On Diophantine approximation by unlike powers of primes

Wenxu Ge; Weiping Li; Tianze Wang

doi:10.1515/math-2019-0045

Artikel Open Access

On Diophantine approximation by unlike powers of primes

Wenxu Ge , Weiping Li und Tianze Wang

Veröffentlicht/Copyright: 30. Mai 2019

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Aus der Zeitschrift Open Mathematics Band 17 Heft 1

Abstract

Suppose that λ₁, λ₂, λ₃, λ₄, λ₅ are nonzero real numbers, not all of the same sign, λ₁/λ₂ is irrational, λ₂/λ₄ and λ₃/λ₅ are rational. Let η real, and ε > 0. Then there are infinitely many solutions in primes p_j to the inequality |λ1p1+λ2p22+λ3p33+λ4p44+λ5p55+η|<(maxpjj)−1/32+ε . This improves an earlier result under extra conditions of λ_j.

Keywords: Diophantine approximation; primes; Davenport-Heilbronn method

MSC 2010: 11D75; 11P32; 11P55

1 Introduction

Given k ≥ 1 and non-zero real numbers λ₁, λ₂, ⋯, λ_s (not all in rational ratio, not all in same sign), we write

F(p)=∑j=1sλjpjk,

where p = (p₁, p₂, …, p_s) with each p_j a prime. Various authors have considered the distribution of values of such forms, see [17, 18] for example.

For k = 1, Vaughan [17] first proved that for any real η, there are infinitely many solutions in primes p_j to the inequlity

|λ1p1+λ2p2+λ3p3+η|<(maxpj)−ξ+ε

with ξ = 1/10. The exponent was subsequently improved by Baker and Harman [1] to ξ = 1/6, Harman [6] to ξ = 1/5 and Matomäki [14] to ξ = 2/9.

For k = 2, Baker and Harman [1] and Harman [7] showed that there are infinitely many solutions in primes p_j to the inequality

|λ1p12+λ2p22+λ3p32+λ4p42+λ5p52+η|<(maxpj)−1/8+ε.

In 2011, Li and Wang [11] proved that there are infinitely many solutions in primes p_j to the inequality

|λ1p1+λ2p22+λ3p32+λ4p42+η|<(maxpj)−1/28+ε.

Later, Languasco and Zaccagnini [9], Liu and Sun [12], and Wang and Yao [20] replaced 1/28 with 1/18, 1/16 and 1/14, respectively.

For k ≥ 3, Vaughan [18] first proved that there are infinitely many solutions in primes p_j to the inequality

|λ1p1k+λ2p2k+⋯+λspsk+η|<(max1≤j≤spj)−σ+ε.

In 2006, Cook and Harman [2] improved the exponent σ.

In 2016, The first author and the second author [3] first established that if λ₁, λ₂, λ₃, λ₄, λ₅ are nonzero real numbers, not all of the same sign and λ₁/λ₂ is irrational, there are infinitely many solutions in primes p_j to the inequality

|λ1p1+λ2p22+λ3p33+λ4p44+λ5p55+η|<(maxpjj)−1/720+ε. (1.1)

Later, Mu [15], Liu [13], Mu and Qu [16] replaced 1720 in (1.1) with 1/180, 5/288 and 5/252 respectively.

In this paper, under some extra conditions of λ_j, we get the following result.

Theorem 1.1

Suppose that λ₁, λ₂, λ₃, λ₄, λ₅ are nonzero real numbers, not all of the same sign, λ₁/λ₂ is irrational, λ₂/λ₄ and λ₃/λ₅ are rational. Let η real, and ε > 0. Then there are infinitely many solutions in primes p_j to the inequality

|λ1p1+λ2p22+λ3p33+λ4p44+λ5p55+η|<(maxpjj)−1/32+ε. (1.2)

In the previous arguments, the key of this problem is the estimates for exponential sums over squares of primes (or for certain double sums if sieve methods are invoked). In [13], Liu used S2(λ2α)≪P21−1/8+ε . In [16], Mu and Qu used sieve method of Harman [7], and got S2∗(λ2α)≪P21−1/7+ε . Using the method of Mu and Qu [16], even if one got the best estimation S2∗(λ2α)≪P21−1/6+ε , 5/252 can only be replaced by 5/216. But in this paper our method don’t depend on the estimates of S₂(λ₂α).

