One Diophantine inequality with unlike powers of prime variables

Abstract

In this paper, we show that if $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, $\lambda _{4}$, $\lambda_{5}$ are nonzero real numbers not all of the same sign, η is real, $0<\sigma<\frac{1}{720}$, and at least one of the ratios $\lambda_{i}/\lambda_{j}$ ($1\leq i< j\leq5$) is irrational, then the inequality $|\lambda_{1}p_{1}+\lambda_{2}p_{2}^{2}+\lambda_{3}p_{3}^{3}+\lambda_{4}p_{4}^{4}+\lambda _{5}p_{5}^{5}+\eta|<(\max_{ 1\leq j\leq5}{p_{j}^{j}})^{-\sigma}$ has infinite solutions with primes $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, $p_{5}$.

1 Introduction

Diophantine inequalities with integer or prime variables have been considered by many scholars. Recently, Yang and Li in [1] proved that the inequality

$$\begin{aligned} \biggl|\lambda_{1}x_{1}^{2}+\lambda_{2}x_{2}^{3}+ \lambda_{3}x_{3}^{4}+\lambda_{4}x_{4}^{5}-p- \frac {1}{2} \biggr|< \frac{1}{2} \end{aligned}$$

has infinite solutions with natural numbers $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$ and prime p. Using the Davenport-Heilbronn method, we establish our result as follows.

Theorem 1.1

Let $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, $\lambda_{4}$, $\lambda_{5}$ be nonzero real numbers not all of the same sign, η is real, $0<\sigma<\frac {1}{720}$, and at least one of the ratios $\lambda_{i}/\lambda_{j}$ ($1\leq i< j\leq5$) is irrational, then the inequality

$$\bigl|\lambda_{1}p_{1}+\lambda_{2}p_{2}^{2}+ \lambda_{3}p_{3}^{3}+\lambda_{4}p_{4}^{4}+ \lambda _{5}p_{5}^{5}+\eta\bigr|< \Bigl(\max _{ 1\leq j\leq5}{p_{j}^{j}}\Bigr)^{-\sigma} $$

has infinite solutions with primes $p_{1}$, $p_{2}$, $p_{3}$, $p_{4}$, $p_{5}$.

2 Notation and outline of the proof

Throughout, we use p to denote a prime number. We denote by δ a sufficiently small positive number and by ε an arbitrarily small positive number, not necessarily the same at different occurrences. Constants, both explicit and implicit, in Landau or Vinogradov symbols may depend on $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, $\lambda_{4}$, $\lambda_{5}$, and η. We write $e(x)=e^{2\pi ix}$. We take X to be the basic parameter, a large real integer. Since at least one of the ratios $\lambda_{i}/\lambda_{j}$ ($1\leq i< j\leq5$) is irrational, without loss of generality we may assume that $\lambda_{1}/\lambda_{2}$ is irrational. For the other cases, the only difference is in the following intermediate region, and we may deal with the same method in Section 4.

Since $\lambda_{1}/\lambda_{2}$ is irrational, there are infinitely many pairs of integers q, a with $|\lambda_{1}/\lambda_{2}-a/q|\leq q^{-2}$, $(a,q)=1$, $q>0$, and $a\neq0$. We choose q to be large in terms of $\lambda_{1}$, $\lambda_{2}$, $\lambda _{3}$, $\lambda_{4}$, $\lambda_{5}$, η and make the following definitions:

$$\begin{aligned}& N=q^{2}, \qquad L=\log N,\qquad 0< \sigma< \frac{\theta}{32}< \frac{1}{720}, \nu =N^{-\sigma}, \qquad \tau=N^{-1+\theta}, \end{aligned}$$

(2.1)

$$\begin{aligned}& P=N^{\theta}L^{-1},\qquad Q=\bigl(|\lambda_{1}|^{-1}+| \lambda_{2}|^{-1}\bigr)N^{1-\theta},\qquad T_{1}=T_{2}^{2}=T_{3}^{3}=T_{4}^{4}=T_{5}^{5}=N^{\frac{1}{3}}. \end{aligned}$$

