Bounded gaps between primes

Boune gaps between primes Yitang Zhang It is prove that Abstract lim inf n p n+1 p n ) < 7 10 7, p n is the n-th prime. Our metho is a refinement of the recent work of Golston, Pintz an Yilirim on the small gaps between consecutive primes. A major ingreient of the proof is a stronger version of the Bombieri-Vinograov theorem that is applicable when the mouli are free from large prime ivisors only see Theorem 2 below), but it is aequate for our purpose. Contents 1. Introuction 2 2. Notation an sketch of the proof 3 3. Lemmas 7 4. Upper boun for S 1 16 5. Lower boun for S 2 22 6. Combinatorial arguments 24 7. The ispersion metho 27 8. Evaluation of S 3 r, a) 29 9. Evaluation of S 2 r, a) 30 10. A truncation of the sum of S 1 r, a) 34 11. Estimation of R 1 r, a; k): The Type I case 39 12. Estimation of R 1 r, a; k): The Type II case 42 13. The Type III estimate: Initial steps 44 14. The Type III estimate: Completion 48 References 55 1

1. Introuction Let p n enote the n-th prime. It is conjecture that lim inf n p n+1 p n ) = 2. While a proof of this conjecture seems to be out of reach by present methos, recently Golston, Pintz an Yilirim [6] have mae significant progress towar the weaker conjecture lim inf p n+1 p n ) <. 1.1) n In particular, they prove that if the primes have level of istribution ϑ = 1/2 + ϖ for an arbitrarily small) ϖ > 0, then 1.1) will be vali see [6, Theorem 1]). Since the result ϑ = 1/2 is known the Bombieri-Vinograov theorem), the gap between their result an 1.1) woul appear to be, as sai in [6], within a hair s breath. Until very recently, the best result on the small gaps between consecutive primes was ue to Golston, Pintz an Yilirim [7] that gives lim inf n p n+1 p n <. 1.2) log pn log log p n ) 2 One may ask whether the methos in [6], combine with the ieas in Bombieri, Frielaner an Iwaniec [1]-[3] which are employe to erive some stronger versions of the Bombieri- Vinograov theorem, woul be goo enough for proving 1.1) see Question 1 on [6, p.822]). In this paper we give an affirmative answer to the above question. We aopt the following notation of [6]. Let H = {h 1, h 2,..., h k0 } 1.3) be a set compose of istinct non-negative integers. We say that H is amissible if ν p H) < p for every prime p, ν p H) enotes the number of istinct resiue classes moulo p occupie by the h i. Theorem 1. Suppose that H is amissible with k 0 3.5 10 6. Then there are infinitely many positive integers n such that the k 0 -tuple contains at least two primes. Consequently, we have {n + h 1, n + h 2,..., n + h k0 } 1.4) lim inf n p n+1 p n ) < 7 10 7. 1.5) The boun 1.5) results from the fact that the set H is amissible if it is compose of k 0 istinct primes, each of which is greater than k 0, an the inequality π7 10 7 ) π3.5 10 6 ) > 3.5 10 6. 2

This result is, of course, not optimal. The conition k 0 3.5 10 6 is also crue an there are certain ways to relax it. To replace the right sie of 1.5) by a value as small as possible is an open problem that will not be iscusse in this paper. 2. Notation an sketch of the proof Notation p a prime number. a, b, c, h, k, l, m integers., n, q, r positive integers. Λq) the von Mangolt function. τ j q) the ivisor function, τ 2 q) = τq). ϕq) the Euler function. µq) the Möbius function. x a large number. L = log x. y, z real variables. ey) = exp{2πiy}. e q y) = ey/q). y the istance from y to the nearest integer. m aq) means m amo q). c/ means a/mo 1) ac 1mo ). q Q means Q q < 2Q. ε any sufficiently small, positive constant, not necessarily the same in each occurrence. B some positive constant, not necessarily the same in each occurrence. A any sufficiently large, positive constant, not necessarily the same in each occurrence. η = 1 + L 2A. κ N the characteristic function of [N, ηn) Z. a summation over reuce resiue classes lmo q). l mo q) C q a) the Ramanujan sum l mo q) e qla). We aopt the following conventions throughout our presentation. The set H given by 1.3) is assume to be amissible an fixe. We write ν p for ν p H); similar abbreviations will be use in the sequel. Every quantity epening on H alone is regare as a constant. For example, the absolutely convergent prouct S = p 1 ν ) p 1 1 ) k0 p p 3

is a constant. A statement is vali for any sufficiently small ε an for any sufficiently large A whenever they are involve. The meanings of sufficiently small an sufficiently large may vary from one line to the next. Constants implie in O or, unless specifie, will epen on H, ε an A at most. We first recall the unerlying iea in the proof of [6, Theorem 1] which consists in evaluating an comparing the sums S 1 = n x λn) 2 2.1) an S 2 = n x k0 λn) is a real function epening on H an x, an { log n if n is prime, θn) = 0 otherwise. The key point is to prove, with an appropriate choice of λ, that i=1 ) θn + h i ) λn) 2, 2.2) S 2 log 3x)S 1 > 0. 2.3) This implies, for sufficiently large x, that there is a n x such that the tuple 1.4) contains at least two primes. In [6] the function λn) mainly takes the form λn) = D is a power of x an Let an 1 k 0 + l 0 )! P n) D µ) log D ) k0 +l 0, l 0 > 0, 2.4) k 0 P n) = n + h j ). j=1 γ;, c) = γn) 1 n x ϕ) n c) n x n,)=1 γn) for, c) = 1, C i ) = {c : 1 c, c, ) = 1, P c h i ) 0mo )} for 1 i k 0. 4

The evaluations of S 1 an S 2 lea to a relation of the form S 2 log 3x)S 1 = k 0 T 2 LT 1 )x + OxL k 0+2l 0 ) + OE) for D < x 1/2 ε, T1 an T2 are certain arithmetic sums see Lemma 1 below), an E = µ) τ 3 )τ k0 1) θ;, c). <D 2 c C i ) 1 i k 0 Let ϖ > 0 be a small constant. If D = x 1/4+ϖ 2.5) an k 0 is sufficiently large in terms of ϖ, then, with an appropriate choice of l 0, one can prove that k 0 T2 LT1 L k 0+2l 0 +1. 2.6) In this situation the error E can be efficiently boune if the primes have level of istribution ϑ > 1/2 + 2ϖ, but one is unable to prove it by present methos. On the other han, for D = x 1/4 ε, the Bombieri-Vinograov theorem is goo enough for bouning E, but the relation 2.6) can not be vali, even if a more general form of λn) is consiere see Sounararajan [12]). Our first observation is that, in the sums T1 an T2, the contributions from the terms with having a large prime ivisor are relatively small. Thus, if we impose the constraint P in 2.4), P is the prouct of the primes less than a small power of x, the resulting main term is still L k 0+2l 0 +1 with D given by 2.5). Our secon observation, which is the most novel part of the proof, is that with D given by 2.5) an with the constraint P impose in 2.4), the resulting error τ 3 )τ k0 1) θ;, c) 2.7) 1 i k 0 <D 2 P c C i ) can be efficiently boune. This is originally ue to the simple fact that if P an is not too small, say > x 1/2 ε, then can be factore as = rq 2.8) with the range for r flexibly chosen see Lemma 4 below). Thus, roughly speaking, the characteristic function of the set { : x 1/2 ε < < D 2, P} may be treate as a well factorable function see Iwaniec [10]). The factorization 2.8) is crucial for bouning the error terms. It suffices to prove Theorem 1 with k 0 = 3.5 10 6 5

which is henceforth assume. Let D be as in 2.5) with Let gy) be given by gy) = ϖ = 1 1168. 1 log D ) k0 +l 0 if y < D, k 0 + l 0 )! y an Write gy) = 0 if y D, D 1 = x ϖ, l 0 = 180. D 0 = exp{l 1/k 0 }, P = p, 2.9) p<d 1 P 0 = p. 2.10) p D 0 In the case P an is not too small, the factor q in 2.8) may be chosen such that q, P 0 ) = 1. This will consierably simplify the argument. We choose λn) = µ)g). 2.11) P n),p) In the proof of Theorem 1, the main terms are not ifficult to hanle, since we eal with a fixe H. This is quite ifferent from [6] an [7], in which various sets H are involve in the argument to erive results like 1.2). By Cauchy s inequality, the error 2.7) is efficiently boune via the following Theorem 2. For 1 i k 0 we have θ;, c) xl A. 2.12) <D 2 P c C i ) The proof of Theorem 2 is escribe as follows. First, applying combinatorial arguments see Lemma 6 below), we reuce the proof to estimating the sum of γ;, c) with certain Dirichlet convolutions γ. There are three types of the convolutions involve in the argument. Write x 1 = x 3/8+8ϖ, x 2 = x 1/2 4ϖ. 2.13) In the first two types the function γ is of the form γ = α β such that the following hol. A 1 ) α = αm)) is supporte on [M, η j 1 M), j 1 19, αm) τ j1 m)l. 6

A 2 ) β = βn)) is supporte on [N, η j 2 N), j 2 19, βn) τ j2 n)l, x 1 < N < 2x 1/2. For any q, r an a satisfying a, r) = 1, the following Siegel-Walfisz assumption is satisfie. n ar) n,q)=1 βn) 1 ϕr) n,qr)=1 A 3 ) j 1 + j 2 20, [MN, η 20 MN) [x, 2x). βn) τ 20 q)nl 200A. We say that γ is of Type I if x 1 < N x 2 ; we say that γ is of Type II if x 2 < N < 2x 1/2. In the Type I an II estimates we combine the ispersion metho in [1] with the factorization 2.8) here r is close to N in the logarithmic scale). Due to the fact that the moulo is at most slightly greater than x 1/2 in the logarithmic scale, after reucing the problem to estimating certain incomplete Kloosterman sums, we nee only to save a small power of x from the trivial estimates; a variant of Weil s boun for Kloosterman sums see Lemma 11 below) will fulfill it. Here the conition N > x 1, which may be slightly relaxe, is essential. We say that γ is of Type III if it is of the form γ = α κ N1 κ N2 κ N3 such that α satisfies A 1 ) with j 1 17, an such that the following hol. A 4 ) N 3 N 2 N 1, MN 1 x 1. A 5 ) [MN 1 N 2 N 3, η 20 MN 1 N 2 N 3 ) [x, 2x). The Type III estimate essentially relies on the Birch-Bombieri result in the appenix to [5] see Lemma 12 below), which is employe by Frielaner an Iwaniec [5] an by Heath-Brown [9] to stuy the istribution of τ 3 n) in arithmetic progressions. This result in turn relies on Deligne s proof of the Riemann Hypothesis for varieties over finite fiels the Weil Conjecture) [4]. We estimate each γ;, c) irectly. However, if one applies the metho in [5] alone, efficient estimates will be vali only for MN 1 x 3/8 5ϖ/2 ε. Our argument is carrie out by combining the metho in [5] with the factorization 2.8) here r is relatively small); the latter will allow us to save a factor r 1/2. In our presentation, all the αm) an βn) are real numbers. 3. Lemmas In this section we introuce a number of prerequisite results, some of which are quote from the literature irectly. Results given here may not be in the strongest forms, but they are aequate for the proofs of Theorem 1 an Theorem 2. Lemma 1. Let ϱ 1 ) an ϱ 2 ) be the multiplicative functions supporte on square-free integers such that ϱ 1 p) = ν p, ϱ 2 p) = ν p 1. 7

