On pairs of equations with unequal powers of primes and powers of 2

1.   Introduction

The Goldbach conjecture is one of the most famous problems, and numerous variations have been derived from it. In the 1950s, Linnik ^[9,10] showed that every sufficiently large even integer can be represented as a sum of two primes and $K$ powers of 2, where $K$ is an absolute constant. In 1975, Gallagher ^[2] established an asymptotic formula for the number of such representations. In 1998, the explicit value of $K$ was first obtained by Liu, Liu, and Wang ^[11]. They showed that $K = 54000$ is acceptable. Afterwards, many mathematicians improved the value of $K$ (see ^{[4,7,8,12,15,16,19]}). The best result so far is due to Pintz and Ruzsa ^[16], who proved that $K = 8$ is acceptable.

In 2017, motivated by the works of Linnik ^[9,10], Liu ^[13] studied a Goldbach–Linnik problem with unequal powers of primes. To be specific, he considered the problem on the representation of the large even integer $N$ in the form

where $p_i$ are prime numbers and $v_j$ are positive integers. He proved that (1.1) is solvable for $k = 41$. Subsequently, the acceptable value of $k$ was successively refined by Lü ^[14], Zhao ^[23], and Zhang ^[20]. Very recently, based on the work of ^[21,22], Zhu ^[24] further improved the result to $k = 17$.

On the other hand, in 2023, Huang ^[5] studied Eq (1.1) in an extended way. He attempted to simultaneously represent pairs of positive even integers $N_1$ and $N_2$ with $N_2 < N_1\ll N_2$, in the form

In ^[5], he proved that the simultaneous equations (1.2) are solvable for $k = 105$. In 2024, Han, Liu, and Yue ^[3] improved the value of $k$ to 36.

In this paper, we shall continue to improve the results of ^[3] and ^[24] and establish the following sharper results:

Theorem 1. For $k = 30$, the simultaneous equations (1.2) are solvable for every sufficiently large positive even integers $N_1$ and $N_2$ satisfying $N_2 < N_1\ll N_2$.

Theorem 2. For $k = 14$, the Eq (1.1) is solvable for every sufficiently large positive even integer $N$.

2.   Notation and outline of the method

In this paper, we assume that $N_1$ and $N_2$ are sufficiently large even integers satisfying $N_2 < N_1\ll N_2$. We fix a positive constant $\eta$ satisfying $\eta\leq 10^{-100}$. Let $\varepsilon$ be an arbitrarily small positive number, and the value of $\varepsilon$ may change from line to line. The letter $p$, with or without a subscript, is reserved for a prime number. As usual, we use $e(\alpha)$ to denote $e^{2\pi i\alpha}$, and $\varphi(n)$ stands for the Euler function. Moreover, we write

In order to apply the circle method, we set

Then we can define

where

Now let

be the weighted number of solutions of (1.2) in $(p_1, \dots, p_{12}, v_1, \dots, v_k)$ with

Then by orthogonality, we have

For $k\geq 30$, we shall prove

where $I$ is the positive constant defined as (3.4).

3.   The lower bound for $\mathcal{R}_1(k, N_1, N_2)$

We first state some auxiliary results. Let

Lemma 3.1. Let $\mathfrak{M}_j$ be defined as (2.3). Then we have

where $\mathfrak{S}(n)\gg 1$ for $n \equiv 0\pmod 2$.

Proof. Let

It follows from [13, Lemma 2.1] that

and $\mathfrak{S}(n)\gg 1$ for $n\equiv 0\pmod 2$. Therefore, it suffices to show that

Write

Then we can deduce from the orthogonality that

where the elementary estimates

are used. Since $u_{k, j}(\lambda) = {\frac{1}{k}}N_j^{\frac{1}{k}}\int_{{\frac{1}{6}}-\eta}^{{\frac{1}{6}}+\eta}e(tN_j\lambda)t^{{\frac{1}{k}}-1}dt = {\frac{1}{k}}N_j^{\frac{1}{k}}g_k(N_j\lambda)$. Then we have

where the bound $g_k(\lambda)\ll |\lambda|^{-1}$ (see [17, Lemma 4.3]) is used. Now (3.1) follows from (3.2) and (3.3).