(2.2)

Let u be a positive real number, we define

$$\begin{aligned}& K_{u}(\alpha)= \biggl(\frac{\sin{\pi u\alpha}}{\pi\alpha} \biggr)^{2}\quad(\alpha \neq0),\qquad K_{u}(0)=u^{2}, \end{aligned}$$

(2.3)

$$\begin{aligned}& F_{k}(\alpha)=\sum_{(\delta N)^{1/k}\leq p\leq N^{1/k}}e\bigl(\lambda _{k}p^{k}\alpha\bigr)\log p,\quad k=1,2,3,4,5, \end{aligned}$$

(2.4)

$$\begin{aligned}& I_{k}(\alpha)= \int_{(\delta N)^{1/k}}^{N^{1/k}}e\bigl(\lambda_{k}y^{k} \alpha\bigr) \,\mathrm{d}y,\quad k=1,2,3,4,5, \end{aligned}$$

(2.5)

$$\begin{aligned}& J_{k}(\alpha)=\mathop{\sum_{|\gamma|\leq T_{k} }}_{\beta\geq\frac{2}{3}}\sum _{\delta N< n\leq N}n^{-1+\rho/k}e(\lambda_{k}\alpha n), \quad k=1,2,3,4,5, \end{aligned}$$

(2.6)

where $\rho=\beta+i\gamma(\beta,\gamma \mbox{ real})$ is a typical non-trivial zero of the Riemann Zeta function.

It follows from (2.3) that

$$ K_{u}(\alpha)\ll\min\bigl(u^{2},| \alpha|^{-2}\bigr),\qquad \int_{-\infty}^{+\infty }e(\alpha y)K_{u}(\alpha) \,\mathrm{d}\alpha=\max\bigl(0,u-|y|\bigr). $$

(2.7)

From (2.7) it is clear that

$$\begin{aligned} J&:= \int_{-\infty}^{+\infty}\prod_{j=1}^{5}F_{j}( \alpha)e(\alpha\eta)K_{\nu }(\alpha)\,\mathrm{d} \alpha \\ &\leq(\log N)^{5}\mathop{\sum_{|\lambda_{1}p_{1}+\lambda_{2}p_{2}^{2}+\lambda _{3}p_{3}^{3}+\lambda_{4}p_{4}^{4}+\lambda_{5}p_{5}^{5}+\eta|< \nu}}_{(\delta N)^{1/k} \leq p_{k}\leq N^{1/k}, k=1,2,3,4,5}1 \\ &=:(\log N)^{5}\mathcal{N}(N). \end{aligned}$$

Thus we have

$$\mathcal{N}(N)\geq(\log N)^{-5}J. $$

To estimate J, we split the range of infinite integration into three sections, traditional named the neighborhood of the origin $\mathfrak{C}=\{\alpha\in\mathbb {R}:|\alpha|\leq\tau\}$, the intermediate region $\mathfrak{D}=\{\alpha\in\mathbb{R}:\tau\leq |\alpha|\leq P\}$, the trivial region $\mathfrak{c}=\{\alpha\in\mathbb {R}:|\alpha|>P\}$.

To prove Theorem 1.1, we shall establish that

$$J(\mathfrak{C})\gg\nu^{2}N^{\frac{77}{60}}, \qquad J(\mathfrak{D})=o \bigl(\nu ^{2}N^{\frac{77}{60}}\bigr),\qquad J(\mathfrak{c})=o\bigl( \nu^{2}N^{\frac{77}{60}}\bigr) $$

in Sections 3, 4, and 5, respectively. Thus

$$\mathcal{N}(N)\gg\nu^{2}(\log N)^{-5}N^{\frac{77}{60}}, $$

and Theorem 1.1 can be established.

