On Roth's theorem concerning a cube and three cubes of primes. Citation Quarterly Journal Of Mathematics, 2004, v. 55 n. 3, p.

Transcription

1 Title On Roth's theorem concerning a cube and three cubes of primes Authors) Ren, X; Tsang, KM Citation Quarterly Journal Of Mathematics, 004, v. 55 n. 3, p Issued Date 004 URL Rights Quarterly Journal of Mathematics. Copyright Oxford University Press.; This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.; This is an electronic version of an article published in [Quarterly Journal Of Mathematics, 004, v. 55 n. 3, p ]

2 ON ROTH S THEOREM CONCERNING A CUBE AND THREE CUBES OF PRIMES by Xiumin Ren 1 and Kai-Man Tsang Department of Mathematics, The University of Hong Kong, Pokfulam, Hong Kong) Abstract In this paper, we prove that with at most ON 171/196+ε ) exceptions, all positive integers up to N are the sum of a cube and three cubes of primes. This improves an earlier result ON 169/170 ) of the first author and the classical result ONL A ) of Roth. 1. Introduction It is conjectured that all sufficiently large integers n satisfying some necessary congruence conditions are the sum of four cubes of primes, i.e. n = p p 3 + p p 3 4. Such a strong conjecture is out of reach at present; but it is reasonable, in view of the following results of Davenport and Hua respectively. Davenport s theorem in [1] asserts that almost all positive integers are the sum of four positive cubes, while a theorem of Hua [5, 6] states that almost all positive integers n with n 0, ± mod 9) are the sum of five cubes of primes. In 1949, Roth [1] proved that almost all positive integers n can be written as n = m 3 + p 3 + p p 3 4, 1.1) where m is a positive integer and p j are primes. To be more precise, we let EN) denote the number of all the integers n not exceeding N which cannot be written as 1.1). Then Roth s theorem actually states that EN) N log A N, where A > 0 is arbitrary. This result can be viewed as an approximation to the above conjecture, and the quality of this approximation is indicated in the upper bound of EN). Roth s theorem has been improved by the first author [11] to EN) N 169/170. The exponent 169/170 was obtained via an approach in which the possible existence of Siegel zero does not have special influence, and hence the Deuring-Heilbronn phenomenon can be avoided. In this paper we inject new ideas into the afore-mentioned approach, and make the following further improvement. Theorem 1.1. For EN) defined as above, we have EN) N 171/196+ε. The new ideas used in this paper include the iterative method and the hybrid mean-value estimate for Dirichlet polynomials of Liu [8], which will be displayed in full details at relevant places in the following sections. An outline of the proof of Theorem 1.1 will be given in. At 1 xmren@maths.hku.hk kmtsang@maths.hku.hk 1

3 this stage, we only point out that our Theorem 1.1 does not depend on the Deuring-Heilbronn phenomenon, and the method of this paper can be successfully applied to a number of additive problems. Notation. As usual, ϕn) and Λn) stand for the function of Euler and von Mangoldt respectively, and dn) is the divisor function. We use χ mod q and χ 0 mod q to denote a Dirichlet character and the principal character modulo q; and Ls, χ) is the Dirichlet L- function. In our statement, N is a large positive integer, and L = log N. The symbol r R means R < r R. The letters ε and A denote positive constants, which are arbitrarily small and arbitrarily large respectively. We use c j to represent absolute positive constants.. Outline of the method For large positive integer N and θ = 5/16 ε, we set P = N θ and Q = NP 1..1) For coprime integers a, q with 1 a q P, we denote by Mq, a) the interval [a/q 1/qQ, a/q + 1/qQ]. These major arcs all lie in [1/Q, 1 + 1/Q] and, since P Q, they are mutually disjoint. Write M for the union of all Mq, a) and define the minor arcs m as the complement of M in [1/Q, 1 + 1/Q]. Let U = N/9) 1/3 and V = U 5/6..) We define and, for W = U or V, T α) = m U em 3 α) Sα, W ) = m W Λm)em 3 α). Define rn) = Λm 1 )Λm )Λm 3 ). n=m m3 4 m 1,m 4 U,m,m 3 V Then rn) = 1+1/Q 1/Q Sα, U)S α, V )T α)e nα)dα = To handle the integral on the major arcs, we need the following. M +..3) m Lemma.1. For all integers n with N/ n N, we have Sα, U)S α, V )T α)e nα)dα = Sn)Jn) + OV U 1 L A )..4) M Here Sn) is the singular series which is defined by 4.1) and it satisfies log log n) c 1 Sn) log n;.5)