Notation: Throughout the paper, the letter δ denotes a sufficiently small, fixed positive number. The letter ε denotes an arbitrarily sufficiently small positive real number. Any statement in which ε occurs holds for each fixed ε > 0. c denotes an absolute constant, not necessarily the same in all occurrences. The letter p, with or without subscript, denotes a prime number. Constants, both explicit and implicit, in Vinogradov symbols may depend on λ₁, λ₂, λ₃, λ₄, λ₅. We write e(x) = exp(2π i, x).

2 Outline of the method

We use the Hardy-Littlewood circle method which first stated by Davenport-Heilbronn. Note that λ₁/λ₂ is irrational and λ₂/λ₄ is rational. Without loss of generality, we assume that |λ₂/λ₄| ≤ 1. Let a/q be a continued fraction convergent to λ₁/λ₂ and put X = q^12/5. Then (λ₂ a)/(λ₄ q) = a′/q′ is a continued fraction convergent to λ₁/λ₄, where (a′, q′) = 1. Thus we have q ≍ q′. Suppose that 0 < τ < 1, and write P_j = X^1/j and 𝓘_j = [δ P_j, P_j] for 1 ≤ j ≤ 5. We define

Kτ(α)=sin⁡πταπα2,Sj(α)=∑p∈Jj(log⁡p)e(αpj).

Then we can easily get

Kτ(α)≪min(τ2,|α|−2),∫RKτ(α)e(αx)dα=max(0,τ−|x|). (2.1)

For any measurable subset 𝔛 of ℝ, we define

J(X):=∫XS1(λ1α)S2(λ2α)S3(λ3α)S4(λ4α)S5(λ5α)Kτ(α)e(ηα)dα. (2.2)

Then by (2.1), we have

J(R)=∑pj∈Jj(log⁡p1)⋯(log⁡p5)∫Re(α(λ1p1+⋯+λ4p44+λ5p55+η))Kτ(α)dα ≤(log⁡X)5∑pj∈Jjmax(0,τ−|λ1p1+⋯+λ4p44+λ5p55+η|) ≤τ(log⁡X)5N(η,X), (2.3)

where 𝓝(η, X) is the number of solutions to the inequality

|λ1p1+⋯+λ4p44+λ5p55+η|<τ,pj∈Jj.

To estimate the integral 𝓙(ℝ), we divide the real line into three parts: the major arc 𝔐, the minor arc 𝔪 and the trivial arc 𝔱, which are defined by

M={α:|α|≤1},m={α:1<|α|≤ξ},t={α:|α|>ξ},

where ξ = τ⁻²X^1/80+ε. By the arguments of section 5 in [15], we have

J(t)=o(τ2X77/60). (2.4)

3 Preliminary lemmas

Lemma 3.1

[19, Theorem 3.1] Suppose that N ≥ 2 and α satisfies

|qα−a|≤q−1,(a,q)=1,q∈N,a∈Z.

Then we have

∑p≤N(log⁡p)e(αp)≪(log⁡N)4(N12q12+N45+Nq−12).

Corollary 3.2

Suppose that X≥Z≥X45+ε and |S₁(α)| > Z. Then there are coprime integers a, q satisfying

1≤q≪(X/Z)2Xε,|qα−a|≪(X/Z)2Xε−1.

Proof

This follows from Lemma 3.1 immediately. □

Lemma 3.3

[8, Theorem 3] Let k ≥ 3 and σ(k) = 1/(3⋅ 2^k−1). Suppose that N ≥ 2 and α satisfies

|qα−a|≤Q−1,(a,q)=1,q∈N,q≤Q,a∈Z,

where Q = N^{(k²−2kσ(k))/(2k−1)}. Then, for any ε > 0,

∑p≤N(log⁡p)e(αpk)≪N1−σ(k)+ε+N1+ε(q+Nk|qα−a|)1/2.

Corollary 3.4

Suppose that P4≥Z≥P41−1/24+ε and |S₄(α)| > Z. Then there are coprime integers a, q satisfying

1≤q≪(P4/Z)2P4ε,|qα−a|≪(P4/Z)2P4ε−4.

