\documentclass[reqno]{amsart} %------------Start of the authours' settings-------------------- \usepackage{amsmath,amssymb,amsthm,amsfonts,mathrsfs} \usepackage{fullpage} \usepackage{xcolor} \usepackage[colorlinks=true, linkcolor=blue, anchorcolor=blue, citecolor=red ]{hyperref} \allowdisplaybreaks % allow equations break between two pages % THEOREMS Environment ------------------ \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem*{conjecture*}{Conjecture} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem*{remark*}{remark} %Shortcuts of commands -------------------- \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \renewcommand{\P}{\mathbb{P}} \newcommand{\E}{\mathbb{E}} \newcommand{\F}{\mathbb{F}} \newcommand{\cM}{\mathcal{M}} \newcommand{\cL}{\mathcal{L}} \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\cS}{\mathcal{S}} \newcommand{\ve}{\varepsilon} \newcommand{\fD}{\mathfrak{D}} \newcommand{\fp}{\mathfrak{p}} \newcommand{\on}{\operatorname} \renewcommand{\Re}{\on{Re}} \renewcommand{\Im}{\on{Im}} \renewcommand{\mod}[1]{\on{mod}#1} \newcommand{\of}[1]{\left(#1\right)} \newcommand{\set}[1]{\left\{#1\right\}} \newcommand{\norm}[1]{\left\Vert#1\right\Vert} \newcommand{\wh}[1]{\widehat{#1}} \newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}} \newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}} \newcommand{\bmat}[1]{\begin{bmatrix}#1\end{bmatrix}} \newcommand{\dmat}[1]{\left\vert\begin{matrix}#1\end{matrix}\right\vert} \newcommand{\smat}[1]{\begin{smallmatrix}#1\end{smallmatrix}} \newcommand{\spmat}[1]{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)} \newcommand{\sbmat}[1]{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]} \usepackage{colonequals} %------------End of the authours' settings---------------------- %------------authour's informaiton---------------- \author{Runbo Li} \address{Department of Mathematics, Imperial College London, London SW7 2AZ, UK} \email{runbo.li26@imperial.ac.uk, runbo.li.carey@gmail.com} %\date{\today} \makeatletter \@namedef{subjclassname@2020}{\textup{}2020 Mathematics Subject Classification} \makeatother %------------Title and keywords------ \title[]{Waring--Goldbach Problem: One square, four fifth powers and fifteen sixth powers} \subjclass[2020]{11P32, 11P55} \keywords{Waring--Goldbach problem, circle method} \begin{document} \begin{abstract} We prove that for every sufficiently large even integer $n$, the equation $$ n = p_1^2 + p_2^5 + p_3^5 + p_4^5 + p_5^5 + p_6^6 + \cdots + p_{20}^6 $$ has a solution with all $p_j$ prime. \end{abstract} \maketitle \tableofcontents \section{Introduction} Let $n$ be a sufficiently large positive integer that satisfies some necessary congruence conditions. Using the Hardy--Littlewood circle method, many mathematicians studied the number of representations of $n$ as the sum of mixed powers of prime numbers. Examples of recent works on this topic include \cite{GMLA2234566}, \cite{GMLA2233rst}, \cite{GMLAOneSquare}, \cite{GMLA2233567}, \cite{Brudern2234567}, \cite{LLZ2244457} and \cite{WangTian2234568}. In this paper, we prove the following result. \begin{theorem}\label{t1} For every sufficiently large even integer $n$, the equation \begin{equation} n = p_1^2 + p_2^5 + p_3^5 + p_4^5 + p_5^5 + p_6^6 + \cdots + p_{20}^6 \end{equation} has a solution with all $p_j$ prime. \end{theorem} \section{Preliminaries} Let $n$ be a sufficiently large positive integer, $P_k = n^{\frac{1}{k}}$ and $L = \log n$. Let $$ \lambda_1 = 1, \quad \lambda_2 = \frac{27}{32}, \quad \lambda_3 = \frac{729}{1024}, \quad \lambda_4 = \frac{19683}{32768}, \quad \lambda_5 = \frac{531441}{1048576}, \quad \lambda_6 = \frac{14348907}{33554432}, \quad \lambda_7 = \frac{387420489}{1073741824}, $$ $$ \lambda_8 = \frac{10460353203}{34359738368}, \qquad \lambda_9 = \frac{9365374360779057135}{36385377104158523392}, \qquad \lambda_{10} = \frac{248229150174961425}{1137043034504953856}, $$ $$ \lambda_{11} = \frac{13208549498258175}{71065189656559616}, \qquad \lambda_{12} = \frac{1418162385496725}{8883148707069952}, \qquad \lambda_{13} = \frac{1234060169123925}{8883148707069952}, $$ $$ \lambda_{14} = \lambda_{15} = \frac{560936440510875}{4441574353534976}, $$ and write $$ \Lambda = \sum_{j = 1}^{15} \lambda_j = \frac{169503075438540046537}{28398343105684701184}, \qquad \Theta = \frac{3}{10} + \frac{\Lambda}{6} = \frac{1103100465143862543341}{851950293170541035520} \approx 1.29479. $$ We define $$ f_k(\alpha) = \sum_{\frac{1}{2} P_k < p \leqslant P_k} e(\alpha p^k) \log p $$ and, for $1 \leqslant j \leqslant 15$, $$ f_{k,j}(\alpha) = \sum_{\frac{1}{2} P_k^{\lambda_j} < p \leqslant P_k^{\lambda_j}} e(\alpha p^k) \log p. $$ We also define $$ F(\alpha) = f_2(\alpha) f_5(\alpha)^4 \prod_{j = 1}^{15} f_{6,j}(\alpha). $$ Then by the orthogonality of additive characters, the integral \begin{equation} \int_{0}^{1} F(\alpha) e(-\alpha n) d \alpha \end{equation} counts the number of solutions of (1) with prime variables lying in certain ranges weighted by a $\log$ factor. Our goal is to show the lower bound \begin{equation} \int_{0}^{1} F(\alpha) e(-\alpha n) d \alpha \gg n^{\Theta} \end{equation} holds for all even $n$. Fix $B \geqslant 1$. Put $$ \mathfrak{M}(q,a) = \left\{\alpha \in [0,\ 1] : \left|\alpha - \frac{a}{q}\right| \leqslant \frac{L^B}{n} \right\}, \qquad \mathfrak{M} = \bigcup_{1 \leqslant q \leqslant L^B} \bigcup_{\substack{0 \leqslant a \leqslant q \\ (a, q) = 1 }} \mathfrak{M}(q,a), \qquad \mathfrak{m} = [0,\ 1] \backslash \mathfrak{M}. $$ Then (3) can be obtained if we get \begin{equation} \int_{\mathfrak{M}} F(\alpha) e(-\alpha n) d \alpha \gg n^{\Theta} \end{equation} and \begin{equation} \int_{\mathfrak{m}} F(\alpha) e(-\alpha n) d \alpha \ll n^{\Theta} L^{-1}. \end{equation} \section{The major arcs} Define $$ S_{k}(q,a) = \sum_{\substack{x = 1 \\ (x,q) = 1 }}^{q} e \left(\frac{a x^k}{q} \right), \qquad