A pair of equations in unlike powers of primes and powers of 2

1 Introduction

As an approximation to Goldbach’s problem, Linnik proved in 1951 [1] under the assumption of the Generalized Riemann Hypothesis (GRH), and later in 1953 [2] unconditionally, that each large even integer N is a sum of two primes p 1 , p 2 and a bounded number of powers of 2, namely

In 2002, Heath-Brown and Puchta [3] applied a rather different approach to this problem and showed that k = 13 and, on the GRH, k = 7 . In 2003, Pintz and Ruzsa [4] established this latter result and announced that k = 8 is acceptable unconditionally. Elsholtz, in an unpublished manuscript, which is yet to appear in print, showed that k = 12 ; this was proved independently by Liu and Lü [5].

In 1999, Liu et al. [6] proved that every large even integer N can be written as a sum of four squares of primes and a bounded number of powers of 2, namely

And Platt and Trudgian [7] got that k = 45 suffices.

In 2001, Liu and Liu [8] proved that every large even integer N can be written as a sum of eight cubes of primes and a bounded number of powers of 2, namely

The acceptable value k = 330 was determined by Platt and Trudgian [7].

In 2011, Liu and Lü [9] considered a hybrid problem of (1.1)–(1.3),

They showed that k = 161 is acceptable and Platt and Trudgian [7] revised it to 156.

As a generalization, recently, Hu and Yang [10] first considered the simultaneous representation of pairs of positive even integers N 2 ≫ N 1 > N 2 , in the form

where k is a positive integer. They proved that the simultaneous Eq. (1.5) is solvable for k = 455 .

The primary purpose of this article is to sharpen this result considerably by establishing the following theorem.

Our proof of Theorem 1.1 uses the Hardy-Littlewood circle method. We make a new estimate of minor arcs and draw on some strategies adopted in the works of Hu and Yang [10] and Kong and Liu [11].

Throughout this article, the letter ϵ denotes a positive constant, which is arbitrarily small but may not the same at different occurrences.

2 The proof of Theorem 1.1

Throughout this article, we assume that N i , i = 1 , 2 are sufficiently large even integers satisfying N 2 ≫ N 1 > N 2 . Then, we set

for i = 1 , 2 .

We define the major arcs M 1 , M 2 and minor arcs C ( M 1 ) , C ( M 2 ) as usual, namely

where i = 1 , 2 and

It follows from the definition of P i and Q i that the arcs M 1 ( a , q ) and M 2 ( a , q ) are mutually disjoint, respectively. We further define

As in [10], let δ = 10 − 4 and

for i = 1 , 2 . We set

and set

for i = 1 , 2 .

Let

be the weighted number of solutions of (1.5) in ( p 1 , … , p 8 , v 1 , … , v k ) with

for j = 1 , 2 , … , k . Then, R ( N 1 , N 2 ) can be written as follows:

We will establish Theorem 1.1 by estimating the term R 1 ( N 1 , N 2 ) , R 2 ( N 1 , N 2 ) , and R 3 ( N 1 , N 2 ) . We need to show that R ( N 1 , N 2 ) > 0 for every pair of sufficiently large positive even integers N 2 ≫ N 1 > N 2 .

Let

for i = 1 , 2 , 3 , 4 . Then, we write

Now, we need to quote two lemmas from Hu and Yang [10, Lemmas 2.3 and 2.4] as follows:

Then, R 2 ( N 1 , N 2 ) is

where we use the trivial bound of G ( α 1 + α 2 ) . We note that

where we use the bound from [10, Lemma 2.5] and [13, Lemma 2.5] as follows:

By using the integral transformation of β = α 1 + α 2 and the periodicity of G ( β ) , we have

By the simple orthogonality and [14, Lemma 4.12], we have

and

From Lemma 2.4 and Cauchy’s inequality, we have

We choose λ = 0.9322 and get

since N 2 ≫ N 1 > N 2 . Similarly,

Then,

Following the lines in [14], we note that

where

Let

where

As in the proof of [14, Lemma 3.2], we treat the case h ≠ 0 and h = 0 separately and obtain

Next, we estimate ∑ h ≠ 0 r 14 ( h ) S 2 ( h ) . Note that

Then, from the proof of [14, Lemma 4.3] we have

where

Define the multiplicative function c ( d ) by

where d is square-free and ( 30 , d ) = 1 . Then,

where the condition ( h ) in ∑ ( h ) denotes that the summation is taken over all ν 1 , … , ν 14 , μ 1 , … , μ 14 satisfying 4 ⩽ ν j , μ j ⩽ L and h = ∑ j = 1 14 ( 2 ν j − μ j ) ≠ 0 .

Let us consider Σ 1 . We have

where ϱ ( q ) denotes the smallest positive integer ϱ , such that 2 ϱ ( q ) ≡ 1 mod q and c 1 is a constant, which we will deal with later. Similarly, Σ 2 ⩽ ( c 2 + ϵ ) L 28 , where c 2 is a constant. Set

and m ( x ) = ∏ e ⩽ x 2 e − 1 . Then, we have

where we use the fact m ( x ) / ϕ ( m ( x ) ) ⩽ e γ log x for x ⩾ 9 . Here, c 3 = ∏ p > 5 1 + 1 c ( p ) 1 + 1 p − 1 2 ⩽ 1.3904 2 can be found in the proof of [14, Lemma 4.1]. With M = 35 , we have

The constant c 2 ⩽ 1.10302 can be handled in the similar way. Then, the numerical computations provide

So we get

Next, by [11, Lemma 2.3] and [9, Lemma 2.5] we have