The integer part of a nonlinear form with integer variables

Kai Lai

doi:10.1186/s13660-015-0874-2

Research
Open Access

The integer part of a nonlinear form with integer variables

Kai Lai¹Email author

Journal of Inequalities and Applications20152015:357

https://doi.org/10.1186/s13660-015-0874-2

Received: 30 August 2015
Accepted: 28 October 2015
Published: 11 November 2015

Abstract

Using the Davenport-Heilbronn method, we show that if $\lambda_{1},\lambda_{2},\ldots,\lambda_{9}$ are positive real numbers, at least one of the ratios $\lambda_{i}/\lambda_{j}$ ($1\leq i< j\leq9$) is irrational, then the integer parts of $\lambda_{1}x_{1}^{3}+\lambda_{2}x_{2}^{3}+\lambda_{3}x_{3}^{4}+\lambda_{4}x_{4}^{4} +\lambda_{5}x_{5}^{5}+\cdots+\lambda_{9}x_{9}^{5}$ are prime infinitely often for natural numbers $x_{1},x_{2},\ldots,x_{9}$.

Keywords

Davenport-Heilbronn method
integer variables
diophantine approximation

1 Introduction

In 2010, Brüdern et al. [1] proved that if $\lambda _{1},\ldots,\lambda_{s}$ are positive real numbers, $\lambda_{1}/\lambda_{2}$ is irrational, all Dirichlet L-functions satisfy the Riemann hypothesis $s\geq \frac{8}{3}k+2$, then the integer parts of

$$\lambda_{1}x^{k}_{1}+\lambda_{2}x^{k}_{2}+ \cdots+\lambda_{s}x^{k}_{s} $$

are prime infinitely often for natural numbers $x_{j}$.

Motivated by [1], using the Davenport-Heilbronn method, we consider the integer part of a nonlinear form with integer variables and mixed powers 3, 4 and 5, and establish one result as follows.

Theorem 1.1

Let $\lambda_{1},\lambda_{2},\ldots,\lambda_{9}$ be positive real numbers, at least one of the ratios $\lambda_{i}/\lambda_{j}$ ($1\leq i< j\leq9$) is irrational. Then the integer parts of

$$\lambda_{1}x_{1}^{3}+\lambda_{2}x_{2}^{3}+ \lambda_{3}x_{3}^{4}+\lambda_{4}x_{4}^{4}+ \lambda _{5}x_{5}^{5}+\cdots+\lambda_{9}x_{9}^{5} $$

are prime infinitely often for natural numbers $x_{1},x_{2},\ldots,x_{9}$.

It is noted that Theorem 1.1 holds without the Riemann hypothesis.

2 Notation

Throughout, we use p to denote a prime number and $x_{j}$ to denote a natural number. We denote by δ a sufficiently small positive number and by ε an arbitrarily small positive number. Constants, both explicit and implicit, in Landau or Vinogradov symbols may depend on $\lambda_{1},\lambda_{2},\ldots,\lambda _{9}$. We write $e(x)=\exp(2\pi i x)$. We use $[x]$ to denote the integer part of real variable x. We take X to be the basic parameter, a large real integer. Since at least one of the ratios $\lambda_{i}/\lambda_{j}$ ($1\leq i< j\leq9$) 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, then 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 $a\neq 0$. We choose q to be large in terms of $\lambda_{1},\lambda_{2},\ldots ,\lambda_{9}$ and make the following definitions.

$$\begin{aligned}[b] &N\asymp X,\qquad L=\log N,\qquad \bigl[N^{1-8\delta} \bigr]=q,\qquad \tau=N^{-1+\delta},\\ &Q= \bigl(|\lambda_{1}|^{-1}+| \lambda_{2}|^{-1} \bigr)N^{1-\delta},\qquad P=N^{6\delta},\qquad T=N^{\frac{1}{3}}. \end{aligned} $$