4 and Jn) is as defined in 4.3) and it satisfies V U 1 Jn) V U 1..6) In the Waring-Goldbach problem, the quality of arithmetical results obtained usually depends on the size of the major arcs. Our major arcs M in this paper are much larger than those in [11], and we need new ideas to control their contribution. The first new idea is a hybrid estimate for Dirichlet polynomials; see Lemma 5.1 below. The second is the iterative procedure in treating I 1,..., I 5 in 4, as illustrated in the proof of Lemma 4.. The minor arcs estimate is taken care of by Theorem 3 in [11]. That it is valid for minor arcs as defined above can be checked easily. Here we record it in the following lemma. Lemma.. Let m be as defined above. Then we have Sα, U) Sα, V ) 4 T α) dα UV 4 P 1/6+ε. m Equipped with Lemmas.1 and., we can now give the proof of Theorem 1.1 immediately. Proof of Theorem 1.1. We start from.3). The contribution of the major arcs is taken care of by Lemma.1. To treat the integral on the minor arcs we apply Bessel s inequality and Lemma. to get N/<n N m m Sα, U) Sα, V ) 4 T α) dα UV 4 P 1/6+ε..7) By a standard argument, we deduce from.7) that for all N/ < n N but at most OU 3 P 1/6+3ε ) exceptions, m V U 1 P ε. Therefore by Lemma.1, for these un-exceptional n, we have rn) = Sn)Jn) + OV U 1 L A ), and these n can be written as 1.1). Let F N) be the number of the exceptional n above. Then we have F N) U 3 P 1/6+3ε N 171/196+ε. The assertion of Theorem 1.1 now follows from EN) = j 0 F N/j ). Now it remains to prove Lemma.1, which will be carried out in the following sections. 3. An explicit expression The purpose of this section is to establish in Lemma 3.1 an explicit expression for the left-hand side of.4). For χ mod q, we define Cχ, a) = q ) am 3 χm)e, 3.1) q m=1 3

5 and write Cq, a) = Cχ 0, a). We also define q ) am S 3 q, a) = e. 3.) q For α = a/q + λ Mq, a), we have q ) ah 3 S α, W ) = e q h=1 h,q)=1 m=1 m W m h mod q) Λm)eλm 3 ) + OL ). By introducing Dirichlet characters to the above sum over m, we can rewrite S α, W ) as Cq, a) ϕq) m W eλm 3 ) + χ mod q Cχ, a) ϕq) Λm)χm) δ χ )eλm 3 ) + OL ). 3.3) m W Here and throughout, δ χ is 1 or 0 according as χ is principal or not. By Lemma 4.8 in [13] one finds that, for W = U or V, m W eλm 3 ) = W W eλu 3 )du + O1). Thus, if we denote by Φλ, W ) the above integral and by Ψχ, λ, W ) the last sum over m in 3.3), then we have S α, W ) = Cq, a) ϕq) Φλ, W ) + χ mod q Cχ, a) ϕq) Ψχ, λ, W ) + OL ) = S 1 λ, W ) + S λ, W ) + OL ), 3.4) say. For T α) we apply Theorem 4.1 in Vaughan [14], to get say. So if we write T α) = S q, a) Φλ, U) + Oq 1/+ε ) = T 1 λ) + Oq 1/+ε ), 3.5) q λ) = {S 1 λ, U) + S λ, U)} {S 1 λ, V ) + S λ, V )} T 1 λ), then 3.4) and 3.5) together with the trivial bounds Sα, W ) W and T α) U show that Sα, U)S α, V )T α) λ) UV q 1/+ε + U V L. Consequently Sα, U)S α, V )T α)e nα)dα M = q P q a=1 a,q)=1 e an q ) 1/qQ λ)e nλ)dλ + OUV P 3/+ε Q 1 + U V P Q 1 L ). 1/qQ The above O-term is OV U 1 L A ), on recalling.1) and.). Now we write 0 λ) = S1λ, V ), 1 λ) = S 1 λ, V )S λ, V ), λ) = Sλ, V ); 4

