<c<1716 and N is sufficiently large. Recently, the range of c in (1.2) was enlarged to 1<c<1411 by Zhu [40]. Kumchev [15] showed that the equation
[mc]+[pc]=N | (1.3) |
is solved for almost all N provided that 1<c<1615, where m is an integer and p is a prime. Afterwards, the range of c in (1.3) was enlarged to 1<c<1711 by Balanzario, Garaev and Zuazua [3].
In 1995, Laporta and Tolev [18] considered the equation
[pc1]+[pc2]+[pc3]=N | (1.4) |
with prime variables p1,p2,p3. Denote the weighted number of solutions of Eq (1.4) by
R(N)=∑[pc1]+[pc2]+[pc3]=N(logp1)(logp2)(logp3). | (1.5) |
They established the following asymptotic formula
R(N)=Γ3(1+1c)Γ(3c)N3c−1+O(N3c−1exp(−log13−δN)) |
for any 0<δ<13 and 1<c<1716. Afterwards, the range of c was enlarged to 1<c<1211 by Kumchev and Nedeva [16], to 1<c<258235 by Zhai and Cao [39], and to 1<c<137119 by Cai [4].
In this paper, we first show a more general result related to (1.5) by proving the following theorem.
Theorem 1.1. Let N be a sufficiently large integer. Then for 1<c<3+3κ−λ3κ+2, we have
R(N)=Γ3(1+1c)Γ(3c)N3c−1+O(N3c−1exp(−log13−δN)) | (1.6) |
for any 0<δ<13, where (κ,λ) is an exponent pair, and the implied constant in the O−symbol depends only on c.
Choosing (κ,λ)=BA2BABABABAB(0,1)=(81242,132242) in Theorem 1.1, we can immediately get the following corollary, which further improves the result of Cai [4].
Corollary 1.2. Under the notations of Theorem 1.1, for 1<c<837727 the asymptotic formula (1.6) follows.
It is easy to verify that the range of c in Corollary 1.2 is larger than one of Cai's result. Our improvement mainly derives from more accurate estimates of exponential sums by combining Van der Corput's method, exponent pairs and some elementary methods. Also the estimates of exponential sums has lots of applications in problems including automorphic forms (e.g., see [8,11,12,20,22,24,29,30,31,32,33,34,35,36,37,38]).
Notation. Throughout the paper, N always denotes a sufficiently large integer. The letter p, with or without subscripts, is always reserved for primes. Let ε∈(0,10−10(3+3κ−λ3κ+2−c)). We denote by {x} and ‖x‖ the fraction part of x and the distance from x to the nearest integer, respectively. Let 1<c<3+3κ−λ3κ+2 and
P=N1c, τ=P1−c−ε, e(x)=e2πix, S(α)=∑p≤P(logp)e(α[pc]). |
To prove Theorem 1.1, we need the following lemmas.
Lemma 2.1 ([9,Lemma 5]). Suppose that zn is a sequence of complex numbers, then we have
|∑N≤n≤2Nzn|2≤(1+NQ)Q∑q=0(1−qQ)Re(∑N≤n≤2N−q¯znzn+q), |
where Re(t) and ¯t denote the real part and the conjugate of the complex number t, respectively.
Lemma 2.2. Suppose that |x|>0 and c>1. Then for any exponent pair (κ,λ) and M≤a<b≤2M, we have
∑a≤n≤be(xnc)≪(|x|Mc)κMλ−κ+M1−c|x|. |
Proof. We can get this lemma from [6,(3.3.4)].
Lemma 2.3 ([2,Lemma 12]). Suppose that t is not an integer and H≥3. Then for any α∈(0,1), we have
e(−α{t})=∑|h|≤Hch(α)e(ht)+O(min(1,1H‖t‖)), |
where
ch(α)=1−e(−α)2πi(h+α). |
Lemma 2.4 ([9,Lemma 3]). Suppose that 3<U<V<Z<X, and {Z}=12, X≥64Z2U, Z≥4U2, V3≥32X. Further suppose that F(n) is a complex valued function such that |F(n)|≤1. Then the sum
∑X≤n≤2XΛ(n)F(n) |
can be decomposed into O(log10X) sums, each of which either of type {I}:
∑M≤m≤2Ma(m)∑N≤n≤2NF(mn) |
with N>Z, where a(m)≪mε and X≪MN≪X, or of type {II}:
∑M≤m≤2Ma(m)∑N≤n≤2Nb(n)F(mn) |
with U≪M≪V, where a(m)≪mε,b(n)≪nε and X≪MN≪X.