Proof

This follows from Lemma 3.3 immediately. □

Lemma 3.5

[7, Lemma 3] Suppose that N ≥ 2 and α satisfies

|qα−a|≤q−1,(a,q)=1,q∈N,a∈Z.

Then, for any ε > 0,

∑p≤N(log⁡p)e(αp2)≪N1+ε1q+1N1/2+qN21/4.

Corollary 3.6

[7, Corollary 1] Suppose that P₂ ≥ Z ≥ P27/8+ε , and that |S₂(α)| > Z. Then there are coprime integers a, q satisfying

1≪q≪(P2/Z)4P2ε,|qα−a|≪(P2/Z)4P2ε−2.

Lemma 3.7

[16, Lemma 3.7] Suppose that

f(α)∈{S1(λ1α)2,S3(λ3α)8,S4(λ4α)16,S2(λ2α)2S3(λ3α)2S5(λ5α)2, S2(λ2α)2S4(λ4α)4,S2(λ2α)2S5(λ5α)6}.

Then we have

∫−11|f(α)|dα≪f(0)X−1+ε; (3.1)

∫R|f(α)|Kτ(α)dα≪τf(0)X−1+ε. (3.2)

We define the multiplicative function w₃(q) by taking

w3(p3u+v)=3p−u−1/2,when u≥0 and v=1;p−u−1,when u≥0 and 2≤v≤3. (3.3)

Lemma 3.8

[21, Lemma 2.3] If α is a real number satisfying that there exist a ∈ ℤ and q ∈ ℕ with (a, q) = 1, 1 ≤ q ≤ P^3/4 and |qα − a| ≤ P^−9/4, then one has

∑P≤x<2Pe(x3α)≪w3(q)P1+P3|α−a/q|,

otherwise, one has ∑P≤x<2Pe(x3α)≪P34+ε .

Lemma 3.9

[21, Lemma 2.1] Let c be a constant. For Q ≥ 2, one has

∑1≤q≤Qd(q)cw3(q)2≪(log⁡Q)A,

where A is a positive constant, d(q) is the divisor function.

4 The major arc

In this section, we give a low bound for the integral on the major arc 𝔐. First, we consider the standard major arc 𝔐^* = {α : |α| ≤ X^{−1+1/12−ε}}. Using the idea due to Harman [7], we get the following lemma (one can also see section 3 of Mu and Qu [16]). One may improve the standard major arc to {α : |α| ≤ X^{−1+2/15−ε}} by using some ideas due to Languasco and Zaccagnini [10] (one can also see [5]). But there is no improvement for our result, because our improvement comes from the minor arc.

Lemma 4.1

We have

J(M∗)≫τ2X77/60. (4.1)

Lemma 4.2

We have

J(M∖M∗)=o(τ2X77/60). (4.2)

Proof

For a given α, by Dirichlet’s theorem in Diophantine approximation, there exist integers a₁, a₂, q₁, q₂ depending on α such that

|q1λ1α−a1|≤X−1+1/100,|q2λ2α−a2|≤X−1+1/100

with (a_j, q_j) = 1 and 1 ≤ q_j ≤ X^1−1/100. Since α ∈ 𝔐 ∖ 𝔐^*, we see that a₁a₂ ≠ 0 and a_j/|α| ≪ q_j. Now we assert that

max(q1,q2)≥X1/100. (4.3)

We will reason by absurdity. Suppose both q₁ and q₂ are less that X^1/100. We have

|a2q1λ1/λ2−a1q2|=a2λ2αq1λ1α−a1−a1λ2αq2λ2α−a2≪X−1+1/50.

Since there is a convergent a/q to λ₁/λ₂ with q = X^5/12. Thus we have

|a2q1λ1/λ2−a1q2|=o(q−1). (4.4)

But

|a2q1|≪q1q2≪X1/50=o(q). (4.5)

This contradicts the definition of q as the denominator of a convergent to λ₁/λ₂ (see Lemma 9 of [1]). Thus one of q₁, q₂ is greater than X^1/100. Then, by Lemmas 3.1 and 3.5, we have

min|S1(λ1α)|,|S2(λ2α)|2≪X1−1/200+ε. (4.6)

Hence, by the arguments of Lemma 4.6 of [3], it is easy to get

J(M∖M∗)=o(τ2X77/60).

