# A Diophantine Problem with Unlike Powers of Primes

Authors: Quanwu Mu; Liyan Xi
Journal: Journal of Mathematics
 (2021)

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## Abstract

Letk be an integer with 4≤k≤6 and η be any real number. Suppose that λ1,λ2,…,λ5 are nonzero real numbers, not all of them have the same sign, and λ1/λ2 is irrational. It is proved that the inequality λ1p1+λ2p22+λ3p33+λ4p44+λ5p5k+η<max1≤j≤5pj−σk has infinitely many solutions in prime variables p1,p2,p3,p4,andp5, where 0<σ4<1/36,0<σ5<4/189, and 0<σ6<1/54. This gives an improvement of the recent results.


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## Body

## 1. Introduction

The determination of the minimals such that the Diophantine equation(1)N=∑i=1sxii+1,is solvable in positive integers x1,…,xs, for all sufficiently large integers N is an interesting problem in additive number theory. In 1951, Roth [1] proved that s=50 is acceptable. This result was subsequently improved by Thanigasalam et al. [2–4], Vaughan and Vaughan [5, 6], Brüdern and Brüdern [7, 8] and Ford and Ford [9, 10]. The best currently known result is due to Ford [10], with s=14. Schwarz [11] suggested to analyze the related Diophantine inequality. The first result was obtained by Brüdern [12], who showed that the values of(2)∑i=122λi+1xii+1,at integer points x1,…,x22 are dense on the real line provided that λ2,…,λ23 are nonzero real numbers and λ2/λ3 is irrational. Thanks to a pruning technique, Brüdern [13] proved that the values taken by(3)∑i=116λi+1xii+1,at integer points x1,…,x16 are dense on the real line if λ2,…,λ17 are nonzero real numbers and at least one of the ratios λi/λj is irrational.Suppose thatx1,…,xs are prime variables, N is a sufficiently large integer, and N+s is even. In 1969, Vaughan proved in his doctoral thesis that (1) is solvable if s=31. Later, Vaughan [5] improved upon his own result by taking s=30 in place of s=31. By calculating, the exponential density more accurately, Shan [14] showed that s=23 is acceptable. In addition, Prachar [15] established that each sufficiently large odd integer N can be represented as(4)N=p1+p22+p33+p44+p55,where p1,p2,…,p5 are prime numbers. As a corollary of [16] in Theorem 1, Ren and Tsang obtained the same result as Prachar. It is of some interest to consider the analogous form for Diophantine inequalities. Let λ1,λ2,…,λ5 be nonzero real numbers, not all of them have the same sign and λ1/λ2 as irrational. In 2016, Ge and Li [17] proved that, for any given real numbers η and σ,0<σ<1/720, there exist infinitely many solutions in prime numbers pj to the inequality(5)λ1p1+λ2p22+λ3p33+λ4p44+λ5p55+η<max1≤j≤5pjj−σ.Letk≥4 be an integer. The first author [18] investigated the solvability of more general Diophantine inequality(6)λ1p1+λ2p22+λ3p33+λ4p44+λ5p5k+η<max1≤j≤5pj−σk,and proved that (6) has infinitely many solutions in prime variables pj for 0<σ4<5/288 and 0<σk<5/6k2k+1 with k≥5. Subsequently, Liu [19] obtained 0<σ5<5/288. In [20], the first author and Qu showed that 0<σ5<5/252 is acceptable. Very recently, this result was improved by Zhu [21], who obtained 0<σ5<1/48. In [22], Gao and Liu gave an improvement ([18] in Theorem 1.2) in case k≥6, and they proved 0<σ6<1/56 particularly.The main purpose of this paper is to sharpen the above results in case4≤k≤6. We obtain the following theorem.Theorem 1.
Letk be an integer with 4≤k≤6 and η be any given real number. Suppose that λ1,λ2,…,λ5 are nonzero real numbers, not all of them are same sign, and λ1/λ2 is irrational. Then, inequality (6) has infinitely many solutions in prime variables p1,p2,…,p5, where 0<σ4<1/36,0<σ5<4/189, and 0<σ6<1/54.The improvement derives not only from the use of the functionρm constructed by Harman and Kumchev (see Section 8 in [23] and Section 5 in [24], for details) but also from some ingredients in [21]. It is worth remarking that Ge et al. [25] obtained 0<σ5<1/32, if the condition “λ1/λ2 is irrational” in Theorem 1 is replaced by “λ1/λ2 is irrational and λ2/λ4 and λ3/λ5 are rational.”Notation. Throughout the paper, ε and δ are arbitrarily small, fixed positive real numbers. Any statement in which ε occurs holds for each positive ε. The implicit constants in O-term, ≪- and ≫-symbols depend at most on λ1,λ2,…,λ5 and ε. The letter p, with or without subscript, is reserved for a prime number. By A≍B, we mean that A≪B and A≫B. For simplicity, we write ℒ=logX and eα=exp2πiα.


