Goldbach–Linnik–Type Problem of Symmetric Mixed Powers of Primes and Powers of Two

Abstract

This article demonstrates that every sufficiently large odd integer can be expressed as the sum of one square of a prime, six cubes of primes, and 23 powers of two. This finding represents an improvement on the previous results of Sinnadurai in 1965 and Hooley in 1981.

1. Introduction and Main Result

It is highly probable that for each $s>1$, every sufficiently large integer can be represented as the sum of one square and s k-th powers of positive integers with $k⩾3$. To be specific, our focus will be on the Diophantine equation

$$N=x^2+y_1^k+y_2^k+\cdots +y_s^k,$$

(1)

where $s,k\in \mathbb{N}$ are natural numbers with $k⩾3$. This family of equations is a subset of the variants of Waring’s problem, which have been extensively studied by various scholars since the early applications of the Hardy–Littlewood method. A purely heuristic application of this method, based solely on major arc analysis, suggests that the number $R_{k,s}\left(N\right)$ of solutions to (1) in natural numbers $x,y_1,\cdots ,y_s$ adheres to the asymptotic relation

$$R_{k,s}\left(N\right)={\displaystyle \frac{\mathsf{\Gamma }\left(\frac{3}{2}\right){\mathsf{\Gamma }}^s(1+\frac{1}{k})}{\mathsf{\Gamma }(\frac{1}{2}+\frac{s}{k})}}{\mathfrak{S}}_{k,s}\left(N\right)N^{{\displaystyle \frac{s}{k}}-{\displaystyle \frac{1}{2}}}(1+o\left(1\right))$$

(2)

as $N\to +\infty$, provided that $s>{\displaystyle \frac{k}{2}}$. Here, the singular series is defined by

$${\mathfrak{S}}_{k,s}\left(N\right)=\sum _{q=1}^{\infty }{\displaystyle \frac{1}{q^{s+1}}}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q\left(\sum _{x=1}^{q}e\left({\displaystyle \frac{a{x}^2}{q}}\right)\right){\left(\sum _{y=1}^{q}e\left({\displaystyle \frac{a{y}^k}{q}}\right)\right)}^{s}e\left(-{\displaystyle \frac{aN}{q}}\right),$$

where $e\left(a\right)=e^{2\pi ia}$. The first analysis of the problem was conducted by Stanley [1] in 1930. Following the methodology established by Hardy and Littlewood [2,3] in their classic series “Partitio Numerorum”, Stanley [1] formulated the asymptotic Formula (2) for $s⩾s_1\left(k\right)$, where

$$s_1\left(3\right)=7,s_1\left(4\right)=14,s_1\left(5\right)=28,s_1\left(k\right)=2^{k-3}(k-2)+O\left(k\right)(k>5).$$

(3)

Later, Sinnadurai [4] verified (2) for $R_{3,6}\left(N\right)$, and Hooley [5] provided a different proof for this result. However, when $k⩾4$, the authors are unaware of any improvements on Stanley’s bounds (3) documented in the literature. Yet, since the 1920s, the theory of Waring’s problem has experienced waves of innovation, resulting in significantly smaller lower bounds for s for which (2) can be demonstrated. The latest developments indicate that for $k⩾3$ and $s⩾2^{k-1}+2$, then for any $\epsilon >0$, one can demonstrate that

$$R_{k,s}\left(N\right)={\displaystyle \frac{\mathsf{\Gamma }\left(\frac{3}{2}\right){\mathsf{\Gamma }}^s(1+\frac{1}{k})}{\mathsf{\Gamma }(\frac{1}{2}+\frac{s}{k})}}{\mathfrak{S}}_{k,s}\left(N\right)N^{{\displaystyle \frac{s}{k}}-{\displaystyle \frac{1}{2}}}+O\left(N^{{\displaystyle \frac{s}{k}}-{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{k·2^{k-1}}}+\epsilon }\right).$$

(4)

Moreover, there exists a function $s_0\left(k\right)$ that satisfies

$$s_0\left(k\right)⩽{\displaystyle \frac{k^2}{2}}logk+O(k^{2}loglogk),$$

and ensures that (4) holds whenever $s⩾s_0\left(k\right)$. Notably, when $k=3$ we may take $s=6$, thereby including the formula $R_{3,6}\left(n\right)$ as already established by Sinnadurai [4] and Hooley [5] within (4).

Given the result of Sinnadurai [4] and Hooley [5], it is reasonable to conjecture that for every sufficiently large odd integer N, the following equation is solvable:

$$N=p^2+p_1^3+p_2^3+p_3^3+p_4^3+p_5^3+p_6^3$$

(5)

where p (with or without a subscript) denotes a prime number. But this conjecture is perhaps out of reach at present. Motivated by this conjecture and the famous works of Linnik [6,7] and Gallagher [8] for representing a sufficiently large even integer N as sums of two primes and powers of 2, we consider the corresponding approximation of (5). For the literature on this topic, one can refer to [9,10,11,12].

In this paper, we focus on the symmetric approximation of the conjecture (5) and establish the following theorem:

Theorem 1.

Every sufficiently large odd integer can be represented as a sum of one square of a prime, six cubes of primes and 23 powers of two.

Notation. 

Throughout this paper, let p always denote a prime number with or without subscripts; $\epsilon$ always denotes an arbitrarily small positive number, which may not be the same at different occurrences. We use $e\left(\alpha \right)$ to represent $e^{2\pi i\alpha }$. Furthermore, $\phi \left(n\right)$ stands for Euler’s function.

2. Outline of the Proof of Theorem 1

Let N be a sufficiently large odd integer and $\eta$ be a fixed positive constant which satisfies $\eta ⩽{10}^{-100}$. Write

$$U_2={\left({\displaystyle \frac{(1-\eta )N}{8}}\right)}^{{\displaystyle \frac{1}{2}}},U_3={\left({\displaystyle \frac{(1-\eta )N}{16}}\right)}^{{\displaystyle \frac{1}{3}}},P_3={\left({\displaystyle \frac{\eta N}{4}}\right)}^{{\displaystyle \frac{1}{3}}},V_3=U_3^{{\displaystyle \frac{5}{6}}},L={\displaystyle \frac{log(N/logN)}{log2}},$$

(6)

and

$$\begin{array}{cc}\hfill & H\left(\alpha \right)=\sum _{1⩽v⩽L}e\left(2^v\alpha \right),F\left(\alpha \right)=\sum _{U_2<p⩽2U_2}(logp)e\left(p^2\alpha \right),S_1\left(\alpha \right)=\sum _{U_3<p⩽2U_3}(logp)e\left(p^3\alpha \right),\hfill \\ \hfill & S_2\left(\alpha \right)=\sum _{P_3<p⩽2P_3}(logp)e\left(p^3\alpha \right),T\left(\alpha \right)=\sum _{V_3<p⩽2V_3}(logp)e\left(p^3\alpha \right).\hfill \end{array}$$