6 and for i = 0, 1,, define I i = q P I 3+i = q P Then we have proved q a=1 a,q)=1 q a=1 a,q)=1 e an q e an q Lemma 3.1. For I j defined as above, we have M Sα, U)S α, V )T α)e nα)dα = ) 1/qQ S 1 λ, U)T 1 λ) i λ)e nλ)dλ, 3.6) 1/qQ ) 1/qQ S λ, U)T 1 λ) i λ)e nλ)dλ. 3.7) 1/qQ 5 I j + OV U 1 L A ). j=0 In the following sections we prove that I 0 produces the main term, while the others contribute to the error term. We need some more notations. 4. Estimation of I j for j = 0, 1,..., 5. S q, a) be as defined in 3.1) and 3.). We define Bn, q, χ 1, χ, χ 3 ) = q a=1 a,q)=1 Let χ 1, χ and χ 3 be characters mod q, Cχ, a) and e an q ) Cχ 1, a)cχ, a)cχ 3, a)s q, a), and write Bn, q) = Bn, q, χ 0, χ 0, χ 0 ), An, q) = Bn, q) ϕ 3 q)q, Sn) = An, q). 4.1) q=1 This Sn) is the singular series appearing in Lemma.1. By Lemma 18 and in [1] and Lemma 4.4 in [11], we see that the singular series is absolutely convergent and satisfies the first inequality in.5). The second inequality in.5) can be established by making use of Lemmas 15 and 16 in [1] as follows: An, q) 1 + c p 1 ) + c p q p n p n1 3 ) ) n c + c p p n1 1 ) log n. 4.) ϕn) Lemma 4.1. Let I 0 be as defined in 3.6). Then for all N/ < n N, we have I 0 = Sn)Jn) + OV U 1 L A ), where Sn) and Jn) are defined in 4.1) and 4.3). 5

7 Proof. By definition we have Define I 0 = q P Bn, q) ϕ 3 q)q Jn) = 1/qQ 1/qQ Φ λ, U)Φ λ, V )e nλ)dλ. Φ λ, U)Φ λ, V )e nλ)dλ. 4.3) Then by Lemma 5. in [11], Jn) is well defined and satisfies.6). Using the elementary estimate we get and therefore, λ 1/qQ ) 1 Φλ, W ) min W, W, 4.4) λ Φ λ, U)Φ λ, V ) dλ V U 4 1/qQ dλ λ V UP ) 1 q, I 0 = Jn) { An, q) + O V UP ) } 1 q An, q). 4.5) q P q P By Lemma 18 in [1], An, q) = Sn) + OP 1/4+ε ), q P so the main term on the right hand-side of 4.5) becomes Jn)Sn) + OV U 1 P 1/4+ε ). To estimate the O-term in 4.5), we use the bound see [1, pp. 77]) An, q) q 3/+ε n, q) 1/, to get q An, q) q 1/+ε n, q) 1/ q P q P d n d ε q 1/+ε P 1/+ε, q P/d and consequently the O-term in 4.5) is OV U 1 P 1/+ε ). This proves Lemma 4.1. Lemma 4.. Let I j, j = 1,,..., 5 be as defined in 3.6) and 3.7). Then we have I j V U 1 L A. To prove Lemma 4., we need the following Lemmas Lemma 4.3. If χ j, j = 1,, 3, are primitive characters mod r j, and r 0 = [r 1, r, r 3 ] is the least common multiple of r 1, r, r 3, then for χ 0 mod q we have Bn, q, χ1 χ 0, χ χ 0, χ 3 χ 0 ) r 0 L. q P r 0 q ϕ 3 q)q 6