## 2. Preliminaries

We apply the Davenport–Heilbronn circle method (see [26] and Chapter 11 in [27]) to prove Theorem 1. Since λ1/λ2 is irrational, there are infinitely many convergents to its continued fraction. Let q be any denominator of a convergent to λ1/λ2. As in [20], let X run through the sequences:(7)X=q7/3.We set(8)ℐ=13X1/2,23X1/2,S2∗α=∑m∈ℐρmeαm2,where the function ρm is defined by 5.2 in [24]. According to [24], ρm is a nontrivial lower bound for the characteristic function of the set of primes in ℐ, and it satisfies(9)ρm≤1,ifmis a prime,0,otherwise.For further properties ofρm, see Lemma 1 and (4.2)–(4.4) in [24]. Let(10)Ij=δX1/j,X1/j,Sjα=∑p∈Ijlogpeαpj.By the prime number theorem, it is easy to show thatSjα≪X1/j. For any fixed τ>0, set Kτα=πα−2sin2πτα for α≠0 and Kτ0=τ2. Clearly, we have(11)Kτα≪minτ2,α−2.A straightforward application of the Cauchy integral formula gives(12)∫−∞+∞exαKταdα=max0,τ−x.Identity (12) is also a corollary of Lemma 4 in [26]. For 4≤k≤6, put(13)Gα=S1λ1αS2∗λ2α∏j=34SjλjαSkλ5αeαηKτα.We write(14)Iτ,η,X=∫XGαdα,for any measurable subset X of ℝ. It follows from (9) and (12) that(15)Iτ,η,ℝ=∑pj∈Ij,j=1,3,4,p5∈Ik,m2∈ℐρm2∏1≤j≤5j≠2logpj×∫−∞+∞eλ1p1+λ2m22+∑j=34λjpjj+λ5p5k+ηαKταdα=∑pj∈Ij,j=1,3,4,p5∈Ik,m2∈ℐρm2∏1≤j≤5j≠2logpj×max0,τ−λ1p1+λ2m22+∑j=34λjpjj+λ5p5k+η≤τℒ4NX,where NX denotes the number of solutions of the inequality(16)λ1p1+λ2p22+λ3p33+λ4p44+λ5p5k+η<τ,with p2∈ℐ,p5∈Ik, and pj∈Ij for j∈1,3,4. In what follows, we take(17)τ=X−1/36+20εifk=4,X−4/189+20εifk=5,X−1/54+20εifk=6,actually. We now divide the real line into three disjoint parts:(18)M=α:α≤X−1/8,m=α:X−1/8<α≤ξ,t=α:α>ξ,where ξ=τ−2X1/16−1/4k+10ε. These 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 toIτ,η,ℝ is from the major arc, and the contribution from the minor arcs and the trivial region can be neglected.