(7)

Define

$$\mathcal{R}(k,N):=\sum _{\begin{array}{c}N=p_1^2+p_2^3+p_3^3+p_4^3+p_5^3+p_6^3+p_7^3+2^{v_1}+\cdots +2^{v_k}\\ U_2<p_1⩽2U_2,U_3<p_2,p_3⩽2U_3,P_3<p_4,p_5⩽2P_3\\ V_3<p_6,p_7⩽2V_3,1⩽v_k⩽L\end{array}}(log{p}_1)(log{p}_2)\cdots (log{p}_7).$$

To apply the circle method, we define

$$Q_1=N^{25/216-2\epsilon },Q_2=N^{191/216+\epsilon }.$$

Then, we define the major arcs $\mathfrak{M}$ and minor arcs $\mathfrak{m}$ as follows

$$\mathfrak{M}=\bigcup _{q⩽Q_1}\bigcup _{\begin{array}{c}1⩽a⩽q\\ (a,q)=1\end{array}}\mathfrak{M}(q,a),\mathfrak{m}=\left[{\displaystyle \frac{1}{Q_2}},1+{\displaystyle \frac{1}{Q_2}}\right]\setminus \mathfrak{M},$$

(8)

where

$$\mathfrak{M}(q,a)=\left[{\displaystyle \frac{a}{q}}-{\displaystyle \frac{1}{q{Q}_2}},{\displaystyle \frac{a}{q}}+{\displaystyle \frac{1}{q{Q}_2}}\right].$$

By the orthogonality of the exponential sum, one has

$$\begin{array}{cc}\hfill \mathcal{R}(k,N)=& {\int }_0^{1}F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha \hfill \\ \hfill =& \left({\int }_{\mathfrak{M}}+{\int }_{\mathfrak{m}}\right)F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha \hfill \\ \hfill =& \mathcal{I}(k,\mathfrak{M},N)+\mathcal{I}(k,\mathfrak{m},N).\hfill \end{array}$$

To prove Theorem 1, we need the two propositions below, which will be proved in Section 3 and Section 4, respectively.

Proposition 1.

Suppose that $k=23$. Define $\mathfrak{M}$ as in (8). Then we have

$$\mathcal{I}(k,\mathfrak{M},N)⩾0.14457265·{\displaystyle \frac{{\eta }^{2/3}{(1-\eta )}^{13/18}{N}^{25/18}}{4^{19/9}}}{L}^k.$$

Proposition 2.

Suppose that $k=23$. Define $\mathfrak{m}$ as in (8). Then we have

$$|\mathcal{I}(k,\mathfrak{m},N)|⩽8.499467692{\lambda }^k{\displaystyle \frac{{\eta }^{2/3}{(1-\eta )}^{13/18}{N}^{25/18}}{4^{19/9}}}{L}^k+O\left({\displaystyle \frac{{\eta }^{2/3}{(1-\eta )}^{13/18}{N}^{25/18}}{4^{19/9}}}{L}^{k-1}\right),$$

where $\lambda =0.83372131691$.

3. Proof of Proposition 1

In this section, we shall investigate the lower bound of $\mathcal{I}(k,\mathfrak{M},N)$. Some preliminaries have to be given first. Let

$$\begin{array}{c}\hfill C_2(q,a)=\sum _{\begin{array}{c}m=1\\ (m,q)=1\end{array}}^{q}e\left({\displaystyle \frac{a{m}^2}{q}}\right),C_3(q,a)=\sum _{\begin{array}{c}m=1\\ (m,q)=1\end{array}}^{q}e\left({\displaystyle \frac{a{m}^3}{q}}\right),\\ \hfill B(n,q)=\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q{C}_2(q,a)C_3^6(q,a)e\left(-{\displaystyle \frac{an}{q}}\right),A(n,q)={\displaystyle \frac{B(n,q)}{{\phi }^7\left(q\right)}}.\end{array}$$

(9)

where $\phi \left(n\right)$ stands for Euler’s function.

Proposition 3.

Let $\mathfrak{M}$ and $\mathfrak{m}$ be defined as in (8), and consider $N/2<n⩽N$. Then we have

$${\int }_{\mathfrak{M}}F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha ={\displaystyle \frac{1}{2·3^6}}\mathfrak{S}\left(n\right)\mathfrak{J}\left(n\right)+O\left(N^{25/18}{L}^{-1}\right),$$

where $\mathfrak{S}\left(n\right)$ is a singular series defined by

$$\mathfrak{S}\left(n\right)=\sum _{q=1}^{\infty }A(n,q).$$

Furthermore, for any odd integer n, $\mathfrak{S}\left(n\right)$ is absolutely convergent and satisfies

$$0<c^{*}⩽\mathfrak{S}\left(n\right)\ll 1$$

with some fixed positive constant $c^{*}$, while

$$\mathfrak{J}\left(n\right)=\sum _{\begin{array}{c}n=m_1+m_2+m_3+m_4+m_5+m_6+m_7\\ U_3^3<m_2,m_3⩽8U_3^3\\ P_3^3<m_4,m_5⩽8P_3^3\\ V_3^3<m_6,m_7⩽8V_3^3\\ U_2^2<m_1⩽4U_2^2\end{array}}{m}_1^{-1/2}{\left(m_2{m}_3{m}_4{m}_5{m}_6{m}_7\right)}^{-2/3}\asymp N^{25/18}.$$

(10)

Proof. 

The proof of Proposition 3 is exactly the same as that of Proposition 2.1 of Zhang and Li [13], hence we omit the details herein. □

For $k⩾1$, we define

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

Lemma 1.

Assume that $(p,a)=1$. Then

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

where ${\mathcal{A}}_k$ is the set of non-principal characters χ modulo p for which ${\chi }^k$ is principal, and $\tau \left(\chi \right)$ signifies the Gauss sum

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

In addition, it holds that $\left|\tau \left(\chi \right)\right|=p^{1/2}$ and $|{\mathcal{A}}_k|=(k,p-1)-1$.

Proof. 

This conclusion is supported by Lemma 4.3 found in Vaughan [14]. □

Lemma 2.

