On the Waring-Goldbach problem for one cube and nine biquadrates

Abstract

Let N be a sufficiently large integer. In this paper, it is proved that with at most O\left(N^{\frac{17}{18}+\epsilon }\right)exceptions, all positive integers satisfying some necessary congruence conditions up to N can be represented in the form p_1^3+p_2^4+p_3^4+p_4^4+p_5^4+p_6^4+p_7^4+p_8^4+p_9^4+p_{10}^4, where p_1,p_2,\dots ,p_{10} are prime numbers.

Keywords

Waring-Goldbach problem / Hardy-Littlewood method / exceptional set

1 Introduction and main results

In the study of number theory and everyday teaching practice, the problem of additive prime number theory involving the sum of low-order powers of natural numbers provides a convenient way to develop and explore new techniques in the Hardy-Littlewood method (see [1‒5]). Specifically, based on the research progress in the Waring problem for cubes and biquadrates, many scholars have achieved fruitful results in the study of the Waring problem concerning the representation of sufficiently large integers as sums of mixed powers of cubes and biquadrates.

For each natural number r, let B(r) denote the smallest positive integer s such that all sufficiently large integers can be represented as the sum of r cubes and s biquadrates. In the study of this problem, when 1\le r\le 3, the estimates of upper bounds for B(r) were first obtained by Kawada and Wooley [8], who provided

B\left(1\right)\le 9,B\left(2\right)\le 8,B\left(3\right)\le 6.

For the cases r=5,6, the results were currently given by Brüdern [4, 5]:

B\left(5\right)\le 3,B\left(6\right)\le 2.

For r=7, according to Linnik’s classical result [10], which states that any sufficiently large integer can be expressed as the sum of cubes of seven positive integers, it follows that B\left(7\right)=0. The conclusion for r=4 cannot be directly obtained from the existing literature. However, according to Brüdern [4], almost all sufficiently large positive integers can be expressed as the sum of three cubes and one biquadrate; combined with the conclusion of Kawada and Wooley [8], which states that sufficiently large even integers can be represented as the sum of one cube and four biquadrates, it can be deduced that B\left(4\right)\le 5. Later, Brüdern and Wooley [6] improved the result for r=2 and obtained B\left(2\right)\le 7.

Based on the above results, it is natural to consider the Waring-Goldbach problem for corresponding powers, but current methods in number theory cannot solve these problems. This paper will consider the problem of representing sufficiently large positive integers, which satisfy certain congruence conditions, as the additive prime of one prime cube and nine prime biquadrates, and will provide an upper bound estimate for the exceptional set corresponding to this representation method.

Theorem 1.1　 Let E(N) denote the number of positive integers n that satisfy the congruence conditions n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right)), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right), do not exceed N and cannot be represented as

(1.1)n=p_1^3+p_2^4+p_3^4+p_4^4+p_5^4+p_6^4+p_7^4+p_8^4+p_9^4+p_{10}^4.

Then, for any \epsilon >0, we have

E\left(N\right)\ll N^{\frac{17}{18}+\epsilon }.

Note　We will use the Hardy-Littlewood method to prove Theorem 1.1. For the estimation of the minor arc integrals, we will adopt the approach of Kawada and Wooley [9] and incorporate the exponent and estimates of Zhao [14]. For the integrals over the major arcs, we will perform a secondary subdivision and estimate each part separately. Detailed steps will be provided in the subsequent sections.

2 Outline of the proof of Theorem 1.1

Let N be a sufficiently large positive integer. According to the dichotomy method, we only need to consider positive integers n that satisfy the congruence conditions n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right) and n\in (\frac{N}{2},N]. To apply the Hardy-Littlewood method, we define an appropriate Farey subdivision of the unit interval.

Let A>0 be a sufficiently large constant, whose value will be determined at the end of the proof. Define

Q_0={\mathrm{l}\mathrm{o}\mathrm{g}}^{A}N,Q_1=N^{\frac{1}{6}},Q_2=N^{\frac{5}{6}},{\mathfrak{I}}_0=\left[-\frac{1}{Q_2},1-\frac{1}{Q_2}\right].

According to Dirichlet’s rational approximation theorem (see Vaughan [12, Lemma 2.1]), each \alpha \in [-\frac{1}{Q_2},1-\frac{1}{Q_2}] can be represented as

(2.1)\alpha =\frac{a}{q}+\lambda ,\left|\lambda \right|\le \frac{1}{q{Q}_2},

where a and q are integers satisfying 1\le a\le q\le Q_2 and (a,q)=1. Define

\mathfrak{M}\left(q,a\right)=\left[\frac{a}{q}-\frac{Q_1}{qN},\frac{a}{q}+\frac{Q_1}{qN}\right],\mathfrak{M}=\bigcup _{1\le q\le Q_1} \bigcup _{\begin{array}{c}1\le a\le q\\ (a,q)=1\end{array}} \mathfrak{M}(q,a),

{\mathfrak{M}}_0\left(q,a\right)=\left[\frac{a}{q}-\frac{Q_0^{200}}{qN},\frac{a}{q}+\frac{Q_0^{200}}{qN}\right],{\mathfrak{M}}_0=\bigcup _{\begin{array}{c}1\le q\le Q_0^{100}\end{array}} \bigcup _{\begin{array}{c}1\le a\le q\\ (a,q)=1\end{array}} {\mathfrak{M}}_0(q,a),

{\mathfrak{m}}_1={\mathfrak{I}}_0\setminus \mathfrak{M},{\mathfrak{m}}_2=\mathfrak{M}\setminus {\mathfrak{M}}_0.

With the above subdivisions, we obtain a Farey subdivision of the unit interval:

(2.2){\mathfrak{I}}_0={\mathfrak{M}}_0\cup {\mathfrak{m}}_1\cup {\mathfrak{m}}_2.

For k\in \left\{3,4\right\}, let

where X_k=(\frac{N}{32}{)}^{\frac{1}{k}}. Define

According to (2.2), we have

\begin{array}{rl}R\left(n\right)& ={\int }_0^1 f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \\ & ={\int }_{-\frac{1}{Q_2}}^{1-\frac{1}{Q_2}} f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \\ & =\left\{{\int }_{{\mathfrak{M}}_0} +{\int }_{{\mathfrak{m}}_1} +{\int }_{{\mathfrak{m}}_2} \right\}{f}_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha .\end{array}

To prove Theorem 1.1, we need the following two propositions.

Proposition 2.1　 For n\in [\frac{N}{2},N], we have

(2.3){\int }_{{\mathfrak{M}}_0} f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha =\mathfrak{G}\left(n\right)\mathfrak{J}\left(n\right)+O\left(\frac{n^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}n}\right),

where \mathfrak{G}\left(n\right) is the singular series defined by (4.1). For positive integers n satisfying the conditions n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right), there exists a constant c^{\ast }>0 such that

(2.4)

In addition, \mathfrak{J}\left(n\right) is defined by Eq. (4.5) and satisfies

\mathfrak{J}\left(n\right)\asymp \frac{n^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{10}n}.