8 The saving of r 0 on the right-hand side will play a key role in our argument, and the quality of the exceptional set will depend on the magnitude of the exponent 5/6. In the next section, we will apply the iterative method of [8] to make use of the full strength of this 5/6, and the reader is referred to the proof of Lemma 4.. Proof of Lemma 4.3. This is a slight modification of Lemma 4.5 in [11]. In fact, by 4.7)-4.10) in [11], we see that q P r 0 q Bn, q, χ1 χ 0, χ χ 0, χ 3 χ 0 ) ϕ 3 q)q r 0 i=0 q P/3 i r 0 q,r 0 )=1 An, q). From this the desired assertion follows by applying 4.) to the last sum. Lemma 4.4. Let g be a positive integer. Then we have r P 1/ 1/rQ [r, g] Ψχ, λ, U) dλ) g U 1/ L c 3. 1/rQ In particular, for g = 1, the above bound can be improved to U 1/ L A. Here indicates that the summation is over all primitive characters mod r. Lemma ) For integer g 1, we have r P [r, g] max Ψχ, λ, V ) λ 1/rQ g V L 17. ) In particular, for g = 1 the above bound can be improved to V L A. The proofs of Lemmas 4.4 and 4.5 will be postponed to the next section. lemmas ready, we can now give the proof of Lemma 4.. Proof of Lemma 4.. We first consider I 5. By definition, With these I 5 = Bn, q, χ 1, χ, χ 3 ) ϕ 3 q)q q P χ 1 mod q χ mod q χ 3 mod q 1/qQ Ψχ 1, λ, V )Ψχ, λ, V )Ψχ 3, λ, U)Φλ, U)e nλ)dλ. 1/qQ Reducing the characters into primitive characters and observing that, for primitive character with r q, Ψχχ 0, λ, W ) Ψχ, λ, W ) = Λm)χχ 0 m) χm))eλm 3 ) δ χχ 0 δ χ ) eλm 3 ) m W m W log p L, p q, p r p j W 7

9 we have I 5 = r 1 P r 3 P χ 1 mod r 1 χ 3 mod r 3 q P [r 1,r,r 3 ] q Bn, q, χ 1 χ 0, χ χ 0, χ 3 χ 0 ) qϕ 3 q) 1/qQ Ψχ 1 χ 0, λ, V )Ψχ χ 0, λ, V )Ψχ 3 χ 0, λ, U)Φλ, U)e nλ)dλ 1/qQ L [r1, r, r 3 ] I5, 4.6) r 1 P r 3 P χ 1 mod r 1 χ 3 mod r 3 by Lemma 4.3. Here with I 5 = 1/[r1,r,r 3 ]Q 1/[r 1,r,r 3 ]Q Ψ χ 1, λ, V )Ψ χ, λ, V )Ψ χ 3, λ, U) Φλ, U) dλ, Ψ χ, λ, W ) = Ψχ, λ, W ) + L. 4.7) By Cauchy s inequality, I5 max λ 1/r 1 Q Ψ χ 1, λ, V ) max λ 1/r Q Ψ χ, λ, V ) ) 1/ 1/r3 Q 1/ 1/Q Ψ χ 3, λ, U)) dλ Φλ, U) dλ). 1/r 3 Q 1/Q Here an application of 4.4) easily gives 1/Q 1/Q On the other hand, by Lemma 4.4 and 4.7), r 3 P χ 3 mod r 3 Φλ, U) dλ U ) 1/r3 [r1, r, r 3 ] Q Ψ χ 3, λ, U)) dλ [r 1, r ] U 1/ L c 3 + L [r 1, r ] U 1/ L c 3. 1/r 3 Q r 3 P χ 3 mod r 3 ) 1/ [r1, r, r 3 ] r 3 Q) 1/ Collecting these estimates, we get from 4.6) that I 5 U 1 L c 3+1 max λ 1/r 1 Q Ψ χ 1, λ, V ) r 1 P χ 1 mod r 1 r P χ mod r [r1, r ] max λ 1/r Q Ψ χ, λ, V ). 4.9) By Lemma 4.5 and 4.7), the last double sum is r 1 V L 17 + L [r1, r ] r 1 V L 17. r P χ mod r 8