Suppose that $k=23$. Let $\Xi (N,k)=\{(1-\eta )N⩽n⩽N:n=N-2^{v_1}-2^{v_2}-\cdots -2^{v_k},1⩽v_1,\cdots ,v_k⩽L\}$. Then, for $N\equiv 1(mod2)$, one has

$$\sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}\mathfrak{S}\left(n\right)⩾0.98815819L^k.$$

(11)

Proof. 

Let $C_2(q,a),C_3(q,a),B(n,q)$ and $A(n,q)$ be defined as in (9). By Lemma 1, for $p⩾5$, $(p,a)=1$ and $p\equiv 2(mod3)$, we have $C_3(p,a)=-1$. In order to obtain an appropriate lower bound for $B(n,p)$ with $p\equiv 2(mod3)$, we define the notation

$$G(\chi ,n)=\sum _{h=1}^p\chi \left(h\right)e\left({\displaystyle \frac{hn}{p}}\right),c_q\left(n\right)=\sum _{\begin{array}{c}h=1\\ (h,q)=1\end{array}}^{q}e\left({\displaystyle \frac{hn}{q}}\right),$$

where $c_q\left(n\right)$ is the Ramanujan sum. By Theorem 7.5.4 in [15], for $p⩾5$, we have

$$C_2(p,a)=S_2(p,a)-1=\chi \left(a\right)S_2(p,1)-1,$$

where $\chi (·)$ is the Legendre symbol $\left({\displaystyle \frac{·}{p}}\right)$. Thus, if $p\equiv 2(mod3)$, we can write $B(n,p)$ as

$$B(n,p)=S_2(p,1)G(\chi ,-n)-c_p(-n).$$

By Theorem 7.5.5 and Theorem 7.5.8 of Hua [15], we have

$$S_2(p,1)=\left\{\begin{array}{cc}\sqrt{p},\hfill & \mathrm{if}p\equiv 1(mod4),\hfill \\ i\sqrt{p},\hfill & \mathrm{if}p\equiv 3(mod4),\hfill \end{array}\right.$$

and

$$\left|G(\chi ,n)\right|=\left\{\begin{array}{cc}\sqrt{p},\hfill & \mathrm{if}p\nmid n,\hfill \\ 0,\hfill & \mathrm{if}p|n,\hfill \end{array}\right.c_p\left(n\right)=\left\{\begin{array}{cc}-1,\hfill & \mathrm{if}p\nmid n,\hfill \\ p-1,\hfill & \mathrm{if}p|n,\hfill \end{array}\right.$$

and thus $B(n,p)⩾-(p-1)$ for $p\equiv 2(mod3)$.

For $p\equiv 1(mod3)$, by Lemma 1, we have

$$\left|B(n,p)\right|=\left|\sum _{a=1}^{p-1}{C}_2(p,a)C_3^6(p,a)e\left(-{\displaystyle \frac{an}{p}}\right)\right|⩽(p-1)(\sqrt{p}+1){(2\sqrt{p}+1)}^6.$$

Combining the above two cases, we obtain the following result:

$$\begin{array}{cc}\hfill \prod _{p⩾300}(1+A(n,p))⩾& \prod _{\begin{array}{c}p⩾300\\ p\equiv 2(mod3)\end{array}}\left(1-{\displaystyle \frac{1}{{(p-1)}^6}}\right)\hfill \\ \hfill & \times \prod _{\begin{array}{c}p⩾300\\ p\equiv 1(mod3)\end{array}}\left(1-{\displaystyle \frac{(\sqrt{p}+1){(2\sqrt{p}+1)}^6}{{(p-1)}^6}}\right).\hfill \end{array}$$

On the one hand, we have the numerical lower bound

$$\begin{array}{cc}\hfill & \prod _{\begin{array}{c}300⩽p⩽p_{10000}\\ p\equiv 2(mod3)\end{array}}\left(1-{\displaystyle \frac{1}{{(p-1)}^6}}\right)\times \prod _{\begin{array}{c}300⩽p⩽p_{10000}\\ p\equiv 1(mod3)\end{array}}\left(1-{\displaystyle \frac{(\sqrt{p}+1){(2\sqrt{p}+1)}^6}{{(p-1)}^6}}\right)\hfill \\ \hfill & ⩾0.9992282,\hfill \end{array}$$

where the $p_r$ denotes the r-th prime. On the other hand, there holds

$$\prod _{p>p_{10000}}\left(1+A(n,p)\right)⩾\prod _{p>p_{10000}}\left(1-{\displaystyle \frac{1}{p^2}}\right)⩾0.99999923688927.$$

Thus,

$$\prod _{p⩾300}(1+A(n,p))⩾0.9992274374=:C_1.$$

Set $\mathcal{P}=\{3,7,13\}$. By noting that $n\equiv 1(mod2),N\equiv 1(mod2)$ and $v_j⩾1\mathrm{with}$$j=1,2,\cdots ,k$, we deduce that

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}\mathfrak{S}\left(n\right)⩾& \left(\underset{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}{min}\prod _{p\notin \mathcal{P}}\left(1+A(n,p)\right)\right)\left(\sum _{n\in \Xi (N,k)}\prod _{p\in \mathcal{P}}(1+A(n,p))\right)\hfill \\ \hfill ⩾& \left(\underset{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}{min}\prod _{\begin{array}{c}p\notin \mathcal{P}\\ 2⩽p<300\end{array}}(1+A(n,p))\right)\left(\underset{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}{min}\prod _{p⩾300}\left(1+A(n,p)\right)\right)\hfill \\ \hfill & \times \left(\sum _{n\in \Xi (N,k)}\prod _{p\in \mathcal{P}}\left(1+A(n,p)\right)\right)\hfill \\ \hfill ⩾& 0.98895427533C_1\sum _{n\in \Xi (N,k)}\prod _{p\in \mathcal{P}}(1+A(n,p))\hfill \\ \hfill ⩾& 0.98819024\sum _{1⩽j⩽q}\sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv j(modq)\end{array}}\prod _{p\in \mathcal{P}}(1+A(n,p))\hfill \\ \hfill =& 0.98819024\sum _{1⩽j⩽q}\left(\prod _{p\in \mathcal{P}}(1+A(j,p))\right)\sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv j(modq)\end{array}}1,\hfill \end{array}$$

(12)

where $q={\prod }_{p\in \mathcal{P}}p$. For an odd q, denote by $\varrho \left(q\right)$ the smallest positive integer $\varrho$ such that $2^{\varrho \left(q\right)}\equiv 1(modq)$. Then the inner sum in (12) becomes

$$\mathcal{S}:=\sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv j(modq)\end{array}}1={\left({\displaystyle \frac{L}{\varrho \left(q\right)}}+O\left(1\right)\right)}^k\sum _{\begin{array}{c}1⩽v_1,\cdots ,v_k⩽\varrho \left(q\right)\\ 2^{v_1}+\cdots +2^{v_k}\equiv N-j(modq)\end{array}}1.$$