## 3. The Major Arc

Our first goal is to show that(19)Iτ,η,M≫τ2X13/12+1/kℒ−1.The 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(20)M1=α:α≤X−1+5/12k−ε,M2=α:X−1+5/12k−ε<α≤X−7/8,M3=α:X−7/8<α≤X−1/8.Then, we haveM=M1∪M2∪M3 and(21)Iτ,η,M=Iτ,η,M1+Iτ,η,M2+Iτ,η,M3.By a similar argument as that in pp. 1656–1657 in [20], we can obtain(22)Iτ,η,M1≫τ2X13/12+1/kℒ−1.To estimate the integralsIτ,η,M2 and Iτ,η,M3, we need the following two lemmas.Lemma 1.
Letj≥2 be an integer. Then, for nonzero real number λ and any ε>0, we have(23)Sjλα≪X1/j1−j⋅41−j+εα−41−jforX−1≤α≤X−1+1/2j,X1j1−1/2⋅41−j+εforX−1+1/2j<α≤X−1/2j.Proof.
It follows from Theorem1 in [28].Lemma 2.
For4≤k≤6, suppose that(24)Fα∈S12λ1α,S38λ3α,S2∗λ2α2S34λ3α,S2∗λ2α2S44λ4α,S2∗λ2α2S56λ5α,S2∗λ2α2S68λ5α,S2∗λ2αS3λ3αSkλ5α2,S2∗λ2α2S42λ4αS54λ5α.
Then, we have(25)∫−11Fαdα≪X−1F01+ε,∫ℝFαKταdα≪τX−1F01+ε.Proof.
See Lemma 3.7 in [20].Whenα∈M2, it follows from (23) that(26)S4λ4α≪X15/64+εα−1/64≪X1/4−5/768k+2ε.Combining this with the Cauchy–Schwarz inequality and Lemma2 gives(27)Iτ,η,M2≪τ2supα∈M2S4λ4α∫M2S1λ1αS2∗λ2αS3λ3αSkλ5αdα≪τ2supα∈M2S4λ4α∫−11S1λ1α2dα1/2×∫−11S2∗λ2αS3λ3αSkλ5α2dα1/2≪τ2X1/4−5/768k+2ε⋅X1+2ε1/2⋅X2/3+2/k+ε1/2≪τ2X13/12+1/k−ε,where (11) is used.Whenα∈M3, (23) implies(28)S4λ4α≪X1/4−1/512+ε.Proceeding as in the proof of (27), we have(29)Iτ,η,M3≪τ2X13/12+1/k−ε.This with (27), (22), and (21) yields (19).


