The integer part of a nonlinear form with integer variables

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 $x_{1},x_{2},\ldots,x_{9}$.

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.

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 $\frak{c}=\{\alpha\in{\mathbb{R}}:|\alpha|>P\}$.

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. □

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

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

 □

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

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

 □

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.