Let ν be a positive real number, we define

$$\begin{aligned} &K_{\nu}(\alpha)=\nu \biggl(\frac{\sin\pi \nu\alpha}{\pi\nu\alpha} \biggr)^{2},\quad\alpha\neq0,\qquad K_{\nu}(0)=\nu, \\ &F_{i}(\alpha)=\sum_{1\leq x\leq X^{\frac{1}{3}}}e \bigl(\alpha x^{3} \bigr), \quad i=1,2, \\ &F_{j}(\alpha)=\sum_{1\leq x\leq X^{\frac{1}{4}}}e \bigl(\alpha x^{4} \bigr),\quad j=3,4, \\ &F_{k}(\alpha)=\sum_{1\leq x\leq X^{\frac{1}{5}}}e \bigl(\alpha x^{5} \bigr), \quad k=5,\ldots,9, \\ &G(\alpha)=\sum_{p\leq N}(\log p)e(\alpha p), \\ &f_{i}(\alpha)=\int_{1}^{X^{\frac{1}{3}}}e \bigl( \alpha x^{3} \bigr)\,dx, \quad i=1,2, \\ &f_{j}(\alpha)=\int_{1}^{X^{\frac{1}{4}}}e \bigl( \alpha x^{4} \bigr)\,dx,\quad j=3,4, \\ &f_{k}(\alpha)=\int_{1}^{X^{\frac{1}{5}}}e \bigl( \alpha x^{5} \bigr)\,dx,\quad k=5,\ldots,9, \\ &g(\alpha)=\int_{1}^{N}e(\alpha x)\,dx. \end{aligned}$$

(2.1)

It follows from (2.1) that

$$\begin{aligned}& K_{\nu}(\alpha)\ll\min \bigl(\nu,\nu^{-1}| \alpha|^{-2} \bigr), \end{aligned}$$

(2.2)

$$\begin{aligned}& \int_{-\infty}^{+\infty}e(\alpha y)K_{\nu}(\alpha) \,d\alpha=\max \bigl(0,1-\nu^{-1}|y| \bigr). \end{aligned}$$

(2.3)

From (2.3) it is clear that

$$\begin{aligned}[b] J &=: \int_{-\infty}^{+\infty}\prod _{i=1}^{9}F_{i}(\lambda_{i} \alpha) G(-\alpha)e \biggl(-\frac{1}{2}\alpha \biggr)K_{\frac{1}{2}}( \alpha)\,d\alpha \\ &\leq \log N\mathop{\sum_{|\lambda_{1}x_{1}^{3}+\lambda_{2}x_{2}^{3}+\lambda_{3}x_{3}^{4}+\lambda_{4}x_{4}^{4} +\lambda_{5}x_{5}^{5}+\cdots+\lambda_{9}x_{9}^{5}-p-\frac{1}{2}|< \frac{1}{2}}}_{ {1\leq x_{1},x_{2}\leq X^{1/3}, 1\leq x_{3},x_{4}\leq X^{1/4},1\leq x_{5},\ldots,x_{9}\leq X^{1/5}, p\leq N}}1 \\ &=: (\log N){\mathcal{N}}(X), \end{aligned} $$

thus

$${\mathcal{N}}(X)\geq(\log N)^{-1}J. $$

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

3 The neighborhood of the origin

Lemma 3.1

If $\alpha=a/q+\beta$, where $(a,q)=1$, then

$$\sum_{1\leq x\leq N^{1/t}}e \bigl(\alpha x^{t} \bigr)=q^{-1}\sum_{m=1}^{q}e \bigl(am^{t}/q \bigr)\int_{1}^{N^{1/t}}e \bigl( \beta y^{t} \bigr)\,dy+O \bigl(q^{1/2+\varepsilon }\bigl(1+N|\beta|\bigr) \bigr). $$

Proof

This is Theorem 4.1 of [2]. □

If $|\alpha|\in\frak{C}$, by Lemma 3.1, taking $a=0$, $q=1$, then

$$ F_{i}(\alpha)=f_{i}(\alpha)+O \bigl(X^{\delta} \bigr), \quad i=1,2,\ldots,9. $$

(3.1)

Lemma 3.2

Let $\rho=\beta+i\gamma$ be a typical zero of the Riemann zeta function, C be a positive constant,

$$I(\alpha)=\sum_{|\gamma|\leq T, \beta\geq \frac{2}{3}}\sum _{n\leq N}n^{\rho-1}e(n\alpha),\qquad J(\alpha)=O \bigl(\bigl(1+| \alpha|N\bigr)N^{\frac{2}{3}}L^{C} \bigr), $$

then

$$\begin{aligned}& G(\alpha)=g(\alpha)-I(\alpha)+J(\alpha), \end{aligned}$$

(3.2)