10 Therefore, I 5 U 1 V L c 3+18 r 1 P χ 1 mod r 1 r 1 max λ 1/r 1 Q Ψ χ 1, λ, V ) V U 1 L A, by applying 4.7) and ) in Lemma 4.5. This proves the lemma for j = 5. We remark that, in the iterative argument above, we have used the saving r 0 to its full strength, as pointed out right after Lemma 4.3. One sees that the iterative argument is crucial for our improvement, since previously in [11] we used the inequality r 0 r 5/18+ε 1 r 5/18+ε r 5/18+ε 3, which is responsible for the weaker bound EN) N 169/170 there. We would also like to add that the key ingredient to fulfil the iterative argument is the hybrid estimate in Lemmas 5.1 and 5. below. To finish the proof of Lemma 4., we need to sketch how to estimate I j for j = 1,..., 4. As an example, we consider I 3. By definition and by reducing the characters into primitive ones, we get by Lemma 4.3 and 4.7) that I 3 L r P 1/rQ r Ψ χ, λ, U) Φλ, U) Φ λ, V ) dλ 1/rQ V L r P ) 1/ 1/rQ 1/Q r Ψ χ, λ, U)) dλ Φλ, U) dλ 1/rQ 1/Q ) 1/ V U 1 L A, by 4.8) and Lemma 4.4 for g = 1. This finishes the proof of Lemma Proof of Lemmas 4.4 and 4.5. We need to do some preparations to establish Lemmas 4.4 and 4.5. positive integer and let M j, j = 1,..., 10, be positive integers satisfying Let M be a large 9 M M 1 M 10 M) and M 6,..., M 10 M) 1/5. 5.1) For any positive integer m, let log m, if j = 1, a j m) = 1, if j =,..., 5, µm), if j = 6,..., ) We define the following functions of a complex variable s: f j s, χ) = m M j a j m)χm) m s, F s, χ) = f 1 s, χ) f 10 s, χ). 5.3) Then we have the following hybrid type estimate for F 1/ + it, χ), which is Lemma.1 in Liu [8]. 9

11 Lemma 5.1. For any T > 0, d 1 and 1 R M, we have r R d r T T ) { 1 R F + it, χ dt d T + R } d 1/ T 1/ M 3/10 + M 1/ L c 4. We also need zero-density estimates in forms stated in the following lemma. Lemma 5.. Let T 1, Q 1 and 1 d Q. Then q Q d q χ mod q Q ) Aσ)1 σ) T Nσ, T, χ) L 14, d where for 1/ σ < 3/4, Aσ) = 3/ σ); and for 3/4 σ 1, Aσ) = 1/5 + ε. Proof. We observe that under the restriction d q, Theorem 1. in Montgomery [9] can be restated as follows: For 1/ σ 4/5 q Q d q χ mod q Q ) 31 σ)/ σ) T Nσ, T, χ) L 9, d and for 4/5 σ 1 q Q d q χ mod q Q ) 1 σ)/σ T Nσ, T, χ) L 14. d This proves the lemma for 1/ σ < 3/4 and 5/6 σ 1. For 3/4 σ 5/6, the desired assertion is included in 1.1) of Huxley [7]. Proof of Lemma 4.4. Note that for primitive character, δ χ = 1 if r = 1, and δ χ = 0 otherwise. Hence 1/rQ [r, g] Ψχ, λ, U) dλ 1 r P = g 1/Q + 1<r P = J 1 + J, 1/Q m U 1/rQ Λm) 1)eλm 3 ) dλ [r, g] 1/rQ 1/rQ m U 10 ) 1/ 1/ Λm)χm)eλm 3 ) dλ 1/ 5.4)

