# On a Diophantine Inequality withs Primes

Authors: Xiaofei Yan; Lu Zhang
Journal: Journal of Mathematics
 (2021)

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## Abstract

Let2<c<δ. In this study, for prime numbers p1,…,ps and a sufficiently large real number N, we prove the Diophantine inequality p1c+⋯+psc−N<N−9/10cδ−c, where δ=294−210s/123−97s and s≥5. When s=5, this result improves a previous result.


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## Body

## 1. Introduction

Suppose thatk≥1 is an integer and c>1 is not an integer. Let ε be a small positive number. The Waring–Goldbach problem is to study the solvability of the Diophantine equality:(1)N=p1k+⋯+psk,in prime numbers p1,…,ps. In [1], the author studied the analog of the Waring–Goldbach problem. For any sufficiently large real number N, let Hc denote the smallest natural number s such that the Diophantine inequality,(2)p1c⋯+psc−N<ε,is solvable in prime numbers p1,…,ps. In [1], the author proved that(3)limsupc⟶∞Hcclogc≤4.In [1], the author also obtained that Hc≤5, for 1<c<3/2. Later, the result was improved in [2–6]. Now, the best result for Hc≤5 is 2<c<52/25 by Li and Cai [3].In [4], the authors first proved that Hc≤4, for 1<c<81/68. Later, the result was improved in [7]. Now, the best result for Hc≤4 is 1<c<1193/889 by Zhang and Li [8]. When s=3 in inequality (2), Tolev [9] obtained the result 1<c<27/26. Afterwards, the range of c was enlarged by several authors in [10–16]. Now, the best result for s=3 is 1<c<43/36 by Cai [12]. When s=2, in inequality (2), Laporta obtained 1<c<15/14 in [17]. Laporta’s result was improved by some authors in [4, 18, 19]. Now, the best result for r=2 is 1<c<59/44 by Li and Cai [19].In this paper, we focus on the Diophantine inequality (2) and prove the following result.Theorem 1.
Let2<c<δ. For any sufficiently large real number N, let ε=N−9/10cδ−c and let B0N,s denote the number of solutions of the Diophantine inequality:(4)p1c+p2c+⋯+psc−N<ε,in prime numbers p1,p2,…,ps, where δ=294−210s/123−97s and s≥5. We can obtain(5)B0N,s≫εNs/c−1logsN.Fors=5 in Theorem 1, we can get better result than Li and Cai [3].Corollary 2.
Under the notations of Theorem1, for 2<c<378/181, we have(6)B0N,5≫εN5/c−1log5N.
For comparison,2<c<378/181=2.0883977,…,52/25=2.08.Notation 1.
In this paper, letδ=294−210s/123−97s and δ1=131−97s/123−97s, s≥5. We also assume(7)2<c<δ,X=3N101/c,η=11000δ−c,κ=X1−c−η,ε=N−9/10cδ−c,χ=Xδ−c,a=9ε10,b=ε10,Sx=∑X/2<p≤Xexpclogp,Ix=∫X/2Xextcdt.
LetΛn denote von Mangoldt’s function. We write ω∼Ω if the range of ω is Ω<ω≤2Ω. aω≪bω means that aω=Obω.