3 The neighborhood of the origin

We let

$$\begin{aligned} B_{k}(\alpha)=F_{k}(\alpha)-I_{k}( \alpha)+J_{k}(\alpha),\quad k=1,2,3,4,5. \end{aligned}$$

(3.1)

We use C to denote a positive absolute constant, not necessarily the same one on each occurrence.

Lemma 3.1

We have

$$\begin{aligned} B_{k}(\alpha)\ll N^{\frac{2}{3k}}L^{C}\bigl(1+| \alpha|N\bigr), \quad k=1,2,3,4,5. \end{aligned}$$

(3.2)

This is Lemma 7 of Vaughan [2].

Lemma 3.2

For $k=1,2,3,4,5$, we have

$$\begin{aligned} &I_{k}(\alpha)\ll N^{\frac{1}{k}}\min \bigl(1,N^{-1}|\alpha|^{-1}\bigr), \end{aligned}$$

(3.3)

$$\begin{aligned} & \int_{-\frac{1}{2}}^{\frac{1}{2}}\bigl|J_{k}(\alpha)\bigr|^{2} \,\mathrm{d}\alpha\ll N^{\frac{2}{k}-1}\exp\bigl(-2L^{-\frac{1}{5}}\bigr), \end{aligned}$$

(3.4)

$$\begin{aligned} & \int_{-\frac{1}{2}}^{\frac{1}{2}}\bigl|I_{k}(\alpha)\bigr|^{2} \,\mathrm{d}\alpha\ll N^{\frac{2}{k}-1}, \end{aligned}$$

(3.5)

$$\begin{aligned} & \int_{-\tau}^{\tau}\bigl|B_{k}(\alpha)\bigr|^{2} \,\mathrm{d}\alpha\ll N^{\frac {2}{k}-1}\exp\bigl(-2L^{-\frac{1}{5}}\bigr), \end{aligned}$$

(3.6)

$$\begin{aligned} & \int_{-\tau}^{\tau}\bigl|F_{k}(\alpha)\bigr|^{2} \,\mathrm{d}\alpha\ll N^{\frac{2}{k}-1}. \end{aligned}$$

(3.7)

Proof

The inequality (3.6) follows from (2.1) and Lemma 3.1. The others are similar to Lemma 8 of Vaughan [2]. □

Lemma 3.3

We have

$$\begin{aligned} \int_{\mathfrak{C}}\Biggl\vert \prod_{i=1}^{5}F_{i}( \alpha)-\prod_{i=1}^{5}I_{i}( \alpha)\Biggr\vert K_{\nu}(\alpha) \,\mathrm{d}\alpha\ll \nu^{2}N^{\frac{77}{60}}\exp\bigl(-L^{-\frac{1}{5}}\bigr). \end{aligned}$$

(3.8)

Proof

Note that

$$\begin{aligned} &\prod_{i=1}^{5}F_{i}(\alpha)- \prod_{i=1}^{5}I_{i}(\alpha) \\ &\quad=\bigl(F_{1}(\alpha)-I_{1}(\alpha)\bigr)\prod _{i=2}^{5}F_{i}(\alpha) +I_{1}(\alpha) \bigl(F_{2}(\alpha)-I_{2}(\alpha) \bigr)\prod_{i=3}^{5}F_{i}(\alpha) \\ &\qquad{}+I_{1}(\alpha)I_{2}(\alpha) \bigl(F_{3}( \alpha)-I_{3}(\alpha)\bigr)F_{4}(\alpha)F_{5}( \alpha) +\prod_{i=1}^{3}I_{i}( \alpha) \bigl(F_{4}(\alpha)-I_{4}(\alpha)\bigr)F_{5}( \alpha) \\ &\qquad{}+\prod_{i=1}^{4}I_{i}( \alpha) \bigl(F_{5}(\alpha)-I_{5}(\alpha)\bigr). \end{aligned}$$

