Elsevier

Journal of Number Theory

Volume 202, September 2019, Pages 220-253
Journal of Number Theory

General Section
On a Diophantine inequality over primes

https://doi.org/10.1016/j.jnt.2019.01.008Get rights and content

Abstract

Let N be a sufficiently large real number. In this paper, it is proved that for 1<c<6/5, the Diophantine inequality|p1c+p2c+p3c+p4cN|<log1N is solvable in prime numbers p1,p2,p3,p4. This result constitutes an improvement upon that of Mu (2015) [15] for the range 1<c<97/81.

Section snippets

Introduction and main result

In 1952, Piatetski-Shapiro [16] considered the following analogue of the Waring–Goldbach problem. Assume that c>1 is not an integer and let ε be a positive number. If r is a sufficiently large integer (depending only on c), then the inequality|p1c+p2c++prcN|<ε has a solution in prime numbers p1,p2,,pr for sufficiently large N. More precisely, if the least r such that (1.1) has a solution in prime numbers for every ε>0 and N>N0(c,ε) is denoted by H(c), then, in [16], Piatetski-Shapiro proved

Notation

Throughout this paper, let p,q,r,s, with or without subscripts, always denote prime numbers; η always denotes an arbitrary small positive constant, which may not be the same at different occurrences; mM means M<r2M. As usual, e(x)=e2πix; f(x)g(x) means that f(x)=O(g(x)); f(x)g(x) means that f(x)g(x)f(x). N is a sufficiently large real number; c is a fixed real number satisfying 1<c<6/5; f(x,y)Δg(x,y) means thati+jxiyjf(x,y)=i+jxiyjg(x,y)(1+O(Δ)) for all pairs (i,j) for which it

Outline of the method

In this section, we shall introduce the idea of the proof of Theorem 1.1. First, we need the following lemma.

Lemma 3.1

Let a,b be real numbers, 0<b<a/4, and let ℓ be a positive integer. There exists a function φ(y) which is k times continuously differentiable and such that{φ(y)=1,for|y|ab,0<φ(y)<1,forab<|y|<a+b,φ(y)=0,for|y|a+b, and its Fourier transformΦ(x)=+e(xy)φ(y)dy satisfies the inequality|Φ(x)|min(2a,1π|x|,1π|x|(2π|x|b)).

Proof

See Piatetski-Shapiro [16] or Segal [18]. 

We writeP(z)=p<zp

Preliminary lemmas

Throughout this paper, we denote by φ(y) the function from Lemma 3.1 with parameters a=9ε/10, b=ε/10 and =[logX]. Also, we denote by Φ(x) the Fourier transform of φ(y).

Lemma 4.1

For 1<c<2, we haveττ|S(x)|2dxX2clogX,ττ|S1(x)|2dxX2clogX,ττ|I0(x)|2dxX2clog1X.

Proof

See Lemma 2.3 of Mu [15]. 

Lemma 4.2

DefineJ(X)=+I4(x)Φ(x)e(Nx)dx. Then for 1<c<2, there holdsJ(X)εX4c.

Proof

See Lemma 2.5 of Mu [15]. 

Lemma 4.3

Let 1<c<2. Then we haveτH|S(x)|2|Φ(x)|dxX1+η,τH|S(x)|4|Φ(x)|dxX4c+3η,τH|S1(x)|2|Φ(x)|dxX1+η.

Proof

See

Exponential sums

In this section, we shall prove the exponential sum estimates which we will need to get the asymptotic formula (3.5).

Lemma 5.1

Let α and β be real, αβ(α1)(β1)(α2)(β2)0,X>0,M,N1, |a(m)|1,|b(n)|1. Then we have(XMN)η|mMnNa(m)b(n)e(XmαnβMαNβ)|(X4M31N34)142+(X6M53N51)166+(X6M46N41)156+(X2M38N29)140+(XM9N6)110+(X2M7N6)110+(X3M43N32)146+(XM6N6)18+M12N+MN12+X12MN.

Proof

See Theorem 9 of Sargos and Wu [17]. 

Lemma 5.2

Let D be a subdomain of the rectangle{(x,y):M<x2M,N<y2N,M>N} such that any line parallel to any

Asymptotic formulas

In this section, let Φ(x) be the Fourier transform of the function φ(y) defined in Lemma 3.1. DefineW0(n)=ττI03(x)Φ(x)e((ncN)x)dx andW1(n)=ττI02(x)I1(x)Φ(x)e((ncN)x)dx.

Lemma 6.1

Suppose that v>0,xx0(v),xvzx. Let ω(x) be the continuous solution of the differential–difference equation{ω(x)=1x,if1x2,(xω(x))=ω(x1),ifx>2. Then for any u(x,2x], we havex<nu(n,P(z))=11=ω(logxlogz)uxlogz+O(xlog2x). In particular, we haveω(x)={1+log(x1)x,forx[2,3],1+log(x1)x+1x2x1log(t1)tdt,forx[3,4]

Proof of Theorem 1.1

We start with (3.11). First, we shall establish the connections among J0(X),J1(X) and J(X). By (3.1), (6.26) and (6.27), we have|x|>τI03(x)I(x)Φ(x)e(Nx)dxετ1x3X3c3log3X1xXc1dxX1c+3η. By prime number theorem, it is easy to getI0(x)=I(x)logX+O(Xlog2X) andI1(x)=I(x)logX112512dθ(1θ)θ+O(Xlog2X). From (7.1) and (7.2), we derive thatJ0(X)=ττI03(x)I(x)Φ(x)e(Nx)dx+O(X1c+3η)=ττ(I(x)logX+O(Xlog2X))3I(x)Φ(x)e(Nx)dx+O(X1c+3η)=1log3XττI4(x)Φ(x)e(Nx)dx+O(X4c(logX)6)=1log3X

Acknowledgment

The authors would like to express the most sincere gratitude to Professor Wenguang Zhai for his valuable advices and constant encouragement.

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