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

(3.3)

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

(3.4)

Proof

Equations (3.2), (3.3), (3.4) can be seen from Lemma 5, (29) and (33) given by Vaughan [3]. □

Lemma 3.3

We have

$$\begin{aligned}& \int_{-\frac{1}{2}}^{\frac{1}{2}}\bigl|f_{i}( \alpha)\bigr|^{2}\,d\alpha \ll X^{-\frac{1}{3}},\quad i=1,2, \\& \int_{-\frac{1}{2}}^{\frac{1}{2}}\bigl|f_{j}( \alpha)\bigr|^{2}\,d\alpha \ll X^{-\frac{1}{2}}, \quad j=3,4, \\& \int_{-\frac{1}{2}}^{\frac{1}{2}}\bigl|f_{k}( \alpha)\bigr|^{2}\,d\alpha \ll X^{-\frac{3}{5}}, \quad k=5,\ldots, 9. \end{aligned}$$

Proof

These results are from Lemma 5 of [3]. □

Lemma 3.4

We have

$$\int_{{\frak{C}}}\Biggl|\prod_{i=1}^{9}F_{i}( \lambda_{i}\alpha) G(-\alpha)-\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha) g(-\alpha)\Biggr|K_{\frac{1}{2}}(\alpha)\,d\alpha\ll X^{\frac{13}{6}}L^{-1}. $$

Proof

It is obvious that

$$\begin{aligned}& F_{i}(\lambda_{i}\alpha)\ll X^{\frac{1}{3}},\qquad f_{i}(\lambda_{i}\alpha)\ll X^{\frac{1}{3}}, \quad i=1,2, \\& F_{j}(\lambda_{j}\alpha)\ll X^{\frac{1}{4}},\qquad f_{j}(\lambda_{j}\alpha)\ll X^{\frac{1}{4}}, \quad j=3,4, \\& F_{k}(\lambda_{k}\alpha)\ll X^{\frac{1}{5}},\qquad f_{k}(\lambda_{k}\alpha)\ll X^{\frac{1}{5}},\quad k=5, \ldots,9, \\& G(-\alpha)\ll N,\qquad g(-\alpha)\ll N, \\& \begin{aligned}[b] &\prod_{i=1}^{9}F_{i}( \lambda_{i}\alpha)G(-\alpha)-\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha)g(-\alpha) \\ &\quad= \bigl(F_{1}(\lambda_{1}\alpha)-f_{1}( \lambda_{1}\alpha) \bigr)\prod_{i=2}^{9}F_{i}( \lambda_{i}\alpha)G(-\alpha)\\ &\qquad{} + \bigl(F_{2}( \lambda_{2}\alpha)-f_{2}(\lambda_{2}\alpha) \bigr) \mathop{\prod_{i=1}}_{{i\neq2}}^{9}F_{i}( \lambda_{i}\alpha)G(-\alpha)+\cdots \\ & \qquad{} + \bigl(F_{9}(\lambda_{9}\alpha)-f_{9}( \lambda_{9}\alpha) \bigr)\prod_{i=1}^{8}f_{i}( \lambda_{i}\alpha)G(-\alpha) +\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha) \bigl(G(-\alpha)-g(-\alpha) \bigr). \end{aligned} \end{aligned}$$