By noting that

$$\mathcal{S}={\left({\displaystyle \frac{L}{\varrho \left(q\right)}}+O\left(1\right)\right)}^k{\displaystyle \frac{1}{q}}\sum _{s=0}^{q-1}e\left({\displaystyle \frac{s(j-N)}{q}}\right){\left(\sum _{1⩽\nu ⩽\varrho \left(q\right)}e\left({\displaystyle \frac{s{2}^{\nu }}{q}}\right)\right)}^k,$$

we have

$$\begin{array}{cc}\hfill \mathcal{S}& ⩾{\left({\displaystyle \frac{L}{\varrho \left(q\right)}}+O\left(1\right)\right)}^k{\displaystyle \frac{1}{q}}\left({\varrho }^k\left(q\right)-(q-1)\underset{1⩽s⩽q-1}{max}|\sum _{1⩽\nu ⩽\varrho \left(q\right)}e\left({\displaystyle \frac{s{2}^{\nu }}{q}}\right){|}^k\right)\hfill \\ \hfill & ={\displaystyle \frac{L^k}{q}}\left(1-(q-1){\left({\displaystyle \frac{1}{\varrho \left(q\right)}}\underset{1⩽s⩽q-1}{max}\left|\sum _{1⩽\nu ⩽\varrho \left(q\right)}e\left({\displaystyle \frac{s{2}^{\nu }}{q}}\right)\right|\right)}^k\right)+O\left(L^{k-1}\right).\hfill \end{array}$$

It should be noted that $q=273$, and hence $\varrho \left(q\right)=12$. It is easy to check that

$$\underset{1⩽s⩽q-1}{max}\left|\sum _{1⩽\nu ⩽\varrho \left(q\right)}e\left({\displaystyle \frac{s{2}^{\nu }}{q}}\right)\right|<6.00001,$$

and

$$(q-1){\left({\displaystyle \frac{1}{\varrho \left(q\right)}}\underset{1⩽s⩽q-1}{max}\left|\sum _{1⩽\nu ⩽\varrho \left(q\right)}e\left({\displaystyle \frac{s{2}^{\nu }}{q}}\right)\right|\right)}^{23}⩽0.000032425.$$

Therefore

$$\mathcal{S}⩾{\displaystyle \frac{(1-0.000032425)L^k}{q}}+O\left(L^{k-1}\right),$$

and

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}\mathfrak{S}\left(n\right)& ⩾0.98819024\sum _{1⩽j⩽q}\left(\prod _{p\in \mathcal{P}}(1+A(j,p))\right){\displaystyle \frac{(1-0.000032425)L^k}{q}}+O\left(L^{k-1}\right)\hfill \\ \hfill & ={\displaystyle \frac{0.98819024(1-0.000032425)L^k}{q}}\prod _{p\in \mathcal{P}}\left(\sum _{j=1}^p(1+A(j,p))\right)+O\left(L^{k-1}\right).\hfill \end{array}$$

Observing that

$$\sum _{j=1}^p(1+A(j,p))=p+{\displaystyle \frac{1}{{(p-1)}^7}}\sum _{a=1}^{p-1}{C}_2(p,a)C_3^6(p,a)\sum _{j=1}^{p}e\left(-{\displaystyle \frac{aj}{q}}\right)=p,$$

one has

$$\sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}\mathfrak{S}\left(n\right)⩾0.98815819L^k+O\left(L^{k-1}\right).$$

(13)

This completes the proof of Lemma 2. □

Lemma 3.

For $(1-\eta )N⩽n⩽N$ and $\mathfrak{J}\left(n\right)$ defined by (10), we have

$$\mathfrak{J}\left(n\right)⩾213.3129454{\displaystyle \frac{{\eta }^{2/3}{(1-\eta )}^{13/18}{N}^{25/18}}{4^{19/9}}}{L}^k.$$

(14)

Proof. 

The sum $\mathfrak{J}\left(n\right)$ is defined over the domain which can be represented as

$$\begin{array}{cc}\hfill \mathcal{D}=& \{(m_1,\cdots ,m_7):U_2^2<m_1⩽4U_2^2,U_3^3<m_2,m_3⩽8U_3^3,P_3^3<m_4,m_5⩽8P_3^3,\hfill \\ \hfill & V_3^3<m_6,m_7⩽8V_3^3,n=m_1+m_2+\cdots +m_7\}.\hfill \end{array}$$

To provide the lower bound for $\mathfrak{J}\left(n\right)$, we define

$$\begin{array}{cc}\hfill {\mathcal{D}}^{*}=& \{(m_1,\cdots ,m_7):U_2^2<m_1⩽{\displaystyle \frac{1-14\eta }{2}}N,U_3^3<m_3⩽7U_3^3+{\displaystyle \frac{3\eta }{2}}N,U_3^3<m_2⩽8U_3^3,\hfill \\ \hfill & m_2+m_3⩽9U_3^3,P_3^3<m_4,m_5⩽8P_3^3,V_3^3<m_6,m_7⩽8V_3^3,\hfill \\ \hfill & m_2=n-m_1-m_3-m_4-m_5-m_6-m_7\}.\hfill \end{array}$$

By the definition of ${\mathcal{D}}^{*}$, one deduces that

$$\begin{array}{c}\hfill U_3^3<n-m_1-m_3-5\eta N⩽n-m_1-m_3-m_4-m_5-m_6-m_7=m_2⩽9U_3^3-m_3=8U_3^3.\end{array}$$

Therefore, ${\mathcal{D}}^{*}$ is a subset of $\mathcal{D}$. By the elementary estimate

$$\sum _{\alpha <m⩽\beta }{m}^{-\nu }⩾(1-\eta ){(1-\nu )}^{-1}({\beta }^{1-\nu }-{\alpha }^{1-\nu }),$$