Then by (2.7), (3.1), Lemma 3.2,

$$\begin{aligned} & \int_{\mathfrak{C}}\Biggl\vert \bigl(F_{1}( \alpha)-I_{1}(\alpha)\bigr)\prod_{i=2}^{5}F_{i}( \alpha)\Biggr\vert K_{\nu}(\alpha)\,\mathrm{d}\alpha \\ &\quad\ll\nu^{2}N^{\frac{47}{60}} \int_{-\tau}^{\tau}\bigl|\bigl(B_{1}( \alpha)-J_{1}(\alpha )\bigr)F_{2}(\alpha)\bigr|\,\mathrm{d}\alpha \\ &\quad\ll\nu^{2}N^{\frac{47}{60}} \biggl( \int_{-\tau}^{\tau}\bigl|\bigl(B_{1}( \alpha)-J_{1}(\alpha)\bigr)\bigr|^{2}\,\mathrm{d}\alpha \biggr)^{\frac{1}{2}} \biggl( \int_{-\tau}^{\tau}\bigl|F_{2}(\alpha)\bigr|^{2} \,\mathrm{d}\alpha \biggr)^{\frac {1}{2}} \\ &\quad\ll\nu^{2}N^{\frac{47}{60}} \biggl( \int_{-\tau}^{\tau}\bigl(\bigl|\bigl(B_{1}( \alpha)\bigr|^{2}+\bigl|J_{1}(\alpha)\bigr)\bigr|^{2}\bigr)\,\mathrm {d}\alpha \biggr)^{\frac{1}{2}} \\ &\quad\ll\nu^{2}N^{\frac{77}{60}}\exp\bigl(-L^{-\frac{1}{5}}\bigr). \end{aligned}$$

The other cases are similar, and the proof of Lemma 3.3 is completed. □

Lemma 3.4

We have

$$\begin{aligned} \int_{|\alpha|>\tau}\Biggl\vert \prod_{i=1}^{5}I_{i}( \alpha)\Biggr\vert K_{\nu}(\alpha )\,\mathrm{d}\alpha \ll \nu^{2}N^{\frac{77}{60}-4\theta}. \end{aligned}$$

(3.9)

It follows from (2.7) and (3.3).

Lemma 3.5

We have

$$\begin{aligned} \int_{-\infty}^{+\infty}\prod_{j=1}^{5}I_{j}( \alpha)e(\alpha\eta)K_{\nu }(\alpha)\,\mathrm{d} \alpha \gg \nu^{2}N^{\frac{77}{60}}. \end{aligned}$$

(3.10)

Proof

To prove (3.10), we write the left side as

$$\begin{aligned} \int_{\delta N}^{N} \int_{(\delta N)^{\frac{1}{2}}}^{N^{\frac{1}{2}}}\cdots \int_{(\delta N)^{\frac{1}{5}}}^{N^{\frac{1}{5}}} \int_{-\infty}^{+\infty} e \Biggl(\alpha \Biggl(\eta+\sum _{j=1}^{5}\lambda_{j}y_{j}^{j} \Biggr) \Biggr)K_{\nu }(\alpha)\,\mathrm{d} \alpha \,\mathrm{d}y_{1} \,\mathrm{d}y_{2}\cdots\,\mathrm{d}y_{5}, \end{aligned}$$

which, by (2.7), is

$$\begin{aligned} \int_{\delta N}^{N} \int_{(\delta N)^{\frac{1}{2}}}^{N^{\frac{1}{2}}}\cdots \int_{(\delta N)^{\frac{1}{5}}}^{N^{\frac{1}{5}}} \max \Biggl(0,\nu-\Biggl\vert \eta+ \sum_{j=1}^{5}\lambda_{j}y_{j}^{j} \Biggr\vert \Biggr) \,\mathrm{d}y_{1}\,\mathrm{d}y_{2}\cdots \,\mathrm{d}y_{5}. \end{aligned}$$