Lemma 2.5. Let f(t) be a real value function and continuous differentiable at least three times on [a,b](1≤a<b≤2a), |f‴(x)|∼Δ>0, then we have
∑a<n≤be(f(n))≪aΔ16+Δ−13. |
Moreover, if 0<c1λ1≤c2λ1, |f″(x)|∼λ1a−1, then we have
∑a<n≤be(f(n))≪a12λ121+λ−11; |
if c2λ1≤12, then we have
∑a<n≤be(f(n))≪λ−11. |
Proof. The first result was proved by Sargos [26]. And the remaining two results were due to Jia [13].
Lemma 2.6 ([23,Lemma 2]). Let M>0, N>0, um>0, υn>0, Am>0, Bn>0 (1≤m≤M,1≤n≤N). Let also Q1 and Q2 be given non-negative numbers, Q1≤Q2. Then there is one q such that Q1≤q≤Q2 and
M∑m=1Amqum+N∑n=1Bnq−υn≪M∑m=1N∑n=1(AυnmBumn)1um+υn+M∑m=1AmQum1+N∑n=1BnQ−υn2. |
Lemma 2.7 ([36,Lemma 5]). Let f(x), g(x) be algebraic functions in [a,b], |f″(x)|∼1R, f‴(x)≪1RU, U≥1, |g(x)|≪G, |g′(x)|≪GU−1. [α,β] is the image of [a,b] under the mapping y=f′(x). nu is the solution of f′(n)=u.
bu={1,α<u<β,12,u=α∈Noru=β∈N. |
Then we have
∑a<n≤bg(n)e(f(n))=∑α<u≤βbug(nu)√|f″(nu)|e(f(nu)−unu+18)+O(Glog(β−α+2)+G(b−a+R)u−1)+O(Gmin(√R,1‖α‖)+Gmin(√R,1‖β‖)). |
Lemma 2.8 ([13,Lemma 3]). Suppose that x∼N, f(x)≪P, and f′(x)≫Δ. Then we have
∑n∼Nmin(D,1‖f(n)‖)≪(P+1)(D+Δ−1)log(2+Δ−1). |
Lemma 2.9. For 0<α<1 and any exponent pair (κ,λ), we have
T(α,X)=∑X<n≤2Xe(α[nc])≪Xκc+λ1+κlogX+XαXc. |
Proof. Throughout the proof of this lemma, we write H=X−κc+1−λ+κ1+κ for convenience. Using Lemma 2.3 we can get
T(α,X)=∑|h|≤Hch(α)∑X<n≤2Xe((h+α)nc)+O((logX)∑X<n≤2Xmin(1,1H||nc||)). |
Then by the expansion
min(1,1H||θ||)=∞∑h=−∞ahe(hθ), |
where
|ah|=min(log2HH, 1|h|, Hh2), |
we have
∑X<n≤2Xmin(1,1H||nc||)≤∞∑h=−∞|ah||∑X<n≤2Xe(hnc)|≪Xlog2HH+∑1≤h≤H1h((hXc)κXλ−κ+XhXc)+∑h≥HHh2((hXc)κXλ−κ+XhXc)≪Xκc+λ1+κlogX, |
where we estimated the sum over n by Lemma 2.2.
In a similar way, we have
∑|h|≤Hch(α)∑X<n≤2Xe((h+α)nc)=c0(α)∑X<n≤2Xe(αnc)+∑1≤h≤Hch(α)∑X<n≤2Xe((h+α)nc)≪Xκc+λ1+κlogX+XαXc. |
Then this lemma follows.