## 4. The Minor Arcs

The next thing to do in the proof is to establish that(30)Iτ,η,m≪τ2X13/12+1/kℒ−2.This work forms the bulk of the present paper. We subdividem into four disjoint parts: m=m1∪m2∪m3∪m4, where(31)m1=α∈m:S1λ1α≤X6/7+2ε,m2=α∈m:S1λ1α>X6/7+2ε,S2∗λ2α≤X3/7+2ε,S3λ3α>X11/36+2ε,m3=α∈m:S1λ1α>X6/7+2ε,S2∗λ2α≤X3/7+2ε,S3λ3α≤X11/36+2ε,m4=α∈m:S1λ1α>X6/7+2ε,S2∗λ2α>X3/7+2ε.Therefore,(32)Iτ,η,m=∑j=14Iτ,η,mj.To prove (30), it suffices to show that Iτ,η,mj≪τ2X13/12+1/kℒ−2 holds for 1≤j≤4.We apply Hölder’s inequality and Lemma2 to estimate Iτ,η,m1. When k=4, we have(33)Iτ,η,m1≪supα∈m1S1λ1α1/4∫ℝS1λ1α2Kταdα3/8×∫ℝS3λ3α8Kταdα1/8∫ℝS2∗λ2α2S4λ4α4Kταdα1/4×∫ℝS2∗λ2α2Skλ5α4Kταdα1/4≪X6/7+2ε1/4τX1+ε3/8τX5/3+ε1/8τX1+ε1/4τX4/k+ε1/4≪τX13/12+1/k−1/28+2ε.Ifk=5, then(34)Iτ,η,m1≪supα∈m1S1λ1α3/16∫ℝS1λ1α2Kταdα13/32×∫ℝS3λ3α8Kταdα3/32∫ℝS2∗λ2α2S4λ4α4Kταdα1/4×∫ℝS2∗λ2α2Skλ5α6Kταdα1/8×∫ℝS2∗λ2αS3λ3αSkλ5α2Kταdα1/8≪X6/7+2ε3/16τX1+ε13/32τX5/3+ε3/32τX1+ε1/4τX6/k+ε1/8τX2/3+2/k+ε1/8≪τX13/12+1/k−3/112+2ε.In casek=6, we obtain(35)Iτ,η,m1≪supα∈m1S1λ1α1/6∫ℝS1λ1α2Kταdα5/12×∫ℝS3λ3α8Kταdα1/12∫ℝS2∗λ2α2S4λ4α4Kταdα1/4×∫ℝS2∗λ2α2Skλ5α8Kταdα1/12×∫ℝS2∗λ2αS3λ3αSkλ5α2Kταdα1/6≪X6/7+2ε1/6τX1+ε5/12τX5/3+ε1/12τX1+ε1/4τX8/k+ε1/12τX2/3+2/k+ε1/6≪τX13/12+1/k−1/42+2ε.It follows from (33)–(35) and (17) that(36)Iτ,η,m1≪τ2X13/12+1/k−ε≪τ2X13/12+1/kℒ−2.In order to establish an upper bound forIτ,η,m2 as small as possible, we need the following lemma.Lemma 3.
(Lemma 3.4 in [21]). Let(37)N=α:X11/36+2ε<S3λ3α≤X1/3.
Then, we have(38)∫NS3λ3α2S4λ4α2Kταdα≪τX1/6+4ε.For4≤k≤6, by the Cauchy–Schwarz inequality, Lemmas 2 and 3, we obtain(39)Iτ,η,m2≪X1/ksupα∈m2S2∗λ2α∫NS3λ3α2S4λ4α2Kταdα1/2×∫ℝS1λ1α2Kταdα1/2≪τX1/k+3/7+1/12+1/2+4ε≪τX13/12+1/k−1/14+5ε,where the trivial upper bound Skλ5α≪X1/k is used. It is easily derived from (17) that(40)Iτ,η,m2≪τ2X13/12+1/k−ε≪τ2X13/12+1/kℒ−2.The upper bound estimation ofIτ,η,m3 plays a crucial role in the proof. The parameter τ, which is given by (17), is determined in this step. When k=4, by Hölder’s inequality and Lemma 2, we have(41)Iτ,η,m3≪supα∈m3S3λ3α∫ℝS1λ1α2Kταdα1/2×∫ℝS2∗λ2α|2S4λ4α4Kταdα1/4×∫ℝS2∗λ2α2Skλ5α4Kταdα1/4≪X11/36+2ετX1+ε1/2τX1+ε1/4τX4/k+ε1/4≪τX13/12+1/k−1/36+4ε.In the case ofk=5, we obtain(42)Iτ,η,m3≪supα∈m3S2∗λ2α1/6supα∈m3S3λ3α1/3×∫ℝS1λ1α2Kταdα1/2∫ℝS3λ3α8Kταdα1/12×∫ℝS2∗λ2α2S4λ4α4Kταdα1/4×∫ℝS2∗λ2α2Skλ5α6Kταdα1/6≪X3/7+2ε1/6X11/36+2ε1/3τX1+ε1/2τX5/3+ε1/12×τX1+ε1/4τX6/k+ε1/6≪τX13/12+1/k−4/189+4ε.Ifk=6, we deduce that(43)Iτ,η,m3≪supα∈m3S3λ3α2/3∫ℝS1λ1α2Kταdα1/2×∫ℝS2∗λ2α2S4λ4α4Kταdα1/4×∫ℝS2∗λ2α2Skλ5α8Kταdα1/12×∫ℝS2∗λ2αS3λ3αSkλ5α2Kταdα1/6≪X11/36+2ε2/3τX1+ε1/2τX1+ε1/4τX8/k+ε1/12τX2/3+2/k+ε1/6≪τX1312+1k−154+4ε.Inequalities (41)–(43) and (17) together give(44)Iτ,η,m3≪τ2X13/12+1/k−ε≪τ2X13/12+1/kℒ−2.In the remainder of this section, we shall be trying to estimateIτ,η,m4. By a familiar dyadic dissection argument, we divide m4 into at most ≪ℒ3 disjoint sets EZ1,Z2,y. For α∈EZ1,Z2,y, we have(45)Z1<S1λ1α≤2Z1,Z2<S2∗λ2α≤2Z2,y<α≤2y,where Z1=2k1X6/7+2ε,Z2=2k2X3/7+2ε, and y=2k3X−1/8 for some nonnegative integers k1,k2,andk3. For the sake of convenience, we take the notation A as a shortcut for EZ1,Z2,y, and let mA stand for the Lebesgue measure of A.Lemma 4.
(Lemma 4.3 in [20]). We have(46)mA≪yX18/7+9εZ1−2Z2−4.Whenk=4, it follows from (11) and Hölder’s inequality that(47)Iτ,η,A≪∫AS1λ1α2S2∗λ2α4Kταdα1/12∫ℝS1λ1α2Kταdα5/12×∫ℝS3λ3α8Kταdα1/8∫ℝS4λ4α16Kταdα1/48×∫ℝSkλ5α16Kταdα1/48∫ℝS2∗λ2α2S4λ4α4Kταdα1/6×∫ℝS2∗λ2α2Skλ5α4Kταdα1/6≪Z12Z24⋅mA⋅minτ2,y−21/12τX1+ε5/12τX5/3+ε1/8×τX3+ε1/48τX12/k+ε1/48τX1+ε1/6τX4/k+ε1/6≪yX18/7+9ε⋅minτ2,y−21/12τ11/12X41/48+11/12k+ε≪τX13/12+1/k−1/28+2ε,where Lemmas 2 and 3 are used.Whenk=5, by the similar argument as in the proof of (47), we obtain(48)Iτ,η,A≪∫AS1λ1α2S2∗λ2α4Kταdα1/16∫ℝS1λ1α2Kταdα7/16×∫ℝS3λ3α8Kταdα1/8∫ℝS2∗λ2α2S4λ4α4Kταdα1/8×∫ℝ|S2∗λ2α2S4λ4α2Skλ5α4Kταdα1/4≪Z12Z24⋅mA⋅minτ2,y−21/16τX1+ε7/16×τX5/3+ε1/8τX1+ε1/8τX1/2+4/k+ε1/4≪yX18/7+9ε⋅minτ2,y−21/16τ15/16X43/48+1/k+ε≪τX13/12+1/k−3/112+2ε.Whenk=6, we have(49)Iτ,η,A≪∫AS1λ1α2S2∗λ2α4Kταdα1/16∫ℝS1λ1α2Kταdα7/16×∫ℝS3λ3α8Kταdα1/8∫ℝS2∗λ2α2S4λ4α4Kταdα1/4×∫ℝS2∗λ2α2Skλ5α8Kταdα1/8≪Z12Z24⋅mA⋅minτ2,y−21/16τX1+ε7/16τX5/3+ε1/8τX1+ε1/4τX8/k+ε1/8≪yX18/7+9ε⋅minτ2,y−21/16τ15/16X43/48+1/k+ε≪τX13/12+1/k−3/112+2ε.Thanks to (17) and (47)–(49), we are led to the conclusion that(50)Iτ,η,m4≪ℒ3⋅maxAIτ,η,A≪τ2X13/12+1/k−ε≪τ2X13/12+1/kℒ−2.This together with (36), (40), (44), and (32) gives (30).