(3.11)

We let $z_{k}=y_{k}^{k}$, $k=1,2,3,4,5$, then the integral (3.11) can be written as

$$\begin{aligned} \frac{1}{120} \int_{\delta N}^{N} \cdots \int_{\delta N}^{N} z_{2}^{-\frac{1}{2}}z_{3}^{-\frac{2}{3}}z_{4}^{-\frac{3}{4}}z_{5}^{-\frac{4}{5}} \max \Biggl(0,\nu-\Biggl\vert \eta+\sum_{j=1}^{5} \lambda_{j}z_{j}\Biggr\vert \Biggr) \,\mathrm{d}z_{1} \cdots\,\mathrm{d}z_{5}. \end{aligned}$$

(3.12)

Since $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, $\lambda_{4}$, and $\lambda_{5}$ are not all of the same sign, we may assume without loss of generality that $\lambda_{1}<0$, $\lambda _{2}>0$. Consider the region

$$\mathcal{B}=\bigl\{ (z_{2},z_{3},z_{4},z_{5}): \delta^{\frac{1}{2}} N\leq z_{2} \leq 2\delta^{\frac{1}{2}} N, \delta N\leq z_{j} \leq2\delta N\ (j=3,4,5)\bigr\} . $$

Then, for δ sufficiently small and large N, whenever $(z_{2},z_{3},z_{4},z_{5})\in\mathcal{B}$ one has

$$2\delta N< -(\lambda_{2}z_{2}+\lambda_{3}z_{3}+ \lambda_{4}z_{4}+\lambda_{5}z_{5})\lambda _{1}^{-1}< \frac{1}{2}N $$

and so every $z_{1}$ with $|\lambda_{1}z_{1}+\cdots+\lambda_{5}z_{5}+\eta|\leq \frac{1}{2}\nu$ satisfies $\delta N< z_{1}< N$. Therefore the integral (3.12) is greater than

$$\begin{aligned} \frac{1}{480}\nu^{2} \int_{\mathcal{B}}z_{2}^{-\frac{1}{2}}z_{3}^{-\frac {2}{3}}z_{4}^{-\frac{3}{4}} z_{5}^{-\frac{4}{5}} \,\mathrm{d}z_{2}\,\mathrm{d}z_{3} \,\mathrm{d}z_{4}\,\mathrm{d}z_{5} \gg\nu^{2}N^{\frac{77}{60}}. \end{aligned}$$

This completes the proof of Lemma 3.5. □

Together with Lemmas 3.3, 3.4, 3.5, we have

$$\begin{aligned} J(\mathfrak{C})= \int_{\mathfrak{C}}\prod_{j=1}^{5}F_{j}( \alpha)e(\alpha\eta )K_{\nu}(\alpha)\,\mathrm{d}\alpha \gg \nu^{2}N^{\frac{77}{60}}. \end{aligned}$$

(3.13)

4 The intermediate region

Lemma 4.1

We have

$$\begin{aligned} & \int_{-\infty}^{+\infty}\bigl|F_{j}(\alpha)\bigr|^{2^{j}}K_{\nu}( \alpha)\,\mathrm{d} \alpha \ll N^{\frac{2^{j}}{j}-1+\varepsilon}, \quad j=2,3,4,5, \end{aligned}$$

(4.1)

$$\begin{aligned} & \int_{-\infty}^{+\infty}\bigl|F_{1}(\alpha)\bigr|^{2}K_{\nu}( \alpha)\,\mathrm{d} \alpha \ll NL. \end{aligned}$$

(4.2)