Lemma 2.10 ([42,Lemma 2.1]). Suppose that f(n) is a real-valued function in the interval [N,N1], where 2≤N<N1≤2N. If 0<c1λ1≤|f′(n)|≤c2λ1≤12, then we have
∑N<n≤N1e(f(n))≪λ−11. |
If |f(j)(n)|∼λ1N−j+1(j=1,2), then we have
∑N<n≤N1e(f(n))≪λ−11+N12λ121. |
If |f(j)(n)|∼λ1N−j+1(j=1,2,3,4,5,6), then we have
∑N<n≤N1e(f(n))≪λ−11+Nλλκ1, |
where (κ,λ) is any exponent pair.
Lemma 2.11 ([9,Lemma 6]). Suppose that 0<a<b≤2a and R is an open convex set in C containing the real segment [a,b]. Suppose further that f(z) is analytic on R. f(x) is real for real x∈R. f″(z)≤M for z∈R. There is a constant k>0 such that f″(x)≤−kM for all real x∈R. Let f′(b)=α and f′(a)=β, and define xυ for each integer υ in the range α<υ<β by f′(xυ)=υ. Then we have
∑a<n≤be(f(n))=e(−18)∑α<υ≤β|f″(xυ)|−12e(f(xυ)−υxυ)+O(M−12+log(2+M(b−a))). |
Lemma 3.1. Let P56≪X≪P, H=X1−(1+2κ)c+λ2+2κ and ch(α) denote complex numbers such that ch(α)≪(1+|h|)−1. Then uniformly for α∈(τ,1−τ), we have
SI=∑|h|∼Hch(α)∑M≤m≤2Ma(m)∑N≤n≤2Ne((h+α)(mn)c)≪X(1+2κ)c+λ2+2κ+2ε | (3.1) |
for any a(m)≪mε, where (κ,λ) is any exponent pair, X≪MN≪X and M≪Y with Y=min{X1,X2,X3,X4,X5,X6,X7,X8},
X1=X152(1+2κ)c+λ2+2κ−c2−112, X2=X5211(1+2κ)c+λ2+2κ−4c11−811,X3=X318(1+2κ)c+λ2+2κ−c8−238, X4=X2(1+2κ)c+2λ1+κ−3,X5=X4(1+2κ)c+4λ1+κ−467,X6=X167(1+2κ)c+λ1+κ−257,X7=X203(1+2κ)c+λ1+κ−343,X8=X73(1+2κ)c+λ1+κ−113. |
Proof. It is easy to deduce that
SI≪Mε∑h∼HKh, |
where Kh=∑m∼M|∑n∼Ne((α+h)(mn)c)|. According to Hölder's inequality, we have
K4h≪M3∑m∼M|∑n∼Ne((α+h)(mn)c)|4. | (3.2) |
Let zn=zn(m,α)=(α+h)(mn)c. Suppose that Q, J are two positive integers such that 1≤Q≤Nlog−1X, 1≤J≤Nlog−1X. For the inner sum in (3.2), applying Lemma 2.1 twice, we can get
K4h≪X4Q2+X4J+X3JQJ∑j=1Q∑q=1|Eq,j|, | (3.3) |
where
Eq,j=∑m∼M∑N<n≤2N−q−je(zn−zn+q+zn+q+j−zn+j). | (3.4) |
Let Δ(nc;q,j)=(n+q+j)c−(n+q)c−(n+j)c+nc, G(m,n)=(α+h)mcΔ(nc;q,j). Then zn−zn+q−zn+j+zn+q+j=G(m,n). Thus we have
Eq,j=∑m∑ne(G(m,n)). | (3.5) |
For any t≠1,0, we have
Δ(nt;q,j)=t(t−1)qjnt−2+O(Nt−3qj(q+j)), | (3.6) |
then
∂G∂n=c(c−1)(c−2)(α+h)qjmcnc−3(1+O(q+jN)) |
and
∂2G∂n2=c(c−1)(c−2)(c−3)(α+h)qjmcnc−4(1+O(q+jN)). | (3.7) |
If c(c−1)(c−2)(α+h)qjMcNc−3≤1100, by Lemma 2.5 we have
∑m∑ne(G(m,n))≪MN3((α+h)qjMcNc)−1. |