Let an We have an T 2 = T 1 = 0 1 T 2 = 0 T 1 = 1 µ 1 2 )ϱ 1 0 1 2 ) g 0 1 )g 0 2 ) 0 1 2 2 1 k 0 + 2l 0 )! 1 k 0 + 2l 0 + 1)! µ 1 2 )ϱ 2 0 1 2 ) ϕ 0 1 2 ) 2 2l0 l 0 g 0 1 )g 0 2 ). ) Slog D) k 0+2l 0 + ol k 0+2l 0 ) 3.1) ) 2l0 + 2 Slog D) k 0+2l 0 +1 + ol k 0+2l 0 +1 ). 3.2) l 0 + 1 Proof. The sum T 1 is the same as the sum T R l 1, l 2 ; H 1, H 2 ) in [6, 7.6)] with H 1 = H 2 = H k 1 = k 2 = k 0 ), l 1 = l 2 = l 0, R = D, so 3.1) follows from [6, Lemma 3]; the sum T 2 is the same as the sum T R l 1, l 2 ; H 1, H 2, h 0 ) in [6, 9.12)] with H 1 = H 2 = H, l 1 = l 2 = l 0, h 0 H, R = D, so 3.2) also follows from [6, Lemma 3]. an Remark. A generalization of this lemma can be foun in [12]. Lemma 2. Let A 1 ) = r,)=1 A 2 ) = r,)=1 µr)ϱ 1 r) gr) r µr)ϱ 2 r) gr). ϕr) Suppose that < D an µ) = 1. Then we have A 1 ) = ϑ 1) S log D ) l0 + OL l0 1+ε ) 3.3) l 0! an A 2 ) = ϑ 2) l 0 + 1)! S log D ) l0 +1 + OL l0+ε ), 3.4) ϑ 1 ) an ϑ 2 ) are the multiplicative functions supporte on square-free integers such that ϑ 1 p) = 1 ν ) 1 p, ϑ 2 p) = 1 ν ) 1 p 1. p p 1 8

Proof. Recall that D 0 is given by 2.10). Since ϱ 1 r) τ k0 r), we have trivially A 1 ) 1 + logd/) ) 2k 0 +l 0, so we may assume D/ > exp{log D 0 ) 2 } without loss of generality. Write s = σ + it. For σ > 0 we have µr)ϱ 1 r) = ϑ r 1+s 1, s)g 1 s)ζ1 + s) k 0 ϑ 1, s) = p It follows that r,)=1 1 ν p p 1+s ) 1, G 1 s) = A 1 ) = 1 2πi 1/L) p 1 ν ) p 1 1 ) k0. p 1+s p 1+s ϑ 1, s)g 1 s) D/) s s ζ1 + s) k 0 s. k 0+l 0 +1 Note that G 1 s) is analytic an boune for σ 1/3. We split the line of integration into two parts accoring to t D 0 an t > D 0. By a well-known result on the zero-free region for ζs), we can move the line segment {σ = 1/L, t D 0 } to { σ = κlog D0 ) 1, t D 0 }, κ > 0 is a certain constant, an apply some stanar estimates to euce that A 1 ) = 1 ϑ 1, s)g 1 s) D/) s s 2πi ζ1 + s) k 0 s k 0+l 0 + OL A ). +1 Note that ϑ 1, 0) = ϑ 1 ) an s =1/L ϑ 1, s) ϑ 1 ) = ϑ 1, s)ϑ 1 ) l µl)ϱ 1 l) 1 l s ). l If s 1/L, then ϑ 1, s) log L) B, so that, by trivial estimation, ϑ 1, s) ϑ 1 ) L ε 1. On the other han, by Cauchy s integral formula, for s 1/L we have It follows that 1 2πi s =1/L G 1 s) S 1/L. ϑ 1, s)g 1 s) D/) s s ζ1 + s) k 0 s k 0+l 0 1 +1 2πi ϑ D/) s s 1)S L l0 1+ε. s =1/L s l 0+1 9

This leas to 3.3). The proof of 3.4) is analogous. We have only to note that A 2 ) = 1 ϑ 2, s)g 2 s) D/) s s 2πi ζ1 + s) k 0 1 s k 0+l 0 +1 with ϑ 2, s) = p an G 2 0) = S. an 1 ν p 1 p 1)p s Lemma 3. We have ϱ 1 )ϑ 1 ) <x 1/4 Proof. ϱ 2 )ϑ 2 ) ϕ) <x 1/4 1/L) ) 1, G 2 s) = p 1 ν ) p 1 1 1 ) 1 k0, p 1)p s p 1+s = 1 + 4ϖ) k 0 S 1 log D) k 0 + OL k0 1 ) 3.5) k 0! = 1 + 4ϖ)1 k 0 S 1 log D) k0 1 + OL k0 2 ), 3.6) k 0 1)! Noting that ϑ 1 p)/p = 1/p ν p ), for σ > 0 we have =1 B 1 s) = p Hence, by Perron s formula, ϱ 1 )ϑ 1 ) <x 1/4 = 1 2πi ϱ 1 )ϑ 1 ) 1+s = B 1 s)ζ1 + s) k 0, ) ν p 1 + 1 1 ) k0. p ν p )p s p 1+s 1/L+iD0 B 1 s)ζ1 + s) k0 x s/4 1/L id 0 s s + OD 1 0 L B ). Note that B 1 s) is analytic an boune for σ 1/3. Moving the path of integration to [ 1/3 id 0, 1/3 + id 0 ], we see that the right sie above is equal to 1 B 1 s)ζ1 + s) k 0 x s/4 s + OD0 1 L B ). 2πi s s =1/L Since, by Cauchy s integral formula, B 1 s) B 1 0) 1/L for s = 1/L, an B 1 0) = ) p 1 1 k0 = S p ν p p p) 1, 10

it follows that ϱ 1 )ϑ 1 ) <x 1/4 = 1 L k 0! S 1 4 ) k0 + OL k 0 1 ). This leas to 3.5) since L/4 = 1 + 4ϖ) 1 log D by 2.5). The proof of 3.6) is analogous. We have only to note that, for σ > 0, =1 ϱ 2 )ϑ 2 ) ϕ)p s = B 2 s)ζ1 + s) k 0 1 with B 2 s) = p 1 + ν ) p 1 1 1 ) k0 1, p ν p )p s p 1+s an B 2 0) = S 1. Recall that D 1 an P are given by 2.9), an P 0 is given by 2.10). Lemma 4. Suppose that > D 2 1, P an, P 0 ) < D 1. For any R satisfying D 2 1 < R <, 3.7) there is a factorization = rq such that D 1 1 R < r < R an q, P 0 ) = 1. Proof. Since is square-free an /, P 0 ) > D 1, we may write /, P 0 ) as n, P 0 ) = p j with D 0 < p 1 < p 2 <... < p n < D 1, n 2. j=1 By 3.7), there is a n < n such that n n +1, P 0 ) p j < R an, P 0 ) p j R. j=1 The assertion follows by choosing an noting that r 1/p n +1)R. n r =, P 0 ) p j, q = j=1 j=1 n j=n +1 Lemma 5. Suppose that 1 i k 0 an µqr) = 1. There is a bijection p j, C i qr) C i r) C i q), c a, b) 11

such that cmo qr) is a common solution to c amo r) an c bmo q). Proof. By the Chinese remainer theorem. The next lemma is a special case of the combinatorial ientity ue to Heath-Brown[8]. Lemma 6 Suppose that x 1/10 x < ηx 1/10. For n < 2x we have 10 ) 10 Λn) = 1) j 1 µm 1 )...µm j ) j j=1 m 1,...,m j x The next lemma is a truncate Poisson formula. n 1...n j m 1...m j =n log n 1. Lemma 7 Suppose that η η η 19 an x 1/4 < M < x 2/3. Let f be a function of C, ) class such that 0 fy) 1, an fy) = 1 if M y η M, fy) = 0 if y / [1 M ε )M, 1 + M ε )η M], f j) y) M j1 ε), j 1, the implie constant epening on ε an j at most. Then we have fm) = 1 ˆfh/)e ah) + Ox 2 ) m a) h <H for any H M 1+2ε, ˆf is the Fourier transform of f, i.e., ˆfz) = fy)eyz) y. Lemma 8. Suppose that 1 N < N < 2x, N N > x ε an c, ) = 1. Then for j, ν 1 we have N n N n c) τ j n) ν N N ϕ) Ljν 1, the implie constant epening on ε, j an ν at most. Proof. See [11, Theorem 1]. The next lemma is essentially) containe in the proof of [5, Theorem 4]. Lemma 9 Suppose that H, N 2, > H an c, ) = 1. Then we have min { H, c n/ 1} N) ε H + N). 3.8) n N n,)=1 12

Proof. We may assume N H without loss of generality. Write {y} = y [y] an assume ξ [1/H, 1/2]. Note that {c n/} ξ if an only if bn cmo ) for some b 0, ξ], an 1 ξ {c n/} if an only if bn cmo ) for some b 0, ξ], Thus, the number of the n satisfying n N, n, ) = 1 an c n/ ξ is boune by τq) ε N 1+ε ξ. q Nξ q ±c) Hence, for any interval I of the form I = 0, 1/H], I = [1 1/H, 1), I = [ξ, ξ ] or I = [1 ξ, 1 ξ] with 1/H ξ < ξ 1/2, ξ 2ξ, the contribution from the terms on the left sie of 3.8) with {c n/} I is ε N 1+ε. This completes the proof. Lemma 10. Suppose that β = βn)) satisfies A 2 ) an R x ε N. Then for any q we have ϱ 2 r) βn) 1 2 βn) ϕr) τq) B N 2 L 100A. r R l mo r) n lr) n,q)=1 n,qr)=1 Proof. Since the inner sum is ϕr) 1 N 2 L B by Lemma 8, the assertion follows by Cauchy s inequality an [1, Theorem 0]. Lemma 11 Suppose that N 1, 1 2 > 10 an µ 1 ) = µ 2 ) = 1. Then we have, for any c 1, c 2 an l, n N n, 1 )=1 n+l, 2 )=1 c1 n e + c ) 2n + l) 1 2 ) 1/2+ε + c 1, 1 )c 2, 2 ) 1, 2 ) 2 N. 3.9) 1 2 1 2 Proof. Write 0 = 1, 2 ), t 1 = 1 / 0, t 2 = 2 / 0 an = 0 t 1 t 2. Let K 1, c 1 ; 2, c 2 ; l, m) = n n, 1 )=1 n+l, 2 )=1 c1 n e + c 2n + l) + mn ). 1 2 We claim that K 1, c 1 ; 2, c 2 ; l, m) 0 Sm, b 1 ; t 1 )Sm, b 2 ; t 2 ) 3.10) for some b 1 an b 2 satisfying b i, t i ) c i, i ), 3.11) 13

Sm, b; t) enotes the orinary Kloosterman sum. Note that 0, t 1 an t 2 are pairwise coprime. Assume that an n t 1 t 2 n 0 + 0 t 2 n 1 + 0 t 1 n 2 mo ) l t 1 t 2 l 0 + 0 t 1 l 2 mo 2 ). The conitions n, 1 ) = 1 an n + l, 2 ) = 1 are equivalent to n 0, 0 ) = n 1, t 1 ) = 1 an n 0 + l 0, 0 ) = n 2 + l 2, t 2 ) = 1 respectively. Letting a i mo 0 ), b i mo t i ), i = 1, 2 be given by so that 3.11) hols, by the relation a 1 t 2 1t 2 c 1 mo 0 ), a 2 t 1 t 2 2 c 2 mo 0 ), b 1 2 0t 2 c 1 mo t 1 ), b 2 2 0t 1 c 2 mo t 2 ), we have Hence, 1 i t i 0 + 0 t i mo 1) c 1 n 1 + c 2n + l) 2 a 1 n 0 + a 2 n 0 + l 0 ) 0 + b 1 n 1 t 1 + b 2n 2 + l 2 ) t 2 mo 1). c 1 n + c 2n + l) + mn 1 2 a 1 n 0 + a 2 n 0 + l 0 ) + mn 0 0 + b 1 n 1 + mn 1 t 1 + b 2n 2 + l 2 ) + mn 2 + l 2 ) t 2 ml 2 t 2 mo 1). From this we euce, by the Chinese remainer theorem, that K 1, c 1 ; 2, c 2 ; l, m) = e t2 ml 2 )Sm, b 1 ; t 1 )Sm, b 2 ; t 2 ) k n<k+ n, 1 =1 n+l, 2 )=1 n 0 n, 0 )=1 n+l 0, 0 )=1 whence 3.10) follows. By 3.10) with m = 0 an 3.11), for any k > 0 we have c1 n e + c ) 2n + l) c 1, 1 )c 2, 2 ) 0. 1 2 14 e 0 a1 n + a 2 n + l 0 ) + mn ),