Then by (3.1), Lemmas 3.2 and 3.3, we have

$$\begin{aligned}& \int_{{\frak{C}}}\Biggl| \bigl(F_{1}(\lambda_{1} \alpha)-f_{1}(\lambda_{1}\alpha) \bigr)\prod _{i=2}^{9} F_{i}(\lambda_{i} \alpha)G(-\alpha)\Biggr|K_{\frac{1}{2}}(\alpha)\,d\alpha \ll N^{-1+\delta}X^{\delta}X^{\frac{11}{6}}N \ll X^{\frac{11}{6}+2\delta}, \\& \int_{{\frak{C}}}\Biggl|\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha) \bigl(G(-\alpha )-g(-\alpha) \bigr)\Biggr|K_{\frac{1}{2}}( \alpha)\,d\alpha \\& \quad\ll X^{\frac{11}{6}} \biggl(\int_{{\frak{C}}}\bigl|f_{1}( \lambda_{1}\alpha)\bigr|^{2}K_{\frac {1}{2}}(\alpha)\,d\alpha \biggr)^{\frac{1}{2}} \biggl(\int_{{\frak{C}}}\bigl|J(-\alpha)-I(- \alpha)\bigr|^{2}K_{\frac{1}{2}}(\alpha )\,d\alpha \biggr)^{\frac{1}{2}} \\& \quad\ll X^{\frac{11}{6}} \biggl(\int_{-\frac{1}{2}}^{\frac{1}{2}}\bigl|f_{1}( \lambda_{1}\alpha )\bigr|^{2}\,d\alpha \biggr)^{\frac{1}{2}} \biggl( \int_{{\frak{C}}}\bigl|J(\alpha)\bigr|^{2}\,d\alpha+\int _{-\frac{1}{2}}^{\frac {1}{2}}\bigl|I(\alpha)\bigr|^{2}\,d\alpha \biggr)^{\frac{1}{2}} \\& \quad\ll X^{\frac{11}{6}}X^{-\frac{1}{6}} \bigl(N\exp \bigl(-L^{\frac{1}{5}} \bigr) \bigr)^{\frac {1}{2}} \\& \quad\ll X^{\frac{13}{6}}L^{-1}. \end{aligned}$$

The other cases are similar, and the proof of Lemma 3.4 is completed. □

Lemma 3.5

We have

$$\int_{|\alpha|>N^{-1+\delta}}\Biggl|\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha) g(-\alpha)\Biggr|K_{\frac{1}{2}}(\alpha)\,d\alpha\ll X^{\frac{13}{6}-\frac{13}{6}\delta}. $$

Proof

It follows from Vaughan [2] that for $\alpha\neq0$,

$$\begin{aligned}& f_{i}(\lambda_{i}\alpha)\ll|\alpha|^{-\frac{1}{3}},\quad i=1,2,\qquad f_{j}(\lambda_{j}\alpha)\ll| \alpha|^{-\frac{1}{4}},\quad j=3,4, \\& f_{k}(\lambda_{k}\alpha)\ll|\alpha|^{-\frac{1}{5}},\quad k=5,\ldots,9,\qquad g(-\alpha)\ll|\alpha|^{-1}. \end{aligned}$$

Thus

$$\int_{|\alpha|>N^{-1+\delta}}\Biggl|\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha )g(-\alpha)\Biggr|K_{\frac{1}{2}}(\alpha)\,d\alpha \ll \int_{|\alpha|>N^{-1+\delta}}|\alpha|^{-\frac{19}{6}}\,d\alpha \ll X^{\frac{13}{6}-\frac{13}{6}\delta}. $$

□

Lemma 3.6

We have

$$\int_{-\infty}^{+\infty}\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha) g(-\alpha)e \biggl(-\frac{1}{2}\alpha \biggr)K_{\frac{1}{2}}(\alpha)\,d\alpha\gg X^{\frac{13}{6}}. $$