The proof of Eq. (2.3) in Proposition 2.1 will be given in Section 4. The properties of the singular series described in (2.4) will be proven in Section 5 (Lemma 5.6(ii)).

Proposition 2.2　 Let Z(N) represent the number of positive integers n\in [\frac{N}{2},N] that satisfy the congruence conditions n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right), for which the following estimate holds

{\sum }_{j=1}^2 \left|{\int }_{{\mathfrak{m}}_j} f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \right|\gg \frac{n^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}n}.

Then, we have

\mathcal{Z}\left(N\right)\ll N^{\frac{17}{18}+\epsilon }.

The proof of Proposition 2.2 will be given in Section 6. In the remaining part of this section, we will focus on using Proposition 2.1 and Proposition 2.2 to prove Theorem 1.1.

Proof of Theorem 1.1　According to Proposition 2.2, with at most O\left(N^{\frac{17}{18}+\epsilon }\right) exceptions, all positive integers n\in [\frac{N}{2},N] that satisfy the congruence conditions n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right) also satisfy

{\sum }_{j=1}^2 \left|{\int }_{{\mathfrak{m}}_j} f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \right|\ll \frac{n^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}n}.

Combining this with Proposition 2.1, we know that with at most O\left(N^{\frac{17}{18}+\epsilon }\right) exceptions, all positive integers n\in [\frac{N}{2},N] that satisfy the congruence conditions n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right) can be expressed as p_1^3+p_2^4+p_3^4+p_4^4+p_5^4+p_6^4+p_7^4+p_8^4+p_9^4+p_{10}^4, where P_1,P_2,\dots ,P_{10} are primes. Using the dichotomy method, we obtain

E\left(N\right)\ll \sum _{0\le \ell \ll \mathrm{l}\mathrm{o}\mathrm{g}N} \mathcal{Z}\left(\frac{N}{2^{\ell }}\right)\ll \sum _{0\le \ell \ll \mathrm{l}\mathrm{o}\mathrm{g}N} {\left(\frac{N}{2^{\ell }}\right)}^{\frac{17}{18}+\epsilon }\ll N^{\frac{17}{18}+\epsilon }.

This completes the proof of Theorem 1.1. □

3 Some preliminary lemmas

Lemma 3.1 Let α be a real number such that satisfies |\alpha -\frac{a}{q}|\le q^{-2} with (a,q)=1. Define \beta =\alpha -\frac{a}{q}. Then we have

f_k\left(\alpha \right)\ll d^{{\delta }_k}\left(q\right)(\mathrm{l}\mathrm{o}\mathrm{g}x{)}^c\left(X_k^{\frac{1}{2}}\sqrt{q(1+N|\beta \left|\right)}+X_k^{\frac{4}{5}}+\frac{X_k}{\sqrt{q(1+N|\beta \left|\right)}}\right),

where {\delta }_k=\frac{1}{2}+\frac{\mathrm{l}\mathrm{o}\mathrm{g}k}{\mathrm{l}\mathrm{o}\mathrm{g}2}, and c is a constant.

Proof　Refer to Ren [11, Theorem 1.1]. □

Lemma 3.2　 Let α be a real number, and there exist a\in \mathbb{Z} and q\in \mathbb{N} satisfying

\left(a,q\right)=1,1\le q\le Q,|q\alpha -a|\le Q^{-1}.

If P^{\frac{1}{2}}\le Q\le P^{\frac{5}{2}}, then the following holds

Proof　Refer to Zhao [14, Lemma 8.5]. □

Lemma 3.3 For \alpha \in {\mathfrak{m}}_1, we have

f_3\left(\alpha \right)\ll N^{\frac{11}{36}+\epsilon }.

Proof　According to Dirichlet’s rational approximation (2.1), for \alpha \in {\mathfrak{m}}_1, there must exist Q_1\le q\le Q_2. Using Lemma 3.2, we get

f_3\left(\alpha \right)\ll X_3^{\frac{11}{12}+\epsilon }+X_3^{1+\epsilon }{Q}_1^{-\frac{1}{6}}\ll N^{\frac{11}{36}+\epsilon }.

This completes the proof of Lemma 3.3. □

When 1\le a\le q and (a,q)=1, define

(3.1)\mathcal{J}(q,a)=\left[\frac{a}{q}-\frac{1}{q{Q}_0},\frac{a}{q_1}+\frac{1}{q{Q}_0}\right],\mathcal{J}={\bigcup }_{1\le q\le Q_0} {\bigcup }_{\begin{array}{c}a=-q\\ (a,q)=1\end{array}}^{2q} \mathcal{J}(q,a).

For \alpha \in {\mathfrak{m}}_2, using Lemma 3.1, we have

(3.2)f_3\left(\alpha \right)\ll \frac{N^{\frac{1}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{c}N}{q^{\frac{1}{2}-\epsilon }(1+N|\lambda |{)}^{\frac{1}{2}}}+N^{\frac{4}{15}+\epsilon }=:V_3\left(\alpha \right)+N^{\frac{4}{15}+\epsilon }.

Thus, we can derive the following lemma.

Lemma 3.4　 Let V_3\left(\alpha \right) be defined by (3.2), then

\begin{array}{rl}{\int }_{\mathcal{J}} \left|V_3\right(\alpha ){|}^2\mathrm{d}\alpha & =\sum _{1\le q\le Q_0} \sum _{(a,q)=1}^{2q} {\int }_{\mathcal{J}(q,a)} |V_3\left(\alpha \right){|}^2\mathrm{d}\alpha \\ & \ll N^{-\frac{1}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{2A}N.\end{array}

Proof　We have

\begin{array}{rl}& {\sum }_{1\le q\le Q_0}{\sum }_{(a,q)=1}^{2q}{\int }_{\mathcal{J}(q,a)}\mid V_3\left(\alpha \right){\mid }^2\mathrm{d}\alpha \\ & \ll {\sum }_{1\le q\le Q_0}{q}^{-1+\epsilon }{\sum }_{\begin{array}{c}a=-q\\ (a,q)=1\end{array}}^{2q}{\int }_{\mid \lambda \mid \le \frac{1}{Q_0}}\frac{N^{\frac{2}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{c}N}{1+N\mid \lambda \mid }\mathrm{d}\lambda \\ & \ll \sum _{1\le q\le Q_0} q^{-1+\epsilon }\sum _{\begin{array}{c}a=-q\\ (a,q)=1\end{array}}^{2q} \left({\int }_{\left|\lambda \right|\le \frac{1}{N}} N^{\frac{2}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{c}N\mathrm{d}\lambda +{\int }_{\frac{1}{N}\le \left|\lambda \right|\le \frac{1}{Q_0}} \frac{N^{\frac{2}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{c}N}{N\left|\lambda \right|}\mathrm{d}\lambda \right)\\ & \ll N^{-\frac{1}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{c}N\cdot \sum _{1\le q\le Q_0} q^{-1+\epsilon }\phi \left(q\right)\\ & \ll N^{-\frac{1}{3}}{Q}_0^{1+\epsilon }{\mathrm{l}\mathrm{o}\mathrm{g}}^{c}N\\ & \ll N^{-\frac{1}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{2A}N.\end{array}