It now suffices to prove 3.9) on assuming N 1. By stanar Fourier techniques, the left sie of 3.9) may be rewritten as um)k 1, c 1 ; 2, c 2 ; l, m) with <m< { N um) min, 1 m, }. 3.12) m 2 By 3.10) an Weil s boun for Kloosterman sums, we fin that the left sie of 3.9) is 0 u0) b 1, t 1 )b 2, t 2 ) + t 1 t 2 ) 1/2+ε um) m, b 1, t 1 ) 1/2 m, b 2, t 2 ) ). 1/2 m 0 This leas to 3.9) by 3.12) an 3.11). Remark. In the case 2 = 1, 3.9) becomes n N n, 1 )=1 e 1 c 1 n) 1/2+ε 1 + c 1, 1 )N 1. 3.13) This estimate is well-known see [2, Lemma 6], for example), an it will fin application some. Lemma 12. Let T k; m 1, m 2 ; q) = l mo q) t 1 mo q) t 2 mo q) e q lt1 l + k)t 2 + m 1 t 1 m 2 t 2 ), is restriction to ll + k), q) = 1. Suppose that q is square-free. Then we have T k; m 1, m 2 ; q) k, q) 1/2 q 3/2+ε. Proof. By [5, 1.26)], it suffices to show that T k; m 1, m 2 ; p) k, p) 1/2 p 3/2. In the case k 0mo p), this follows from the Birch-Bombieri result in the appenix to [5] the proof is straightforwar if m 1 m 2 0mo p)); in the case k 0mo p), this follows from Weil s boun for Kloosterman sums. 15

4. Upper boun for S 1 Recall that S 1 is given by 2.1) an λn) is given by 2.11). The aim of this section is to establish an upper boun for S 1 see 4.20) below). Changing the orer of summation we obtain S 1 = µ 1 )g 1 )µ 2 )g 2 ) 1. 1 P 2 P n x P n) 0[ 1, 2 ]) By the Chinese remainer theorem, for any square-free, there are exactly ϱ 1 ) istinct resiue classes mo ) such that P n) 0mo ) if an only if n lies in one of these classes, so the innermost sum above is equal to It follows that T 1 = 1 P ϱ 1 [ 1, 2 ]) x + Oϱ 1 [ 1, 2 ])). [ 1, 2 ] 2 P S 1 = T 1 x + OD 2+ε ), 4.1) µ 1 )g 1 )µ 2 )g 2 ) ϱ 1 [ 1, 2 ]). [ 1, 2 ] Note that ϱ 1 ) is supporte on square-free integers. Substituting 0 = 1, 2 ) an rewriting 1 an 2 for 1 / 0 an 2 / 0 respectively, we euce that T 1 = 0 P 1 P 2 P We nee to estimate the ifference T 1 T 1. We have Σ 1 = Σ 31 = 0 x 1/4 x 1/4 < 0 <D 1 µ 1 2 )ϱ 1 0 1 2 ) 0 1 2 g 0 1 )g 0 2 ). 4.2) T 1 = Σ 1 + Σ 31, µ 1 2 )ϱ 1 0 1 2 ) g 0 1 )g 0 2 ), 0 1 2 2 1 µ 1 2 )ϱ 1 0 1 2 ) g 0 1 )g 0 2 ). 0 1 2 2 In the case 0 > x 1/4, 0 1 < D, 0 2 < D an µ 1 2 ) = 1, the conitions i P, i = 1, 2 are reunant. Hence, T 1 = Σ 2 + Σ 32, 16

It follows that Σ 2 = Σ 32 = Σ 3 = 0 x 1/4 0 P x 1/4 < 0 <D 0 P x 1/4 < 0 <D 0 P 1, 0 )=1 2, 0 )=1 1 P 2 P 1 µ 1 2 )ϱ 1 0 1 2 ) 0 1 2 g 0 1 )g 0 2 ), µ 1 2 )ϱ 1 0 1 2 ) g 0 1 )g 0 2 ). 0 1 2 2 T 1 T 1 Σ 1 + Σ 2 + Σ 3, 4.3) 1 µ 1 2 )ϱ 1 0 1 2 ) g 0 1 )g 0 2 ). 0 1 2 2 First we estimate Σ 1. By Möbius inversion, the inner sum over 1 an 2 in Σ 1 is equal to ϱ 1 0 ) µ 1 )ϱ 1 1 )µ 2 )ϱ 1 2 ) ) g 0 1 )g 0 2 ) µq) 0 1 2 It follows that = ϱ 1 0 ) 0 q, 0 )=1 Σ 1 = µq)ϱ 1 q) 2 0 x 1/4 q, 0 )=1 q 2 A 1 0 q) 2. ϱ 1 0 )µq)ϱ 1 q) 2 q 1, 2 ) 0 q 2 A 1 0 q) 2. 4.4) The contribution from the terms with q D 0 above is D 1 0 L B. Thus, substituting 0 q =, we euce that Σ 1 = ϱ 1 )ϑ ) A 1 ) 2 + OD0 1 L B ), 4.5) <x 1/4 D 0 By the simple bouns which follows from 3.3), ϑ ) = 0 q= 0 <x 1/4 q<d 0 µq)ϱ 1 q). q A 1 ) L l 0 log L) B 4.6) ϑ ) log L) B 17

an ϱ 1 ) L k 0+1/k 0 1, 4.7) x 1/4 <x 1/4 D 0 the contribution from the terms on the right sie of 4.5) with x 1/4 < x 1/4 D 0 is ol k 0+2l 0 ). On the other han, assuming µ) = 1 an noting that for < x 1/4 we have q µq)ϱ 1 q) q so that, by 3.3), ϑ )A 1 ) 2 = 1 l 0!) 2 S2 ϑ 1 ) log D Inserting this into 4.5) we obtain Σ 1 = 1 l 0!) 2 S2 Together with 3.5), this yiels Σ 1 ϑ ) = ϑ 1 ) 1 + O ) τ k0 +1)D0 1, ϱ 1 )ϑ 1 ) x 1/4 = ϑ 1 ) 1, 4.8) ) 2l0 + O τ k0+1)d 1 0 L B) + OL 2l 0 1+ε ). log D ) 2l0 + ol k 0+2l 0 ). 4.9) δ 1 k 0!l 0!) 2 Slog D)k 0+2l 0 + ol k 0+2l 0 ), 4.10) δ 1 = 1 + 4ϖ) k 0. Next we estimate Σ 2. Similar to 4.4), we have Σ 2 = 0 x 1/4 0 P q, 0 )=1 q P A 1) = r,)=1 r P ϱ 1 0 )µq)ϱ 1 q) 2 A 0 q 1 2 0 q) 2. µr)ϱ 1 r)gr). r In a way similar to the proof of 4.5), we euce that Σ 2 = <x 1/4 D 0 P ϱ 1 )ϑ ) A 1) 2 + OD0 1 L B ). 4.11) 18

Assume P. By Möbius inversion we have Noting that A 1) = by 3.3) we euce that r,)=1 µr)ϱ 1 r)gr) r P = q r,p ) D 1 p<d µq) = ϱ 1 q) A 1 q), q q P p. ϑ 1 q) = 1 + OD 1 1 ) if q P an q < D, 4.12) A 1) 1 l 0! Sϑ 1) log D ) l0 ϱ 1 q) + OL l0 1+ε ). 4.13) q q P q<d If q P an q < D, then q has at most 292 prime factors. In aition, by the prime number theorem we have It follows that D 1 p<d 1 p = log 293 + OL A ). 4.14) q P q<d ϱ 1 q) q 1 + 292 ν=1 log 293)k 0 ) ν ν! + OL A ) = δ 2 + OL A ), say. Inserting this into 4.13) we obtain A 1) δ 2 l 0! Sϑ 1) log D ) l0 + OL l0 1+ε ). Combining this with 4.11), in a way similar to the proof of 4.9) we euce that Σ 2 δ2 2 l 0!) 2 S2 Together with 3.5), this yiels ϱ 1 )ϑ 1 ) <x 1/4 log D ) 2l0 + ol k 0+2l 0 ). Σ 2 δ 1δ 2 2 k 0!l 0!) 2 Slog D)k 0+2l 0 + ol k 0+2l 0 ). 4.15) 19

We now turn to Σ 3. In a way similar to the proof of 4.5), we euce that Σ 3 = x 1/4 <<D ϑ) = 0 q= x 1/4 < 0 0 P ϱ 1 ) ϑ) A 1 ) 2, 4.16) µq)ϱ 1 q). q By 4.6) an 4.7), we fin that the contribution from the terms with x 1/4 < x 1/4 D 0 in 4.16) is ol k 0+2l 0 ). Now assume that x 1/4 D 0 < < D, µ) = 1 an P. Noting that the conitions 0 an x 1/4 < 0 together imply 0 P, by 4.8) we obtain ϑ) = 0 q= x 1/4 < 0 µq)ϱ 1 q) q Together with 3.3), this yiels ϑ)a 1 ) 2 = 1 l 0!) 2 S2 ϑ 1 ) log D Combining these results with 4.16) we obtain Σ 3 = 1 l 0!) 2 S2 x 1/4 D 0 <<D P By 4.12), 4.14) an 3.5) we have x 1/4 <<D P ϱ 1 )ϑ 1 ) Together with 4.17), this yiels <D = ϑ 1 ) 1 + O ) τ k0 +1)D0 1. ) 2l0 + O τ k0+1)d 1 0 L B) + OL 2l 0 1+ε ). ϱ 1 )ϑ 1 ) D 1 p<d ϱ 1 )ϑ 1 ) log D ) 2l0 + ol k 0+2l 0 ). 4.17) ϱ 1 p)ϑ 1 p) p p,p ) 1 <D/p ϱ 1 )ϑ 1 ) log 293)δ 1 k 0 1)! S 1 log D) k 0 + ol k 0 ) Σ 3 log 293)δ 1 k 0 1)!l 0!) 2 Slog D)k 0+2l 0 + ol k 0+2l 0 ). 4.18) 20

Since 1 k 0!l 0!) = 1 k0 + 2l 0 2 k 0 + 2l 0 )! it follows from 4.3), 4.10), 4.15) an 4.18) that T 1 T1 2l0 κ 1 k 0 + 2l 0 )! l 0 k 0 ) 2l0 l 0 ), ) Slog D) k 0+2l 0 + ol k 0+2l 0 ), 4.19) ) κ 1 = δ 1 1 + δ2 2 k0 + 2l 0 + log 293)k 0 ). Together with 3.1), this implies that T 1 1 + κ ) 1 2l0 Slog D) k 0+2l 0 + ol k 0+2l 0 ). k 0 + 2l 0 )! l 0 Combining this with 4.1), we euce that S 1 1 + κ ) 1 2l0 Sxlog D) k 0+2l 0 + oxl k 0+2l 0 ). 4.20) k 0 + 2l 0 )! l 0 We conclue this section by giving an upper boun for κ 1. By the inequality an simple computation we have n! > 2πn) 1/2 n n e n ) log 1 + δ2 2 293)k0 ) 292 2 + log 293)k 0 < 2 < 1 292! 292π 185100)584 k 0 an k0 + 2l 0 It follows that This gives k 0 ) < 2k2l 0 0 2l 0 )! < 1 180π 26500) 360. log κ 1 < 3500000 log 293 + 584 log185100) + 360 log26500) < 1200. 292 κ 1 < exp{ 1200}. 4.21) 21

5. Lower boun for S 2 Recall that S 2 is given by 2.2). The aim of this section is to establish a lower boun for S 2 on assuming Theorem 2 see 5.6) below), which together with 4.20) leas to 2.3). We have S 2 = θn)λn h i ) 2 + Ox ε ). 5.1) 1 i k 0 n x Assume that 1 i k 0. Changing the orer of summation we obtain µ 1 )g 1 )µ 2 )g 2 ) n x θn)λn h i ) 2 = 1 P 2 P n x P n h i ) 0[ 1, 2 ]) θn). Now assume µ) = 1. To hanle the innermost sum we first note that the conition P n h i ) 0 mo ) an n, ) = 1 is equivalent to n cmo ) for some c C i ). Further, for any p, the quantity C i p) is equal to the number of istinct resiue classes mo p) occupie by the h i h j with h j h i mo p), so C i p) = ν p 1. This implies C i ) = ϱ 2 ) by Lemma 5. Thus the innermost sum above is equal to c C i [ 1, 2 ]) n x n c[ 1, 2 ]) θn) = ϱ 2[ 1, 2 ]) ϕ[ 1, 2 ]) θn) + n x c C i [ 1, 2 ]) θ; [ 1, 2 ], c). Since the number of the pairs { 1, 2 } such that [ 1, 2 ] = is equal to τ 3 ), it follows that θn)λn h i ) 2 = T 2 θn) + OE i ), 5.2) n x T 2 = 1 P which is inepenent of i, an 2 P E i = <D 2 P n x µ 1 )g 1 )µ 2 )g 2 ) ϱ 2 [ 1, 2 ]) ϕ[ 1, 2 ]) τ 3 )ϱ 2 ) By Cauchy s inequality an Theorem 2 we have c C i ) θ;, c). E i xl A. 5.3) 22