v_{k,r}(\beta) = \frac{1}{k} \sum_{2^{-k} n^{r} < m \leqslant n^{r}} m^{1/k-1} e(\beta m), \qquad v_k(\beta) = v_{k,1}(\beta), $$ $$ U(q,a) = S_{2}(q,a) S_{5}(q,a)^4 S_{6}(q,a)^{15}, \qquad w(\beta) = v_2(\beta) v_5(\beta)^4 \prod_{j = 1}^{15} v_{6, \lambda_j}(\beta). $$ \begin{lemma}\label{l31} Let $q, a$ be integers such as $1 \leqslant q \leqslant L^B$, $0 \leqslant a \leqslant q$ and $(a, q) = 1$. Then for $\alpha \in \mathfrak{M}(q,a)$, we have $$ \sum_{\frac{1}{2} P_k^{r} < p \leqslant P_k^{r}} e(\alpha p^k) \log p = \frac{1}{\varphi(q)} S_{k}(q,a) v_{k,r}(\alpha - a/q) + O\left(P_k^{r} \exp(-c \sqrt{L}) \right) $$ for some constant $c > 0$. \end{lemma} \begin{proof} This lemma follows from [\cite{Hua}, Lemma 6] and partial summation. \end{proof} We shall use this lemma to give an approximation of $F(\alpha)$ using functions $S_{k}(q,a)$ and $v_{k,r}$ defined above. By Lemma~\ref{l31}, we have \begin{equation} F(\alpha) = \frac{1}{\varphi(q)^{20}} U(q,a) w(\alpha - a/q) + O\left(n^{1 + \Theta} L^{-4B} \right). \end{equation} An integration over $\mathfrak{M}$ gives \begin{equation} \int_{\mathfrak{M}} F(\alpha) e(-\alpha n) d \alpha = \sum_{q \leqslant L^B} \frac{A_n(q)}{\varphi(q)^{20}} \int_{-L^B/n}^{L^B/n} w(\beta) e(- \beta n) d \beta + O\left(n^{\Theta} L^{-B} \right), \end{equation} where $$ A_n(q) = \sum_{\substack{a = 1 \\ (a, q) = 1}}^{q} U(q,a) e(-an/q). $$ We first show that \begin{equation} \int_{-L^B/n}^{L^B/n} w(\beta) e(- \beta n) d \beta \gg n^{\Theta}. \end{equation} \begin{lemma}\label{l32} ([\cite{VaughanCircleBOOK}, Lemma 6.2]). Suppose that $|\beta| \leqslant \frac{1}{2}$. Then, $$ v_k(\beta) \ll n^{\frac{1}{k}}(1 + n|\beta|)^{-1}. $$ \end{lemma} By Lemma~\ref{l32} we know that \begin{equation} w(\beta) \ll n^{1 + \Theta}(1 + n|\beta|)^{-2}. \end{equation} By (9), we get \begin{align} \nonumber &\ \int_{-L^B/n}^{L^B/n} w(\beta) e(- \beta n) d \beta \\ \nonumber =&\ \int_{-1/2}^{1/2} w(\beta) e(- \beta n) d \beta + O\left(\int_{L^B/n}^{1/2} |w(\beta)| d \beta \right) \\ \nonumber =&\ \int_{-1/2}^{1/2} w(\beta) e(- \beta n) d \beta + O\left(n^{\Theta} L^{-B} \right) \\ \nonumber =&\ \frac{1}{2 \cdot 5^4 \cdot 6^{15}} \sum_{\substack{m_1 + \cdots + m_{20} = n \\ 2^{-2} n < m_1 \leqslant n \\ 2^{-5} n < m_2, m_3, m_4, m_5 \leqslant n \\ 2^{-6} n^{\lambda_{j-5}} < m_j \leqslant n^{\lambda_{j-5}} (6 \leqslant j \leqslant 20) }} m_1^{-\frac{1}{2}} (m_2 m_3 m_4 m_5)^{-\frac{4}{5}} \prod_{j = 6}^{20} m_j^{-\frac{5}{6}} + O\left(n^{\Theta} L^{-B} \right) \\ \gg&\ \sum_{\substack{m_1 + \cdots + m_{20} = n \\ \frac{1}{4} n < m_1 \leqslant \frac{1}{2} n \\ \frac{1}{32} n < m_2, m_3, m_4 \leqslant \frac{3}{64} n \\ \frac{1}{64} n^{\lambda_{j-5}} < m_j \leqslant \frac{3}{128} n^{\lambda_{j-5}} (6 \leqslant j \leqslant 20) }} m_1^{-\frac{1}{2}} (m_2 m_3 m_4 m_5)^{-\frac{4}{5}} \prod_{j = 6}^{20} m_j^{-\frac{5}{6}} \quad \gg n^{\Theta}. \end{align} Next, we want to deal with the sum \begin{equation} \sum_{q \leqslant L^B} \frac{A_n(q)}{\varphi(q)^{20}}. \end{equation} Define the series $$ \mathfrak{S}(n) = \sum_{q = 1}^{\infty} \frac{A_n(q)}{\varphi(q)^{20}}. $$ \begin{lemma}\label{l33} ([\cite{Hua}, Lemma 5]). For $k \geqslant 2$, we have $$ S_{k}(q,a) \ll q^{\frac{1}{2} + \varepsilon}. $$ \end{lemma} \begin{lemma}\label{l34} ([\cite{VaughanCircleBOOK}, Lemma 4.3]). For $(a, p) = 1$, we have $$ |S_{k}(p,a)| \leqslant ((k, p-1) - 1) \sqrt{p} + 1. $$ \end{lemma} \begin{lemma}\label{l35} ([\cite{Hua}, Lemma 4]). Let $p^{n} \mid k$, $p^{n+1} \nmid k$ and \begin{align*} l = \begin{cases} n + 2, & p = 2, \\ n + 1, & p \neq 2. \\ \end{cases} \end{align*} If $t > l$, then $$ S_{k}(p^t,a) = 0. $$ \end{lemma} By Lemma~\ref{l33}, we know that $U(q,a) \ll q^{10 + \varepsilon}$ for $(a, q) = 1$, and hence $A_n(q) \ll q^{11 + \varepsilon}$ uniformly in $n$. Thus, $\mathfrak{S}(n)$ converges absolutely, and \begin{equation} \sum_{q \leqslant L^B} \frac{A_n(q)}{\varphi(q)^{20}} = \mathfrak{S}(n) + O\left(L^{-B}\right). \end{equation} Since the complete exponential sums are multiplicative, for $(q_1, q_2) = 1$ we have $$ A_n(q_1 q_2) = A_n(q_1) A_n(q_2). $$ Moreover, By Lemma~\ref{l35} and the fact that $A_n(4) = S_{5}(4,a) = 0$, we have $A_n(q) = 0$ unless $q$ is square-free. Hence \begin{equation} \mathfrak{S}(n) = \prod_{p} \mathfrak{S}_p(n) = \prod_{p} \left(1 + \frac{A_n(p)}{(p-1)^{20}} \right). \end{equation} By the orthogonality of additive characters modulo $p$, we have \begin{equation} \mathfrak{S}_p(n) = \frac{p M_n(p)}{(p-1)^{20}}, \end{equation} where $M_n(p)$ denote the number of solutions $(x_1, \ldots, x_{20}) \in \left( \mathbb{F}_p^{\times}\right)^{20}$ of \begin{equation} x_1^2 + x_2^5 + x_3^5 + x_4^5 + x_5^5 + \sum_{j=6}^{20} x_j^6 \equiv n (\bmod p). \end{equation} When $p = 2$, every unit is $1$, and $$ x_1^2 + x_2^5 + x_3^5 + x_4^5 + x_5^5 + \sum_{j=6}^{20} x_j^6 \equiv 0 (\bmod 2). $$ Hence $n \equiv 0 (\bmod 2)$ is necessary. When $p = 3$, every square and sixth power is $1$, and every fifth power is $\pm 1$. It is easy to see that the sum of four fifth powers contains all residue classes modulo $3$, and (15) has a solution. When $p = 5$, we have $x^5 = x$ and $x^6 = \pm 1$. Fix $x_1 = 1$. Since $\mathbb{F}_5^{\times} + \mathbb{F}_5^{\times} = \mathbb{F}_5$, the sum of two fifth powers contains all residue classes modulo $5$, and (15) has a solution. When $p = 7$, we have $x^6 = 1$. Fix $x_1, x_2, x_3 = 1$. Since $\mathbb{F}_7^{\times} + \mathbb{F}_7^{\times} = \mathbb{F}_7$, the sum of two fifth powers contains all residue classes modulo $7$, and (15) has a solution. When $p \geqslant 11$, we define $\mathcal{A}_k = \mathcal{A}_k(p) = \{x^k : x \in \mathbb{F}_p\}$. It is known that $|\mathcal{A}_k(p)| = \frac{p-1}{(k, p-1)}$. Applying the Cauchy--Davenport theorem repeatedly, we have \begin{align} \nonumber |\mathcal{A}_2 + 4 \mathcal{A}_5 + 15 \mathcal{A}_6| \geqslant&\ \min(p, |\mathcal{A}_2| + 4 |\mathcal{A}_5| + 15 |\mathcal{A}_6| - 19) \\ \nonumber \geqslant&\ \min\left(p, \frac{p-1}{2} + \frac{4(p-1)}{5} + \frac{15(p-1)}{6} - 19\right) \\ \nonumber \geqslant&\ \min\left(p, \frac{19p - 114}{5}\right). \end{align} Since $p \geqslant 11$, we have $\frac{19p - 114}{5} > p$, and thus $\mathcal{A}_2 + 4 \mathcal{A}_5 + 15 \mathcal{A}_6 = \mathbb{F}_p$. Now (15) has a solution for every prime $p \geqslant 11$. Combining above cases, we know that $M_n(p) \geqslant 1$ if $n$ is even, and also $\mathfrak{S}(n) > 0$ by (13). Now, by Lemma~\ref{l34} and trivial estimates, we have \begin{align} \nonumber \left|\frac{A_n(p)}{(p-1)^{20}} \right| \leqslant&\ \frac{p(\sqrt{p}+1) (4\sqrt{p}+1)^4 (5\sqrt{p}+1)^{15}}{(p-1)^{20}} \\ \nonumber =&\ \frac{p^{11}}{(p-1)^{20}} \left(1+\frac{1}{\sqrt{p}}\right) \left(4+\frac{1}{\sqrt{p}}\right)^4 \left(5+\frac{1}{\sqrt{p}}\right)^{15} \\ \nonumber \leqslant&\ \frac{p^{11}}{(p/2)^{20}} \left(1+\frac{1}{\sqrt{p}}\right) \left(4+\frac{1}{\sqrt{p}}\right)^4 \left(5+\frac{1}{\sqrt{p}}\right)^{15} \\ \nonumber \leqslant&\ \frac{2^{20} \cdot 2 \cdot 5^4 \cdot 6^{15}}{p^{9}} \end{align} for $p > 10^{10}$. Thus, we get \begin{equation} \prod_{p > 10^{10}} \left(1 + \frac{A_n(p)}{(p-1)^{20}} \right) \geqslant \prod_{p > 10^{10}} \left(1 - \frac{2^{21} \cdot 5^4 \cdot 6^{15}}{p^{9}} \right) > c \end{equation} with some $c > 0$. By (13), (14), (16) and $M_n(p) \geqslant 1$, we have \begin{equation} \mathfrak{S}(n) > c \prod_{p \leqslant 10^{10}} \frac{p M_n(p)}{(p-1)^{20}} > c \prod_{p \leqslant 10^{10}} p^{-19} \gg 1. \end{equation} Now, by (7), (10), (12) and (17), the proof of (4) is completed. \section{The minor arcs} We first recall some mean value estimates proved by Andrade \cite{GMLAOneSquare} and Zhao \cite{Zhao_WaringGoldbach6}. Note that we can remove the logarithmic weights and the restriction to primes in $f_k(\alpha)$ and $f_{k,j}(\alpha)$ at the cost of a power of $L$, and deal with the classical Weyl sums. \begin{lemma}\label{l41} We have $$ \int_{0}^{1} \left|f_2(\alpha)\right|^2 \left|f_5(\alpha)\right|^6 d \alpha \ll n^{\frac{6}{5} + \varepsilon}. $$ \end{lemma} \begin{proof} Taking $k_1 = k_2 = k_3 = 5$ in [\cite{GMLAOneSquare}, Lemma 4], we get Lemma~\ref{l41}. \end{proof} \begin{lemma}\label{l42} We have $$ \int_{0}^{1} \left| \prod_{j=1}^{15} f_{6,j}(\alpha)\right|^2 d \alpha \ll P_6^{\Lambda + \varepsilon}. $$ \end{lemma} \begin{proof} Taking $t = 1$ and $U = P_6$ in [\cite{Zhao_WaringGoldbach6}, Lemma 6.1], we get Lemma~\ref{l42}. \end{proof} Write $\tau = 2^{-100}$. Define $$ \mathcal{E} = \{\alpha \in [0,\ 1] : |f_5(\alpha)| \leqslant P_5^{(\Lambda - 4)/2 - 10 \tau} \}. $$ Using H\"{o}lder's inequality, together with Lemma~\ref{l41}, Lemma~\ref{l42} and the definition of $\mathcal{E}$, we get \begin{align} \nonumber \int_{\mathcal{E}} |F(\alpha)| d \alpha \leqslant&\ \left(\int_{0}^{1} \left|f_2(\alpha)\right|^2 \left|f_5(\alpha)\right|^6 