This completes the proof of Lemma 3.4. □

4 Proof of Proposition 2.1

In this section, we will focus on proving Proposition 2.1. First, we introduce some notations. For a Dirichlet character χ mod q and k\in \{3,4\}, define

C_k(\chi ,a)=\sum _{h=1}^q \overline{\chi \left(h\right)}e\left(\frac{a{h}^k}{q}\right),C_k(q,a)=C_k({\chi }^0,a),

where χ⁰ is the principal character modulo q. Let {\chi }_3,{\chi }_4^{\left(1\right)},{\chi }_4^{\left(2\right)},\dots ,{\chi }_4^{\left(9\right)} be Dirichlet characters modulo q. Define

B(n,q,{\chi }_3,{\chi }_4^{\left(1\right)},{\chi }_4^{\left(2\right)},\dots ,{\chi }_4^{\left(9\right)})=\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q \left(C_3({\chi }_3,a)\prod _{i=1}^9 C_4({\chi }_4^{\left(i\right)},a)\right)e\left(-\frac{an}{q}\right),

B(n,q)=B(n,q,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0,{\chi }^0),

and

(4.1)A(n,q)=\frac{B(n,q)}{{\phi }^{10}\left(q\right)},\mathfrak{S}\left(n\right)=\sum _{q=1}^{\mathrm{\infty }} A(n,q).

Lemma 4.1　 For (a,q)=1 and any Dirichlet character χ mod q, we have

\left|C_k\right(\chi ,a\left)\right|\le 2q^{\frac{1}{2}}{d}^{{\beta }_k}\left(q\right),

where {\beta }_k=\frac{\mathrm{l}\mathrm{o}\mathrm{g}k}{\mathrm{l}\mathrm{o}\mathrm{g}2}.

Proof　Refer to Vinogradov [13, Chapter VI, Problem 14]. □

Lemma 4.2　 Let C_k(q,a) be defined as above. Then we have

(4.2){\sum }_{q\le x} \frac{\left|B\right(n,q\left)\right|}{{\phi }^{10}\left(q\right)}\ll \mathrm{l}\mathrm{o}\mathrm{g}x.

Proof　According to Lemma 4.1, we have

B(n,q)\ll \sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q \left|C_3(q,a)C_4^9(q,a)\right|\ll q^5\phi \left(q\right)d^{20}\left(q\right).

Therefore, the left side of (4.2) is

\ll \sum _{q\le x} \frac{q^5\phi \left(q\right)d^{20}\left(q\right)}{{\phi }^{10}\left(q\right)}\ll \sum _{q\le x} \frac{(\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}q{)}^9{d}^{20}(q)}{q^4}\ll (\mathrm{l}\mathrm{o}\mathrm{g}\mathrm{l}\mathrm{o}\mathrm{g}x{)}^9\sum _{q\le x} \frac{d^{20}\left(q\right)}{q^4}\ll \mathrm{l}\mathrm{o}\mathrm{g}x.

This completes the proof of Lemma 4.2. □

To handle the integrals over the major arcs, we express f_k\left(\alpha \right) in the following form:

For the inner sum on the right-hand side of the above equation, applying the Siegel-Walfisz theorem, we obtain

Thus, we have

f_k\left(\alpha \right)=\frac{C_k(q,a)}{\phi \left(q\right)}{v}_k\left(\lambda \right)+O\left(X_k{\mathrm{e}}^{-c\sqrt{\mathrm{l}\mathrm{o}\mathrm{g}N}}\right),

and further obtain

f_3\left(\alpha \right)f_4^9\left(\alpha \right)=\frac{C_3(q,a)C_4^9(q,a)}{{\phi }^{10}\left(q\right)}{v}_3\left(\lambda \right)v_4^9\left(\lambda \right)+O\left(N^{\frac{31}{12}}{\mathrm{e}}^{-c\sqrt{\mathrm{l}\mathrm{o}\mathrm{g}N}}\right).

Thus, the integral over the major arcs can be expressed as

(4.3)\begin{array}{rl}& {\int }_{{\mathfrak{M}}_0}{f}_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \\ & =\sum _{1\le q\le Q_0^{100}} \sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q e\left(-\frac{an}{q}\right){\int }_{-\frac{Q_0^{200}}{qN}}^{\frac{Q_0^{200}}{qN}} \left(\frac{C_3(q,a)C_4^9(q,a)}{{\phi }^{10}(q)}{v}_3(\lambda )v_4^9(\lambda )\right)\\ & +O\left(N^{\frac{31}{12}}{\mathrm{e}}^{-c\sqrt{\mathrm{l}\mathrm{o}\mathrm{g}N}}\right)\left)e\right(-n\lambda )\mathrm{d}\lambda \\ & =\sum _{1\le q\le Q_0^{100}} \frac{B(n,q)}{{\phi }^{10}\left(q\right)}{\int }_{-\frac{Q_0^{200}}{qN}}^{\frac{Q_0^{200}}{qN}} v_3\left(\lambda \right)v_4^9\left(\lambda \right)e(-n\lambda )\mathrm{d}\lambda +O\left(N^{\frac{19}{12}}{e}^{-c\sqrt{\mathrm{l}\mathrm{o}\mathrm{g}N}}\right).\end{array}

It’s noted that

v_k\left(\lambda \right)={\int }_{X_k^k}^{(2X_k{)}^k} \frac{x^{\frac{1}{k}-1}e\left(\lambda x\right)}{\mathrm{l}\mathrm{o}\mathrm{g}x}\mathrm{d}x.

Thus, the innermost integral in (4.3) can be expressed as

(4.4){\int }_{-\frac{Q_0^{200}}{qN}}^{\frac{Q_0^{200}}{qN}} \left({\int }_{X_3^3}^{(2X_3{)}^3} \frac{x^{-\frac{2}{3}}e\left(\lambda x\right)}{\mathrm{l}\mathrm{o}\mathrm{g}x}\mathrm{d}x\right){\left({\int }_{X_4^4}^{(2X_4{)}^4} \frac{x^{-\frac{3}{4}}e\left(\lambda x\right)}{\mathrm{l}\mathrm{o}\mathrm{g}x}\mathrm{d}x\right)}^{9}e(-n\lambda )\mathrm{d}\lambda .