12 say. By Gallagher s lemma see [3, Lemma 1]), 1/Q ) Λm) 1)eλm 3 1 ) dλ Q where 1/Q m U By the trivial bound one derives Let Y 0 = maxv 1/3, U), = ) 1 8U 3 Q U 3 Q v<m 3 v+q m U X 0 = minv + Q) 1/3, U). Y 0 <m X 0 Λm) 1) X 0 Y 0 )L U QL, J 1 g U 1/ L. Λm) 1) dv Λm) 1) dv, Y 0 <m X 0 We now turn to J. Applying Gallagher s lemma as before, the integral in J is ) 1 8U 3 Λm)χm) dv. 5.5) rq U 3 rq v<m 3 v+rq m U Y = Y v) = maxv 1/3, U), X = Xv) = minv + rq) 1/3, U). 5.6) Then the sum in 5.5) can be rewritten as Λm)χm). 5.7) Y <m X Now we recall Heath-Brown s identity see [4, Lemma 1]) for k = 5, which states that for m M, Λm) = 5 j=1 ) 5 1) j 1 log m 1 )µm j+1) µm j). j m 1 m j =m m j+1,...,m j M) 1/5 Putting this in 5.7), the sum is written into a linear combination of OL 10 ) terms, each of which is of the form ΣX; M) = m 1 M 1 m 10 M 10 Y <m 1 m 10 X a 1 m 1 )χm 1 ) a 10 m 10 )χm 10 ), where a i m) are given by 5.), and M j are positive integers satisfying 5.1) with M = U. Here M denotes the vector M 1, M,..., M 10 ). Therefore we get from 5.4) and 5.5) that J [r, g] 1 { 8U 3 1/ ΣX; M) ) dv}. 5.8) rq 1<r P U 3 rq 11 M

13 Applying Perron s summation formula see for example [13, Lemma 3.1]) and then shifting the contour to the left, we have ΣX; M) = 1 1+1/L+iT πi 1+1/L it { = 1 1/ it πi 1+1/L it F s, χ) Xs Y s ds + O s 1/+iT + + 1/ it 1+1/L+iT 1/+iT ) XL T } F s, χ) Xs Y s ds + O s XL where F s, χ) is as defined in 5.3), and T is a parameter satisfying T X. The integral on the two horizontal segments above is bounded by in view of the trivial estimate Thus, max F σ ± it, χ) Xσ 1/ σ 1+1/L T 10 F σ ± it, χ) = f j σ ± it, χ) L j=1 UL T, 10 j=1 M 1 σ j U 1 σ L. ΣX; M) = 1 T ) 1 X 1/+it F π T + it, χ Y 1/+it dt + O 1/ + it Moreover, one sees from the estimate UL T ). T ), X 1/+it Y 1/+it 1/ + it = X Y u 1/+it du X 1/ Y 1/ U 5/ rq); and the trivial bound X 1/+it Y 1/+it 1/ + it U 1/ t that X 1/+it Y 1/+it 1/ + it ) rq U 1/ min, U 5/ t r U 1/ min P, 1 ). t Thus, by taking T = U, we obtain ΣX; M) ru 1/ P +U 1/ t P/r P/r< t U 1 ) 1 F + it, χ dt ) 1 F + it, χ dt t + OL ).