Proof

By (2.7), we have

$$\begin{aligned} & \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)\bigr|^{4}K_{\nu}( \alpha)\,\mathrm{d} \alpha \\ &\quad= \sum_{(\delta N)^{\frac{1}{2}}\leq p_{1},p_{2},p_{3},p_{4} \leq N^{\frac{1}{2}}} \prod _{i=1}^{4}\log p_{i} \max\bigl(0,\nu-\bigl| \lambda_{2}\bigl(p_{1}^{2}+p_{2}^{2}-p_{3}^{2}-p_{4}^{2} \bigr)\bigr|\bigr) \\ &\quad\ll L^{4}\sum_{(\delta N)^{\frac{1}{2}}\leq p_{1},p_{2},p_{3},p_{4} \leq N^{\frac{1}{2}}} \max\bigl(0, \nu-\bigl|\lambda_{2}\bigl(p_{1}^{2}+p_{2}^{2}-p_{3}^{2}-p_{4}^{2} \bigr)\bigr|\bigr). \end{aligned}$$

Since N is large, $|\lambda_{2}(p_{1}^{2}+p_{2}^{2}-p_{3}^{2}-p_{4}^{2})|<\nu$ if and only if $p_{1}^{2}+p_{2}^{2}=p_{3}^{2}+p_{4}^{2}$. Thus, by Hua’s inequality,

$$\begin{aligned} \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)\bigr|^{4}K_{\nu}( \alpha)\,\mathrm{d} \alpha \ll\nu N^{1+\varepsilon}. \end{aligned}$$

The proofs of the cases $j=3,4,5$ and (4.2) are similar. □

Lemma 4.2

We have

$$\begin{aligned} \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)\bigr|^{2}\bigl|F_{4}( \alpha)\bigr|^{4}K_{\nu}(\alpha )\,\mathrm{d} \alpha \ll\nu N^{1+\varepsilon}. \end{aligned}$$

(4.3)

Proof

By (2.7), we have

$$\begin{aligned} & \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)\bigr|^{2}\bigl|F_{4}( \alpha)\bigr|^{4}K_{\nu }(\alpha)\,\mathrm{d} \alpha \\ &\quad\ll L^{6}\mathop{\sum_{(\delta N)^{\frac{1}{2}}\leq p_{1},p_{2} \leq N^{\frac{1}{2}}}} _{(\delta N)^{\frac{1}{4}}\leq p_{3},p_{4},p_{5},p_{6} \leq N^{\frac{1}{4}}} \max\bigl(0,\nu-\bigl|\lambda_{2}\bigl(p_{1}^{2}-p_{2}^{2} \bigr)-\lambda _{4}\bigl(p_{3}^{4}+p_{4}^{4}-p_{5}^{4}-p_{6}^{4} \bigr)\bigr|\bigr) \\ &\quad\ll\nu L^{6}R(N), \end{aligned}$$

where $R(N)$ is the number of the solutions of the equation

$$\begin{aligned} &\lambda_{2}\bigl(p_{1}^{2}-p_{2}^{2} \bigr)=\lambda_{4}\bigl(p_{3}^{4}+p_{4}^{4}-p_{5}^{4}-p_{6}^{4} \bigr), \\ &(\delta N)^{\frac{1}{2}}\leq p_{1},p_{2} \leq N^{\frac{1}{2}}, \qquad (\delta N)^{\frac{1}{4}}\leq p_{3},p_{4},p_{5},p_{6} \leq N^{\frac{1}{4}}. \end{aligned}$$

Then we have

$$R(N)\ll N^{\frac{1}{2}}\mathop{\sum_{(\delta N)^{\frac{1}{4}}\leq p_{3},p_{4},p_{5},p_{6} \leq N^{\frac{1}{4}}}}_{p_{3}^{4}+p_{4}^{4}-p_{5}^{4}-p_{6}^{4}= 0}1+ \mathop{\sum _{(\delta N)^{\frac{1}{4}}\leq p_{3},p_{4},p_{5},p_{6} \leq N^{\frac {1}{4}}}}_{p_{3}^{4}+p_{4}^{4}-p_{5}^{4}-p_{6}^{4}\neq0}d\bigl(\bigl|p_{3}^{4}+p_{4}^{4}-p_{5}^{4}-p_{6}^{4}\bigr| \bigr), $$

where $d(n)$ is the divisor function. Now (4.3) follows from [3], (2.1). □

Lemma 4.3

([4])