□

5 The minor arc

First, we divide the minor arc 𝔪 into four parts. Let 𝔪′ = 𝔪₁ ∪ 𝔪₂ ∪ 𝔪₃, and 𝔪₄ = 𝔪 ∖ 𝔪′, where

m1={α∈m:|S1(λ1α)|≤X1−1/6+ε},m2={α∈m:|S1(λ1α)|>X1−1/6+ε;|S2(λ2α)|>X1/2−1/16+ε},m3={α∈m:|S1(λ1α)|>X1−1/6+ε;|S4(λ4α)|>X1/4−1/96+ε}.

Now, we begin to estimate the integral on 𝔪_j respectively. First, it is easy to see that

Lemma 5.1

We have

J(m2)≪τX77/60−1/32+ε. (5.2)

Proof

We use the method of Harman [7]. We divide 𝔪₂ into disjoint sets such that for α ∈ 𝓐(Z₁, Z₂, y), we have

Z1≤|S1(λ1α)|<2Z1orZ2≤|S2(λ2α)|<2Z2ory≤|α|<2y,

where Z₁ = X^1−1/6+ε2^t₁, Z₂ = X^{1/2−1/16+ε}2^t₂, y = 2^s for some positive integers t₁, t₂, s. Thus, by Corollaries 3.2 and 3.6, there exist two pairs of coprime integers (a₁, q₁), (a₂, q₂) with a₁a₂ ≠ 0 and

1≤q1≪(X/Z1)2Xε,|q1λ1α−a1|≪(X/Z1)2Xε−1;1≤q2≪(X1/2/Z2)4Xε,|q2λ2α−a2|≪(X1/2/Z2)4Xε−1.

Then for any α ∈ 𝓐(Z₁, Z₂, y), we have |a_j/α| ≪ q_j.

Let 𝓐′ = 𝓐(Z₁, Z₂, y, Q₁, Q₂) be the subset of 𝓐(Z₁, Z₂, y) for which q_j ∼ Q_j. Then, by a familiar argument (see P. 147 of [17] for example),

|a2q1λ1λ2−a1q2|=|a2(q1λ1α−a1)+a1(a2−q2λ2α)λ2α|≪Q2(X/Z1)2Xε−1+Q1(X1/2/Z2)4Xε−1≪X3+2εZ12Z24≪X−5/12−4ε.

Also

|a2q1|≪yQ1Q2.

Note that q = X^5/12. We have

∥a2q1λ1λ2∥≤14q,q1∼Q1,a2≍yQ2, (5.3)

since X is sufficiently large. Then by the pigeon-hole principle and the Legendres law of best approximation for continued fractions, the above inequality (5.7) have ≪ yQ₁Q₂q⁻¹ solutions of |a₂q₁| (see Lemma 9 of [1]). Clearly, each value of |a₂q₁| corresponds to ≪ X^ε values of a₁, a₂, q₁, q₂ by the well-known bound on the divisor function. Hence, we conclude that

μ(A′)≪XεyQ1Q2qmin(X/Z1)2Xε−1Q1−1,(X1/2/Z2)4Xε−1Q2−1 ≪XεyQ1Q2qX1+εZ1Z22Q11/2Q21/2≪X1+2εyQ11/2Q21/2qZ1Z22≪X3+3εyqZ12Z24, (5.4)

where μ(𝓐′) is the Lebesgue measure of 𝓐′. Thus we have

J(A′)≪Z1Z2X1/3+1/4+1/5μ(A′)min(τ2,y−2)≪τX227/60+3εqZ1Z23≪τX77/60−1/16+ε.