From now we always suppose that c(c−1)(c−2)(α+h)qjMcNc−3≥1100. By Lemma 2.7 we have
∑N<n≤2N−j−qe(G(m,n))=e(18)∑α<υ<β|∂2G∂n2(m,nυ)|−12e(G(m,nυ)−υnυ)+R(m,q,j), |
where
∂G∂n(m,nυ)=υ,β=∂G∂n(m,N),α=∂G∂n(m,2N−q−j),R=N4[(h+α)qjXc]−1,υ∼(h+α)qjMcNc−3,R(m,q,j)=O(logX+RN−1+min(√R,max(1‖α‖,1‖β‖))). | (3.8) |
By Lemma 2.8, the contribution of R(m,q,j) to E(q,j) is
≪MlogX+MRN−1+∑m∼Mmin(√R,1‖α‖)+∑m∼Mmin(√R,1‖β‖)≪MlogX+X3−c[(h+α)qjM2]−1+[(h+α)qj]12MXc2−1logX. | (3.9) |
Then we only need to deal with the following exponential sum
∑m∼M∑α<υ<β|∂2G∂n2(m,nυ)|−12e(G(m,nυ)−υnυ)=∑υ∑m∈Iυ|∂2G∂n2(m,nυ)|−12e(G(m,nυ)−υnυ), |
where Iυ is a subinterval of [M,2M]. For a fixed υ, we define Δλ′=Δ(nλ′υ;q,j), where λ′ is an arbitrary real number. We take the derivative of m in (3.8) and get
n′υ=−cΔc−1(c−1)mΔc−2. | (3.10) |
It follows from (3.7) that
ddm(∂2G∂n2(m,nυ))=c2(h+α)mc−1Δc−2((c−1)Δ2c−2−(c−2)Δc−1Δc−3). |
Recalling (3.6), we can get
ddm(∂2G∂n2(m,nυ))=c2(c−1)(c−2)(c−3)(h+α)qjmc−1nc−4υ(1+O(q+jN)). |
Thus for m, |∂2G∂n2(m,nυ)|−12 is monotonic. Let g(m)=G(m,nυ(m))−υnυ(m). Then we have
g′(m)=c(α+h)mc−1Δc,g″(m)=c(α+h)c−1(c−1)2ΔcΔc−2−c2Δ2c−1m2−cΔc−2=c(α+h)(c−1)g1(m)−g2(m)g0(m),g‴(m)=c(α+h)(c−1)(g′1−g′2)g0−g′0(g1−g2)g20, |
where
g′1=((c−1)2cΔc−1Δc−2+(c−1)2(c−2)ΔcΔc−3)n′υ(m),g′2=2c2(c−1)Δc−1Δc−2n′υ(m),g′0=(2−c)m1−c(c−1)Δc−2((c−1)Δ2c−2+cΔc−1Δc−3). |
From the above formulas we can obtain
g‴(m)∼(h+α)qjM−1Xc−2. |
Using Lemma 2.5 and partial summation we can get
∑m∼M∑υ|∂2G∂n2(m,nυ)|−12e(G(m,nυ)−υnυ)≪(M((h+α)qjM−1Xc−2)16+((h+α)qjM−1Xc−2)−13)×(h+α)qjMcNc−3((h+α)qjMcNc−4)−12≪((h+α)qj)23M116X2c3−43+((h+α)qj)16M43Xc6−13. | (3.11) |
By (3.5), (3.9) and (3.11), we have
Eq,jlog−1X≪M+((h+α)qjM2)−1X3−c+((h+α)qj)12MXc2−1+((h+α)qj)23M116X2c3−43+((h+α)qj)16M43Xc6−13. | (3.12) |
Inserting (3.12) into (3.3), we obtain
K4hlog−1X≪X4Q2+X4J+MX3+((h+α)QJM2)−1X6−c+((h+α)QJ)12MXc2+2+((h+α)QJ)23M116X2c3+53+((h+α)QJ)16M43Xc6+83. |
Then choosing optimal J∈[0,Nlog−1X] and Q∈[0,Nlog−1X] and using Lemma 6 twice we can get
Khlog−3X≪B(h), |
where
B(h)=X56+(α+h)114M17Xc14+57+(α+h)112M1148Xc12+1724+(α+h)130M415Xc30+1115+X34M14+(α+h)−14X1−c4+X2328M18+X2532M732+X1720M340+X1114M314. |
Recalling the definitions of H and Y, we have
SIlog−3X≪MεHB(H)≪X(1+2κ)c+λ2+2κ+2ε, |
and Lemma 3.1 is proved.