It follows from 5.1)-5.3) an the prime number theorem that Similar to 4.2), we may rewrite T 2 as T 2 = 0 P 1 P S 2 = k 0 T 2 x + OxL A ). 5.4) 2 P µ 1 2 )ϱ 2 0 1 2 ) g 0 1 )g 0 2 ). ϕ 0 1 2 ) In a way much similar to the proof of 4.19), from the secon assertions of Lemma 2 an Lemma 3 we euce that ) T 2 T2 κ 2 2l0 + 2 < Slog D) k 0+2l 0 +1 + ol k 0+2l 0 +1 ), 5.5) k 0 + 2l 0 + 1)! l 0 + 1 ) κ 2 = δ 1 1 + 4ϖ)1 + δ2 2 k0 + 2l 0 + 1 + log 293)k 0 ). k 0 1 Together with 3.2), this implies that T 2 1 κ 2 k 0 + 2l 0 + 1)! 2l0 + 2 l 0 + 1 ) Slog D) k 0+2l 0 +1 + ol k 0+2l 0 +1 ). Combining this with 5.4), we euce that S 2 k ) 01 κ 2 ) 2l0 + 2 Sxlog D) k 0+2l 0 +1 + oxl k 0+2l 0 +1 ). 5.6) k 0 + 2l 0 + 1)! l 0 + 1 We are now in a position to prove Theorem 1 on assuming Theorem 2. By 4.20), 5.6) an the relation 4 L = log D, 1 + 4ϖ we have S 2 log 3x)S 1 ωsxlog D) k 0+2l 0 +1 + oxl k 0+2l 0 +1 ), 5.7) ω = k 01 κ 2 ) k 0 + 2l 0 + 1)! which may be rewritten as 1 2l0 ω = k 0 + 2l 0 )! Note that l 0 ) 2l0 + 2 l 0 + 1 ) 22l0 + 1) l 0 + 1 κ 2 = k 0k 0 + 2l 0 + 1)1 + 4ϖ) κ 1 2l 0 + 1)2l 0 + 2) 23 41 + κ 1 ) 1 + 4ϖ)k 0 + 2l 0 )! 2l0 l 0 ), k 0 1 κ 2 ) k 0 + 2l 0 + 1 41 + κ ) 1). 1 + 4ϖ < 10 8.

Thus, by 4.21), both of the constants κ 1 an κ 2 are extremely small. It follows by simple computation that ω > 0. 5.8) Finally, from 5.7) an 5.8) we euce 2.3), whence Theorem 1 follows. Remark. The bouns 4.19) an 5.5) are crue an there may be some ways to improve them consierably. It is even possible to evaluate T 1 an T 2 irectly. Thus one might be able to show that 2.3) hols with a consierably smaller k 0. 6. Combinatorial arguments The rest of this paper is evote to proving Theorem 2. sections we assume that 1 i k 0. Write In this an the next six D 2 = x 1/2 ε. On the left sie of 2.12), the contribution from the terms with D 2 is xl A by the Bombieri-Vinograov Theorem. Recalling that D 1 an P 0 are given by 2.9) an 2.10) respectively, by trivial estimation, for D 2 < < D 2 we may also impose the constraint, P 0 ) < D 1, an replace θn) by Λn). Thus Theorem 2 follows from the following Λ;, c) xl A. 6.1) D 2 <<D 2 P,P 0 )<D 1 c C i ) The aim of this section is to reuce the proof of 6.1) to showing that γ;, c) xl 41A 6.2) D 2 <<D 2 P,P 0 )<D 1 c C i ) for γ being of Type I, II or III. Let L be given by Ln) = log n. By Lemma 6, for n x we have Λn) = Λ 1 n) 10 ) 10 Λ 1 = 1) j 1 µκ Mj )... µκ M1 ) κ Nj )... Lκ N1 ). j j=1 M j,...,m 1,N j,...,n 1 Here M j,..., M 1, N j,..., N 1 1 run over the powers of η satisfying M t x 1/10, 6.3) [M j...m 1 N j...n 1, η 20 M j...m 1 N j...n 1 ) [x, 2x) φ. 6.4) 24

Let Λ 2 have the same expression as Λ 1 but with the constraint 6.4) replace by [M j...m 1 N j...n 1, η 20 M j...m 1 N j...n 1 ) [x, 2x). 6.5) Since Λ 1 Λ 2 is supporte on [η 20 x, η 20 x] [2η 20 x, 2η 20 x] an Λ 1 Λ 2 )n) τ 20 n)l, by Lemma 8 we have Λ 1 Λ 2 ;, c) xl A. Further, let D 2 <<D 2 P,P 0 )<D 1 c C i ) 10 ) 10 Λ 3 = 1) j 1 log N 1 ) µκ Mj )... µκ M1 ) κ Nj )... κ N1 ) 6.6) j j=1 M j,...,m 1,N j,...,n 1 with M j,..., M 1, N j,..., N 1 satisfying 6.3) an 6.5). Since Λ 2 Λ 3 )n) τ 20 n)l 2A, by Lemma 8 we have Λ 2 Λ 3 ;, c) xl A. D 2 <<D 2 P,P 0 )<D 1 c C i ) Now assume that 1 j j 10. Let γ be of the form γ = log N j )µκ Mj )... µκ M1 ) κ Nj )... κ N1 ). with M j,..., M 1, N j,..., N 1 satisfying 6.3) an 6.5), an N j... N 1. We claim that either the estimate γ;, c) x1 ϖ+ε 6.7) trivially hols for < D 2 an c, ) = 1, or γ is of Type I, II or III. Write M t = x µt an N t = x νt. We have 0 µ t 1 10, 0 ν j... ν 1, 1 µ j +... + µ 1 + ν j +... + ν 1 < 1 + log 2 L. In the case 3/8 + 8ϖ < ν 1 1/2, γ is of Type I or II by choosing β = κ N1 ; in the case 1/2 < ν 1 1/2 + 3ϖ, γ is of Type II by choosing α = κ N1 ; in the case 1/2 + 3ϖ < ν 1, the estimate 6.7) trivially hols. Since ν 1 2/5 if j = 1, 2, it remains to eal with the case j 3, ν 1 3 8 + 8ϖ. 25

Write ν = µ j +... + µ 1 + ν j +... + ν 4 the partial sum ν j +... + ν 4 is voi if j = 3). In the case ν + ν 1 3/8 + 8ϖ, γ is obviously of Type III. Further, if ν has a partial sum, say ν, satisfying 3 8 + 8ϖ < ν + ν 1 < 5 8 8ϖ, then γ is of Type I or II. For example, if 3 8 + 8ϖ < µ j +... + µ 1 + ν 1 1 2, we choose β = µκ Mj )... µκ M1 ) κ N1 ); if 1 2 < µ j +... + µ 1 + ν 1 < 5 8 8ϖ, we choose α = µκ Mj )... µκ M1 ) κ N1 ). It now suffices to assume that an every partial sum ν of ν satisfies either ν + ν 1 5 8ϖ, 6.8) 8 ν + ν 1 3 8 + 8ϖ or ν + ν 1 5 8 8ϖ. Let ν 1 be the smallest partial sum of ν such that ν 1 + ν 1 5/8 8ϖ the existence of ν 1 follows from 6.8), an there may be more than one choice of ν 1), an let ν be a positive term in ν 1. Since ν 1 ν is also a partial sum of ν, we must have so that ν 1 ν + ν 1 3/8 + 8ϖ, ν 1 4 16ϖ. This implies that ν must be one of the ν t, t 4 that arises only if j 4). In particular we have ν 4 1/4 16ϖ. Now, the conitions 1 4 16ϖ ν 4 ν 3 ν 2 ν 1, ν 4 + ν 3 + ν 2 + ν 1 < 1 + log 2 L together imply that 1 2 32ϖ ν 3 + ν 4 < 1 2 + log 2 2L. It follows that γ is of Type I or II by choosing β = κ N3 κ N4. It shoul be remarke, by the Siegel-Walfisz theorem, that for all the choices of β above the Siegel-Walfisz assumption in A 2 ) hols. Noting that the sum in 6.6) contains OL 40A ) terms, by the above iscussion we conclue that 6.2) implies 6.1). 26

7. The ispersion metho In this an the next three sections we treat the Type I an II estimates simultaneously via the methos in [1, Section 3-7]. We henceforth assume that γ = α β satisfies A 1 ), A 2 ) an A 3 ). Recall that x 1 an x 2 are given by 2.13). We shall apply Lemma 4 with if γ is of Type I x 1 < N x 2 ), an R = x ε N 7.1) R = x 3ϖ N 7.2) if γ is of Type II x 2 < N < 2x 1/2 ). Note that D 2 1 < R < D 2. By Lemma 4 an Lemma 5, the proof of 6.2) is reuce to showing that R /D 1 <r<r µr) a C i r), for µqr) = a, r) = b, q) = 1, D 2 /r<q<d 2 /r q P q,rp 0 )=1 γ; r, a; q, b) = It therefore suffices to prove that Bγ; Q, R) := µr) r R subject to the conitions a C i r) n ar) n bq) q Q q P q,rp 0 )=1 b C i q) γn) 1 ϕqr) b C i q) γ; r, a; q, b) xl 41A, n,qr)=1 γn). γ; r, a; q, b) xl 43A, 7.3) x ϖ R < R < R 7.4) an 1 2 x1/2 ε < QR < x 1/2+2ϖ, 7.5) which are henceforth assume. For notational simplicity, in some expressions the subscript i will be omitte even though they epen on it. In what follows we assume that r R, µr) = 1, a C i r). 7.6) 27

Let cr, a; q, b) be given by if an cr, a; q, b) = sgn γ; r, a; q, b) q Q, q P, q, rp 0 ) = 1, b C i q), cr, a; q, b) = 0 Changing the orer of summation we obtain γ; r, a; q, b) = q Q q P q,rp 0 )=1 b C i q) otherwise. m,r)=1 αm)dr, a; m), Dr, a; m) = q,m)=1 b cr, a; q, b) mn ar) mn bq) βn) 1 ϕqr) n,qr)=1 ) βn). It follows by Cauchy s inequality that Bγ; Q, R) 2 MRL B r R µr) a C i r) m,r)=1 fm)dr, a; m) 2, 7.7) fy) is as in Lemma 7 with η = η 19. We have fm)dr, a; m) 2 = S 1 r, a) 2S 2 r, a) + S 3 r, a), 7.8) m,r)=1 S j r, a), j = 1, 2, 3 are efine by S 1 r, a) = m,r)=1 fm) q,m)=1 cr, a; q, b) b mn ar) mn bq) βn)) 2, S 2 r, a) = q 1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) ϕq b 1 q 2 r) 2 b 2 βn 1 )βn 2 ) fm), n 1 n 2,q 2 r)=1 mn 1 ar) mn 1 b 1 q 1 ) m,q 2 )=1 28