we derive that

$$\begin{array}{cc}\hfill \mathfrak{J}\left(n\right)⩾& \sum _{(m_1,\cdots ,m_7)\in {\mathcal{D}}^{\ast }}{m}_1^{-1/2}{\left(m_2{m}_3{m}_4{m}_5{m}_6{m}_7\right)}^{-2/3}\hfill \\ \hfill ⩾& \sum _{U_2^2<m_1⩽4U_2^2}{m}_1^{-1/2}\times \sum _{P_3^3<m_4,m_5⩽8P_3^3}{\left(m_4{m}_5\right)}^{-2/3}\times \sum _{V_3^3<m_6,m_7⩽8V_3^3}{\left(m_6{m}_7\right)}^{-2/3}\hfill \\ \hfill & \times \sum _{U_3^3<m_3⩽7U_3^3+3\eta N/2}{\left((9U_3^3-m_3)m_3\right)}^{-2/3}\hfill \\ \hfill ⩾& {(1-\eta )}^5·2·3^4·\left({\left({\displaystyle \frac{1-14\eta }{2}}N\right)}^{1/2}-{\left({\displaystyle \frac{1-\eta }{8}}N\right)}^{1/2}\right)P_3^2{V}_3^2\hfill \\ \hfill & \times \sum _{U_3^3<m_3⩽7U_3^3+3\eta N/2}{\left((9U_3^3-m_3)m_3\right)}^{-2/3}\hfill \\ \hfill ⩾& {(1-\eta )}^5·2·3^4·(1-\eta )\left(2^{-1/2}-8^{-1/2}\right)N^{1/2}{P}_3^2{V}_3^2\hfill \\ \hfill & \times \sum _{U_3^3<m_3⩽7U_3^3+3\eta N/2}{\left((9U_3^3-m_3)m_3\right)}^{-2/3}.\hfill \end{array}$$

(15)

Regarding the sum on the right side of the above inequality, applying partial summation yields the following:

$$\begin{array}{cc}\hfill \sum _{U_3^3<m_3⩽7U_3^3+3\eta N/2}{\left((9U_3^3-m_3)m_3\right)}^{-2/3}⩾& (1-\eta ){\int }_{U_3^3}^{7U_3^3+{\displaystyle \frac{3\eta }{2}}N}{\left((9U_3^3-u)u\right)}^{-2/3}\mathrm{d}u\hfill \\ \hfill ⩾& (1-\eta )U_3^{-1}{\int }_1^7{\left((9-t)t\right)}^{-2/3}\mathrm{d}t\hfill \\ \hfill ⩾& 0.931080434U_3^{-1},\hfill \end{array}$$

which, combined with (15) and the identity $N={(1-\eta )}^{-1}16U_3^3$, implies that

$$\begin{array}{cc}\hfill \mathfrak{J}\left(n\right)⩾& {(1-\eta )}^5·2·3^4·(1-\eta )\left(2^{-1/2}-8^{-1/2}\right)·0.931080434U_3^{-1}{N}^{1/2}{P}_3^2{V}_3^2\hfill \\ \hfill ⩾& 213.3129454·P_3^2{U}_3^{1/2}{V}_3^2.\hfill \end{array}$$

This completes the proof of Lemma 3. □

For $k=23$, it should be noted that

$$H^k\left(\alpha \right)e(-N\alpha )=\sum _{n\in \Xi (N,k)}e(-n\alpha ),$$

which combined with (11) and (14) yields

$$\begin{array}{cc}\hfill I(k,\mathfrak{M},N)=& \sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}{\int }_{\mathfrak{M}}F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)e(-n\alpha )\mathrm{d}\alpha \hfill \\ \hfill =& \sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}\left({\displaystyle \frac{1}{2·3^6}}\mathfrak{S}\left(n\right)\mathfrak{J}\left(n\right)+O\left(N^{25/18}{L}^{-1}\right)\right)\hfill \\ \hfill ⩾& {\displaystyle \frac{1}{1458}}\times 213.3129454\times P_3^2{U}_3^{1/2}{V}_3^2\times \sum _{\begin{array}{c}n\in \Xi (N,k)\\ n\equiv 1(mod2)\end{array}}\mathfrak{S}\left(n\right)+O\left(N^{25/18}{L}^{k-1}\right)\hfill \\ \hfill ⩾& 0.14457265·P_3^2{U}_3^{1/2}{V}_3^2{L}^k.\hfill \end{array}$$

This completes the proof of Proposition 1.

4. Proof of Proposition 2

In this section, we present the upper bound estimate for $\mathcal{I}(k,\mathfrak{m},N)$. To achieve this, we further divide the minor arc $\mathfrak{m}$. We propose the following definition:

$$\mathcal{E}\left(\lambda \right)=\left\{\alpha \in (0,1]:\left|H\right(\alpha \left)\right|⩾\lambda L\right\}.$$

(16)

Then, we have

$$\begin{array}{cc}\hfill I(k,\mathfrak{m},N)=& {\int }_{\mathfrak{m}}F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha \hfill \\ \hfill =& \left({\int }_{\mathfrak{m}\setminus \mathcal{E}\left(\lambda \right)}+{\int }_{\mathfrak{m}\cap \mathcal{E}\left(\lambda \right)}\right)F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha \hfill \\ \hfill =& I(k,\mathfrak{m}\setminus \mathcal{E}(\lambda ),N)+I(k,\mathfrak{m}\cap \mathcal{E}(\lambda ),N).\hfill \end{array}$$

Lemma 4.

Suppose that α is a real number, and that there exist integers $a\in \mathbb{Z}$ and $q\in \mathbb{N}$ satisfying

$$(a,q)=1,1⩽q⩽N^{3/4},|q\alpha -a|⩽N^{-3/4}.$$

Then, we have

$$F\left(\alpha \right)\ll U_2^{1-1/8+\epsilon }+{\displaystyle \frac{U_2^{1+\epsilon }}{\sqrt{q(1+N|\alpha -a/q\left|\right)}}}.$$

Proof. 

See Theorem 3 of Kumchev [16]. □

Lemma 5.

For $\alpha \in \mathfrak{m}$, we have

$$F\left(\alpha \right)\ll N^{191/432+\epsilon }.$$

Proof. 

By Dirichlet’s lemma on rational approximation (e.g., see Lemma 2.1 of Vaughan [14]), each real number $\alpha \in \mathfrak{m}$ can be written as

$$\alpha ={\displaystyle \frac{a}{q}}+\lambda ,(a,q)=1,\left|\lambda \right|⩽{\displaystyle \frac{1}{q{\mathcal{Q}}_0}},$$

with $1⩽q⩽{\mathcal{Q}}_0=N^{3/4}$. Since $\alpha \in \mathfrak{m}$, it follows that $q>Q_1$ or $q⩽Q_1=N^{25/216-2\epsilon }$ with $\left|\lambda \right|>{\left(q{Q}_2\right)}^{-1}$. Therefore, one always has

$$q(1+N|\lambda \left|\right)>min\left(Q_1,N{Q}_2^{-1}\right)\gg N^{25/216-2\epsilon }.$$

By Lemma 4, we obtain

$$\underset{\alpha \in \mathfrak{m}}{sup}\left|F\left(\alpha \right)\right|\ll U_2^{1-1/8+\epsilon }+{\displaystyle \frac{U_2^{1+\epsilon }}{\sqrt{N^{25/216-2\epsilon }}}}\ll N^{191/432+\epsilon },$$

which completes the proof of Lemma 5. □

For $\mathcal{A}\subseteq (U_3,2U_3]\cap \mathbb{N}$, where N denotes the set of natural numbers, and we define

$$g\left(\alpha \right)=g_{\mathcal{A}}\left(\alpha \right)=\sum _{n\in \mathcal{A}}(logn)e\left(n^3\alpha \right),$$

and the multiplicative function $w_3\left(q\right)$ by

$$w_3\left(q\right)=\left\{\begin{array}{cc}3p^{-u-1/2},\hfill & \mathrm{if}u⩾0,v=1,\hfill \\ p^{-u-1},\hfill & \mathrm{if}u⩾0,2⩽v⩽3,\hfill \end{array}\right.$$

we then have the following lemma:

Lemma 6.