Lemma 3.2. Let P56≪X≪P, H=X1−(1+2κ)c+λ2+2κ, F=(h+α)Xcand ch(α) denote complex numbers such that ch(α)≪(1+|h|)−1. Then uniformly for α∈(τ,1−τ), we have
SII=∑|h|∼Hch(α)∑M≤m≤2Ma(m)∑N≤n≤2Nb(n)e((h+α)(mn)c)≪X(1+2κ)c+λ2+2κ+2ε, | (3.13) |
for any a(m)≪mε, b(n)≪nε, where (κ,λ) is any exponent pair, X≪MN≪X and
max{X3−(1+2κ)c+λ1+κF−1,X4−2(1+2κ)c+2λ1+κ,X26−533(1+2κ)c+λ1+κF539}≪M≪X(1+2κ)c+λ1+κ−1. |
Proof. Taking Q=[X2−(1+2κ)c+λ1+κlog−1X], then we have Q=o(N). By Cauchy's inequality and Lemma 2.1, we have
|SII|2≪∑m∼M|a(m)|2∑m∼M|∑n∼Nb(n)e(f(mn))|2≪M2N2log2A+2BXQ+MNlog2AXQQ∑q=1Eq, | (3.14) |
where Eq=∑n∼N|b(n+q)b(n)||∑m∼Me(G(mn))| and G(m,n)=G(m,n,q)=(h+α)mcΔ(n,q;c), Δ(n,q;c)=(n+q)c−nc.
If |∂G∂m|≤103Mq−2, by Lemma 2.10 we have
Eq≪∑n∼N|b(n+q)b(n)|(MNFq+(FqMN)12M12)≪∑n∼N(|b(n+q)|2+|b(n)|2)(MNFq+Mq)≪MNqlog2BX | (3.15) |
noting that M≫XF.
Now we suppose |∂G/∂m|>103Mq−2. By Lemma 11 we get
∑m∼Me(G(m,n))≪MN1/2(Fq)1/2|∑r1(n)≤r≤r2(n)φ(n,r)e(s(r,n))|+logX+MN1/2(Fq)−1/2, |
where s(r,n)=G(M(r,n),n)−rm(r,n), φ(r,n)=(Fq)12MN12|∂2G(m(r,n),n)∂m2|−12 and
r1(n)=∂G∂m(M,n), r2(n)=∂G∂m(2M,n). |
Thus we have
Eq≪MN1/2(Fq)1/2∑n∼N|b(n+q)b(n)||∑r1(n)<r≤r2(n)φ(n,r)e(s(r,n))|+Nlog2B+1X+MN3/2(Fq)−1/2log2BX. | (3.16) |
So it suffices to bound the sum
Σ1=∑n∼N|b(n+q)b(n)||∑r1(n)<r≤r2(n)φ(n,r)e(s(r,n))|. |
Let T=[Fq3/M2N] and R=Fq/MN. By Cauchy's inequality and Lemma 2.1 again we get
Σ12≪∑n∼N|b(n+q)b(n)|2∑n∼N|∑r1(n)<r≤r2(n)φ(n,r)e(s(r,n))|2≪N2R2log4BXT+NRlog4BXTΣ2, | (3.17) |
where
Σ2=T∑t=1|∑n∼N∑r1(n)<r≤r2(n)−tφ(n,r+t)φ(n,r)e(s(r+t)−s(r,n))| |
and where we used the estimate
∑n∼N|b(n+q)b(n)|2≪∑n∼N(|b(n+q)|4+|b(n)|4)≪Nlog4BX. |
It is easy to check that 10<T=o(R).