S 3 r, a) = cr, a; q 1, b 1 )cr, a; q 2, b 2 ) ϕq q 1 b 1 q 1 r)ϕq 2 r) 2 b 2 βn 1 )βn 2 ) fm). n 1,q 1 r)=1 n 2,q 2 r)=1 m,q 1 q 2 r)=1 By 7.7) an 7.8), the proof of 7.3) is reuce to showing that S1 r, a) 2S 2 r, a) + S 3 r, a) ) xnr 1 L 87A 7.9) r a on assuming A B. Here we have omitte the constraints given in 7.6) for notational simplicity, so they have to be remembere in the sequel. 8. Evaluation of S 3 r, a) In this section we evaluate S 3 r, a). We shall make frequent use of the trivial boun ˆfz) M. 8.1) By Möbius inversion an Lemma 7, for q j Q, j = 1, 2 we have m,q 1 q 2 r)=1 fm) = ϕq 1q 2 r) q 1 q 2 r ˆf0) + Ox ε ). 8.2) This yiels S 3 r, a) = ˆf0) q 1 b 1 q 2 n 1,q 1 r)=1 n 2,q 2 r)=1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) ϕq 1 )ϕq 2 )ϕr) b 2 βn 1 )βn 2 ) + Ox ε N 2 R 2 ). ϕq 1 q 2 ) q 1 q 2 r In view of 2.10), if q 1 q 2, P 0 ) = 1, then either q 1, q 2 ) = 1 or q 1, q 2 ) > D 0. Thus, on the right sie above, the contribution from the terms with q 1, q 2 ) > 1 is, by 8.1) an trivial estimation, xnd0 1 R 2 L B. It follows that Xr, a) = q 1 b 1 S 3 r, a) = ˆf0)Xr, a) + OxND 1 0 R 2 L B ), 8.3) q 2,q 1 )=1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q 1 q 2 rϕr) b 2 n 1,q 1 r)=1 n 2,q 2 r)=1 βn 1 )βn 2 ). 29

The aim of this section is to show that 9. Evaluation of S 2 r, a) S 2 r, a) = ˆf0)Xr, a) + OxND 1 0 R 2 L B ). 9.1) Assume cr, a; q 1, b 1 )cr, a; q 2, b 2 ) 0. Let νmo q 1 r) be a common solution to ν amo r), ν b 1 mo q 1 ). Substituting mn 1 = n an applying Lemma 8 we obtain βn 1 ) fm) n 1 mn 1 ar) mn 1 b 1 q 1 ) m,q 2 )=1 τ 20 n) xlb q n<2x 1 r. n νq 1 r) It follows that the contribution from the terms with q 1, q 2 ) > 1 in S 2 r, a) is so that, S 2 r, a) = q 1 b 1 q 2,q 1 )=1 n 1 n 2,q 2 r)=1 xnd 1 0 R 2 L B, cr, a; q 1, b 1 )cr, a; q 2, b 2 ) ϕq 2 r) b 2 βn 1 )βn 2 ) fm) + OxND0 1 R 2 L B ). 9.2) mn 1 ar) mn 1 b 1 q 1 ) m,q 2 )=1 Note that the innermost sum in 9.2) is voi unless n 1, q 1 r) = 1. For µq 1 q 2 r) = 1 an q 2, P 0 ) = 1 we have q 2 ϕq 2 ) = 1 + Oτq 2)D0 1 ), an, by Lemma 8, n 1,q 1 r)=1 βn 1 ) mn 1 ar) mn 1 b 1 q 1 ) m,q 2 )>1 fm) τ 20 n) τ 20q 2 )xl B. q n<2x 1 rd 0 n νq 1 r) n,q 2 )>1 Thus the relation 9.2) remains vali if the constraint m, q 2 ) = 1 in the innermost sum is remove an the enominator ϕq 2 r) is replace by q 2 ϕr). Namely we have S 2 r, a) = cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q q 2 ϕr) 1 b 1 q 2,q 1 )=1 b 2 9.3) βn 1 )βn 2 ) fm) + OxND0 1 R 2 L B ). n 1 n 2,q 2 r)=1 mn 1 ar) mn 1 b 1 q 1 ) 30

By Lemma 7, for n 1, q 1 r) = 1 we have mn 1 ar) mn 1 b 1 q 1 ) fm) = 1 q 1 r h <H 2 ) h ˆf e q1 r hµ) + Ox 2 ), q 1 r an µmoq 1 r) is a common solution to Inserting this into 9.3) we euce that H 2 = 4QRM 1+2ε, µn 1 amo r), µn 1 b 1 mo q 1 ). 9.4) S 2 r, a) = ˆf0)Xr, a) + R 2 r, a) + OxND 1 0 R 2 L B ), 9.5) R 2 r, a) = q 1 b 1 n 1,q 1 r)=1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q 1 q 2 rϕr) b 2 ) h βn 1 ) ˆf e q1 r hµ). q 1 r q 2,q 1 )=1 1 h <H 2 n 2,q 2 r)=1 ) βn 2 ) The proof of 9.1) is now reuce to estimating R 2 r, a). First we note that the secon inequality in 7.5) implies H 2 x 1/2+2ϖ+2ε N < 2x 2ϖ+2ε, 9.6) since M 1 x 1 N here an in what follows, we use the secon inequality in 7.5) only). This implies that R 2 r, a) = 0 if γ is of Type I. Now assume that γ is of of Type II. Noting that by 9.4), we have R r, a; n) = µ q 1 r aq 1n 1 + b 1rn 1 mo 1) r q 1 R 2 r, a) N 1+ε R 2 q,n)=1 b cr, a; q, b) q n N n,r)=1 1 h <H 2 31 R r, a; n), 9.7) ) h ahqn ˆf e bhrn ). qr r q

To estimate the sum of R r, a; n) we observe that R r, a; n) 2 = q,n)=1 b q,n)=1 1 h <H 2 1 h <H 2 cr, a; q, b)cr, a; q, b ) qq b ) h h ˆf ˆf qr q r ) ah q h q) n e bhrn ) + b h rn. r q q It follows, by changing the orer of summation an applying 8.1), that M 2 n N n,r)=1 R r, a; n) 2 q cr, a; q, b)cr, a; q, b ) b q b qq Wr, a; q, b; q, b ; h, h ), 1 h <H 2 1 h <H 2 9.8) Wr, a; q, b; q, b ; h, h ) = n N n,qq r)=1 ah q h q) n e bhrn ) + b h rn. r q q Since M 1 N 1, by the secon inequality in 7.4) an 7.2) we have H 2 Q 1 x 3ϖ+ε. 9.9) It follows that, on the right sie of 9.8), the contribution from the terms with h q = hq is NQ 2 τhq) x 3ϖ+ε N. 9.10) 1 h<h 2 q<2q Now assume that cr, a; q, b)cr, a; q, b ) 0, 1 h < H 2, 1 h < H 2 an h q hq. Letting = [q, q ]r, we have ah q h q) r bh r q + b h r q c mo 1) for some c with It follows by the estimate 3.13) that c, r) = h q h q, r). Wr, a; q, b; q, b ; h, h ) 1/2+ε + c, )N. 9.11) 32

Since N > x 2, by the first inequality in 7.4), 7.2) an 2.13) we have R 1 < x 4ϖ N 1 < x 1/2+8ϖ. 9.12) Together with 7.5), this implies that Q x 10ϖ. 9.13) By 9.13) an 7.5) we have 1/2 Q 2 R) 1/2 x 1/4+6ϖ. On the other han, noting that h q h q h q hq )qq mo r), we have Together with 9.6), 9.12) an 9.13), this yiels c, ) c, r)[q, q ] [q, q ]H 2 Q. 9.14) c, )N H 2 NQR 1 x 16ϖ+ε. Combining these estimates with 9.11) we euce that Wr, a; q, b; q, b ; h, h ) x 1/4+7ϖ. Together with 9.6), this implies that, on the right sie of 9.8), the contribution from the terms with h q hq is x 1/4+12ϖ which is sharper than the right sie of 9.10). Combining these estimates with 9.8) we conclue that R r, a; n) 2 x 1 3ϖ+ε M. n N n,r)=1 This yiels, by Cauchy s inequality, R r, a; n) x 1 3ϖ/2+ε. n N n,r)=1 Inserting this into 9.7) we obtain R 2 r, a) x 1 ϖ NR 2 9.15) which is sharper than the O term in 9.5). The relation 9.1) follows from 9.5) an 9.15) immeiately. 33

10. A truncation of the sum of S 1 r, a) We are unable to evaluate each S 1 r, a) irectly. However, we shall establish a relation of the form S 1 r, a) = ˆf0)Xr, a) + R1 r, a) ) + OxNR 1 L 87A ) 10.1) r a r a with R 1 r, a) to be specifie below in 10.10). In view of 8.3) an 9.1), the proof of 7.9) will be reuce to estimating R 1 r, a). By efinition we have S 1 r, a) = cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q 1 b 1 q 2 b 2 βn 1 )βn 2 ) fm). 10.2) n 1 n 2 n 1 r) mn 1 ar) mn 1 b 1 q 1 ) mn 2 b 2 q 2 ) Let Ur, a; q 0 ) enote the sum of the terms in 10.2) with q 1, q 2 ) = q 0. Clearly we have Ur, a; q 0 ) = 0 unless q 0 < 2Q, q 0 P, q 0, rp 0 ) = 1, which are henceforth assume. We first claim that Ur, a; q 0 ) xnd 0 R) 1 L B. 10.3) Assume that, for j = 1, 2, r a q 0 >1 q j Q, q j P, q j, rp 0 ) = 1, b j C i q j ) an q 1, q 2 ) = q 0. Write q 1 = q 1 /q 0, q 2 = q 2 /q 0. By Lemma 5, there exist t 1, t 2 C i q 0 ), b 1 C i q 1) an b 2 C i q 2) such that b j t j mo q 0 ), b j b jmo q j). Note that the conitions mn 1 t 1 mo q 0 ) an mn 2 t 2 mo q 0 ) together imply that t 2 n 1 t 1 n 2 mo q 0 ). 10.4) Thus the innermost sum in 10.2) is voi if 10.4) fails to hol for any t 1, t 2 C i q 0 ). On the other han, if 10.4) hols for some t 1, t 2 C i q 0 ), the innermost sum in 10.2) may be rewritten as fm) mn 1 a 1 q 0 r) mn 1 b 1 q 1 ) mn 2 b 2 q 2 ) 34

a 1 mo q 0 r) is a common solution to a 1 amo r) an a 1 t 1 mo q 0 ). Hence, changing the orer of summation we obtain cr, a; q 1, b 1 )cr, a; q 2, b 2 ) βn 1 )βn 2 ) fm) b 2 b 1 t 1 C i q 0 ) t 2 C i q 0 ) n 1 n 1 n 2 n 1 r) t 1 n 2 t 2 n 1 q 0 ) n 2 n 1 r) βn 1 )βn 2 ) J n, q ) = b j C iq ) b j nq ) mn 1 a 1 q 0 r) 1. mn 1 ar) mn 1 b 1 q 1 ) mn 2 b 2 q 2 ) fm)j mn 1, q 1)J mn 2, q 2), This yiels, by summing over q 1 an q 2 with q 1, q 2 ) = q 0 an changing the orer of summation, Ur, a; q 0 ) βn 1 )βn 2 ) t 1 C i q 0 ) t 2 C i q 0 ) mn 1 a 1 q 0 r) X n) = n 1 n 2 n 1 r) t 1 n 2 t 2 n 1 q 0 ) fm)x mn 1 )X mn 2 ), q Q/q 0 µq ) J i n, q ). 10.5) We may assume that n 1, q 0 r) = 1, since the innermost sum in 10.5) is voi otherwise. Let a 2 mo q 0 r) be a common solution to a 2 a 1 mo r) an a 2 t 2 mo q 0 ). In the case n 2 n 1 mo r), t 1 n 2 t 2 n 1 mo q 0 ), the conition mn 1 a 1 mo q 0 r) is equivalent to mn 2 a 2 mo q 0 r). Thus the innermost sum in 10.5) is fm)x mn 1 ) 2 + fm)x mn 2 ) 2. mn 1 a 1 q 0 r) mn 2 a 2 q 0 r) Since it follows that J n, q ) = { 1 if q P n h i ), q, n) = 1 0 otherwise, X n) q P n h i ) q,n)=1 µq ). 10.6) 35