Using the basic estimate

\begin{array}{rl}{v}_k\left(\lambda \right)& ={\int }_{X_k^k}^{(2X_k{)}^k} \frac{x^{\frac{1}{k}-1}e\left(\lambda x\right)}{\mathrm{l}\mathrm{o}\mathrm{g}x}\mathrm{d}x\\ & \ll \frac{N^{\frac{1}{k}-1}}{\mathrm{l}\mathrm{o}\mathrm{g}N}min\left(N,\frac{1}{\left|\lambda \right|}\right),\end{array}

we know that if we extend the integration range in (4.4) to [-\frac{1}{2},\frac{1}{2}], the resulting error is

\ll {\int }_{\frac{Q_0^{200}}{qN}}^{\frac{1}{2}} \frac{N^{-\frac{89}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{10}N}\cdot \frac{\mathrm{d}\lambda }{{\lambda }^{10}}\ll \frac{N^{-\frac{89}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{10}N}\cdot \frac{q^9{N}^9}{Q_0^{1800}}\ll \frac{N^{\frac{19}{12}}}{(\mathrm{l}\mathrm{o}\mathrm{g}N{)}^{500A}}\ll \frac{n^{\frac{19}{12}}}{(\mathrm{l}\mathrm{o}\mathrm{g}n{)}^{500A}}.

Thus, we obtain

{\int }_{-\frac{Q_0^{200}}{qN}}^{\frac{Q_0^{200}}{qN}} v_3\left(\lambda \right)v_4^9\left(\lambda \right)e(-n\lambda )\mathrm{d}\lambda =\mathfrak{J}\left(n\right)+O\left(\frac{n^{\frac{19}{12}}}{(\mathrm{l}\mathrm{o}\mathrm{g}n{)}^{500A}}\right),

where

(4.5)\begin{array}{rl}\mathfrak{J}\left(n\right)& ={\int }_{-\frac{1}{2}}^{\frac{1}{2}} \left({\int }_{X_3^3}^{(2X_3{)}^3} \frac{x^{-\frac{2}{3}}e\left(\lambda x\right)}{\mathrm{l}\mathrm{o}\mathrm{g}x}\mathrm{d}x\right){\left({\int }_{X_4^4}^{(2X_4{)}^4} \frac{x^{-\frac{3}{4}}e\left(\lambda x\right)}{\mathrm{l}\mathrm{o}\mathrm{g}x}\mathrm{d}x\right)}^{9}e(-n\lambda )\mathrm{d}\lambda \\ & ={\int }_{X_3^3}^{(2X_3{)}^3} {\int }_{X_4^4}^{(2X_4{)}^4} \cdots {\int }_{X_4^4}^{(2X_4{)}^4} {\int }_{-\frac{1}{2}}^{\frac{1}{2}} \frac{x_1^{-\frac{2}{3}}(x_2{x}_3{x}_4{x}_5{x}_6{x}_7{x}_8{x}_9{x}_{10}{)}^{-\frac{3}{4}}}{(\mathrm{l}\mathrm{o}\mathrm{g}{x}_1)(\mathrm{l}\mathrm{o}\mathrm{g}{x}_2)\cdots (\mathrm{l}\mathrm{o}\mathrm{g}{x}_{10})}\\ & \times e\left(\right(x_1+x_2+x_3+\cdots +x_{10}-n\left)\lambda \right)\mathrm{d}\lambda \mathrm{d}{x}_1\mathrm{d}{x}_2\cdots \mathrm{d}{x}_{10}\\ & \asymp \frac{X_3^{-2}{X}_4^{-27}}{(\mathrm{l}\mathrm{o}\mathrm{g}N{)}^{10}}{\int }_{X_3^3}^{(2X_3{)}^3} {\int }_{X_4^4}^{(2X_4{)}^4} {\int }_{X_4^4}^{(2X_4{)}^4} \cdots {\int }_{X_4^4}^{(2X_4{)}^4} {\int }_{-\frac{1}{2}}^{\frac{1}{2}} \\ & \times e\left(\right(x_1+x_2+\cdots +x_{10}-n\left)\lambda \right)\mathrm{d}\lambda \mathrm{d}{x}_1\mathrm{d}{x}_2\cdots \mathrm{d}{x}_{10}\\ & \asymp \frac{X_3^{-2}{X}_4^{-27}}{(\mathrm{l}\mathrm{o}\mathrm{g}N{)}^{10}}{N}^9\asymp \frac{N^{\frac{19}{12}}}{(\mathrm{l}\mathrm{o}\mathrm{g}N{)}^{10}}\asymp \frac{n^{\frac{19}{12}}}{(\mathrm{l}\mathrm{o}\mathrm{g}n{)}^{10}}.\end{array}

In summary, according to (4.5) and Lemma 4.2, Eq. (4.3) can be simplified to

{\int }_{{\mathfrak{M}}_0}{f}_3\left(\alpha \right)f_4^9\left(\alpha \right)e\left(-n\alpha \right)\mathrm{d}\alpha =\mathfrak{G}\left(n\right)\mathfrak{J}\left(n\right)+O\left(\frac{n^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}n}\right),

which is Eq. (2.3). This completes the proof of Proposition 2.1.

5 Properties of the singular series

In this section, we will focus on presenting several properties of the singular series, which play a crucial role in Proposition 2.1.

Lemma 5.1　 Let p be a prime that satisfies p^{\alpha }\parallel k. For (a,p)=1, if \ell \ge \gamma \left(p\right), then C_k(p^{\ell },a)=0, where

\gamma \left(p\right)=\{\begin{array}{l}\alpha +2,\text{i}\text{f}p\ne 2\text{o}\text{r}p=2,\alpha =0;\\ \alpha +3,\text{i}\text{f}p=2,\alpha >0.\end{array}

Proof　See Hua [7, Lemma 8.3]. □

For k\ge 1, define

S_k(q,a)=\sum _{m=1}^q e\left(\frac{a{m}^k}{q}\right).

Lemma 5.2　 Let (p,a)=1. Then we have

S_k\left(p,a\right)=\sum _{\chi \in {\mathcal{A}}_k}\overline{\chi \left(a\right)}\tau \left(\chi \right),

where {\mathcal{A}}_k denotes the set of non-principal characters χ modulo p such that χ^k is the principal character. Furthermore, \tau \left(\chi \right) denotes the Gauss sum

\sum _{m=1}^p \chi \left(m\right)e\left(\frac{m}{p}\right).

Additionally, we have \left|\tau \right(\chi \left)\right|=p^{\frac{1}{2}} and \left|{\mathcal{A}}_k\right|=(k,p-1)-1.