Recall that s(r,n)=G(m(r,n),n)−rm(r,n), where m(r,n) denotes the solution of
∂G∂m(m,n)=r. |
It is easy to deduce that
∂s∂r(r,n)=∂G∂m∂m∂r−m(r,n)−r∂m∂r=−m(r,n). |
So we can obtain
H(n):=Hr,t,q(n)=s(r+t,n)−s(r,n)=∫r+tr∂s∂u(u,n)du=−∫r+trm(u,n)du, |
which implies that |H(j)|∼tMN−j, (j=0,1,2,3,4,5,6). Denote by I(r,t) the interval N<n≤2N, r1(n)<n≤r2(n)−t. Then we have
Σ2≪T∑t=1∑r∼R|∑n∈I(r,t)φ(n,r+t)φ(n,r)e(s(r+t,n)−s(r,n))|. |
Thus using partial summation, we get
Σ2≪T∑t=1∑r∼R(tMN)κNλ≪RMκNλ−κT1+κ≪NR | (3.18) |
with the exponent pair (κ,λ), if we note that M≫X26−533(1+2κ)c+λ1+κF539. From (3.15)–(3.18) we get that for any 1≤q≤Q,
Eq≪MNlog2B+1Xq+Nlog2B+1X+MN32(Fq)−12log2BX. | (3.19) |
Now this lemma follows from inserting (3.19) into (3.14).
Lemma 3.3. For τ≤α≤1−τ, we have
S(α)≪P(1+2κ)c+λ2+2κ+4ε, |
where (κ,λ) is any exponent pair.
Proof. Throughout the proof of this lemma, we write H=X1−(1+2κ)c+λ2+2κ for convenience. By a dissection argument we only need to prove that
∑X<n≤2XΛ(n)e(α[nc])≪X(1+2κ)c+λ2+2κ+3ε | (3.20) |
holds for P56≤X≤P and τ≤α≤1−τ. According to Lemma 2.3, we have
∑X<n≤2XΛ(n)e(α[nc])=∑|h|≤Hch(α)∑X<n≤2XΛ(n)e((h+α)nc)+O((logX)∑X<n≤2Xmin(1,1H‖nc‖)). | (3.21) |
By the expansion
min(1,1H‖nc‖)=∞∑h=−∞ahe(hnc), |
where
|ah|≤min(log2HH,1|h|,Hh2), |
we get
∑X<n≤2Xmin(1,1H‖nc‖)≤∞∑h=−∞ah|∑X<n≤2Xe(hnc)|≪Xlog2HH+∑1≤h≤H1h((hXc)12+XhXc)+∑h≥HHh2((hXc)12+XhXc)≪X(1+2κ)c+λ2+2κlogX, | (3.22) |
where we estimated the sum over n by Lemma 2.2 with the exponent pair (12,12).
Let R=max{X3−(1+2κ)c+λ1+κF−1,X4−2(1+2κ)c+2λ1+κ,X26−533(1+2κ)c+λ1+κF539}. Recall the definition of Y in Lemma 3.1. Let U=R, V=X(1+2κ)c+λ1+κ−1, Z=[XY−1]+12. By Lemma 4 with F(n)=e((h+α)nc), then we reduce the estimate of
∑|h|≤Hch(α)∑X<n≤2XΛ(n)e((h+α)nc) |
to the estimates of sums of type I
S′I=∑|h|≤Hch(α)∑M<m≤2Ma(m)∑N<n≤2Ne((h+α)(mn)c), N>Z, |
and sums of type II
S′II=∑|h|≤Hch(α)∑M<m≤2Ma(m)∑N<n≤2Nb(n)e((h+α)(mn)c), U<M<V, |
where a(m)≪mε, b(n)≪nε, X≪MN≪X. By Lemma 3.1, we get
S′I≪X(1+2κ)c+λ2+2κ+2ε. | (3.23) |
By Lemma 3.2, we get
S′II≪X(1+2κ)c+λ2+2κ+3ε. | (3.24) |
From (3.23) and (3.24) we can obtain
∑|h|≤Hch(α)∑X<n≤2XΛ(n)e((h+α)nc)≪X(1+2κ)c+λ2+2κ+3ε. | (3.25) |
Now (3.20) follows from (3.21), (3.22) and (3.25). Thus we complete the proof of this Lemma.