14 Note that the right hand side is independent of v. Thus by inserting this in 5.8), we obtain say. Clearly J U QP ) 1 +U Q 1 1<r P 1<r P [r, g] M r 1 [r, g] M +U 3/ Q 1 L 1 r 1 [r, g] = J 1 + J + J 3, J 3 U 3/ Q 1 L 1 g 1<r P 1<r P To estimate J 1, one notes that [r, g] = rgr, g) 1. Thus J 1 U 1 g M g U 1 L M r R d r d g t P/R d 5/6 ε By Lemma 5.1, the last double sum is t P/r P/r< t U ) 1 F + it, χ dt ) 1 F + it, χ dt t r 1 g U 3/ P L 1 g U 1/. 1<r P d r d 5/6 ε max d R P R d g ) 1 F + it, χ dt. r t P/r T ) L max 1 1 T P/R F r R T + it, χ dt d r } L c 4+1 {P R/d + P R/d) 1/ U 3/10 + U 1/, ) 1 F + it, χ dt and hence J 1 g U 1 L c 4+1 P 7/6+ε + P 1/ U 3/10 + U 1/ ) g U 1/ L c 4+1, on noting that P U /5. Now it remains to estimate J. We have J g U 1 P L M r R d r P/R t U 13 d 5/6 ε max d R P R 11/6+ε d g ) 1 F + it, χ dt t. 5.9)

15 The last double sum can be estimated via Lemma 5.1 again, which gives 1 T ) L max 1 P/R<T U T F + it, χ dt L c 4+1) Inserting this in 5.9), we get r R d r T {R d 1 + R 3/ dp ) 1/ U 3/10 + RP 1 U 1/ }. J g U 1 P L c 4+1) P 1/6+ε + P 1/ U 3/10 + P 1 U 1/ ) g U 1/ L c 4+1. The first assertion of Lemma 4.4 now follows by letting c 3 = c Now we consider the special case of g = 1. We write 1 r P 1/ 1/rQ r Ψχ, λ, U) dλ) = H 1 + H, 1/rQ where H 1, H denote contributions from those with r L B and L B < r P, respectively with B = 10A. To estimate H, we follow the argument in estimating J but let g = 1 and add the restriction r > L B. We will get H U 1/ L A. To estimate H 1, we apply Gallager s lemma as before to get H 1 r L B r 1 rq where X, Y are as defined by 5.6). m x 8U 3 U 3 rq Y <m X γ T Λm)χm) δ χ ) dv Now we apply the explicit formula see [, 17, 9)-10); 19, 4)-9)]) Λm)χm) = δ χ x x ρ ) xlog qxt ) ρ + O, 5.10) T where < T x is a parameter and ρ = σ + iγ is a typical nontrivial zero of the Dirichlet L-function Ls, χ). Let T = P L B. Then Y <m X Λm)χm) δ χ ) Xρ Y ρ + OUP 1 L B+ ) ρ γ P L B U rq) U σ 1 + OUP 1 L B+ ). γ P L B 1/, Hence H 1 U 1/ r L B γ P L B U σ 1 + U 1/ L A. 14

16 By Satz VIII.6. of Prachar [10] and Siegel s theorem see [, 1]), there exists a positive constant c 5 such that for r L B, Ls, χ) is zero-free in the region σ 1 c 5 / max{log r, log 4/5 x}, t x. Let ηn) = c 5 log 4/5 N. Then by integrating by parts and Lemma 5. with d = 1, we have H 1 U 1/ L 15 max 1/ σ 1 ηn) L4B P ) 1/5+ε)1 σ) U σ 1 + U 1/ L A U 1/ L A, since P U 5/1 ε. Proof of Lemma ) By integrating by parts and noticing λ V 3 < 1, we have Ψχ, λ, V ) = V V max V u V eλu 3 )d V <m u V <m u Λm)χm) δ χ ) Λm)χm) δ χ ). 5.11) By applying 5.10) with T = x/ = V to 5.11), we get r P r P [r, g] g V L d P d g max λ 1/rQ Ψχ, λ, V ) [r, g] 1 + γ ) 1 V β + g P L max 0<T <V 1 + T ) 1 r R d r := K + g V L A, say. By making use of Lemma 5., we have r R d r γ T V β 1 γ V d 5/6 ε max d R P R L 14 R T/d ) A1/)/ V 1/ + L 15 1 γ T 1/ V β 1 + g V L A R T/d ) Aσ)1 σ) V σ 1 dσ, where for 1/ σ < 3/4, Aσ) = 3/ σ); while for 3/4 σ 1, Aσ) = 1/5 + ε. Since Aσ)1 σ) 1 for 1/ σ 1, we see that the total power of T in K is negative. Thus K g V L 17 max 1/ σ 1 d P d g d 5/6 ε 15 max d R P R R /d ) Aσ)1 σ) V σ )