Proof　See Vaughan [12, Lemma 4.3]. □

Lemma 5.3　 For (p,n)=1, we have

(5.1)\left|{\sum }_{a=1}^{p-1} \frac{S_3(p,a)S_4^9(p,a)}{p^6}e\left(-\frac{an}{p}\right)\right|\le 39366p^{-\frac{9}{2}}.

Proof　Let’s denote the left-hand side of Eq. (5.1) by S. According to Lemma 5.2, we have

\mathcal{S}=\frac{1}{p^{10}}\sum _{a=1}^{p-1}\left(\sum _{{\chi }_3\in {\mathcal{A}}_3}\overline{{\chi }_3\left(a\right)}\tau \left({\chi }_3\right)\right){\left(\sum _{{\chi }_4\in {\mathcal{A}}_4}\overline{{\chi }_4\left(a\right)}\tau \left({\chi }_4\right)\right)}^{9}e\left(-\frac{an}{p}\right).

If \left|{\mathcal{A}}_k\right|=0 for some k\in \left\{3,4\right\}, then S=0. If this is not the case, then

\begin{array}{rl}S=& \frac{1}{p^{10}}\sum _{{\chi }_3\in {\mathcal{A}}_3} \sum _{{\chi }_4^{\left(1\right)}\in {\mathcal{A}}_4} \sum _{{\chi }_4^{\left(2\right)}\in {\mathcal{A}}_4} \cdots \sum _{{\chi }_4^{\left(9\right)}\in {\mathcal{A}}_4} \tau \left({\chi }_3\right)\tau \left({\chi }_4^{\left(1\right)}\right)\tau \left({\chi }_4^{\left(2\right)}\right)\cdots \tau \left({\chi }_4^{\left(9\right)}\right)\\ & \times \sum _{a=1}^{p-1} {\chi }_3\left(a\right){\chi }_4^{\left(1\right)}\left(a\right){\chi }_4^{\left(2\right)}\left(a\right)\cdots {\chi }_4^{\left(9\right)}\left(a\right)e\left(-\frac{an}{p}\right).\end{array}

From Lemma 5.2, we know that the number of terms in the outer sums does not exceed \left(\right(3,p-1)-1)\times \left(\right(4,p-1)-1{)}^9\le 2\times 3^9=39366. For each of these terms, we have

\left|\begin{array}{c}\tau \left({\chi }_3\right)\tau \left({\chi }_4^{\left(1\right)}\right)\tau \left({\chi }_4^{\left(2\right)}\right)\cdots \tau \left({\chi }_4^{\left(9\right)}\right)\end{array}\right|=p^5.

Since each of these terms

\overline{{\chi }_3\left(a\right){\chi }_4^{\left(1\right)}\left(a\right){\chi }_4^{\left(2\right)}\left(a\right)\cdots {\chi }_4^{\left(9\right)}\left(a\right)}

is a Dirichlet character χ (mod p), the inner sum is

\sum _{a=1}^{p-1}\chi \left(a\right)e\left(-\frac{an}{p}\right)=\overline{\chi \left(-n\right)}\sum _{a=1}^{p-1}\chi \left(-an\right)e\left(-\frac{an}{p}\right)=\overline{\chi \left(-n\right)}\tau \left(\chi \right).

Noting that for the principal character χ⁰ mod p, we always have \tau \left({\chi }^0\right)=-1, so

|\begin{array}{c}\overline{\chi \left(-n\right)}\tau \left(\chi \right)\end{array}|\le p^{\frac{1}{2}}.

Combining all these discussions, we obtain:

\left|\mathcal{S}\right|\le \frac{1}{p^{10}}\cdot 39366\cdot p^5\cdot p^{\frac{1}{2}}=39366p^{-\frac{9}{2}}.

This completes the proof of Lemma 5.3. □

Lemma 5.4　 Let \mathcal{L}(p,n) denote the number of solutions to the following congruence equation

x_1^3+x_2^4+x_3^4+\cdots +x_{10}^4\equiv n\left(\mathrm{m}\mathrm{o}\mathrm{d}p\right),1\le x_1,x_2,\dots ,x_{10}\le p-1.

Then, for n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right), we have \mathcal{L}(p,n)>0.

Proof　Using the orthogonality of exponential functions, we know that

p\cdot \mathcal{L}(p,n)=\sum _{a=1}^p C_3(p,a)C_4^9(p,a)e\left(-\frac{an}{p}\right)=(p-1{)}^{10}+E_p,

where

E_p=\sum _{a=1}^{p-1} C_3(p,a)C_4^9(p,a)e\left(-\frac{an}{p}\right).

According to Lemma 5.2, we have

\left|E_p\right|\le (p-1)(2\sqrt{p}+1)(3\sqrt{p}+1{)}^9.

It is easy to verify that holds forp\ge 23. Hence, we know that \mathcal{L}(p,n)>0 when p\ge 23. Forp=2,3,5,7,11,13,17,19, under the congruence conditions n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right), we can directly verify one by one that \mathcal{L}(p,n)>0. □

Lemma 5.5　 Multiplicativity of A(n,q) with respect to q.

Proof　According to the definition of A(n,q) in Eq. (4.1), we only need to prove that B(n,q) is multiplicative with respect to q. Let q=q_1{q}_2 that satisfies (q_1,q_2)=1. Then we have

(5.2)\begin{array}{rl}B(n,q_1{q}_2)& =\sum _{(a,q_1{q}_2)=1}^{q_1{q}_2} C_3(q_1{q}_2,a)C_4^9(q_1{q}_2,a)e\left(-\frac{an}{q_1{q}_2}\right)\\ & =\sum _{\begin{array}{c}{a}_1=1\\ (a_1,q_1)=1\end{array}}^{q_1} \sum _{\begin{array}{c}{a}_2=1\\ (a_2,q_2)=1\end{array}}^{q_2} C_3(q_1{q}_2,a_1{q}_2+a_2{q}_1)C_4(q_1{q}_2,a_1{q}_2\\ & +a_2{q}_1)e\left(-\frac{a_{1}n}{q_1}\right)e\left(-\frac{a_{2}n}{q_2}\right).\end{array}

When (q_1,q_2)=1, for k\in \left\{3,4\right\}, we have

(5.3)\begin{array}{rl}{C}_k(q_1{q}_2,a_1{q}_2+a_2{q}_1)& =\sum _{\begin{array}{c}m=1\\ (m,q_1{q}_2)=1\end{array}}^{q_1{q}_2} e\left(\frac{(a_1{q}_2+a_2{q}_1)m^k}{q_1{q}_2}\right)\\ & =\sum _{\stackrel{m_1=1}{(m_1,q_1)=1}}^{q_1} \sum _{\stackrel{m_2=1}{(m_2,q_2)=1}}^{q_2} e\left(\frac{(a_1{q}_2+a_2{q}_1)(m_1{q}_2+m_2{q}_1{)}^k}{q_1{q}_2}\right)\\ & =\sum _{\begin{array}{c}{m}_1=1\\ (m_1,q_1)=1\end{array}}^{q_1} e\left(\frac{a_1(m_1{q}_2{)}^k}{q_1}\right)\sum _{\begin{array}{c}{m}_2=1\\ (m_2,q_2)=1\end{array}}^{q_2} e\left(\frac{a_2(m_2{q}_1{)}^k}{q_2}\right)\\ & =C_k(q_1,a_1)C_k(q_2,a_2).\end{array}