Assume that j = 1, 2, 1 µ k 0 an µ i. Write n jµ = n j n j, h µ h i ), h jµ = h µ h i n j, h µ h i ). Noting that the conitions p mn j +h µ h i ) an p n j together imply that p mn jµ +h jµ), by 10.6) we have X mn j ) τmn jµ + h jµ) τmn jµ + h jµ) k0 1. 1 µ k 0 µ i 1 µ k 0 µ i Since n jµ, h jµ) = n jµ, q 0 r) = 1, it follows by Lemma 8 that mn j a j q 0 r) fm)x mn j ) 2 MLB q 0 r + xε/3 here the term x ε/3 is necessary when q 0 r > x ε/4 M). Combining these estimates with 10.5) we euce that ) ML B Ur, a; q 0 ) q 0 r + xε/3 βn 1 ) βn 2 ). n 1,q 0 r)=1 Using Lemma 8 again, we fin that the innermost sum is It follows that NLB q 0 r + xε/3. t 1 C i q 0 ) t 2 C i q 0 ) ) xnl Ur, a; q 0 ) ϱ 2 q 0 ) 2 B q 0 r) + x1+ε/2 + x ε N. 2 q 0 r This leas to 10.3), since NR 1 > x ε an NQR x 1/2+2ϖ N x 1 ϖ NR 1 n 2 n 1 r) t 1 n 2 t 2 n 1 q 0 ) by 7.5), 7.1), 7.2) an the secon inequality in 7.4). We now turn to Ur, a; 1). Assume µq 1 q 2 r) = 1. In the case n 1, q 1 r) = n 2, q 2 r) = 1, the innermost sum in 10.2) is, by Lemma 7, equal to 1 q 1 q 2 r h <H 1 ˆf h q 1 q 2 r ) e q1 q 2 r µh) + Ox 2 ), H 1 = 8Q 2 RM 1+2ε 10.7) 36

an µmo q 1 q 2 r) is a common solution to µn 1 amo r), µn 1 b 1 mo q 1 ), µn 2 b 2 mo q 2 ). 10.8) It follows that X r, a) = q 1 b 1 Ur, a; 1) = ˆf0)X r, a) + R 1 r, a) + O1), 10.9) q 2,q 1 )=1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q 1 q 2 r b 2 n 1,q 1 r)=1 n 2 n 1 r) n 2,q 2 )=1 βn 1 )βn 2 ) an R 1 r, a) = q 1 b 1 n 1,q 1 r)=1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q 1 q 2 r b 2 βn 1 )βn 2 ) q 2,q 1 )=1 n 2 n 1 r) n 2,q 2 )=1 1 h <H ) h ˆf e q1 q q 1 q 2 r 2 r µh). 10.10) By 10.2), 10.3) an 10.9) we conclue that ˆf0)X r, a) + R 1 r, a) ) + OxND 0 R) 1 L B ). r a S 1 r, a) = r a In view of 8.1), the proof of 10.1) is now reuce to showing that X r, a) Xr, a)) N 2 R 1 L 87A. 10.11) We have X r, a) Xr, a) = q 1 r a b 1 q 2,q 1 )=1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q 1 q 2 r b 2 Vr; q 1, q 2 ) with Vr; q 1, q 2 ) = n 1,q 1 r)=1 n 2 n 1 r) n 2,q 2 )=1 βn 1 )βn 2 ) 1 ϕr) n 1,q 1 r)=1 n 2,q 2 r)=1 βn 1 )βn 2 ) which is inepenent of a. It follows that X r, a) Xr, a)) 1 R r a q 1 Q q 2 Q 37 ϱ 2 q 1 q 2 ) q 1 q 2 r R r,q 1 q 2 )=1 ϱ 2 r) Vr; q 1, q 2 ). 10.12)

Noting that Vr; q 1, q 2 ) = l mo r n lr) n,q 1 )=1 βn) 1 ϕr) n,q 1 r)=1 ) βn) n lr) n,q 2 )=1 βn) 1 ϕr) n,q 2 r)=1 ) βn), by Cauchy s inequality, the conition A 2 ) an Lemma 10, we fin that the innermost sum in 10.12) is τq 1 q 2 ) B N 2 L 100A, whence 10.11) follows. A combination of 8.3), 9.1) an 10.1) leas to S1 r.a) 2S 2 r, a) + S 3 r, a) ) = R 1 r, a) + OxNR 1 L 87A ). 10.13) r a r a Note that µ q 1 q 2 r aq 1q 2 n 1 + b 1q 2 rn 1 + b 2q 1 rn 2 mo 1) r q 1 q 2 by 10.8). Hence, on substituting n 2 = n 1 + kr, we may rewrite R 1 r, a) as R 1 r, a) = 1 r k <N/R R 1 r, a; k), 10.14) R 1 r, a; k) = q 1 b 1 q 2,q 1 )=1 n,q 1 r)=1 n+kr,q 2 )=1 cr, a; q 1, b 1 )cr, a; q 2, b 2 ) q 1 q 2 b 2 1 h <H 1 βn)βn + kr)e hξr, a; q 1, b 1 ; q 2, b 2 ; n, k)). ) h ˆf q 1 q 2 r with ξr, a; k; q 1, b 1 ; q 2, b 2 ; n) = aq 1q 2 n r + b 1q 2 rn q 1 + b 2q 1 rn + kr) q 2. Recall that, in the Type I an II cases, we have reuce the proof of 6.2) to proving 7.9) at the en of Section 7. Now, by 10.13) an 10.14), the proof of 7.9) is in turn reuce to showing that R 1 r, a; k) xl 88A for k < NR 1. In fact, we shall prove the sharper boun R 1 r, a; k) x 1 ϖ/2 10.15) 38

in the next two sections. We conclue this section by showing that the gap between 10.15) an some trivial bouns is not too large. It trivially follows from 8.1) that On the other han, in view of 2.13), since R 1 r, a; k) x 1+ε H 1. H 1 x ε QR) 2 MN) 1 NR 1, an, by the first inequality in 7.4), 7.1) an 7.2), { NR 1 x ϖ+ε if x 1 < N x 2 < x 4ϖ if x 2 < N < 2x 1/2, it follows from 7.5) that H 1 { x 5ϖ+2ε if x 1 < N x 2 x 8ϖ+ε if x 2 < N < 2x 1/2. 10.16) 10.17) Thus, in orer to prove 10.15), we nee only to save a small power of x from the trivial estimate. The bouns 10.16) an 10.17) will fin application in the next two sections. for 11. Estimation of R 1 r, a; k): The Type I case In this an the next sections we assume that k < NR 1, an abbreviate R 1, cq 1, b 1 ), cq 2, b 2 ) an ξq 1, b 1 ; q 2, b 2 ; n) R 1 r, a; k), cr, a; q 1, b 1 ), cr, a; q 2, b 2 ) an ξr, a, k; q 1, b 1 ; q 2, b 2 ; n) respectively, with the aim of proving 10.15). The variables r, a an k may also be omitte some else for notational simplicity. The proof is somewhat analogous to the estimation of R 2 r, a) in Section 9; the main tool we nee is Lemma 11. Assume that x 1 < N x 2 an R is as in 7.1). We have R 1 N ε q 1 cq 1, b 1 ) q 1 b 1 n N n,q 1 r)=1 Fq 1, b 1 ; n), 11.1) 39

Fq 1, b 1 ; n) = 1 h <H 1 q 2,q 1 n+kr))=1 cq 2, b 2 ) q 2 b 2 ) h ˆf e hξq 1, b 1 ; q 2, b 2 ; n)). q 1 q 2 r In what follows assume cq 1, b 1 ) 0. To estimate the sum of Fq 1, b 1 ; n) we observe that, similar to 9.8), M 2 n N n,q 1 r)=1 Fq 1, b 1 ; n) 2 Gh, h ; q 1, b 1, q 2, b 2 ; q 2, b 2) = q 2,q 1 )=1 q 2,q 1)=1 1 h <H 1 n N n,q 1 r)=1 n+kr,q 2 q 2 )=1 b 2 b 2 cq 2, b 2 )cq 2, b 2) q 2 q 2 1 h <H 1 Gh, h ; q 1, b 1, q 2, b 2 ; q 2, b 2), e h ξq 1, b 1 ; q 2, b 2; n) hξq 1, b 1 ; q 2, b 2 ; n) ). 11.2) The conition N x 2 is essential for bouning the terms with h q 2 = hq 2 in 11.2). By 7.5) we have H 1 Q 1 x ε QR)MN) 1 N x 2ϖ+ε. It follows that, on the right sie of 11.2), the contribution from the terms with h q 2 = hq 2 is NQ 2 τhq) B x 2ϖ+ε N. 11.3) 1 h<h 1 q Q Now assume that cq 2, b 2 )cq 2, b 2) 0, q 2 q 2, q 1 ) = 1 an h q 2 hq 2. We have h ξq 1, b 1 ; q 2, b 2; n) hξq 1, b 1 ; q 2, b 2 ; n) h q 2 h q 2 )aq 1 n r + h q 2 h q 2 )b 1 rn q 1 Letting 1 = q 1 r an 2 = [q 2, q 2], we may write + h b 2q 1 rn + kr) q 2 hb 2q 1 rn + kr) q 2 mo 1). for some c 1 with an h q 2 h q 2 )a q 1 r h b 2q 1 r q 2 + h q 2 h q 2 )b 1 r q 1 c 1 1 mo 1) c 1, r) = h q 2 h q 2, r), hb 2q 1 r q 2 c 2 2 mo 1) 40

for some c 2, so that h ξq 1, b 1 ; q 2, b 2; n) hξq 1, b 1 ; q 2, b 2 ; n) c 1 n 1 Since 1, 2 ) = 1, it follows by Lemma 11 that + c 2n + kr) 2 mo 1). Gh, h ; q 1, b 1, q 2, b 2 ; q 2, b 2) 1 2 ) 1/2+ε + c 1, 1 )N 1. 11.4) We appeal to the conition N > x 1 that gives, by 10.16), Together with 7.5), this yiels R 1 < x ϖ+ε N 1 < x 3/4 15ϖ+ε N. 11.5) 1 2 ) 1/2 Q 3 R) 1/2 x 3/4+3ϖ R 1 x 12ϖ+ε N. A much sharper boun for the secon term on the right sie of 11.4) can be obtaine. In a way similar to the proof of 9.14), we fin that c 1, 1 ) c 1, r)q 1 H 1 Q 2. It follows by 10.17), 7.5) an the first inequality in 11.5) that c 1, 1 ) 1 H 1 QR)R 2 x 1/2+9ϖ+4ε N 2 x 1/4 6ϖ. Here we have use the conition N > x 1 again. Combining these estimates with 11.4) we euce that Gh, h ; q 1, b 1, q 2, b 2 ; q 2, b 2) x 12ϖ+ε N. Together with 10.17), this implies that, on the right sie of 11.2), the contribution from the terms with h q 2 hq 2 is x 12ϖ+ε H 2 1N x 2ϖ+5ε N which has the same orer of magnitue as the right sie of 11.3) essentially. Combining these estimates with 11.2) we obtain Fh; q 1, b 1 ; n) 2 x 1 2ϖ+5ε M. n N n,q 1 r)=1 This yiels, by Cauchy s inequality, Fh; q 1, b 1 ; n) x 1 ϖ+3ε. 11.6) n N n,q 1 r)=1 The estimate 10.15) follows from 11.1) an 11.6) immeiately. 41