Proof

From (2.3) one has

$$\begin{aligned} & \int_{-\infty}^{+\infty}\prod _{i=1}^{9}f_{i}(\lambda_{i} \alpha) g(-\alpha)e \biggl(-\frac{1}{2}\alpha \biggr)K_{\frac{1}{2}}( \alpha)\,d\alpha \\ &\quad= \int_{1}^{X^{\frac{1}{3}}}\int_{1}^{X^{\frac{1}{3}}} \int_{1}^{X^{\frac {1}{4}}}\int_{1}^{X^{\frac{1}{4}}} \int_{1}^{X^{\frac{1}{5}}}\cdots\int_{1}^{X^{\frac{1}{5}}} \int_{1}^{N}\int_{-\infty}^{+\infty}e \biggl(\alpha \biggl(\lambda_{1}x_{1}^{3}+ \lambda_{2}x_{2}^{3}+\lambda_{3}x_{3}^{4}+ \lambda_{4}x_{4}^{4} \\ & \qquad{} +\lambda_{5}x_{5}^{5}+\cdots+ \lambda_{9}x_{9}^{5}-x-\frac{1}{2} \biggr) \biggr) K_{\frac{1}{2}}(\alpha)\,d\alpha \,dx \,dx_{9}\cdots \,dx_{5}\,dx_{4}\,dx_{3}\,dx_{2} \,dx_{1} \\ &\quad= \frac{1}{450{,}000}\int_{1}^{X}\cdots\int _{1}^{X} \int_{1}^{N} \int_{-\infty}^{+\infty} x_{1}^{-\frac{2}{3}}x_{2}^{-\frac {2}{3}}x_{3}^{-\frac{3}{4}} x_{4}^{-\frac{3}{4}}x_{5}^{-\frac{4}{5}}\cdots x_{9}^{-\frac{4}{5}}e \Biggl(\alpha \Biggl(\sum _{i=1}^{9}\lambda_{i} x_{i}-x- \frac{1}{2} \Biggr) \Biggr) \\ & \qquad{}\cdot K_{\frac{1}{2}}(\alpha)\,d\alpha \,dx \,dx_{9}\cdots \,dx_{1} \\ &\quad= \frac{1}{450{,}000}\int_{1}^{X}\cdots\int _{1}^{X}\int_{1}^{N}x_{1}^{-\frac {2}{3}}x_{2}^{-\frac{2}{3}}x_{3}^{-\frac{3}{4}} x_{4}^{-\frac{3}{4}}x_{5}^{-\frac{4}{5}}\cdots x_{9}^{-\frac{4}{5}} \\ & \qquad{} \cdot\max \Biggl(0,\frac{1}{2}-\Biggl|\sum _{i=1}^{9}\lambda_{i} x_{i}-x- \frac {1}{2}\Biggr| \Biggr)\,dx \,dx_{9}\cdots \,dx_{1}. \end{aligned}$$

Let $|\sum_{i=1}^{9}\lambda_{i} x_{i}-x-\frac{1}{2}|\leq\frac{1}{4}$, then $\sum_{i=1}^{9}\lambda_{i} x_{i}-\frac{3}{4}\leq x\leq \sum_{i=1}^{9}\lambda_{i} x_{i}-\frac{1}{4}$. Based on

$$\sum_{i=1}^{9}\lambda_{i} x_{i}-\frac{3}{4}>1,\qquad \sum_{i=1}^{9} \lambda_{i} x_{i}-\frac{1}{4}< N, $$

one may take

$$\lambda_{j}X \Biggl(8\sum_{i=1}^{9} \lambda_{i} \Biggr)^{-1} \leq x_{j} \leq \lambda_{j}X \Biggl(4\sum_{i=1}^{9} \lambda_{i} \Biggr)^{-1},\quad j=1,\ldots,9, $$

hence

$$\int_{-\infty}^{+\infty}\prod_{i=1}^{9}f_{i}( \lambda_{i}\alpha) g(-\alpha)e \biggl(-\frac{1}{2}\alpha \biggr)K_{\frac{1}{2}}(\alpha)\,d\alpha \geq\frac{1}{3{,}600{,}000}\prod _{j=1}^{9}\lambda_{j} \Biggl(8\sum _{i=1}^{9}\lambda_{i} \Biggr)^{-9}X^{\frac{13}{6}}. $$

This completes the proof of Lemma 3.6. □

4 The intermediate region

Lemma 4.1

We have

$$\begin{aligned}& \int_{-\infty}^{+\infty}\bigl|F_{i}( \lambda_{i}\alpha)\bigr|^{8}K_{\frac{1}{2}}(\alpha )\,d\alpha \ll X^{\frac{5}{3}+\frac{1}{3}\varepsilon}, \quad i=1,2, \end{aligned}$$

(4.1)

$$\begin{aligned}& \int_{-\infty}^{+\infty}\bigl|F_{j}( \lambda_{j}\alpha)\bigr|^{16}K_{\frac {1}{2}}(\alpha)\,d\alpha \ll X^{3+\frac{1}{4}\varepsilon}, \quad j=3,4, \end{aligned}$$

(4.2)

$$\begin{aligned}& \int_{-\infty}^{+\infty}\bigl|F_{k}( \lambda_{k}\alpha)\bigr|^{32}K_{\frac {1}{2}}(\alpha)\,d\alpha \ll X^{\frac{27}{5}+\frac{1}{5}\varepsilon}, \quad k=5,\ldots,9, \end{aligned}$$

(4.3)

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

(4.4)