Lemma 3.2. Let

Suppose that ${\frac{n}{N_j}} = 1+O({\frac{1}{\log N_j}})$. Then we have

Proof. For (i), it is easy to see that

For (ii), write $t_6 = s-\sum\limits_{i = 1}^{5}t_i$, then we have

where

Note that $\int_{-R}^{R}e(\lambda u)d\lambda = {\frac{\sin2\pi Ru}{\pi u}}$. Thus

Since $\phi(s)$ is bounded and $1-6\eta < {\frac{n}{N_j}} = 1+O({\frac{1}{\log N_j}}) < 1+6\eta$. Hence, we can deduce from the Fourier's integral theorem (see [1, P.22]) that

For $\phi({\frac{n}{N_j}})$, by the trivial estimate, we can obtain

Let $t_i = u_i+{\frac{1}{6}}$, then we have

where $I$ is defined as (3.4) and

From the condition $1-{\frac{n}{N_j}} = O({\frac{1}{\log N_j}})$, we can obtain

Now by combining (3.5)–(3.9), we obtain

Lemma 3.3. Suppose that $(a, p) = 1$. Then we have

Proof. It follows easily from [18, Lemma 4.3].

Lemma 3.4. We have

Proof. For $p = 17$ or $23\leq p\leq 199$, we can directly calculate $\min\limits_{1\leq n\leq p}(1+A(n, p))$ by computer and obtain that

Thus

For $199 < p\leq 10^5$, if $p\equiv 2\pmod 3$ and $(a, p) = 1$, then we can deduce from Lemma 3.3 (i) and (ii) that

If $p\equiv 1\pmod 3$, then by Lemma 3.3 (i), we have

Combining (3.11) and (3.12), we can deduce from numerical calculation that

For $p > 10^5$, it follows from [13, Section 3, P.443] that

Now we can conclude from (3.10) and (3.13)–(3.14) that

Lemma 3.5. Let

Then for $N_1 \equiv N_2\equiv 0\pmod 2$, we have

Proof. According to [13,(3.3)], we have

Set $C = 0.9792$ and $\mathcal{P} = \{3, 5, 7, 11, 13, 19\}$. Then by applying Lemma 3.4, we obtain

where the obvious fact $1+A(n, 2) = 2$ for $n \equiv 0\pmod 2$ is used. Write $q = \prod\limits_{ p\in\mathcal{P}}p = 285285$, $t = N_2-N_1$. Thus

Let $S$ denote the innermost sum in (3.17). Since $N_1 \equiv N_2\equiv 0\pmod 2$, we have

Let $\rho(q)$ denote the smallest positive integer $\rho$ such that $2^{\rho}\equiv 1\pmod q.$ Thus

When $q = 285285$ and $k\geq 30$, with the help of a computer, we can verify that $\rho(q) = 180$ and

Therefore, from (3.18) and (3.19), we have

Inserting (3.20) into (3.17), we obtain

Note that $q = \prod\limits_{p\in \mathcal{P}}p$, we can deduce from the Chinese remainder theorem that

By the numerical calculation, we can obtain

Thus, we can conclude from (3.21)–(3.23) that

The proof of (ii) is similar. From (3.16)–(3.17), we have

The innermost sum can be estimated using the same method as in (3.18)–(3.20), except by replacing $k\geq 30$ with $k\geq 14$. By numerical calculation, we have

Similar to (3.22), we have

Moreover, by applying the bound $\sum\limits_{1 \leq j \leq p} e(-\frac{a j}{p}) = 0$ for $a\neq 0\pmod p$, we can obtain

Now by combining (3.25)–(3.28), we can derive that

Proposition 3.1. Suppose that $k\geq 30$ and $N_1 \equiv N_2\equiv 0\pmod 2$. Then we have

Proof. Note that $H^k(\alpha_1+\alpha_2)e(-N_1\alpha_1-\alpha_2N_2) = \sum\limits_{(n_1, n_2)\in \Xi(N_1, N_2, k)\atop{n_1\equiv n_2\equiv 0\pmod 2}}e(-n_1\alpha_1)e(-n_2\alpha_2)$, then by Lemma 3.1, Lemma 3.2 (ii) and Lemma 3.5 (i), we have

where the trivial bound $\sum\limits_{(n_1, n_2)\in \Xi(N_1, N_2, k)\atop{n_1\equiv n_2\equiv 0\pmod 2}}1\ll L^k$ is used.

4.   The upper bound for $\mathcal{R}_2(k, N_1, N_2)$

In this section, we will give the upper bound for $\mathcal{R}_2(k, N_1, N_2)$. For this purpose, we need to introduce a further division of the minor arcs $\mathfrak{m}$. Let

Then we have

In order to estimate $\mathcal{R}_3(k, N_1, N_2, u)$, we define

Lemma 4.1. We have

Proof. See [22, Lemma 3.1].