Substituting (5.3) into (5.2) gives

\begin{array}{rl}B(n,q_1{q}_2)& =\sum _{\begin{array}{c}{a}_1=1\\ (a_1,q_1)=1\end{array}}^{q_1} C_3(q_1,a_1)C_4^9(q_1,a_1)e\left(-\frac{a_{1}n}{q_1}\right)\\ & \sum _{\begin{array}{c}{a}_2=1\\ (a_2,q_2)=1\end{array}}^{q_2} C_3(q_2,a_2)C_4^9(q_2,a_2)e\left(-\frac{a_{2}n}{q_2}\right)\\ & =B(n,q_1)B(n,q_2).\end{array}

This completes the proof of Lemma 5.5. □

Lemma 5.6　 Let A(n,q) be defined by Eq. (4.1). Then

(i) we have

(5.4)\sum _{q>Z} A(n,q)\ll Z^{-\frac{7}{2}+\epsilon }d\left(n\right).

(ii) there exists an absolute constant c^{\ast }>0, such that for n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right),

Proof　According to Lemma 5.5, we know that B(n,q) is multiplicative with respect to q. Hence, we have

(5.5)B(n,q)=\prod _{p^t\parallel q} B(n,p^t)=\prod _{p^t\parallel q} \sum _{\begin{array}{c}a=1\\ (a,p)=1\end{array}}^{p^t} C_3(p^t,a)C_4^9(p^t,a)e\left(-\frac{an}{p^t}\right).

From Eq. (5.5) and Lemma 5.1, we see that B(n,q)=\prod _{p\parallel q} B(n,p) or 0, depending on whether q is square-free. Thus, we obtain

(5.6)\sum _{q=1}^{\mathrm{\infty }} A(n,q)=\sum _{\begin{array}{c}q=1\\ q\text{s}\text{q}\text{u}\text{a}\text{r}\text{e}\text{-}\text{f}\text{r}\text{e}\text{e}\end{array}}^{\mathrm{\infty }} A(n,q).

Let

\mathcal{R}(p,a):=C_3(p,a)C_4^9(p,a)-S_3(p,a)S_4^9(p,a).

Then,

(5.7)\begin{array}{rl}A(n,p)=& \frac{1}{(p-1{)}^{10}}\sum _{a=1}^{p-1} S_3(p,a)S_4^9(p,a)e\left(-\frac{an}{p}\right)\\ & +\frac{1}{(p-1{)}^{10}}\sum _{a=1}^{p-1} \mathcal{R}(p,a)e\left(-\frac{an}{p}\right).\end{array}

By applying Lemma 4.1 and noting that S_k(p,a)=C_k(p,a)+1, we have S_k(p,a)\ll p^{\frac{1}{2}} and thus \mathcal{R}(p,a)\ll p^{\frac{9}{2}}. Therefore, the second term on the right side of Eq. (5.7) is \le c_1{p}^{-\frac{9}{2}}. On the other hand, according to Lemma 5.3, the first term on the right side of Eq. (5.7) is \le 2^{10}\cdot 39366p^{-\frac{9}{2}}=40310784p^{-\frac{9}{2}}. Let c_2=c_1+40310784. Then we obtain that when p\nmid n,

(5.8)\left|A\right(n,p\left)\right|\le c_2{p}^{-\frac{9}{2}}.

In addition, if we directly apply Lemma 4.1, we get

\begin{array}{rl}\left|B(n,p)\right|& =\left|\sum _{a=1}^{p-1} C_3(p,a)C_4^9(p,a)e\left(-\frac{an}{p}\right)|\le \sum _{a=1}^{p-1} \left|C_3(p,a)C_4^9(p,a)\right|\right|\\ & \le (p-1)\cdot 2^{10}\cdot p^5\cdot 786432=805306368p^5(p-1),\end{array}

from which we get

(5.9)\left|\begin{array}{c}A(n,p)\end{array}\right|=\frac{\left|B\right(n,p\left)\right|}{{\phi }^{10}\left(p\right)}\le \frac{805306368p^5}{(p-1{)}^9}\le \frac{2^9\cdot 805306368p^5}{p^9}=\frac{412316860416}{p^4}.

Let c_3=max(c_2,412316860416). Then for square-free q, we have

\begin{array}{rl}\left|A\right(n,q\left)\right|& =\left(\prod _{\begin{array}{c}p|q\\ p\nmid n\end{array}} \left|A\right(n,p\left)\right|\right)\left(\prod _{\begin{array}{c}p|q\\ p|n\end{array}} \left|A\right(n,p\left)\right|\right)\\ & \le \left(\prod _{\begin{array}{c}p|q\\ p\nmid n\end{array}} \left(c_3{p}^{-\frac{9}{2}}\right)\right)\left(\prod _{\begin{array}{c}p|q\\ p|n\end{array}} \left(c_3{p}^{-4}\right)\right)\\ & =c_3^{\omega \left(q\right)}\left(\prod _{p|q} p^{-\frac{9}{2}}\right)\left(\prod _{p\left|\right(n,q)} p^{\frac{1}{2}}\right)\ll q^{-\frac{9}{2}+\epsilon }(n,q{)}^{\frac{1}{2}}.\end{array}

Therefore, from Eq. (5.6), we obtain

\begin{array}{rl}\sum _{q>Z} \left|A\right(n,q\left)\right|& \ll \sum _{q>Z} q^{-\frac{9}{2}+\epsilon }(n,q{)}^{\frac{1}{2}}=\sum _{d|n} \sum _{q>\frac{Z}{d}} (dq{)}^{-\frac{9}{2}+\epsilon }{d}^{\frac{1}{2}}=\sum _{d|n} d^{-4+\epsilon }\sum _{q>\frac{Z}{d}} q^{-\frac{9}{2}+\epsilon }\\ & \ll \sum _{d|n} d^{-4+\epsilon }{\left(\frac{Z}{d}\right)}^{-\frac{7}{2}+\epsilon }=Z^{-\frac{7}{2}+\epsilon }\sum _{d|n} d^{-\frac{1}{2}}\\ & \ll Z^{-\frac{7}{2}+\epsilon }d\left(n\right),\end{array}

which proves Eq. (5.4) and demonstrates the absolute convergence of \mathfrak{S}\left(n\right).