Suppose that $(a,q)=1$, $|\alpha -a/q|\leq q^{-2}$, then

$$\begin{aligned} \sum_{1\leq p \leq X}(\log p) e(p\alpha) \ll(\log X)^{5}\bigl(X^{1/2}q^{1/2}+X^{4/5}+Xq^{-1/2} \bigr). \end{aligned}$$

Lemma 4.4

([5])

Suppose that $(a,q)=1$, $|\alpha -a/q|\leq q^{-2}$, $\phi(x) =\alpha x^{k}+\alpha_{1}x^{k-1}+\cdots+\alpha_{k-1}x+\alpha_{k}$ ($k\geq2$), then

$$\begin{aligned} \sum_{1\leq p \leq X}(\log p) e\bigl(\phi(p)\bigr) \ll X^{1+\varepsilon}\bigl(q^{-1}+X^{-1/2}+qX^{-k} \bigr)^{4^{1-k}}. \end{aligned}$$

Lemma 4.5

For $\tau<|\alpha|\leq P$, we have

$$\begin{aligned} V(\alpha):=\min\bigl(F_{1}(\alpha),F_{2}( \alpha)^{2}\bigr) \ll N^{1-\frac{\theta }{2}+\varepsilon}. \end{aligned}$$

Proof

Let $\tau<|\alpha|\leq P$, we choose $a_{j}$, $q_{j}$ ($j=1,2$) so that $|\lambda_{j}\alpha-a_{j}/q_{j}|\leq Q^{-1}q_{j}^{-1}$ with $(a_{j},q_{j})=1$ and $1\leq q_{j} \leq Q$. By the method of Davenport and Heilbronn (see Lemma 11 of [6]), we have $\max(q_{1},q_{2})\geq P$. Then Lemma 4.5 follows from Lemmas 4.3 and 4.4. □

Lemma 4.6

We have

$$\begin{aligned} J(\mathfrak{D})= \int_{\mathfrak{D}}\prod_{j=1}^{5}F_{j}( \alpha)e(\alpha\eta )K_{\nu}(\alpha)\,\mathrm{d}\alpha \ll\nu^{2} N^{\frac{77}{60}-(\frac{\theta}{32}-\sigma)+\varepsilon}. \end{aligned}$$

(4.4)

Proof

By Lemmas 4.1, 4.2, 4.5, and Hölder’s inequality, we have

$$\begin{aligned} & \int_{\mathfrak{D}}\Biggl\vert \prod_{j=1}^{5}F_{j}( \alpha)e(\alpha\eta)K_{\nu }(\alpha)\Biggr\vert \,\mathrm{d}\alpha \\ &\quad\ll V(\alpha)^{\frac{1}{16}} \int_{-\infty}^{+\infty}\Biggl\vert \bigl(F_{1}( \alpha)^{\frac {15}{16}}F_{2}(\alpha) +F_{1}( \alpha)F_{2}(\alpha)^{\frac{7}{8}} \bigr) \prod _{j=3}^{5}F_{j}(\alpha)\Biggr\vert K_{\nu}(\alpha)\,\mathrm{d}\alpha \\ &\quad\ll V(\alpha)^{\frac{1}{16}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{1}(\alpha)\bigr|^{2}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{15}{32}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)\bigr|^{4}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{1}{8}} \\ &\qquad{} \times \biggl( \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)^{2}F_{4}( \alpha)^{4}\bigr| K_{\nu}(\alpha)\,\mathrm{d}\alpha \biggr)^{\frac{1}{4}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{3}(\alpha)\bigr|^{8}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{1}{8}} \\ &\qquad{} \times \biggl( \int_{-\infty}^{+\infty}\bigl|F_{5}(\alpha)\bigr|^{32} K_{\nu}(\alpha)\,\mathrm{d}\alpha \biggr)^{\frac{1}{32}} +V(\alpha)^{\frac{1}{16}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{1}(\alpha)\bigr|^{2}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{1}{2}}\\ &\qquad{}\times \biggl( \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)\bigr|^{4}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{3}{32}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)^{2}F_{4}( \alpha)^{4}\bigr| K_{\nu}(\alpha)\,\mathrm{d}\alpha \biggr)^{\frac{1}{4}}\\ &\qquad{} \times \biggl( \int_{-\infty}^{+\infty}\bigl|F_{3}(\alpha)\bigr|^{8}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{1}{8}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{5}(\alpha)\bigr|^{32} K_{\nu}(\alpha)\,\mathrm{d}\alpha \biggr)^{\frac{1}{32}} \\ &\quad\ll\nu N^{\frac{77}{60}-\frac{\theta}{32}+\varepsilon} \ll\nu^{2} N^{\frac{77}{60}-(\frac{\theta}{32}-\sigma)+\varepsilon}. \end{aligned}$$