17 Let σ 0 = 8 + 1ε)/13 + 6ε), which is the solution of Aσ)1 σ) = 5/6 ε. Then K g V L 17 max 1/ σ σ 0 P Aσ)1 σ) V σ 1 +g V L 17 max σ 0 <σ 1 P Aσ)1 σ) V σ 1. The first maximum is OL A ) if P V 1/h ε where h = max 1/ σ σ0 {Aα) 5/61 σ)} = 7/3. Similarly the second maximum is O1) if P V 1/h where h = max σ0 σ 1 Aα) = 1/5 + ε. This proves 1). ) We write r P r max Ψχ, λ, V ) = Z 1 + Z, λ 1/rQ where Z 1 and Z denote contributions from those with r L B and L B < r P, respectively. By similar arguments as those leading to 5.1) but with g = 1, d = 1 and the restriction that R L B, we get Z V L 17 max max 1/ σ 1 L B R P R Aσ)1 σ) V σ 1 V L 17 max 1/ σ σ 0 P Aσ)1 σ) V σ 1 V L A, +V L 17 max σ 0 <σ 17/18 P Aσ)1 σ) V σ 1 + V L B/+17 since P V 5/1 ε. This finishes the estimate for Z. Now we turn to Z 1. By 5.11), we have Z 1 r L B max V u V V <m u Λm)χm) δ χ ). Now the desired estimate follows by applying Siegel-Walfisz theorem in the form of the bound: For a primitive character, r log B X Λm)χm) δ χ ) X exp cb) ) log X. m X This proves the lemma. Acknowledgement. The authors are indebted to the referee for valuable suggestions that simplify parts of the argument and bring improvements to the paper. The original version of the paper has the result that EN) N 53/54+ɛ. The improvement obtained in the present Theorem 1.1 is due to an idea of the referee. The first author is supported by a Post-Doctoral Fellowship of The University of Hong Kong. 16

18 References [1] H. Davenport, On Waring s problem for cubes, Acta Math ), [] H. Davenport, Multiplicative Number Theory, nd edition, Springer Verlag, Berlin [3] P.X. Gallagher, A large sieve density estimate near σ = 1, Invent. Math ), [4] D.R. Heath-Brown, Prime numbers in short intervals and a generalized Vaughan s identity, Can. J. Math ), [5] L.K. Hua, Some results in additive prime number theory, Quart. J. Math ), [6] L.K. Hua, Additive theory of prime numbers, Science press, Beijing 1957; English version, American Mathematical Society, Providence [7] M.N. Huxley, Large values of Dirichlet polynomials, III, Acta Arith ), [8] J.Y. Liu, On Lagrange s theorem with prime variables, to appear in Quart. J. Math. [9] H.L. Montgomery, Topics in Multiplicative number theory, LNM 7, Springer-Verlag, Berlin. Heidelberg. New York [10] K. Prachar, Primzahlverteilung, Springer Verlag, Berlin [11] X. Ren, The exceptional set in Roth s theorem concerning a cube and three cubes of primes, Quart. J. Math ), [1] K.F. Roth, On Waring s problem for cubes, Proc. London Math. Soc. ) ), [13] E.C. Titchmarsh Revised by D.R. Heath-Brown), The theory of the Riemann zeta-function, nd edition, Oxford University Press, Oxford [14] R.C. Vaughan, The Hardy-Littlewood method, nd edition, Cambridge University Press, Cambridge