Let $\mathcal{M}$ denote the union of the intervals $\mathcal{M}(q,a)$ for $1⩽a⩽q⩽P_3^{3/4}$ and $(a,q)=1$, where

$$\mathcal{M}(q,a)=\left\{\alpha :|q\alpha -a|⩽P_3^{-9/4}\right\}.$$

Let $G\left(\alpha \right)$ and $h\left(\alpha \right)$ be integrable functions of period one. Then, for any measurable set $\mathfrak{m}\subseteq (0,1]$, we have

$${\int }_{\mathfrak{m}}g\left(\alpha \right)G\left(\alpha \right)h\left(\alpha \right)\mathrm{d}\alpha \ll N^{1/3}{\mathcal{J}}_0^{1/4}{\left({\int }_{\mathfrak{m}}{\left|G\left(\alpha \right)\right|}^2\mathrm{d}\alpha \right)}^{1/4}{\mathcal{J}}^{1/2}\left(\mathfrak{m}\right)+N^{7/24+\epsilon }\mathcal{J}\left(\mathfrak{m}\right),$$

where

$$\mathcal{J}\left(\mathfrak{m}\right)={\int }_{\mathfrak{m}}\left|G\left(\alpha \right)h\left(\alpha \right)\right|\mathrm{d}\alpha ,{\mathcal{J}}_0=\underset{\beta \in (0,1]}{sup}{\int }_{\mathcal{M}}{\displaystyle \frac{w_3^2\left(q\right){\left|h(\alpha +\beta )\right|}^2}{(1+P_3^3{|\alpha -a/q|)}^2}}\mathrm{d}\alpha .$$

Proof. 

See Lemma 3.1 of Zhao [17]. □

In order to give appropriate an upper bound estimate of $\mathcal{I}(k,\mathfrak{m}\cap \mathcal{E}(\lambda ),N)$, one needs to consider the measure of the set $\mathcal{E}\left(\lambda \right)$. We have the following lemma:

Lemma 7.

Let

$$H_h\left(\alpha \right)=\sum _{n=0}^{h-1}e\left(2^n\alpha \right),\mathcal{F}(\xi ,h)={\displaystyle \frac{1}{2^h}}\sum _{r=0}^{2^h-1}exp\left(\xi ·\Re \left(H_h\left(2^{-h}r\right)\right)\right).$$

Then, it holds that

$$\mathrm{meas}\left(\mathcal{E}\left(\lambda \right)\right)⩽N^{-E\left(\lambda \right)},$$

where

$$E\left(\lambda \right)={\displaystyle \frac{\xi \lambda }{log2}}-{\displaystyle \frac{log\mathcal{F}(\xi ,h)}{hlog2}}-{\displaystyle \frac{\epsilon }{log2}}$$

for any $h\in \mathbb{N},\xi >0$ and $\epsilon >0$.

Proof. 

See the arguments on pp. 562–565 of Heath–Brown and Puchta [18]. □

Lemma 8.

Let $E\left(\lambda \right)$ be defined as in Lemma 7. Then, one has

$$E\left(0.83372131691\right)>{\displaystyle \frac{2}{3}}+{10}^{-10}.$$

Proof. 

Taking $\xi =1.0835751541553,h=40$ as in Lemma 7, we deduce the desired conclusion immediately. □

Lemma 9.

Let $F\left(\alpha \right),S_1\left(\alpha \right),S_2\left(\alpha \right)$ and $T\left(\alpha \right)$ be defined as in (7). Then we have

$$\begin{array}{cc}\hfill \left(\mathrm{i}\right)& {\int }_0^1\left|S_1^4\left(\alpha \right)T^4\left(\alpha \right)\right|\mathrm{d}\alpha ⩽7.3909U_3{V}_3^4;\hfill \\ \hfill \left(\mathrm{ii}\right)& {\int }_0^1\left|F^2\left(\alpha \right)T^4\left(\alpha \right)\right|\mathrm{d}\alpha \ll N^{10/9+\epsilon };\hfill \\ \hfill \left(\mathrm{iii}\right)& {\int }_0^1\left|F^2\left(\alpha \right)S_i^2\left(\alpha \right)T^2\left(\alpha \right)\right|\mathrm{d}\alpha \ll N^{11/9+\epsilon },(i=1,2).\hfill \end{array}$$

Proof. 

For (i), one can see Lemma 2.6 of Liu [19]. For (ii) and (iii), one can see Lemma 2.3 of Liu [20]. □

Lemma 10.

Let $F\left(\alpha \right),S_2\left(\alpha \right)$ be defined as in (7). Then we have

$$\begin{array}{c}\hfill {\int }_0^1\left|F^2\left(\alpha \right)S_2^4\left(\alpha \right)\right|\mathrm{d}\alpha \ll 9.77431043494196P_3^4.\end{array}$$

Proof. 