12. Estimation of R 1 r, a; k): The Type II case Assume that x 2 < N < 2x 1/2 an R is as in 7.2). We have R 1 N ε Kn), 12.1) n N n,r)=1 Kn) = q 1,n)=1 b 1 q 2,q 1 n+kr))=1 cq 1, b 1 )cq 2, b 2 ) q 1 q 2 b 2 1 h <H 1 ) h ˆf e hξq 1, b 1 ; q 2, b 2 ; n)). q 1 q 2 r Let # stan for a summation over the 8-tuples q1, b 1 ; q 2, b 2 ; q 1, b 1; q 2, b 2) with q 1, q 2 ) = q 1, q 2) = 1. To estimate the sum of Kn) we observe that, similar to 9.8), M 2 n N n,r)=1 Kn) 2 # cq 1, b 1 )cq 2, b 2 )cq 1, b 1)cq 2, b 2) q 1 q 2 q 1q 2 1 h <H 1 1 h <H 1 Mh, h ; q 1, b 1 ; q 2, b 2 ; q 1, b 1; q 2, b 2), 12.2) Mh, h ; q 1, b 1 ; q 2, b 2 ; q 1, b 1; q 2, b 2) = eh ξq 1, b 1; q 2, b 2; n) hξq 1, b 1 ; q 2, b 2 ; n)). n N Here is restriction to n, q1 q 1r) = n + kr, q 2 q 2) = 1. Similar to 9.9), we have H 1 Q 2 x 3ϖ+ε. Hence, on the right sie of 12.2), the contribution from the terms with h q 1 q 2 = hq 1q 2 is NQ 4 τhqq ) B x 3ϖ+ε N. 12.3) 1 h<h 1 q Q q Q Note that the bouns 9.12) an 9.13) are vali in the present situation. Since R is near to x 1/2 in the logarithmic scale an Q is small, it can be shown via Lemma 11 that the terms with h q 1 q 2 hq 1q 2 on the right sie of 12.2) make a small contribution in comparison with 12.3). Assume that cq 1, b 1 )cq 2, b 2 )cq 1, b 1)cq 2, b 2) 0, q 1, q 2 ) = q 1, q 2) = 1, h q 1 q 2 hq 1q 2. 42

We have with h ξq 1, b 1; q 2, b 2; n) hξq 1, b 1 ; q 2, b 2 ; n) s n r + t 1 n q 1 + t 1 n q 1 + t 2n + kr) q 2 + t 2n + kr) q 2 mo 1) 12.4) s ah q 1q 2 hq 1 q 2 )mo r), t 1 b 1 hq 2 rmo q 1 ), t 1 b 1h q 2rmo q 1), t 2 b 2 hq 1 rmo q 2 ), t 2 b 2h q 1rmo q 2). Letting 1 = [q 1, q 1]r, 2 = [q 2, q 2], we may rewrite 12.4) as h ξq 1, b 1; q 2, b 2; n) hξq 1, b 1 ; q 2, b 2 ; n) c 1 n 1 + c 2n + kr) 2 mo 1) for some c 1 an c 2 with It follows by Lemma 11 that c 1, r) = h q 1q 2 hq 1 q 2, r). Mh, h ; q 1, b 1 ; q 2, b 2 ; q 1, b 1; q 2, b 2) 1 2 ) 1/2+ε + c 1, 1 ) 1, 2 ) 2 N 1. 12.5) By 7.5) an 9.13) we have 1 2 ) 1/2 Q 4 R) 1/2 x 1/4+16ϖ. On the other han, we have 1, 2 ) q 1 q 1, q 2 q 2) Q 2, since q 2 q 2, r) = 1, an, similar to 9.14), c 1, 1 ) c 1, r)[q 1, q 1] [q 1, q 1]H 1 Q 2. It follows by 10.16), 9.13) an the first inequality in 9.12) that c 1, 1 ) 1, 2 ) 2 N 1 H 1 NQ 6 R 1 x 72ϖ. Combining these estimates with 12.5) we euce that Mh, h ; q 1, b 1 ; q 2, b 2 ; q 1, b 1; q 2, b 2) x 1/4+16ϖ+ε. Together with 10.16), this implies that, on the right sie of 12.2), the contribution from the terms with h q 1 q 2 hq 1q 2 is x 1/4+16ϖ+ε H 2 1 x 1/4+33ϖ 43

which is sharper than the right sie of 12.3). Combining these estimates with 12.2) we obtain Kn) 2 x 1 3ϖ+ε M. n N n,r)=1 This yiels, by Cauchy s inequality, n N n,r)=1 Kn) x 1 ϖ. 12.6) The estimate 10.15) follows from 12.1) an 12.6) immeiately. 13. The Type III estimate: Initial steps Assume that γ = α κ N3 κ N2 κ N1 is of Type III. Our aim is to prove that for any an c satisfying γ;, c) x1 ε/2 13.1), c) = 1, x 1/2 ε < < x 1/2+2ϖ, P,, P 0 ) < D 1, which are henceforth assume. This leas to 6.2). We first erive some lower bouns for the N j from A4) an A5). We have ) 1/2 x N 1 N 2 x 5/16 4ϖ, 13.2) MN 1 an x N 3 x 1/4 16ϖ M x 1/4 16ϖ. 13.3) MN 1 N 2 Let f be as in Lemma 7 with η = η an with N 1 in place of M. Note that the function κ N1 f is supporte on [N1, N 1 ] [ηn 1, ηn 1 + ] with N 1 ± = 1 ± N1 ε )N 1. Letting γ = α κ N3 κ N2 f, we have ϕ) n,)=1 γ γ )n) x 1 ε/2, an n c) γ γ )n) L N 1 q N 1 q,)=1 1 l<3x/q lq c) τ 19 l) + L 44 ηn 1 q ηn + 1 q,)=1 1 l<3x/q lq c) τ 19 l) x1 ε/2.

It therefore suffices to prove 13.1) with γ replace by γ. In fact, we shall prove the sharper boun γ ;, c) x1 ϖ/3 In a way similar to the proof of 8.2) we obtain This yiels, by 13.2), 1 ϕ) n,)=1 n,)=1 γ n) = ˆf0) fn) = ϕ) ˆf0) + Ox ε ). m,)=1 n 3 N 3 n 2 N 2 n 3,)=1 n 2,)=1 αm) + O 1 x 3/4 ). 13.4) Here an in what follows, n N stans for N n < ηn. On the other han, we have γ n) = αm) fn 1 ). n c) m,)=1 n 3 N 3 n 2 N 2 n 3,)=1 n 2,)=1 The innermost sum is, by Lemma 7, equal to It follows that γ ;, c) = 1 m M m,)=1 1 h <H n 3 N 3 n 3,)=1 mn 3 n 2 n 1 c) ) ˆfh/)e chmn3 n 2 + Ox 2 ), n 2 N 2 n 2,)=1 H = N 1+2ε 1. αm) 1 h <H ) ˆfh/)e chmn3 n 2 + O 1 x 3/4 ). The proof of 13.4) is therefore reuce to showing that ) ˆfh/)e ahn3 n 2 x 1 ϖ/2+2ε M 1 13.5) 1 h<h for any a with a, ) = 1. n 3 N 3 n 2 N 2 n 3,)=1 n 2,)=1 45

On substituting 1 = /h, ) an applying Möbius inversion, the left sie of 13.5) may be rewritten as ) ˆfh/ 1 )e 1 ahn3 n 2 1 1 h<h n 3 N 3 n 2 N 2 h, 1 )=1 n 3,)=1 n 2,)=1 = 1 2 = b 3 2 b 2 2 µb 3 )µb 2 ) It therefore suffices to show that 1 h<h h, 1 )=1 n 3 N 3 n 3, 1 )=1 n 2 N 2 n 2, 1 )=1 for any 1, b, N 3, an N 2 satisfying 1 h<h h, 1 )=1 n 3 N 3 /b 3 n 3, 1 )=1 n 2 N 2 /b 2 n 2, 1 )=1 ˆfh/ 1 )e 1 ahb3 b 2 n 3 n 2 ), H = 1 N 1+2ε 1. 13.6) ˆfh/ 1 )e 1 bhn3 n 2 ) x 1 ϖ/2+ε M 1 13.7) 1, b, 1 ) = 1, 1 N 3 N 3 N 3, which are henceforth assume. Note that 13.2) implies 1 N 2 N 2 N 2, 13.8) H x 3/16+6ϖ+ε. 13.9) In view of 13.6), the left sie of 13.7) is voi if 1 N1 1 2ε, so we may assume 1 > N1 1 2ε. By the trivial boun ˆfz) N 1, 13.10) an 3.13), we fin that the left sie of 13.7) is HN 3 N 1 1/2+ε 1 + 1 1 N 2 ) 3/2+ε 1 N 2ε 1 N 3. In the case 1 x 5/12 6ϖ, the right sie is x 1 ϖ+3ε M 1 by A4) an 2.13). This leas to 13.7). Thus we may further assume 1 > x 5/12 6ϖ. 13.11) We appeal to the Weyl shift an the factorization 2.8) with 1 in place of. By Lemma 4, we can choose a factor r of 1 such that x 44ϖ < r < x 45ϖ. 13.12) 46

Write N 1, k) = 1 h<h h, 1 )=1 n 3 N 3 n 3, 1 )=1 n 2 N 2 n 2 +hkr, 1 )=1 ˆfh/ 1 )e 1 bhn2 + hkr)n 3 ), so that the left sie of 13.7) is just N 1, 0). Assume k > 0. We have with Q i 1, k) = N 1, k) N 1, 0) = Q 1 1, k) Q 2 1, k), 13.13) 1 h<h h, 1 )=1 n 3 N 3 n 3, 1 )=1 l I i h) l, 1 )=1 ˆfh/ 1 )e 1 bhln3 ), i = 1, 2, I 1 h) = [ ηn 2, ηn 2 + hkr ), I 2 h) = [ N 2, N 2 + hkr ). To estimate Q i 1, k) we first note that, by Möbius inversion, Q i 1, k) = ˆfh/t)e t bhln3 ). st= 1 µs) 1 h<h/s n 3 N 3 l I i h) n 3, 1 )=1 l, 1 )=1 The inner sum is voi unless s < H. Since H 2 = o 1 ) by 13.9) an 13.11), it follows, by changing the orer of summation, that Q i 1, k) ˆfh/t)e t bhln3 ), st= 1 t>h n 3 N 3 n 3, 1 )=1 l I i H) l, 1 )=1 h J i s,l) J i s, l) is a certain interval of length < H an epening on s an l. Noting that, by integration by parts, z ˆfz) min { } N1 2, z 2 N1 ε, by partial summation an 13.10) we obtain ˆfh/t)e t bhln3 ) N1 1+ε min { H, bln 3 /t 1}. h J i s,l) It follows that Q i 1, k) N1 1+ε t 1 t>h l I i H) l, 1 )=1 n 3 <2N 3 n 3, 1 )=1 min { H, bln 3 /t 1}. Since H = on 3 ) by 13.3) an 13.9), the innermost sum is N3 1+ε by Lemma 9. In view of 13.6), this leas to Q i 1, k) 1+ε 1 krn 3. 13.14) 47

We now introuce the parameter K = [x 1/2 48ϖ N 1 N 2 ] 13.15) which is x 1/8 56ϖ by 13.2). By A5) an the secon inequality in 13.12), we see that the right sie of 13.14) is x 1 ϖ+ε M 1 if k < 2K. Hence, by 13.13), the proof of 13.7) is reuce to showing that 1 K N 1, k) x 1 ϖ/2+ε M 1. 13.16) k K 14. The Type III estimate: Completion The aim of this section is to prove 13.16) that will complete the proof of Theorem 2. We start with the relation hn 2 + hkr) l + kr mo 1 ) for h, 1 ) = n 2 + hkr, 1 ) = 1, l hn 2 mo 1 ). Thus we may rewrite N 1, k) as N 1, k) = ) νl; 1 ) e 1 bl + kr)n3 with l mo 1 ) l+kr, 1 )=1 νl; 1 ) = hn 2 l 1 ) n 3 N 3 n 3, 1 )=1 ˆfh/ 1 ). Here is restriction to 1 h < H, h, 1 ) = 1 an n 2 N 2. It follows by Cauchy s inequality that 2 N 1, k) P 1 P 2, 14.1) P 1 = l mo 1 ) k K νl; 1 ) 2, P 2 = l mo 1 ) k K l+kr, 1 )=1 n N 3 n, 1 )=1 The estimation of P 1 is straightforwar. By 13.10) we have e 1 bl + kr)n ) 2. P 1 N 2 1 #{h 1, h 2 ; n 1, n 2 ) : h 2 n 1 h 1 n 2 mo 1 ), 1 h i < H, n i N 2}. 48

The number of the 4-tuples h 1, h 2 ;, n 1, n 2 ) satisfying the above conitions is 2 τm)). l mo 1 ) Since HN 2 1+ε 1 by 13.6), it follows that 1 m<2hn 2 m l 1 ) The estimation of P 2 is more involve. We claim that P 1 1+ε 1 N 2 1. 14.2) P 2 1 x 3/16+52ϖ+ε K 2. 14.3) Write 1 = rq. Note that N 3 r x1/6 69ϖ 14.4) by 13.8), 13.11), 13.3) an the secon inequality in 13.12). Since e 1 bl + kr)n) = e 1 bl + kr)nr + s)) + Or), n N 3 n, 1 )=1 it follows that Hence, k K n N 3 l+kr, 1 )=1 n, 1 )=1 Ul) = 0 s<r s,r)=1 0 s<r n N 3 s,r)=1 /r nr+s,q)=1 k K l+kr, 1 )=1 P 2 e 1 bl + kr)n ) = Ul) + OKr), l mo 1 ) n N 3 /r nr+s,q)=1 e 1 bl + kr)rn + s) ). Ul) 2 + 1 Kr) 2. 14.5) The secon term on the right sie is amissible for 14.3) by the secon inequality in 13.12). On the other han, we have Ul) 2 = V k 2 k 1 ; s 1, s 2 ), 14.6) V k; s 1, s 2 ) = l mo 1 ) n 1 N 3 /r n 1 r+s 1,q)=1 k 1 K k 2 K n 2 N 3 /r l mo 1 ) n 2 r+s 2,q)=1 0 s 1 <r 0 s 2 <r s 1,r)=1 s 2,r)=1 49 e 1 bln1 r + s 1 ) bl + kr)n 2 r + s 2 ) ).