Summing over all possible values of Z₁, Z₂, y, Q₁, Q₂, we conclude that

J(m2)≪τX77/60−1/32+ε. (5.5)

□

Lemma 5.2

We have

J(m3)≪τX77/60−1/32+ε. (5.6)

Proof

The proof is similar to that of lemma 5.1, we only give a brief proof. We divide 𝔪₃ into disjoint sets such that for α ∈ 𝓐(Z₁, Z₂, y), we have

Z1≤|S1(λ1α)|<2Z1orZ2≤|S4(λ4α)|<2Z2ory≤|α|<2y,

where Z₁ = X^1−1/6+ε2^t₁, Z₂ = X^{1/4−1/96+ε}2^t₂, y = 2^s for some positive integers t₁, t₂, s. Thus, by Corollaries 3.2 and 3.4, there exist two pairs of coprime integers (a₁, q₁), (a₂, q₂) with a₁a₂ ≠ 0 and

1≤q1≪(X/Z1)2Xε,|q1λ1α−a1|≪(X/Z1)2Xε−1;1≤q2≪(X1/4/Z2)2Xε,|q2λ4α−a2|≪(X1/4/Z2)2Xε−1.

Let 𝓐′ = 𝓐(Z₁, Z₂, y, Q₁, Q₂) be the subset of 𝓐(Z₁, Z₂, y) for which q_j ∼ Q_j. Then,

|a2q1λ1λ4−a1q2|≪X3/2+2εZ12Z22≪X−31/48−2ε.

Also

|a2q1|≪yQ1Q2.

Since q′ ≍ q = X^5/12, we have

∥a2q1λ1λ4∥≤14q′,q1∼Q1,a2≍yQ2. (5.7)

Hence, we conclude that

μ(A′)≪XεyQ1Q2q′min(X/Z1)2Xε−1Q1−1,(X1/4/Z2)2Xε−1Q2−1 ≪XεyQ1Q2q′X1/4+εZ1Z2Q11/2Q21/2≪X1/4+2εyQ11/2Q21/2q′Z1Z2≪X3/2+3εyq′Z12Z22. (5.8)

Thus by Lemma 3.7, we have

J(A′)≪∫A′|S1(λ1α)S4(λ4α)|2Kτ(α)dα1/2∫R|S2(λ2α)S3(λ3α)S5(λ5α)|2Kτ(α)dα1/2≪τX16/15+ε1/2min(τ2,y−2)Z12Z22X3/2+3εyq′Z12Z221/2≪τX77/60+2ε(q′)1/2≪τX77/60−5/24+2ε.

Summing over all possible values of Z₁, Z₂, y, Q₁, Q₂, we conclude that

J(m3)≪τX77/60−1/32+ε.

□

Lemma 5.3

We have

J(m4)≪τX77/60−1/32+ε. (5.9)

Proof

We use the method of the first author and Zhao [4]. First, by Cauchy’s inequality, we get

J(m4)≪∫R|S1(λ1α)|2Kτ(α)dα1/2I(2)1/2≪(τX1+ε)1/2I(2)1/2, (5.10)

where

I(t)=∫m4|S2(λ2α)2S3(λ3α)tS4(λ4α)2S5(λ5α)2|Kτ(α)dα. (5.11)

Then we have

I(2)=∑p∈J3(log⁡p)∫m4e(αλ3p3)S3(−λ3α)|S2(λ2α)2S4(λ4α)2S5(λ5α)2|Kτ(α)dα≤(log⁡X)∑n∈J3∫m4e(αλ3n3)S3(−λ3α)|S2(λ2α)2S4(λ4α)2S5(λ5α)2|Kτ(α)dα.

Then, by Cauchy’s inequality, we get

I(2)≪P31/2(log⁡X)L1/2, (5.12)

where

L=∑n∈J3∫m4e(αλ3n3)S3(−λ3α)|S2(λ2α)2S4(λ4α)2S5(λ5α)2|Kτ(α)dα2

For the sum 𝓛, we have

L=∑n∈J3∫m4∫m4|S2(λ2α)2S4(λ4α)2S5(λ5α)2S2(λ2β)2S4(λ4β)2S5(λ5β)2|S3(−λ3α)S3(λ3β)e(λ3n3(α−β))Kτ(α)Kτ(β)dαdβ≤∫m4|S2(λ2β)2S4(λ4β)2S5(λ5β)2S3(λ3β)F(β)|Kτ(β)dβ, (5.13)

where

F(β)=∫m4|S2(λ2α)2S4(λ4α)2S5(λ5α)2S3(−λ3α)T(λ3(α−β))|Kτ(α)dα (5.14)

and

T(x)=∑n∈J3e(xn3).