We can deduce from the arguments on page 417 of [21] that

$${\int }_0^1\left|F^2\left(\alpha \right)S_2^4\left(\alpha \right)\right|\mathrm{d}\alpha ⩽(8+o\left(1\right))·\mathfrak{S}·\tilde{\mathfrak{I}},$$

(17)

where

$$\mathfrak{S}=\sum _{q=1}^{\infty }\mathcal{A}\left(q\right),\mathcal{A}\left(q\right)={\displaystyle \frac{1}{{\phi }^6\left(q\right)}}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q|C_2^2(q,a)C_3^4(q,a)|,$$

$$\tilde{\mathfrak{I}}={\displaystyle \frac{1}{2^2·3^4}}\sum _{\begin{array}{c}{m}_1+m_2+m_3=n_1+n_2+n_3\\ P_3^3<m_2,m_3,n_2,n_3⩽8P_3^3\\ U_2^2<m_1,n_1⩽4U_2^2\end{array}}{\left(m_1{n}_1\right)}^{-1/2}{\left(m_2{m}_3{n}_2{n}_3\right)}^{-2/3}.$$

From Lemma 8.3 of Hua [22], we know that $\mathcal{A}\left(q\right)$ is multiplicative and $\mathcal{A}\left(p^j\right)=0$ for $j⩾2$. Therefore, we have

$$\mathfrak{S}=\prod _p\left(1+\mathcal{A}\left(p\right)\right).$$

By Lemma 1, we have

$$1+\mathcal{A}\left(p\right)⩽1+{\displaystyle \frac{{(\sqrt{p}+1)}^2{(2\sqrt{p}+1)}^4}{{(p-1)}^5}}.$$

Then, we obtain

$$\prod _{p⩾301}(1+\mathcal{A}\left(p\right))<1.0093278341.$$

By simple calculation, it is easy to see that

$$\prod _{2⩽p⩽300}(1+\mathcal{A}\left(p\right))<3.49275743638.$$

Thus, we obtain

$$\mathfrak{S}=\left(\prod _{2⩽p⩽300}\left(1+\mathcal{A}\left(p\right)\right)\right)\left(\prod _{p⩾301}\left(1+\mathcal{A}\left(p\right)\right)\right)<3.52533729815.$$

(18)

For $\tilde{\mathfrak{I}}$, by noting that

$$m_1=n_1+n_2+n_3-m_2-m_3⩾n_1+2P_3^3-16P_3^3⩾\left(1-{\displaystyle \frac{29}{8}}\eta \right)n_1,$$

and

$$\sum _{P_3^3<m⩽8P_3^3}{m}^{-2/3}\sim 3P_3,\sum _{U_2^2<m⩽4U_2^2}{m}^{-1}\sim 2log2,$$

one deduces that

$$\begin{array}{cc}\hfill \tilde{\mathfrak{I}}⩽& {\displaystyle \frac{1}{2^2·3^4}}\sum _{\begin{array}{c}{m}_1+m_2+m_3=n_1+n_2+n_3\\ P_3^3<m_2,m_3,n_2,n_3⩽8P_3^3\\ U_2^2<m_1,n_1⩽4U_2^2\end{array}}{\left(1-{\displaystyle \frac{29}{8}}\eta \right)}^{-1/2}{n}_1^{-1}{\left(m_2{m}_3{n}_2{n}_3\right)}^{-2/3}\hfill \\ \hfill ⩽& {\displaystyle \frac{1}{2^2·3^4}}\left(1+{\displaystyle \frac{29}{8}}\eta \right)·(2log2)·{\left(3P_3\right)}^4⩽\left({\displaystyle \frac{log2}{2}}+2\eta \right)P_3^4.\hfill \end{array}$$

(19)

Combining (17), (18) and (19), we have

$$\begin{array}{c}\hfill {\int }_0^1|F^2\left(\alpha \right)S_2^4\left(\alpha \right)|\mathrm{d}\alpha ⩽9.77431044P_3^4.\end{array}$$

This completes the proof of Lemma 10. □

It follows from Lemma 9, Lemma 10 and the Cauchy–Schwarz inequality that

$$\begin{array}{cc}\hfill {\int }_0^1|F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)|\mathrm{d}\alpha ⩽& {\left({\int }_0^1|F^2\left(\alpha \right)S_2^4\left(\alpha \right)|\mathrm{d}\alpha \right)}^{1/2}{\left({\int }_0^1|S_1^4\left(\alpha \right)T^4\left(\alpha \right)|\mathrm{d}\alpha \right)}^{1/2}\hfill \\ \hfill ⩽& {\left(9.77431044P_3^4\right)}^{1/2}·{\left(7.3909U_3{V}_3^4\right)}^{1/2}\hfill \\ \hfill ⩽& 8.499467692P_3^2{U}_3^{1/2}{V}_3^2,\hfill \end{array}$$

which, combined with the definition of $\mathcal{E}\left(\lambda \right)$ in (16), yields:

$$\begin{array}{cc}\hfill & \left|{\int }_{\mathfrak{m}\setminus \mathcal{E}\left(\lambda \right)}F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha \right|\hfill \\ \hfill ⩽& {\left(\lambda L\right)}^k·{\int }_0^1\left|F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)\right|\mathrm{d}\alpha ⩽8.499467692{\lambda }^k{P}_3^2{U}_3^{1/2}{V}_3^2{L}^k.\hfill \end{array}$$

(20)

Lemma 11.

For $\lambda =0.83372131691$, we have

$${\int }_{\mathfrak{m}\cap \mathcal{E}\left(\lambda \right)}F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha \ll P_3^2{U}_3^{1/2}{V}_3^2{L}^{k-1}.$$

(21)

Proof. 

It follows from Hölder’s inequality that

$$\begin{array}{cc}\hfill & {\int }_{\mathfrak{m}\cap \mathcal{E}\left(\lambda \right)}F\left(\alpha \right)S_1^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)H^k\left(\alpha \right)e(-N\alpha )\mathrm{d}\alpha \hfill \\ \hfill \ll & L^k{\left({\int }_0^1\left|S_1^4\left(\alpha \right)T^4\left(\alpha \right)\right|\mathrm{d}\alpha \right)}^{1/4}{\left({\int }_0^1|S_2\left(\alpha \right){|}^8\mathrm{d}\alpha \right)}^{3/16}\hfill \\ \hfill & \times {\left({\int }_{\mathfrak{m}}\left|F\left(\alpha \right){|}^2\right|S_2\left(\alpha \right)\left|\right|S_1\left(\alpha \right){|}^2|T\left(\alpha \right){|}^2\mathrm{d}\alpha \right)}^{1/2}{\left({\int }_{\mathcal{E}\left(\lambda \right)}1\mathrm{d}\alpha \right)}^{1/16}.\hfill \end{array}$$

(22)