Proof

By (2.2) and Hua’s inequality, for $i=1,2$, we have

$$\begin{aligned} & \int_{-\infty}^{+\infty}\bigl|F_{i}( \lambda_{i}\alpha)\bigr|^{8}K_{\frac{1}{2}}(\alpha )\,d\alpha \\ &\quad\ll \sum_{m=-\infty}^{+\infty}\int _{m}^{m+1}\bigl|F_{i}(\lambda_{i} \alpha )\bigr|^{8}K_{\frac{1}{2}}(\alpha)\,d\alpha \\ &\quad\ll \sum_{m=0}^{1}\int _{m}^{m+1}\bigl|F_{i}(\lambda_{i} \alpha)\bigr|^{8}\,d\alpha +\sum_{m=2}^{+\infty}m^{-2} \int_{m}^{m+1}\bigl|F_{i}( \lambda_{i}\alpha )\bigr|^{8}\,d\alpha \\ &\quad\ll X^{\frac{5}{3}+\frac{1}{3}\varepsilon}+X^{\frac{5}{3}+\frac {1}{3}\varepsilon}\sum_{m=2}^{+\infty}m^{-2} \\ &\quad\ll X^{\frac{5}{3}+\frac{1}{3}\varepsilon}. \end{aligned}$$

The proofs of (4.2)-(4.4) are similar to (4.1). □

Lemma 4.2

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

$$\sum_{x=1}^{M}e \bigl(\phi(x) \bigr)\ll M^{1+\varepsilon } \bigl(q^{-1}+M^{-1}+qM^{-k} \bigr)^{2^{1-k}}. $$

Proof

This is Lemma 2.4 (Weyl’s inequality) of Vaughan [2]. □

Lemma 4.3

For every real number $\alpha\in\frak{D}$, let $W(\alpha)=\min(|F_{1}(\lambda_{1}\alpha)|,|F_{2}(\lambda_{2}\alpha)|)$, then

$$W(\alpha)\ll X^{\frac{1}{3}-\frac{1}{4}\delta+\frac{1}{3}\varepsilon}. $$

Proof

For $\alpha\in\frak{D}$ and $i=1,2$, we choose $a_{i}$, $q_{i}$ such that

$$ |\lambda_{i}\alpha-a_{i}/q_{i}|\leq q_{i}^{-1}Q^{-1} $$

(4.5)

with $(a_{i},q_{i})=1$ and $1\leq q_{i}\leq Q$.

Firstly, we note that $a_{1}a_{2}\neq0$. Secondly, if $q_{1},q_{2}\leq P$, then

$$\biggl|a_{2}q_{1}\frac{\lambda_{1}}{\lambda_{2}}-a_{1}q_{2}\biggr| \leq \biggl|\frac{a_{2}/q_{2}}{\lambda_{2}\alpha}q_{1}q_{2} \biggl( \lambda_{1}\alpha-\frac{a_{1}}{q_{1}} \biggr)\biggr|+ \biggl|\frac{a_{1}/q_{1}}{\lambda_{2}\alpha}q_{1}q_{2} \biggl(\lambda_{2}\alpha-\frac{a_{2}}{q_{2}} \biggr)\biggr| \ll PQ^{-1}< \frac{1}{2q}. $$

We recall that q was chosen as the denominator of a convergent to the continued fraction for $\lambda_{1}/\lambda_{2}$. Thus, by Legendre’s law of best approximation, we have $|q'\frac{\lambda_{1}}{\lambda_{2}}-a'|>\frac{1}{2q}$ for all integers $a'$, $q'$ with $1\leq q'< q$, thus $|a_{2}q_{1}|\geq q=[N^{1-8\delta}]$. However, from (4.5) we have $|a_{2}q_{1}|\ll q_{1}q_{2}P \ll N^{18\delta}$, this is a contradiction. We have thus established that for at least one i, $P< q_{i}\ll Q$. Hence Lemma 4.2 gives the desired inequality for $W(\alpha)$. □

Lemma 4.4

We have

$$\int_{\frak{D}}\prod_{i=1}^{9}F_{i}( \lambda_{i}\alpha) G(-\alpha)e \biggl(-\frac{1}{2}\alpha \biggr)K_{\frac{1}{2}}(\alpha)\,d\alpha \ll X^{\frac{13}{6}-\frac{1}{16}\delta+\varepsilon}. $$