## 2. Some Lemmas

In order to prove our theorem, we need the following lemmas.Lemma 1.
Letr be a positive integer. There exists a function ϕy which is r=logX times continuously differentiable and satisfies(8)ϕy=1,fory≤a−b,0<ϕy<1,fora−b<y<a+b,ϕy=0,fory≥a+b,and its Fourier transformation,(9)Φx=∫−∞+∞e−xyϕydy,satisfies(10)Φx≤min2a,1πx,1πxr2πxbr.Proof.
This can be found in Piatetski-Shapiro [1].Lemma 2.
Letal be a sequence of complex numbers; then, for L,Q≥1, we have(11)∑L<l≤2Lal2≤2+LQ∑q<Q1−qQ∑L≤l+q,l−q≤2Lal+qal−q¯,where z¯ denotes the conjugate of the complex number z.Proof.
This is Lemma2 of Fouvry and Iwaniec [20].Lemma 3.
We have(12)∫−∞∞Isxe−xNΦxdx≫εXs−c.Proof.
Similar to Lemma6 of [16], we have(13)H≔∫−∞∞Isxe−xNΦxdx=∫X/2X⋯∫X/2X∫−∞∞ext1c+⋯+tsc−NΦxdxdt1⋯dts=∫X/2X⋯∫X/2XΦt1c+⋯+tsc−Ndt1⋯dts≥∫X/2X⋯∫X/2X∗dt1⋯dts≥∫μXλXdt1⋯∫μXλXdts−1∫ℳdts,where ∗ means that t1c+⋯+tsc−N<4ε/5, and λ and μ satisfy(14)12<121/c<μ<λ<122−12c1/c<1,ℳ=X2,X∩N−4ε5−t1c−⋯−ts−1c1/c,N+4ε5−t1c−⋯−ts−1c1/c.
We have(15)H≫ε∫μXλX⋯∫μXλXξt1,…,ts−11/c−1dt1⋯dts−1,by the mean-value theorem, where ξt1,…,ts−1≍Xc. Therefore, we can obtain(16)H≫εXs−c,which proves Lemma 3.Lemma 4.
(i)
∫−κκSx2dx≪X2−clog3X(ii)
∫−κκIx2dx≪X2−clog3XProof.
This is Lemma 7 of Tolev [16].Lemma 5.
Let0≤k≤1/2≤l≤1 and let B>0. For any exponent pair k,l and Ω≤Ω′≤10Ω, we have(17)∑Ω≤ω≤Ω′eBωc≪BΩckΩl−k+ΩBΩc.Proof.
This is Lemma 3 of Li and Cai [3].Lemma 6.
Forx<κ, we have(18)Sx=Ix+OXexp−log1/5X.Proof.
This is Lemma 14 of Tolev [16].Lemma 7.
Forκ<x<χ, we have(19)∫κ<x<χS4xΦxdx≪X4−c+X2X3η.Proof.
This is (50) of Zhai and Cao [5].Lemma 8.
Letαandβ be real numbers such that α≠0,1,2andβ≠0,1,2,3. Set(20)TΩ,Γ=∑Ω<ω≤2Ω∑Γ<γ≤2ΓaωbγeFωαγβΩαΓβ,where F≫ΩΓ2,Γ≥Ω≥1, and aω,bγ≤1. Then, we have(21)TΩ,Γ≪F1/16Ω7/8Γ13/16+F1/26Ω93/104Γ23/26+F−1/128Ω467/512Γ65/64+Ω65/72ΓΩΓη.Proof.
This is Theorem1 of Baker and Weingartner [6].Lemma 9.
Let3<Q<K<W<X. Suppose that W−1/2∈ℕ,X≥64W2UQ,W≥4Q2,K3≥32X. Assume further that Fn is a complex-valued function such that Fn≤1. Then, the sum,(22)∑X<n≤2XΛnFn,can be written in Olog10X sums.Proof.
This is Lemma 3 of Heath-Brown [21].