Taking

$$g\left(\alpha \right)=S_1\left(\alpha \right),h\left(\alpha \right)=T\left(\alpha \right),G\left(\alpha \right)={\left|F\left(\alpha \right)\right|}^2\left|S_2\left(\alpha \right)\right|S_1(-\alpha )T(-\alpha )$$

in Lemma 6, we deduce that

$$\begin{array}{cc}\hfill & {\int }_{\mathfrak{m}}|F\left(\alpha \right){|}^2|S_2\left(\alpha \right)||S_1\left(\alpha \right){|}^2|T\left(\alpha \right){|}^2\mathrm{d}\alpha \hfill \\ \hfill \ll & N^{{\displaystyle \frac{1}{3}}}{\mathcal{J}}_0^{{\displaystyle \frac{1}{4}}}{\left({\int }_{\mathfrak{m}}{|F\left(\alpha \right)|}^4|S_1{\left(\alpha \right)|}^2|S_2{\left(\alpha \right)|}^2{|T\left(\alpha \right)|}^2\mathrm{d}\alpha \right)}^{{\displaystyle \frac{1}{4}}}{\left({\int }_{\mathfrak{m}}{|F\left(\alpha \right)|}^2|S_1\left(\alpha \right)||S_2{\left(\alpha \right)||T\left(\alpha \right)|}^2\mathrm{d}\alpha \right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +N^{{\displaystyle \frac{7}{24}}+\epsilon }{\int }_{\mathfrak{m}}{|F\left(\alpha \right)|}^2|S_1\left(\alpha \right)||S_2{\left(\alpha \right)||T\left(\alpha \right)|}^2\mathrm{d}\alpha ,\hfill \end{array}$$

(23)

where

$${\mathcal{J}}_0=\underset{\beta \in (0,1]}{sup}\sum _{q⩽P_3^{3/4}}\sum _{\begin{array}{c}a=1\\ (a,q)=1\end{array}}^q{\int }_{\mathcal{M}(q,a)}{\displaystyle \frac{w_3^2\left(q\right){\left|h(\alpha +\beta )\right|}^2}{(1+P_3^3{|\alpha -a/q|)}^2}}\mathrm{d}\alpha$$

with

$$\mathcal{M}(q,a)=\left\{\alpha :|q\alpha -a|⩽P_3^{-9/4}\right\}.$$

It follows from analogous arguments, referring to Lemma 6.2 of Zhang and Li [13], that

$${\mathcal{J}}_0\ll V_3^2{P}_3^{-3}{(logN)}^c\ll N^{-4/9+\epsilon }.$$

(24)

By Lemma 2.5 of Vaughan [14], Lemma 5, Lemma 9 and Hölder’s inequality, one derives that

$$\begin{array}{cc}\hfill & {\int }_{\mathfrak{m}}\left|F\left(\alpha \right){|}^4\right|S_1\left(\alpha \right){|}^2\left|S_2\left(\alpha \right){|}^2\right|T\left(\alpha \right){|}^2\mathrm{d}\alpha \hfill \\ \hfill \ll & \left(\underset{\alpha \in \mathfrak{m}}{sup}|F\left(\alpha \right){|}^3\right){\left({\int }_0^1|S_1\left(\alpha \right){|}^8\mathrm{d}\alpha \right)}^{1/4}{\left({\int }_0^1|S_2\left(\alpha \right){|}^8\mathrm{d}\alpha \right)}^{1/4}{\left({\int }_0^1\left|F^2\left(\alpha \right)T^4\left(\alpha \right)\right|\mathrm{d}\alpha \right)}^{1/2}\hfill \\ \hfill \ll & {\left(N^{191/432+\epsilon }\right)}^3·{\left(N^{5/3+\epsilon }\right)}^{1/4}·{\left(N^{5/3+\epsilon }\right)}^{1/4}·{\left(N^{10/9+\epsilon }\right)}^{1/2}\ll N^{391/144+\epsilon }.\hfill \end{array}$$

(25)

In addition, by (iii) of Lemma 9 and Cauchy’s inequality, one also has

$$\begin{array}{cc}\hfill & {\int }_{\mathfrak{m}}\left|F\left(\alpha \right){|}^2\right|S_1\left(\alpha \right)\left|\right|S_2\left(\alpha \right)\left|\right|T\left(\alpha \right){|}^2\mathrm{d}\alpha \hfill \\ \hfill \ll & {\left({\int }_0^1\left|F^2\left(\alpha \right)S_1^2\left(\alpha \right)T^2\left(\alpha \right)\right|\mathrm{d}\alpha \right)}^{1/2}{\left({\int }_0^1\left|F^2\left(\alpha \right)S_2^2\left(\alpha \right)T^2\left(\alpha \right)\right|\mathrm{d}\alpha \right)}^{1/2}\ll N^{11/9+\epsilon }.\hfill \end{array}$$

(26)

It follows from (23)–(26) that

$${\int }_{\mathfrak{m}}\left|F\left(\alpha \right){|}^2\right|S_2\left(\alpha \right)\left|\right|S_1\left(\alpha \right){|}^2|T\left(\alpha \right){|}^2\mathrm{d}\alpha \ll N^{109/72+\epsilon },$$

(27)

which, combined with (22), Lemma 8, (i) of lemma 9 and Lemma 2.5 of Vaughan [14], yields (21). This completes the proof of Lemma 11. □

From (20) and Lemma 11, we derive the result of Proposition 2.

5. Proof of Theorem 1

Taking $\lambda =0.83372131691,k=23$ in Proposition 1 and Proposition 2, we can deduce that

$$\begin{array}{cc}\hfill \mathcal{R}(k,N)⩾& \mathcal{I}(k,\mathfrak{M},N)-\left|\mathcal{I}\right(k,\mathfrak{m},N\left)\right|\hfill \\ \hfill ⩾& (0.14457265-8.499467692\times 0.{83372131691}^k)P_3^2{U}_3^{1/2}{V}_3^2{L}^k\hfill \\ \hfill ⩾& 0.01489P_3^2{U}_3^{1/2}{V}_3^2{L}^k.\hfill \end{array}$$

This completes the proof of Theorem 1.

6. Conclusions

In this paper, we have established an asymptotic formula for the Goldbach–Linnik type problem concerning the symmetric mixed powers of primes and powers of 2. Specifically, we have shown that every sufficiently large odd integer can be represented as the sum of one prime square, six cubes of primes, and 23 powers of two. This result extends the classical Goldbach conjecture by incorporating higher powers and mixed types, offering a broader understanding of the additive properties of primes and their powers.

Our findings contribute to analytic number theory by providing deeper insights into the behavior of primes in additive problems. The innovative techniques developed, such as the application of the circle method, pave the way for future research. The symmetry observed in the mixed powers of primes suggests an underlying structure that enhances our understanding of prime distribution.

Practical advantages include potential applications in cryptographic algorithms [23] and computational number theory, aiding efficient algorithm design [24]. This study underscores the importance of cross-disciplinary approaches in advancing mathematical theories.

In conclusion, our work addresses a significant problem in additive number theory and lays the groundwork for future research. It highlights the potential for discovering new mathematical phenomena through the interplay of primes and their powers, enriching our understanding of fundamental mathematical structures.