 □

5 The trivial region

Lemma 5.1

Let $G(\alpha)=\sum e(\alpha f(x_{1},\ldots,x_{m}))$, where f is any real function and the summation is over any finite set of values of $x_{1},\ldots,x_{m}$. Then, for any $A>4$, we have

$$\begin{aligned} \int_{|\alpha|>A}\bigl|G(\alpha)\bigr|^{2}K_{\nu}(\alpha) \,\mathrm{d}\alpha \leq\frac{16}{A} \int_{-\infty}^{+\infty}\bigl|G(\alpha)\bigr|^{2}K_{\nu}( \alpha )\,\mathrm{d}\alpha. \end{aligned}$$

This is Lemma 2 of [7].

Lemma 5.2

We have

$$\begin{aligned} J(\mathfrak{c})= \int_{\mathfrak{c}}\prod_{j=1}^{5}F_{j}( \alpha)e(\alpha\eta )K_{\nu}(\alpha)\,\mathrm{d}\alpha \ll\nu^{2} N^{\frac{77}{60}-(\theta-\sigma)+\varepsilon}. \end{aligned}$$

Proof

By Lemmas 4.1, 4.2, 5.1, and Hölder’s inequality, we have

$$\begin{aligned} & \int_{\mathfrak{c}} \Biggl|\prod_{j=1}^{5}F_{j}( \alpha)e(\alpha\eta)K_{\nu }(\alpha) \Biggr|\,\mathrm{d}\alpha \\ &\quad\ll\frac{1}{P} \int_{-\infty}^{+\infty} \prod_{j=1}^{5}\bigl|F_{j}( \alpha)\bigr|K_{\nu}(\alpha)\,\mathrm{d}\alpha \\ &\quad\ll\frac{1}{P}\max\bigl(\bigl|F_{5}(\alpha)\bigr|\bigr) \biggl( \int_{-\infty}^{+\infty}\bigl|F_{1}(\alpha)\bigr|^{2}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{1}{2}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)\bigr|^{4}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{1}{8}} \\ &\qquad{}\times \biggl( \int_{-\infty}^{+\infty}\bigl|F_{2}(\alpha)^{2}F_{4}( \alpha)^{4}\bigr| K_{\nu}(\alpha)\,\mathrm{d}\alpha \biggr)^{\frac{1}{4}} \biggl( \int_{-\infty}^{+\infty}\bigl|F_{3}(\alpha)\bigr|^{8}K_{\nu}( \alpha)\,\mathrm {d}\alpha \biggr)^{\frac{1}{8}} \\ &\quad\ll\nu N^{\frac{77}{60}-\theta+\varepsilon} \ll\nu^{2} N^{\frac{77}{60}-(\theta-\sigma)+\varepsilon}. \end{aligned}$$

 □