Here is restriction to l, 1 ) = l + kr, 1 ) = 1. To hanle the right sie of 14.6) we first note that if l l 1 r + l 2 qmo 1 ), then the conition ll + kr), 1 ) = 1 is equivalent to l 1 l 1 + k), q) = l 2, r) = 1. In this situation, by the relation 1 r 1 q + q mo 1) r we have ln 1 r + s 1 ) l + kr)n 2 r + s 2 ) 1 r2 l 1 n 1 r + s 1 ) r 2 l 1 + k)n 2 r + s 2 ) q + q2 s 1 s 2 l 2 s 2 s 1 ) r mo 1). Thus the innermost sum in the expression for V k; s 1, s 2 ) is, by the Chinese remainer theorem, equal to C r s 2 s 1 ) e q br2 ln 1 r + s 1 ) br 2 l + k)n 2 r + s 2 ) ). It follows that W k; s 1, s 2 ) = n 1 N 3 /r n 1 r+s 1,q)=1 l mo q) ll+k),q)=1 V k; s 1, s 2 ) = W k; s 1, s 2 )C r s 2 s 1 ), 14.7) n 2 N 3 /r l mo q) n 2 r+s 2,q)=1 e q br2 ln 1 r + s 1 ) br 2 l + k)n 2 r + s 2 ) ). Here is restriction to ll + k), q) = 1. By virtue of 14.7), we estimate the contribution from the terms with k 1 = k 2 on the right sie of 14.6) as follows. For n 1 r + s 1, q) = n 2 r + s 2, q) = 1 we have l mo q) e q br2 ln 1 r + s 1 ) br 2 ln 2 r + s 2 ) ) = C q n 1 n 2 )r + s 1 s 2 ). On the other han, since N 3 x 1/3, by 13.11) an the secon inequality in 13.12) we have N 3 1 x 1/12+6ϖ r 1. 14.8) This implies N 3/r = oq), so that n N 3 /r C q nr + m) q 1+ε 50

for any m. It follows that W 0; s 1, s 2 ) q 1+ε r 1 N 3. Inserting this into 14.7) an using the simple estimate C r s 2 s 1 ) r 2+ε, we euce that 0 s 1 <r 0 s 2 <r 0 s 1 <r 0 s 2 <r s 1,r)=1 s 2,r)=1 V 0; s 1, s 2 ) 1+ε 1 N 3. It follows that the contribution from the terms with k 1 = k 2 on the right sie of 14.6) is 1+ε 1 KN 3 which is amissible for 14.3), since K 1 N 3 x 1/2+48ϖ N 1 1 x 3/16+52ϖ by 13.15) an 13.2). The proof of 14.3) is therefore reuce to showing that V k 2 k 1 ; s 1, s 2 ) 1 x 3/16+52ϖ+ε K 2. 14.9) k 1 K k 2 K k 2 k 1 0 s 1 <r s 1,r)=1 0 s 2 <r s 2,r)=1 In view of 14.4) an 14.8), letting n = min{n : n N 3/r}, n = max{n : n N 3/r}, we may rewrite W k; s 1, s 2 ) as W k; s 1, s 2 ) = n 1 q n 1 r+s 1,q)=1 n 2 q l mo q) n 2 r+s 2,q)=1 F y) is a function of C 2 [0, 1] class such that F n 1 /q)f n 2 /q) e q br2 ln 1 r + s 1 ) br 2 l + k)n 2 r + s 2 ) ), F y) = 1 0 F y) 1, if [ n F y) = 0 if y / an such that the Fourier coefficient κm) = 1 0 n q y n q, q 1 2q,, n q + 1 2q F y)e my) y 51 ],

satisfies { 1 κm) κ m) := min r, 1 m, Here we have use 14.8). By the Fourier expansion of F y) we obtain with W k; s 1 ; s 2 ) = Y k; m 1, m 2 ; s 1, s 2 ) = m 1 = m 2 = } q. 14.10) m 2 κm 1 )κm 2 )Y k; m 1, m 2 ; s 1, s 2 ), 14.11) n 1 q n 2 q l mo q) n 1 r+s 1,q)=1 n 2 r+s 2,q)=1 e q δl, k; m1, m 2 ; n 1, n 2 ; s 1, s 2 ) ) δl, k; m 1, m 2 ; n 1, n 2 ; s 1, s 2 ) = br 2 ln 1 r + s 1 ) br 2 l + k)n 2 r + s 2 ) + m 1 n 1 + m 2 n 2. Moreover, if n j r + s j t j mo q), then n j rt j s j )mo q), so that m 1 n 1 + m 2 n 2 rm 1 t 1 + m 2 t 2 ) rm 1 s 1 + m 2 s 2 ) mo q). Hence, on substituting n j r + s j = t j, we may rewrite Y k; m 1, m 2 ; s 1, s 2 ) as Zk; m 1, m 2 ) = Y k; m 1, m 2 ; s 1, s 2 ) = Zk; m 1, m 2 )e q rm1 s 1 + m 2 s 2 ) ), 14.12) t 1 mo q) t 2 mo q) l mo q) It follows from 14.7), 14.11) an 14.12) that 0 s 1 <r 0 s 2 <r s 1,r)=1 s 2,r)=1 V k; s 1, s 2 ) = Jm 1, m 2 ) = m 1 = m 2 = 0 s 1 <r 0 s 2 <r s 1,r)=1 s 2,r)=1 e q br2 lt 1 br 2 l + k)t 2 + rm 1 t 1 + m 2 t 2 ) ). κm 1 )κm 2 )Zk; m 1, m 2 )Jm 1, m 2 ), 14.13) e q rm1 s 1 + m 2 s 2 ) ) C r s 2 s 1 ). We now appeal to Lemma 12. By simple substitution we have Zk; m 1, m 2 ) = T k, bm 1 r 3, bm 2 r 3 ; q), 52

so Lemma 12 gives Zk; m 1, m 2 ) k, q) 1/2 q 3/2+ε, the right sie being inepenent of m 1 an m 2. On the other han, we have the following estimate that will be prove later m 1 = m 2 = κ m 1 )κ m 2 ) Jm 1, m 2 ) r 1+ε. 14.14) Combining these two estimates with 14.13) we obtain V k; s 1, s 2 ) k, q) 1/2 q 3/2+ε r 1+ε. This leas to 14.9), since 0 s 1 <r 0 s 2 <r s 1,r)=1 s 2,r)=1 by the first inequality in 13.12), an q 1/2 = 1 /r) 1/2 < x 1/4 21ϖ = x 3/16+52ϖ k 1 K k 2 k 1, q) k 2 K k 2 k 1 1/2 q ε K 2, whence 14.3) follows. The estimate 13.16) follows from 14.1)-14.3) immeiately, since N 1 x 3/8+8ϖ M 1, 1 < x 1/2+2ϖ, 31 32 + 36ϖ = 1 ϖ 2. It remains to prove 14.14). The left sie of 14.14) may be rewritten as 1 r In view of 14.10), we have m 1 = m 2 = 0 k<r κ m 1 )κ m 2 + k) Jm 1, m 2 + k). m= κ m) L, an κ m + k) κ m) for 0 k < r, since r < q by 13.11) an the secon inequality in 13.12). Thus, in orer to prove 14.14), it suffices to show that Jm 1, m 2 + k) r 2+ε 14.15) 0 k<r 53

for any m 1 an m 2. Substituting s 2 s 1 = t an applying Möbius inversion we obtain Jm 1, m 2 ) = C r t) e q rm2 t + m 1 + m 2 )s) ) t <r s I t ss+t),r)=1 C r t) t <r r 1 r s I t ss+t) 0r 1 ) e q rm1 + m 2 )s ), 14.16) I t is a certain interval of length < r an epening on t. For any t an square-free r 1, there are exactly τr 1 /t, r 1 )) istinct resiue classes mo r 1 ) such that ss + t) 0 mo r 1 ) if an only if s lies in one of these classes. On the other han, if r = r 1 r 2, then e q rm1 + m 2 )s ) min { r 2, r 2 m 1 + m 2 )/q 1} s I t s ar 1 ) for any a. Hence the inner sum on the right sie of 14.16) is τr) r 2 r min { r 2, r 2 m 1 + m 2 )q 1} which is inepenent of t. Together with the simple estimate C r t) τr)r, t <r this yiels Jm 1, m 2 ) τr) 2 r r 2 r min { r 2, r 2 m 1 + m 2 )/q 1}. It follows that the left sie of 14.15) is τr) 2 r r 1 r 2 =r 0 k 1 <r 1 Assume r 2 r. By the relation 0 k 2 <r 2 min { r 2, r 2 m 1 + m 2 + k 1 r 2 + k 2 )/q 1 }. 14.17) r 2 q q + 1 mo 1), r 2 qr 2 54

for 0 k < r 2 we have r 2 m + k) q r 2m q qk 1 + O r 2 q ) mo 1). This yiels 0 k<r 2 min { r 2, r 2 m + k)/q 1 } r 2 L 14.18) for any m. The estimate 14.15) follows from 14.17) an 14.18) immeiately. Acknowlegments. The author is grateful to the referees for carefully reaing the manuscript an making useful suggestions. References [1] E. Bombieri, J. B. Frielaner an H. Iwaniec, Primes in arithmetic progressions to large mouli, Acta Math. 1561986), 203-251. [2] E. Bombieri, J. B. Frielaner an H. Iwaniec, Primes in arithmetic progressions to large mouli II, Math. Ann. 2771987), 361-393. [3] E. Bombieri, J. B. Frielaner an H. Iwaniec, Primes in arithmetic progressions to large mouli III, J. Am. Math. Soc. 21989), 215-224. [4] P. Deligne, La conjecture e Weil I, Publ. Math. IHES. 431974), 273-307. [5] J. B. Frielaner an H. Iwaniec, Incomplete Kloosterman sums an a ivisor problem, Ann. Math. 1211985), 319-350. [6] D. A. Golston, J. Pintz an C. Y. Yilirim, Primes in Tuples I, Ann. Math. 1702009), 819-862. [7] D. A. Golston, J. Pintz an C. Y. Yilirim, Primes in Tuples II, Acta Math. 2042010), 1-47. [8] D. R. Heath-Brown, Prime numbers in short intervals an a generalize Vaughan ientity, Cana. J. Math. 341982), 1365-1377. [9] D. R. Heath-Brown, The ivisor function 3 n) in arithmetic progressions, Acta Arith. 471986), 29-56. [10] H. Iwaniec, A new form of the error term in the linear sieve, Acta Arith. 371980), 307-320. 55

[11] P. Shiu, A Brun-Titchmarsh theorem for multiplicative functions, J. Reine Angew. Math. 3131980), 161-170. [12] K. Sounararajan, Small gaps between prime numbers: the work of Golston-Pintz- Yilirim, Bull. Amer. Math. Soc. 442007), 1-18. Yitang Zhang Department of Mathematics an Statistics University of New Hampshire Durham, NH 03824 E-mail: yitangz@unh.eu 56