To prove Lemma 5.6(ii), according to Lemma 5.5, we have

(5.10)\begin{array}{rl}S\left(n\right)& =\prod _p \left(1+\sum _{t=1}^{\mathrm{\infty }} A(n,p^t)\right)=\prod _p (1+A(n,p\left)\right)\\ & =\left(\prod _{p\le c_3} (1+A(n,p))\right)\left(\prod _{p>c_{3}p\nmid n} (1+A(n,p))\right)\left(\prod _{\genfrac{}{}{0pt}{}{p>c_3}{p\nmid n}} (1+A(n,p))\right).\end{array}

According to (5.8), we have

(5.11)\prod _{\genfrac{}{}{0pt}{}{p>c_3}{p\nmid n}} \left(1+A(n,p)\right)\ge \prod _{p>c_3} \left(1-\frac{c_3}{p^{\frac{9}{2}}}\right)\ge c_4>0.

From (5.9), we have

(5.12)\prod _{\genfrac{}{}{0pt}{}{p>c_3}{p|n}} \left(1+A(n,p)\right)\ge \prod _{p>c_3} \left(1-\frac{c_3}{p^4}\right)\ge c_5>0.

On the other hand, it is easy to see that

1+A(n,p)=\frac{p\cdot \mathcal{L}(p,n)}{{\phi }^{10}\left(p\right)}.

According to Lemma 5.4, when n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), and n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right), \mathcal{L}(p,n)>0 holds for all prime numbers p. Therefore, 1+A(n,p)>0, and thus

(5.13)\prod _{p\le c_3} \left(1+A(n,p)\right)\ge c_6>0.

Combining equations (5.10) to (5.13) and setting c^{\ast }=c_4{c}_5{c}_6>0, we conclude that \mathfrak{S}\left(n\right)\ge c^{\ast }>0.

Furthermore, from (5.8) and (5.9), we have

\mathfrak{S}\left(n\right)\le \prod _{p\nmid n} \left(1+\frac{c_3}{p^{\frac{9}{2}}}\right)\cdot \prod _{p\mid n} \left(1+\frac{c_3}{p^4}\right)\ll 1.

This completes the proof of Lemma 5.6. □

6 Proof of Proposition 2.2

In this section, we will present the proof of Proposition 2.2. We denote by {\mathcal{Z}}_j\left(N\right) the set of positive integers n satisfying n\in [\frac{N}{2},N], n\equiv 0\left(\mathrm{m}\mathrm{o}\mathrm{d}2\right), n\equiv \pm 1\left(\mathrm{m}\mathrm{o}\mathrm{d}3\right), n\not\equiv 4\left(\mathrm{m}\mathrm{o}\mathrm{d}5\right) and such that

(6.1)\left|{\int }_{{\mathfrak{m}}_j} f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \right|\gg \frac{n^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}n}.

For convenience, we denote the cardinality of {\mathcal{Z}}_j\left(N\right) by z_j. Furthermore, for n\in {\mathcal{Z}}_j\left(N\right), we define the complex number {\xi }_j\left(n\right) via the equation

(6.2)\left|{\int }_{{\mathfrak{m}}_j} f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \right|={\xi }_j\left(n\right){\int }_{{\mathfrak{m}}_j}{f}_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha .

And if n\notin {\mathcal{Z}}_j\left(N\right), then {\xi }_j\left(n\right)=0. Clearly, \left|{\xi }_j\right(n\left)\right|=1 whenever {\xi }_j\left(n\right) is not equal to 0. Therefore, we have

(6.3)\sum _{n\in {\mathcal{Z}}_j\left(N\right)} {\xi }_j\left(n\right){\int }_{{\mathfrak{m}}_j} f_3\left(\alpha \right)f_4^9\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha ={\int }_{{\mathfrak{m}}_j} f_3\left(\alpha \right)f_4^9\left(\alpha \right){\mathcal{K}}_j\left(\alpha \right)\mathrm{d}\alpha ,

where the exponential sum {\kappa }_j\left(\alpha \right) is defined as

{\mathcal{K}}_j\left(\alpha \right)=\sum _{n\in {\mathcal{Z}}_j\left(N\right)} {\xi }_j\left(n\right)e(-n\alpha ).

For j=1,2, let

I_j={\int }_{{\mathfrak{m}}_j} f_3\left(\alpha \right)f_4^9\left(\alpha \right){\mathcal{K}}_j\left(\alpha \right)\mathrm{d}\alpha .

According to Eqs. (6.1)‒(6.3), we have

(6.4)I_j\gg \sum _{n\in {\mathcal{Z}}_j\left(N\right)} \frac{n^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}n}\gg \frac{{\mathcal{Z}}_j{N}^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}N},j=1,2.

According to Kawada and Wooley [9, Lemma 2.1] (where k=4 and j=2), we have

(6.5){\int }_0^1 \left|f_4^4\right(\alpha \left){\mathcal{K}}_j^2\right(\alpha \left)\right|\mathrm{d}\alpha \ll {\mathcal{Z}}_j{N}^{\frac{3}{4}}+{\mathcal{Z}}_j^{\frac{3}{2}}{N}^{\frac{1}{2}+\epsilon }

for j=1,2. Using the Cauchy inequality, VauGhan [12, Lemma 2.5], Lemma 3.3, and Eq. (6.5), we obtain

(6.6)\begin{array}{rl}{I}_1& \ll \left(\underset{\alpha \in {\mathfrak{m}}_1}{sup} {\left|f_3\left(\alpha \right)\right|}^{\frac{1}{2}}\right){\left({\int }_0^1 {\left|f_3\left(\alpha \right)\right|}^8\mathrm{d}\alpha \right)}^{\frac{1}{16}}{\left({\int }_0^1 {\left|f_4\left(\alpha \right)\right|}^{16}\mathrm{d}\alpha \right)}^{\frac{\gamma }{16}}\\ & {\left({\int }_0^1 \left|f_4^4\right(\alpha \left){\mathcal{K}}_1^2\right(\alpha \left)\right|\mathrm{d}\alpha \right)}^{\frac{1}{2}}\\ & \ll {\left(N^{\frac{11}{36}+\epsilon }\right)}^{\frac{1}{2}}{\left(N^{\frac{5}{3}+\epsilon }\right)}^{\frac{1}{16}}{\left(N^{3+\epsilon }\right)}^{\frac{7}{16}}{\left(Z_1{N}^{\frac{3}{4}}+Z_1^{\frac{3}{2}}{N}^{\frac{1}{2}+\epsilon }\right)}^{\frac{1}{2}}\\ & \ll N^{\frac{113}{72}+\epsilon }({\mathcal{Z}}_1^{\frac{1}{2}}{N}^{\frac{3}{8}}+{\mathcal{Z}}_1^{\frac{3}{4}}{N}^{\frac{1}{4}})\ll {\mathcal{Z}}_1^{\frac{1}{2}}{N}^{\frac{35}{18}+\epsilon }+{\mathcal{Z}}_1^{\frac{3}{4}}{N}^{\frac{131}{72}+\epsilon }.\end{array}

From Eqs. (6.4) and (6.6), we get

\begin{array}{c}{\mathcal{Z}}_1{N}^{\frac{19}{12}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{-11}N\ll I_1\ll {\mathcal{Z}}_1^{\frac{1}{2}}{N}^{\frac{35}{18}+\epsilon }+{\mathcal{Z}}_1^{\frac{3}{4}}{N}^{\frac{131}{72}+\epsilon },\end{array}

from which we deduce

(6.7){\mathcal{Z}}_1\ll N^{\frac{17}{18}+\epsilon }.