Let Mβ(r,b)={α∈m4:|rλ3(α−β)−b|≤P3−9/4} . Then the set 𝓜_β(r, b) ≠ ∅ forces that

|b+rλ3β|≤|rλ3(α−β)−b|+|rλ3α|≤P3−9/4+r|λ3|ξ.

Let B={b∈Z:|b+rλ3β|≤P3−9/4+r|λ3|ξ} . We divide the set 𝓑 into two sets 𝓑₁ = {b ∈ ℤ : |b + rλ₃β| ≤ r |λ₃|τ⁻¹} and 𝓑₂ = 𝓑 ∖ 𝓑₁. Let

Mβ=⋃1≤r≤P33/4⋃b∈B(b,r)=1Mβ(r,b).

Then by Lemma 3.8, we have

F(β)≪P3∫Mβ|S2(λ2α)2S4(λ4α)2S5(λ5α)2S3(−λ3α)|w3(r)Kτ(α)1+P33|λ3(α−β)−b/r|dα+P33/4+εI(1), (5.15)

where w₃(r) is defined as in (3.3). Note that |S₂(λ₂α)| ≤ P₂X^−1/16+ε and |S₄(λ₄α)| ≤ P₄X^−1/96+ε for α ∈ 𝔪₄. Then, by Cauchy’s inequality, we get

where

J(β)=∫Mβ|S5(λ5α)2|w3(r)2Kτ(α)(1+P33|λ3(α−β)−b/r|)2dα. (5.17)

Now we begin to estimate the integral 𝔍(β). First, we divide it into two parts.

J(β)=∑1≤r≤P33/4∑b∈B(b,r)=1∫Mβ(r,b)|S5(λ5α)2|w3(r)2Kτ(α)(1+P33|λ3(α−β)−b/r|)2dα=J1(β)+J2(β), (5.18)

where

Jj(β)=∑1≤r≤P33/4∑b∈Bj(b,r)=1∫Mβ(r,b)|S5(λ5α)2|w3(r)2Kτ(α)(1+P33|λ3(α−β)−b/r|)2dα. (5.19)

For the first part, we have

J1(β)≪τ2∑1≤r≤P33/4w3(r)2∑b∈B1(b,r)=1∫|rλ3γ|<P3−9/4|S5(λ5(β+γ)+bλ5/(rλ3))|2(1+P33|λ3γ|)2dγ ≪τ2∑1≤r≤P33/4w3(r)2∫|rλ3γ|<P3−9/4U(B1∗)(1+P33|λ3γ|)2dγ,

where

U(B1∗)=∑b∈B1∗|S5(λ5(β+γ)+bλ5/(rλ3))|2,

and

B1∗={b∈Z:−rv([|λ3|v−1τ−1]+1)<b+rλ3β≤rv([|λ3|v−1τ−1]+1)}.

Since λ₅/λ₃ is rational, we take λ₅/λ₃ = u/v with u, v ∈ ℤ and (u, v) = 1. We take r1=r(u,r) . Then we have

U(B1∗)=∑p1,p2∈J5∑b∈B1∗e((λ5(β+γ)+bλ5/(rλ3))(p15−p25))≤∑p1,p2∈J5∑b∈B1∗eburv(p15−p25)=2rv([|λ3|v−1τ−1]+1)∑p1,p2∈J5u(p15−p25)≡0(modrv)1≪rτ−1∑p1,p2∈J5p15≡p25(modr1v)1≪rτ−1P52(r1v)−2∑1≤b1,b2≤r1v;(b1b2,r1v)=1b15≡b25(modr1v)1≪rτ−1P52(r1v)−1∑1≤b≤r1vb5≡1(modr1v)1≪τ−1P52d(r)c.