## 3. The Estimate ofSx

In this section, we draw our attention to the estimate of exponential sums, which also has lots of applications (e.g., see [22–35]). Suppose that aω≪ωη and bγ≪γη; then, we estimate the exponential sums in the following two forms. Type I:(23)∑ω∼Ω,γ∼ΓΩΓ∼X∑aωFωγ,and Type II:(24)∑ω∼Ω,n∼LΩΓ∼X∑aωbγFωγ.Lemma 10.
For complex number sequencesaω and bγ, suppose that aω≪1,bn≪1, κ≤x≤χ, X16/97s−123≪Γ≪X438−178s/369−291s,ΩΓ≍X; then, we have(25)SII=∑ω∼Ω∑γ∼Γaωbγexωγc≪Xδ1+η.Proof.
By Cauchy’s inequality and Lemma2 with M=X16/97s−123−η, we obtain(26)SII2≪∑γ∼Γbγ2∑γ∼Γ∑ω∼Ωaωexωγc2≪Γ∑γ∼ΓΩM∑0≤m<M1−mM×∑ω∼Ωaω+maω−m¯exωcγ+mc−γ−mc≪ΓΩM∑γ∼ΓΩ1+η+∑1≤m<M1−mM×∑ω∼Ωaω+maω−m¯exωcγ+mc−γ−mc≪X2M+XM∑1≤m≤M∑γ∼Γ∑ω∼Ωexωcγ+mc−γ−mc≪X2M+XMS∗,where(27)S∗=∑1≤m≤M∑γ∼Γ∑ω∼Ωefω,γ,m,with fω,γ,m=xωcγ+mc−γ−mc. By Lemma 5, we choose the exponent pair 1/14,11/14. Then, we can obtain(28)S∗≪∑1≤m≤M∑Γ≤γ≤2ΓmxXc−11/14XΓ11/14+1mxXc−1≪ΓM15/14χ1/14Xc−1/14X11/14Γ−11/14+Γκ−1X1−clogM≪X.
Now, Lemma10 follows from (26) and (28).Lemma 11.
Letaω be a sequence of complex numbers. Suppose that aω≪1, κ≤x≤χ, and Ω≪X6029−2231s/3075−2425s; then, we have(29)SI=∑ω∼Ω∑γ∼Γaωexωγc≪Xδ1+η.Proof.
IfΩ≪X41−27s/123−97s, by Lemma 5, we choose the exponent pair 1/6,2/3. Then, we obtain(30)SI≪∑ω≤2ΩxXc1/6Xω1/2+XxXcω≪Xδ1+η.
IfX41−27s/123−97s≪Ω≪X438−178s/369−291s, then, by Lemma 10, we have(31)SI≪Xδ1+η.
IfX438−178s/369−291s≪Ω≪X6029−2231s/3075−2425s , by Lemma 8 with ω,γ=ω,γ or ω,γ=γ,ω, we have(32)Sx≪Xδ1+η.
Now, the proof of Lemma11 is completed.Lemma 12.
Forκ≤x≤χ, we have(33)Sx≪Xδ1+4η.Proof.
We can obtain(34)S∗∗=∑n∼XΛnexnc≪Xδ1+3η.
Let(35)Q=X16/97s−123,K=X438−178s/369−291s,W=X194s+2954/2425s−3075+12.
LetFγ=exγc in Lemma 9; we reduce the estimation of S∗∗ to the estimations of type I sums:(36)∑ω∼Ω,γ∼ΓΩΓ∼X∑aωFωγ,Γ>W,and type II sums:(37)∑ω∼Ω,γ∼ΓΩΓ∼X∑aωFωγ,Q<Γ<K,and estimate (34) follows from Lemmas 10 and 11.