d \alpha \right)^{\frac{1}{2}} \left(\int_{0}^{1} \left| \prod_{j=1}^{15} f_{6,j}(\alpha)\right|^2 d \alpha \right)^{\frac{1}{2}} \sup_{\alpha \in \mathcal{E}}|f_5(\alpha)| \\ \ll&\ n^{\frac{3}{5} + \varepsilon} P_6^{\Lambda/2 + \varepsilon} P_5^{(\Lambda - 4)/2 - 10 \tau} \ll n^{\theta - \tau}, \end{align} where $$ \theta = \frac{441062789418431385223}{340780117268216414208} \approx 1.29427. $$ Note that $\Theta - \theta > 0.0005$. \section{Proof of Theorem 1.1} \begin{lemma}\label{l51} Suppose that $X^{\frac{1}{24}} \leqslant P \leqslant X^{\frac{9}{20}}$. Then either we have $$ \sum_{X < p \leqslant 2 X} e(\alpha p^5) \ll X^{\frac{47}{48} + \varepsilon}, $$ or there exist integers $q$ and $a$ such that $1 \leqslant q \leqslant P$, $(a, q)=1$ and $|q \alpha - a| \leqslant P X^{-5}$. \end{lemma} \begin{proof} Taking $k = 5$ in [\cite{KumchevWooley}, Lemma 2.2], we get Lemma~\ref{l51}. \end{proof} Now suppose that $\alpha \in [0,\ 1] \backslash \mathcal{E}$. We have $|f_5(\alpha)| > P_5^{(\Lambda - 4)/2 - 10 \tau}$ in this case. Taking $X = \frac{1}{2} P_5$ and $P = \frac{1}{32} n^{\frac{1}{119}}$ in Lemma~\ref{l51}, we find that the first case will never occur since $\frac{\Lambda - 4}{2} - \frac{47}{48} > 0.005$. Hence, we must have \begin{equation} [0,\ 1] \backslash \mathcal{E} \subset \mathfrak{R} = \bigcup_{1 \leqslant q \leqslant \frac{1}{32} n^{\frac{1}{119}}} \bigcup_{\substack{0 \leqslant a \leqslant q \\ (a, q)=1}} \mathfrak{R}(q,a), \end{equation} where \begin{equation} \mathfrak{R}(q,a) = \{\alpha \in [0,\ 1] : |q \alpha - a| \leqslant n^{-\frac{118}{119}} \}. \end{equation} Define a function $\Upsilon : \mathfrak{R} \rightarrow [0,\ 1]$ by $$ \Upsilon(\alpha) = (q + n |q \alpha - a|)^{-1}. $$ Taking $k = 2, 5$ in [\cite{KumchevMichigan}, Theorem 2] and by partial summation, we have \begin{equation} f_2(\alpha) \ll P_2 L^C \Upsilon(\alpha)^{\frac{1}{2} - \varepsilon} \end{equation} and \begin{equation} f_5(\alpha) \ll P_5 L^C \Upsilon(\alpha)^{\frac{1}{2} - \varepsilon} \end{equation} for some constant $C>0$. By (21), (22) and trivial upper bounds for $f_{6,j}(\alpha)$ ($1 \leqslant j \leqslant 15$), we have \begin{equation} F(\alpha) = f_2(\alpha) f_5(\alpha)^4 \prod_{j = 1}^{15} f_{6,j}(\alpha) \ll n^{1 + \Theta} L^{20C} \Upsilon(\alpha)^{\frac{49}{20}}. \end{equation} Integrating over $\mathfrak{R} \backslash \mathfrak{M}$ gives \begin{equation} \int_{\mathfrak{R} \backslash \mathfrak{M}} |F(\alpha)| d \alpha \ll n^{1 + \Theta} L^{20C} \int_{\mathfrak{R} \backslash \mathfrak{M}} |\Upsilon(\alpha)|^{\frac{49}{20}} d \alpha \ll n^{\Theta} L^{20C - \frac{9}{20}B + 1}. \end{equation} Choosing $B = 200 C + 100$ in (24) and by the fact that $\mathfrak{m} \subset \mathcal{E} \cup \mathfrak{R} \backslash \mathfrak{M}$, the proof of (5) is completed. From (4) and (5) we get (3), and the proof of Theorem~\ref{t1} is completed. \bibliographystyle{plain} \bibliography{bib} \end{document}