A Diophantine Problem with Unlike Powers of Primes

Let k be an integer with 4 ≤ k ≤ 6 and η be any real number. Suppose that λ 1 , λ 2 , ... , λ 5 are nonzero real numbers, not all of them have the same sign, and λ 1 / λ 2 is irrational. It is proved that the inequality | λ 1 p 1 + λ 2 p 22 + λ 3 p 33 + λ 4 p 44 + λ 5 p k 5 + η | < ( max 1 ≤ j ≤ 5 p j ) − σ ( k ) has inﬁnitely many solutions in prime variables p 1 ,p 2 , p 3 , p 4 , and p 5 , where 0 < σ ( 4 ) < 1/36 , 0 < σ ( 5 ) < 4/189, and 0 < σ ( 6 ) < 1/54. This gives an improvement of the recent results.


Introduction
e determination of the minimal s such that the Diophantine equation is solvable in positive integers x 1 , . . . , x s , for all sufficiently large integers N is an interesting problem in additive number theory. In 1951, Roth [1] proved that s � 50 is acceptable. is result was subsequently improved by anigasalam et al. [2][3][4], Vaughan and Vaughan [5,6], Brüdern and Brüdern [7,8] and Ford and Ford [9,10]. e best currently known result is due to Ford [10], with s � 14. Schwarz [11] suggested to analyze the related Diophantine inequality. e first result was obtained by Brüdern [12], who showed that the values of at integer points (x 1 , . . . , x 22 ) are dense on the real line provided that λ 2 , . . . , λ 23 are nonzero real numbers and λ 2 /λ 3 is irrational. anks to a pruning technique, Brüdern [13] proved that the values taken by 16 i�1 λ i+1 x i+1 i , at integer points (x 1 , . . . , x 16 ) are dense on the real line if λ 2 , . . . , λ 17 are nonzero real numbers and at least one of the ratios λ i /λ j is irrational.

Preliminaries
We apply the Davenport-Heilbronn circle method (see [26] and Chapter 11 in [27]) to prove eorem 1. Since λ 1 /λ 2 is irrational, there are infinitely many convergents to its continued fraction. Let q be any denominator of a convergent to λ 1 /λ 2 . As in [20], let X run through the sequences: We set where the function ρ(m) is defined by 5.2 in [24]. According to [24], ρ(m) is a nontrivial lower bound for the characteristic function of the set of primes in I, and it satisfies For further properties of ρ(m), see Lemma 1 and (4.2)-(4.4) in [24]. Let By the prime number theorem, it is easy to show that S j (α) A straightforward application of the Cauchy integral formula gives Identity (12) is also a corollary of Lemma 4 in [26]. For 4 ≤ k ≤ 6, put for any measurable subset X of R. It follows from (9) and (12) that 2 Journal of Mathematics where N(X) denotes the number of solutions of the inequality with p 2 ∈ I, p 5 ∈ I k , and p j ∈ I j for j ∈ 1, 3, 4 In what follows, we take actually. We now divide the real line into three disjoint parts: ese sets are called the major arc, the minor arcs, and the trivial regions, respectively.
In the following sections, we shall prove that the dominant contribution to I(τ, η, R) is from the major arc, and the contribution from the minor arcs and the trivial region can be neglected.

The Major Arc
Our first goal is to show that e proof of (19) is quite similar to that given in Section 3 in [20]. For completeness of exposition, we briefly present the proof procedure below. Let By a similar argument as that in pp. 1656-1657 in [20], we can obtain To estimate the integrals I(τ, η, M 2 ) and I(τ, η, M 3 ), we need the following two lemmas. Lemma 1. Let j ≥ 2 be an integer. en, for nonzero real number λ and any ε > 0, we have Proof. It follows from eorem 1 in [28].

The Minor Arcs
e next thing to do in the proof is to establish that is work forms the bulk of the present paper. We subdivide m into four disjoint parts: erefore, To prove (30), it suffices to show that |I(τ, η, m j )| ≪ τ 2 X (13/12)+(1/k) L − 2 holds for 1 ≤ j ≤ 4.
We apply Hölder's inequality and Lemma 2 to estimate |I(τ, η, m 1 )|. When k � 4, we have In case k � 6, we obtain (36) In order to establish an upper bound for |I(τ, η, m 2 )| as small as possible, we need the following lemma.

Lemma 3 (Lemma 3.4 in [21]). Let
(37) For 4 ≤ k ≤ 6, by the Cauchy-Schwarz inequality, Lemmas 2 and 3, we obtain Journal of Mathematics where the trivial upper bound S k (λ 5 α) ≪ X (1/k) is used. It is easily derived from (17) that (40) e upper bound estimation of |I(τ, η, m 3 )| plays a crucial role in the proof. e parameter τ, which is given by (17), is determined in this step. When k � 4, by Hölder's inequality and Lemma 2, we have In the case of k � 5, we obtain If k � 6, we deduce that Inequalities (41)-(43) and (17) together give In the remainder of this section, we shall be trying to estimate |I(τ, η, m 4 )|. By a familiar dyadic dissection argument, we divide m 4 into at most ≪L 3 disjoint sets E(Z 1 , Z 2 , y). For α ∈ E(Z 1 , Z 2 , y), we have where Z 1 � 2 k 1 X (6/7)+2ε , Z 2 � 2 k 2 X (3/7)+2ε , and y � 2 k 3 X − (1/8) for some nonnegative integers k 1 , k 2 , and k 3 . For the sake of convenience, we take the notation A as a shortcut for E(Z 1 , Z 2 , y), and let m(A) stand for the Lebesgue measure of A.

Data Availability
No data were used to support this study.

Conflicts of Interest
e authors declare that they have no conflicts of interest.