## 4. Proof of the theorem

Let(38)BN,s=∑X/2<p1,…,ps≤Xp1c+⋯+psc−N<εlogp1⋯logps,B1N,s=∑X/2<p1,…,ps≤Xlogp1⋯logpsϕp1c+⋯+psc−N.By the definition ofϕy in Lemma 1, we have(39)BN,s≥B1N,s.By the inverse Fourier transformation formula, we obtain(40)B1N,s=∫−∞∞Ssxe−xNΦxdx=D1N+D2N+D3N,where(41)D1N=∫−κκSsxe−xNΦxdx,D2N=∫κ<x≤χSsxe−xNΦxdx,D3N=∫x>χSsxe−xNΦxdx.Let(42)H1N=∫−κκIsxe−xNΦxdx,HN=∫−∞∞Isxe−xNΦxdx.Then, we have(43)D1N=HN+H1N−HN+D1N−H1N.From Lemma10 in [36], we have Ix≪X1−c|x|−1. Thus, by Lemma 1, we have(44)H1N−HN≪∫x>κIxsΦxdx≪Xs−sc∫x>κΦxx−sdx≪εXs−scκ−s+1≪εXs−clogX.It follows from Lemmas1, 4, and 6 that(45)D1N−H1N≪∫−κκSsx−IsxΦxdx≪εmaxx≤κSx−Ix∫−κκ|Ss−1x|+Is−1xdx≪εXs−3maxx≤κSx−Ix∫−κκ|S2x|+I2xdx≪εXs−1−clog3Xmaxx≤κSx−Ix≪εXs−cexp−log1/5X.By Lemma3 and (43)–(45), we obtain(46)D1N≫εXs−c.Let(47)Ux=∑X/2≤n≤Xexnc.We have(48)D2N=∑X/2≤p≤Xlogp∫κ<x≤χexpcSs−1xΦxe−xNdx≤∑X/2≤p≤Xlogp∫κ<x≤χexpcSs−1xΦxe−xNdx≤logX∑X/2≤n≤X∫κ<x≤χexncSs−1xΦxe−xNdx.It follows from (48) and Cauchy’s inequality that(49)D2N2≤Xlog2X∑X/2≤n≤X∫κ<x≤χexncSs−1xΦxe−xNdx2≤Xlog2X∫κ<y≤χSs−1yΦye−yN¯dy×∫κ<x≤χSs−1xΦxe−xNUx−ydx≤X−194s2+1135s−1187/123−97slog2X∫κ<y≤χS4yΦydy×∫κ<x≤χS4xΦxUx−ydx.By Lemma5 with the exponent pair 13/84,55/84 (see in [37]), we obtain(50)Ux≪minxXc13/84X1/2+XxXc,X.It follows from (50) that(51)∫κ<x≤χSx4ΦxUx−ydx≪∫κ<x≤χx−y≤X−cSx4ΦxUx−ydx+∫κ<x≤χX−c<x−y≤2χSx4ΦxUx−ydx≪X∫κ<x≤χx−y≤X−cSx4Φxdx+∫κ<x≤χX−c<x−y≤2χSx4Φxx−yXc13/84X1/2+Xx−yXcdx≪εX647−485s/123−97s+16η∫x−y≤X−cdx+χXc13/84X1/2∫κ<x≤χSx4Φxdx+εX647−485s/123−97s−c+16η∫X−c<x−y≤2χ1x−ydx≪εX647−485s/123−97s−c+17η+X107−81s/123−97s∫κ<x≤χSx4Φxdx,where Lemma 12 is used.Now, by (49) and (51), we obtain(52)D2N2≪X−194s2+1135s−1187/123−97slog2X∫κ<x≤χSy4Φydy×εX647−485s/123−97s−c+17η+X107−81s/123−97s∫κ<x≤χSx4Φxdx≪εX−194s2+650s−540/123−97s−c+18η∫κ<x≤χSy4Φydy+X−194s2+1054s−1080/123−97s∫κ<x≤χSx4Φxdx2≪εX−194s2+456s−294/123−97s−c+19η+X−194s2+666s−588/123−97s+4η,where Lemma 7 is used. From (52), we obtain(53)D2N≪ε1/2X−97s2+228s−147/123−97s−c/2+10η+X−97s2+333s−294/123−97s+2η≪εXs−clogX.It follows from Lemma1 that(54)D3N≪∫x>χSsxΦxdx≪Xs∫x>χ1xr2πxbrdx≪Xsr2πbχr≪1.By (40), (46), (53), and (54),(55)B1N,s≫εXs−c.It follows from (39) and (55) that(56)B0N,s≥BN,slogsX≥B1N,slogsX≫εNs/c−1logsN.Now, by (56), the proof of the theorem is completed.



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*Source: 3303476-2021-11-24.xml*