Lemma 4.2. We have

Proof. It follows from the standard major arcs techniques in the Waring-Goldbach problem.

Lemma 4.3. We have

Proof. Let

Then from [22, Proof of Lemma 4.1, P.417] with $k = 4$, we have

For $J_j^*$, it follows from the same argument leading to (3.2) and (3.3) that

Now by substituting (4.3) into (4.2), we obtain

Lemma 4.4. We have

Proof. By Lemmas 4.2 and 4.3, we have

Lemma 4.5. Let $I$ be defined as (3.4). Then we have

Proof. The proof of Lemma 4.5 is similar to that of Lemma 3.2 (ii). Let $t_6 = t_1-t_2+t_3-t_4+t_5-s$, then we can obtain

where

Applying the Fourier integral theorem, we can obtain

Write $t_i = u_i+{\frac{1}{6}}$. Thus

Making the change of variables $u_2 = -s_2, \, \, \, u_4 = -s_4$, we find that

Now the desired result follows from (4.5)–(4.8).

Proposition 4.1. We have

Proof. Note that $|H(\alpha_1+\alpha_2)| < uL$ for $(\alpha_1, \alpha_2)\in \mathfrak{m}\backslash \mathcal{E}(u)$. Then we have

It is easy to see that $\mathfrak{m}_1\subseteq \mathfrak{m}^*_1, \mathfrak{m}_2\bigcup (\mathfrak{M}_2\backslash\mathfrak{M}^*_2)\subseteq \mathfrak{m}^*_2$ and $\mathfrak{m}_2\bigcap (\mathfrak{M}_2\backslash\mathfrak{M}^*_2) = \emptyset$. Hence

From (4.9)–(4.10) and Lemmas 4.1–4.5, we can obtain

Lemma 4.6. We have

Proof. See [24, Lemma 4.5 and Lemma 4.1].

Lemma 4.7. Let

Write $meas(\mathcal{E}^*(u))$ for the measure of the set $\mathcal{E}^*(u)$. Then we have

Proof. See [6, Lemma 5 and (3.10)].

Proposition 4.2. Let $u = 0.83372131685$. Then we have

Proof. For brevity, we write

From the definition of $\mathfrak{m}$ and $\mathcal{E}{(u)}$, we have

where the trivial bound $H(\alpha_1+\alpha_2)\ll L$ is used. By applying Hölder's inequality, Hua's inequality and Lemma 4.6 (i), (ii), we can obtain

Thus

where we used the integral transformation $\beta = \alpha_1+\alpha_2$. Note that $H(\beta)$ is of period one. Hence

On substituting (4.15) into (4.14), we obtain

where Lemma 4.7 and Lemma 4.6 (iii) are used. In a similar manner, we have

Since $N_2 < N_1\ll N_2$, then we can conclude from (4.12) and (4.16)–(4.17) that

Now we turn to give the estimate for $|\mathcal{R}_2(k, N_1, N_2)|$. Suppose that $k\geq 30$. Then by combining (4.1) and Propositions 4.1–4.2 with $u = 0.83372131685$, we have

5.   Proof of Theorem 1

When $k\geq 30$, by combining Proposition 3.1, (2.4), and (4.18), we can obtain

where Lemma 3.2 (i) is used in the last step. Now the proof of Theorem 1 is completed.

6.   Proof of Theorem 2

We sketch the proof of Theorem 2, since the idea of the proof is similar to that of Theorem 1. We only give the changes that are necessary for our Theorem 2. Let

Suppose that $k\geq 14$ and $u = 0.83372131685$. Then from the orthogonality, we have

Applying Lemmas 3.1–3.2 and Lemma 3.5 (ii), we have

Since $\mathfrak{m}_1\backslash\mathcal{E}^*{(u)}\subset\mathfrak{m}_1^*$, we can deduce from Lemma 4.1 and Lemmas 4.4–4.5 that

Moreover, by Hölder's inequality, Hua's inequality, and Lemmas 4.6–4.7, we have

From (6.1)–(6.4), we can conclude that

Now the proof of Theorem 2 is complete.

Use of Generative-AI tools declaration

The author declares that he has not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The author thanks the referees for their time and comments. The author would like to express the most sincere gratitude to Professor Yingchun Cai for his valuable advice and constant encouragement.

Conflict of interest

The author declares that there is no conflict of interest regarding the publication of this paper.