Proof

By Lemmas 4.1, 4.3 and Hölder’s inequality, we have

$$\begin{aligned} & \int_{{\frak{D}}}\prod_{i=1}^{9}\bigl|F_{i}( \lambda_{i}\alpha)G(-\alpha )\bigr|K_{\frac{1}{2}}(\alpha)\,d\alpha \\ &\quad\ll \max_{\alpha\in{\frak{D}}}\bigl|W(\alpha)\bigr|^{\frac{1}{4}} \biggl( \biggl( \int_{-\infty}^{+\infty}\bigl|F_{1}( \lambda_{1}\alpha)\bigr|^{8} \biggr)^{\frac{1}{8}} \biggl(\int _{-\infty}^{+\infty}\bigl|F_{2}(\lambda_{2} \alpha)\bigr|^{8} \biggr)^{\frac{3}{32}} \\ &\qquad{} + \biggl(\int_{-\infty}^{+\infty}\bigl|F_{1}( \lambda_{1}\alpha)\bigr|^{8} \biggr)^{\frac{3}{32}} \biggl(\int _{-\infty}^{+\infty}\bigl|F_{2}(\lambda_{2} \alpha)\bigr|^{8} \biggr)^{\frac{1}{8}} \biggr) \\ & \qquad{} \cdot \biggl(\prod_{j=3}^{4} \int_{-\infty}^{+\infty}\bigl|F_{j}( \lambda_{j} \alpha)\bigr|^{16} K_{\frac{1}{2}}(\alpha)\,d\alpha \biggr)^{\frac{1}{16}} \biggl(\prod_{k=5}^{9} \int_{-\infty}^{+\infty}\bigl|F_{k}(\lambda_{k} \alpha )\bigr|^{32}K_{\frac{1}{2}}(\alpha)\,d\alpha \biggr) ^{\frac{1}{32}} \\ &\qquad{} \cdot \biggl(\int_{-\infty}^{+\infty}\bigl|G(- \alpha)\bigr|^{2}K_{\frac{1}{2}}(\alpha )\,d\alpha \biggr)^{\frac{1}{2}} \\ &\quad\ll \bigl(X^{\frac{1}{3}-\frac{1}{4}\delta+\frac{1}{3}\varepsilon} \bigr)^{\frac{1}{4}} \bigl(X^{\frac{5}{3}+\frac{1}{3}\varepsilon} \bigr)^{\frac{7}{32}} \bigl(X^{3+\frac{1}{4}\varepsilon} \bigr)^{\frac{1}{8}} \bigl(X^{\frac{27}{5}+\frac{1}{5}\varepsilon} \bigr)^{\frac{5}{32}}(N L)^{\frac {1}{2}} \\ &\quad\ll X^{\frac{13}{6}-\frac{1}{16}\delta+\varepsilon}. \end{aligned}$$

□

5 The trivial region

Lemma 5.1

(Lemma 2 of [4])

Let $V(\alpha)=\sum e(\alpha f(x_{1},\ldots,x_{m}))$, 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

$$\int_{|\alpha|>A}\bigl|V(\alpha)\bigr|^{2}K_{\nu}(\alpha) \,d\alpha \leq\frac{16}{A}\int_{-\infty}^{\infty}\bigl|V( \alpha)\bigr|^{2} K_{\nu}(\alpha)\,d\alpha. $$

Lemma 5.2

We have

$$\int_{\frak{c}}\prod_{i=1}^{9}F_{i}( \lambda_{i}\alpha) G(-\alpha)e \biggl(-\frac{1}{2}\alpha \biggr)K_{\frac{1}{2}}(\alpha)\,d\alpha \ll X^{\frac{13}{6}-6\delta+\varepsilon}. $$