Thus, by Lemma 3.9, we have

J1(β)≪τP52∑1≤r≤P33/4w3(r)2d(r)c∫|rλ3γ|<P3−9/41(1+P33|λ3γ|)2dγ ≪τP52X−1∑1≤r≤P33/4w3(r)2d(r)c≪τP52X−1+ε. (5.20)

Now, we begin to estimate 𝔍₂(β). First, without loss of generality we need only consider the set

B2′={b∈Z:r|λ3|τ−1<b+rλ3β≤P3−9/4+r|λ3|ξ}

which falls in the set

B2∗={b∈Z:rvκ1<b+rλ3β≤rvκ2},

where κ₁ = [|λ₃|v⁻¹τ⁻¹] and κ₂ = [|λ₃|v⁻¹ξ] + 2. Then we have

J2(β)≪∑1≤r≤P33/4∑b∈B2∗∫Mβ(r,b)|S5(λ5α)|2w3(r)2Kτ(α)(1+P33|λ3(α−β)−b/r|)2dα ≪∑1≤r≤P33/4w3(r)2∑b∈B2∗∫Mβ(r,b)|S5(λ5α)|2|α|−2(1+P33|λ3(α−β)−b/r|)2dα ≪∑1≤r≤P33/4w3(r)2∑κ1≤k<κ21(k−1)2∑rvk<b+rλ3β≤rv(k+1)∫Mβ(r,b)|S5(λ5α)|2(1+P33|λ3(α−β)−b/r|)2dα ≪∑1≤r≤P33/4w3(r)2∑κ1≤k<κ21(k−1)2∫|rλ3γ|<P3−9/4U(Ck)(1+P33|λ3γ|)2dγ,

where 𝓒_k = {b ∈ ℤ : rvk < b + rλ₃β ≤ rv(k + 1)}. On the other hand, similar to the above estimate of U(B1∗) , we have U(Ck)≪P52d(r)c . Thus we have

J2(β)≪P25X−1∑1≤r≤P33/4w3(r)2d(r)c∑κ1≤k<κ21(k−1)2≪τP25X−1+ε. (5.21)

Combining (5.15)-(5.21), we have

F(β)≪τ1/2P2P3P4P5X−1/2−7/96+εI(2)1/2+P33/4+εI(1) (5.22)

uniformly for β ∈ ℝ.

Hence, by (5.12), (5.13) and (5.22), we have

I(2)≪P37/8I(1)+τ1/4(P2P32P4P5)1/2X−1/4−7/192+εI(1)1/2I(2)1/4. (5.23)

By Hölder’s inequality and Lemma 3.7, we have

Thus we have

I(2)≪P37/8I(2)1/3(τP22P31/2P42P52X−1+ε)2/3+I(2)5/12τ7/12(P2P3P4P5)7/6X−7/12−7/192+ε.

Then this implies

I(2)≪τP22P31/2P42P52X−1+εP321/16+τ(P2P3P4P5)2X−1−1/16+ε≪τX47/30−1/16+ε. (5.24)

Then (5.9) follows from (5.10) and (5.24) immediately. □

6 Completion of the proof of Theorem 1.1

We take τ = X^1/32+2ε. Combining (2.4), (5.1) and Lemmas 4.1, 4.2, 5.1, 5.2, 5.3, we deduce that 𝓙(ℝ) ≫ τ² X^77/60. Thus by (2.3), we have

N(η,X)≫τX77/60(log⁡X)−5.

Note that max(pjj)≍X , so τ≍max(pjj)−1/32+2ε . Then we see that the following inequality

|λ1p1+⋯+λ4p44+λ5p55+η|<max(pjj)−1/32+2ε

has τ X^77/60(log X)⁻⁵ solutions in primes p_j. Since X = q^12/5 and λ₁/λ₂ is irrational, there are infinitely many pairs of integers q, a. This implies that the last inequality has infinitely many solutions in primes p_j.

Funding: The first author is partially supported by the National Natural Science Foundation of China (Grant No. 11871193). The second and third authors are partially supported by the National Natural Science Foundation of China (Grant No. 11471112).
Availability of data and materials: Not applicable.
Competing interests: The authors declare that they have no competing interests.
Author’s contributions: All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.
Consent for publication: Not applicable

Acknowledgement

We thank the referees for their time and comments.

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Received: 2018-11-21

Accepted: 2019-04-05

Published Online: 2019-05-30

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https://doi.org/10.1515/math-2019-0045

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Diophantine approximation; primes; Davenport-Heilbronn method

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