## 5. The Trivial Regions

Finally, it only remains to treatIτ,η,t. Suppose that r and j are positive integers with r≤j. For any ξ∈Xε,+∞ and nonzero real λ, we have (see (5.1) and (5.2) in [20])(51)∫ξ+∞Sjλα2rKταdα≪ξ−1X2r−r/j+ε,∫ξ+∞S2∗λ2α4Kταdα≪ξ−1X1+ε.It follows from Hölder’s inequality that(52)Iτ,η,t≪∫ξ+∞Skλ5αS2∗λ2α∏1≤j≤4j≠2SjλjαKταdα≪∫ξ+∞S1λ1α2Kταdα1/2∫ξ+∞S2∗λ2α4Kταdα1/4×∫ξ+∞S3λ3α8Kταdα1/8∫ξ+∞S4λ4α16Kταdα1/16×∫ξ+∞Skλ5α16Kταdα1/16≪ξ−1X1+ε1/2ξ−1X1+ε1/4ξ−1X5/3+ε1/8ξ−1X3+ε1/16ξ−1X12/k+ε1/16≪ξ−1X55/48+3/4k+2ε.Recalling thatξ=τ−2X1/16−1/4k+10ε and inserting this expression into (52) yields(53)Iτ,η,t≪τ2X13/12+1/k−ε≪τ2X13/12+1/kℒ−2.


## 6. Completion of the Proof

We are now in a position to get the desired conclusion. It should be noted that(54)Iτ,η,ℝ=Iτ,η,M+Iτ,η,m+Iτ,η,t.From this and (19), (30), and (53), we infer that Iτ,η,ℝ≫τ2X13/12+1/kℒ−1. Hence, by (15),(55)NX≫τX13/12+1/kℒ−5.This implies inequality (16) has ≫τX13/12+1/kℒ−5 solutions in quintuples of primes p1,p2,…,p5 with p2∈ℐ,p5∈Ik, and pj∈Ij, for j∈1,3,4. Notice that λ1/λ2 is irrational, q is any denominator of a convergent to λ1/λ2 and X=q7/3. By substituting (17) into (55), we deduce that NX⟶+∞ as q⟶+∞. In view of(56)max1≤j≤5pj≍X,and (17), we obtain the required range of σk in Theorem 1. This completes the proof of Theorem 1.



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