Proof

By Lemmas 5.1, 4.1 and Schwarz’s inequality, we have

$$\begin{aligned} & \int_{\frak{c}}\prod_{i=1}^{9}F_{i}( \lambda_{i}\alpha) G(-\alpha)e \biggl(-\frac{1}{2}\alpha \biggr)K_{\frac{1}{2}}(\alpha)\,d\alpha \\ &\quad\ll \int_{\frak{c}}\Biggl|\prod_{i=1}^{9}F_{i}( \lambda_{i}\alpha)G(-\alpha )\Biggr|K_{\frac{1}{2}}(\alpha)\,d\alpha \\ &\quad\ll \frac{1}{P}\int_{-\infty}^{+\infty}\Biggl|\prod _{i=1}^{9}F_{i}(\lambda _{i}\alpha) G(-\alpha)\Biggr|K_{\frac{1}{2}}(\alpha)\,d\alpha \\ &\quad\ll N^{-6\delta}\bigl|F_{1}(\lambda_{1} \alpha)\bigr|^{\frac{1}{4}} \biggl(\biggl(\int_{-\infty}^{+\infty}\bigl|F_{1}( \lambda_{1}\alpha)\bigr|^{8} \biggr)^{\frac{3}{32}} \biggl(\int _{-\infty}^{+\infty}\bigl|F_{2}(\lambda_{2} \alpha)\bigr|^{8} \biggr)^{\frac{1}{8}}\biggr) \\ & \qquad{} \cdot \biggl(\prod_{j=3}^{4} \int_{-\infty}^{+\infty}\bigl|F_{j}( \lambda_{j} \alpha)\bigr|^{16} K_{\frac{1}{2}}(\alpha)\,d\alpha \biggr)^{\frac{1}{16}} \biggl(\prod_{k=5}^{9} \int_{-\infty}^{+\infty}\bigl|F_{k}(\lambda_{k} \alpha )\bigr|^{32}K_{\frac{1}{2}}(\alpha)\,d\alpha \biggr) ^{\frac{1}{32}} \\ & \qquad{} \cdot \biggl(\int_{-\infty}^{+\infty}\bigl|G(- \alpha)\bigr|^{2}K_{\frac{1}{2}}(\alpha )\,d\alpha \biggr)^{\frac{1}{2}} \\ &\quad\ll N^{-6\delta} \bigl(X^{\frac{1}{3}} \bigr)^{\frac{1}{4}} \bigl(X^{\frac{5}{3}+\frac{1}{3}\varepsilon} \bigr)^{\frac{7}{32}} \bigl(X^{3+\frac{1}{4}\varepsilon} \bigr)^{\frac{1}{8}} \bigl(X^{\frac{27}{5}+\frac{1}{5}\varepsilon} \bigr)^{\frac{5}{32}}(N L)^{\frac {1}{2}} \\ &\quad\ll X^{\frac{13}{6}-6\delta+\varepsilon}. \end{aligned}$$

□

6 The proof of Theorem 1.1

From Lemmas 3.4, 3.5 and 3.6 we conclude that $J({\frak{C}})\gg X^{\frac {13}{6}}$. From Lemma 4.4 it follows that $J({\frak{D}})=o(X^{\frac{13}{6}})$. From Lemma 5.2 we have $J({\frak{c}})=o(X^{\frac{13}{6}})$. Thus

$$J\gg X^{\frac{13}{6}},\qquad {\mathcal{N}}(X)\gg X^{\frac{13}{6}}L^{-1}, $$

namely, under conditions of Theorem 1.1,

$$ \biggl|\lambda_{1}x_{1}^{3}+\lambda_{2}x_{2}^{3}+ \lambda_{3}x_{3}^{4}+\lambda_{4}x_{4}^{4}+ \lambda _{5}x_{5}^{5}+\cdots+\lambda_{9}x_{9}^{5} -p-\frac{1}{2}\biggr|< \frac{1}{2} $$

(6.1)

has infinitely many solutions in positive integers $x_{1},x_{2},\ldots,x_{9}$ and prime p. It is evident from (6.1) that

$$p< \lambda_{1}x_{1}^{3}+\lambda_{2}x_{2}^{3}+ \lambda_{3}x_{3}^{4}+\lambda_{4}x_{4}^{4}+ \lambda _{5}x_{5}^{5}+\cdots+\lambda_{9}x_{9}^{5}< p+1, $$

and hence

$$\bigl[\lambda_{1}x_{1}^{3}+\lambda_{2}x_{2}^{3}+ \lambda_{3}x_{3}^{4}+\lambda_{4}x_{4}^{4}+ \lambda _{5}x_{5}^{5}+\cdots+\lambda_{9}x_{9}^{5} \bigr]=p. $$

The proof of Theorem 1.1 is complete.

Declarations

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)

College of Computer and Information Engineering, Henan University of Economics and Law, Zhengzhou, 450011, P.R. China

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