Next, we estimate an upper bound for z_2. According to Eq. (3.2), we have

(6.8)I_2\ll {\int }_{{\mathfrak{m}}_2} \left|V_3\right(\alpha \left)f_4^9\right(\alpha \left){\mathcal{K}}_2\right(\alpha \left)\right|\mathrm{d}\alpha +N^{\frac{4}{15}+\epsilon }{\int }_{{\mathfrak{m}}_2} \left|f_4^9\right(\alpha \left){\mathcal{K}}_2\right(\alpha \left)\right|\mathrm{d}\alpha =:I_{21}+I_{22}.

For \alpha \in {\mathfrak{m}}_2, either or holds. Thus, according to Lemma 3.1, we have

(6.9)\begin{array}{rl}\underset{\alpha \in {\mathfrak{m}}_2}{sup} \left|f_4\right(\alpha \left)\right|& \ll X_4^{\frac{1}{2}+\epsilon }\sqrt{q\left(1+N|\alpha -\frac{a}{q}|\right)}+X_4^{\frac{4}{5}+\epsilon }+\frac{X_4(\mathrm{l}\mathrm{o}\mathrm{g}N{)}^c}{{\left(q(1+N|\alpha -{\displaystyle \frac{a}{q}}|)\right)}^{\frac{1}{2}-\epsilon }}\\ & \ll X_4^{\frac{1}{2}+\epsilon }{Q}_1^{\frac{1}{2}}+X_4^{\frac{4}{5}+\epsilon }+\frac{X_4(\mathrm{l}\mathrm{o}\mathrm{g}N{)}^c}{(Q_0^{100}{)}^{\frac{1}{2}-\epsilon }}\ll \frac{N^{\frac{1}{4}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{40A}N}.\end{array}

Noting that {\mathfrak{m}}_2\subseteq I, where I is defined in Eq. (3.1), and using the Cauchy inequality, it is evident that {\mathcal{K}}_2\left(\alpha \right)\ll {\mathcal{Z}}_2. From Hua [7, Theorem 4, p.19], Lemma 3.4 and (6.9), we obtain

(6.10)\begin{array}{rl}{I}_{21}& \ll {\mathcal{Z}}_2\cdot \underset{\alpha \in {\mathfrak{m}}_2}{sup} \left|f_4\right(\alpha \left)\right|\cdot {\left({\int }_0^1 \left|f_4\right(\alpha ){|}^{16}\mathrm{d}\alpha \right)}^{\frac{1}{2}}{\left({\int }_{\mathcal{J}} \left|V_3\right(\alpha ){|}^2\mathrm{d}\alpha \right)}^{\frac{1}{2}}\\ & \ll {\mathcal{Z}}_2\cdot \left(\frac{N^{\frac{1}{4}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{40A}N}\right)\cdot (N^3{\mathrm{l}\mathrm{o}\mathrm{g}}^{c}N{)}^{\frac{1}{2}}\cdot (N^{-\frac{1}{3}}{\mathrm{l}\mathrm{o}\mathrm{g}}^{2A}N{)}^{\frac{1}{2}}\\ & \ll \frac{{\mathcal{Z}}_2{N}^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{20A}N},\end{array}

where the parameter A is chosen to be sufficiently large for the above upper bound to hold. Furthermore, using the Cauchy inequality, Eq. (6.5), and Hua [7, Theorem 4, p.19], we obtain

(6.11)\begin{array}{rl}{I}_{22}& \ll N^{\frac{4}{15}+\epsilon }\left({\int }_0^1 \right|f_4{\left(\alpha ){|}^{16}\mathrm{d}\alpha \right)}^{\frac{3}{8}}{\left({\int }_0^1 |f_4\left(\alpha \right){|}^8\mathrm{d}\alpha \right)}^{\frac{1}{8}}{\left({\int }_0^1 |f_4^4\left(\alpha \right){\mathcal{K}}_2^2\left(\alpha \right)|\mathrm{d}\alpha \right)}^{\frac{1}{2}}\\ & \ll N^{\frac{4}{15}+\epsilon }{\left(N^{3+\epsilon }\right)}^{\frac{3}{8}}{\left(N^{\frac{5}{4}+\epsilon }\right)}^{\frac{1}{8}}{\left({\mathcal{Z}}_2{N}^{\frac{3}{4}}+{\mathcal{Z}}_2^{\frac{3}{2}}{N}^{\frac{1}{2}+\epsilon }\right)}^{\frac{1}{2}}\\ & \ll N^{\frac{743}{480}+\epsilon }\left({\mathcal{Z}}_2^{\frac{1}{2}}{N}^{\frac{3}{8}}+{\mathcal{Z}}_2^{\frac{3}{4}}{N}^{\frac{1}{4}}\right)\\ & \ll {\mathcal{Z}}_2^{\frac{1}{2}}{N}^{\frac{923}{480}+\epsilon }+{\mathcal{Z}}_2^{\frac{3}{4}}{N}^{\frac{863}{480}+\epsilon }.\end{array}

Combining Eqs. (6.4), (6.8), (6.10), and (6.11), we obtain

\frac{{\mathcal{Z}}_2{N}^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{11}N}\ll I_2=I_{21}+I_{22}\ll \frac{{\mathcal{Z}}_2{N}^{\frac{19}{12}}}{{\mathrm{l}\mathrm{o}\mathrm{g}}^{20A}N}+{\mathcal{Z}}_2^{\frac{1}{2}}{N}^{\frac{923}{480}+\epsilon }+{\mathcal{Z}}_2^{\frac{3}{4}}{N}^{\frac{863}{480}+\epsilon },

which implies

(6.12){\mathcal{Z}}_2\ll N^{\frac{103}{120}+\epsilon }.

From Eqs. (6.7) and (6.12), we have

\mathcal{Z}\left(N\right)\ll {\mathcal{Z}}_1+{\mathcal{Z}}_2\ll N^{\frac{17}{18}+\epsilon }.

This completes the proof of Proposition 2.2. □

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