It is easy to see that
R(N)=∫1−τ−τS3(α)e(−αN)dα=∫τ−τS3(α)e(−αN)dα+∫1−ττS3(α)e(−αN)dα=R1(N)+R2(N). | (4.1) |
Following the argument of Laporta and Tolev [18,pages 928–929], we can get that
R1(N)=Γ3(1+1c)Γ(3c)N3c−1+O(N3c−1exp(−log13−δN)) | (4.2) |
for 1<c<32 and any 0<δ<13, where the implied constant in the O−symbol depends only on c.
Let
S(α,X)=∑X<p≤2Xe(α[pc])logp, T(α,X)=∑X<n≤2Xe(α[nc]). |
We can get
R2(N)=∫1−ττS3(α)e(−αN)dα≪(logX)maxP56≤X≤0.5P|∫1−ττS2(α)S(α,X)e(−αN)dα|+P116log2P, | (4.3) |
where we used
∫1−ττ|S2(α)|dα≪∫10|S2(α)|dα≪Plog2P. | (4.4) |
Now, we start to estimate the absolute value on the right hand in (4.3). By Cauchy's inequality we have
|∫1−ττS2(α)S(α,X)e(−αN)dα|=|∑X<p≤2X(logp)∫1−ττS2(α)e(α[pc]−αN)dα|≤∑X<p≤2X(logp)|∫1−ττS2(α)e(α[pc]−αN)dα|≪(logX)∑X<n≤2X|∫1−ττS2(α)e(α[nc]−αN)dα|≪X12(logX)(∑X<n≤2X|∫1−ττS2(α)e(α[nc]−αN)dα|2)12=X12(logX)(∫1−ττ¯S2(β)e(−βN)dβ∫1−ττS2(α)T(α−β,X)e(−αN)dα)12≪X12(logX)(∫1−ττ|S(β)|2dβ∫1−ττ|S(α)|2|T(α−β,X)|dα)12. | (4.5) |
Then we have
∫1−ττ|S(α)|2|T(α−β,X)|dα≪∫τ<α<1−τ|α−β|≤X−c|S(α)|2|T(α−β,X)|dα+∫τ<α<1−τ|α−β|>X−c|S(α)|2|T(α−β,X)|dα. | (4.6) |
By Lemma 3.3, we have
∫τ<α<1−τ|α−β|≤X−c|S(α)|2|T(α−β,X)|dα≪Xmaxα∈(τ,1−τ)|S(α)|2∫|α−β|≤X−c1dα≪X1−cP(1+2κ)c+λ1+κ+8ε, | (4.7) |
where we used the trivial bound T(α,X)≪X. By Lemma 2.9, Lemma 3.3 and (4.4), we get
∫τ<α<1−τ|α−β|>X−c|S(α)|2|T(α−β,X)|dα≪∫τ<α<1−τ|α−β|>X−c|S(α)|2(Xκc+λ1+κlogX+X|α−β|Xc)dα≪Xκc+λ1+κ(logX)∫1−ττ|S(α)|2dα+maxα∈(τ,1−τ)|S(α)|2∫|α−β|>X−cX|α−β|Xcdα≪Xκc+λ1+κPlog3P+X1−cP(1+2κ)c+λ1+κ+9ε. | (4.8) |
Thus, combining (4.6)–(4.8) we obtain
∫1−ττ|S(α)|2|T(α−β,X)|dα≪Xκc+λ1+κPlog3P+X1−cP(1+2κ)c+λ1+κ+9ε. | (4.9) |
By (4.3), (4.5) and (4.9), we can obtain
R2(N)≪P3−c−ε. | (4.10) |
Now putting (4.1), (4.2) and (4.10) into together, we have
R(N)=Γ3(1+1c)Γ(3c)N3c−1+O(N3c−1exp(−log13−δN)) |
follows for any 0<δ<13, where the implied constant in the O−symbol depends only on c. Thus we complete the proof of Theorem 1.1.
The authors would like to thank the referees for their many useful comments. This work is supported by National Natural Science Foundation of China (Grant Nos. 11801328 and 11771256).
The authors declare no conflict of interest.
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