Bounded gaps between primes


 Madeline Cain
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1 Boune gaps between primes Yitang Zhang It is prove that Abstract lim inf n p n+1 p n ) < , p n is the nth 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 BombieriVinograov 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 Lower boun for S Combinatorial arguments The ispersion metho Evaluation of S 3 r, a) Evaluation of S 2 r, a) A truncation of the sum of S 1 r, a) Estimation of R 1 r, a; k): The Type I case Estimation of R 1 r, a; k): The Type II case The Type III estimate: Initial steps The Type III estimate: Completion 48 References 55 1
2 1. Introuction Let p n enote the nth 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 BombieriVinograov 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 nonnegative 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 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 ) < ) 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 π ) π ) >
3 This result is, of course, not optimal. The conition k 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
4 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 > ) 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
5 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 ) 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 BombieriVinograov 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 =
6 which is henceforth assume. Let D be as in 2.5) with Let gy) be given by gy) = ϖ = 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
7 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 SiegelWalfisz 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 BirchBombieri result in the appenix to [5] see Lemma 12 below), which is employe by Frielaner an Iwaniec [5] an by HeathBrown [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 squarefree integers such that ϱ 1 p) = ν p, ϱ 2 p) = ν p 1. 7
8 Let an We have an T 2 = T 1 = 0 1 T 2 = 0 T 1 = 1 µ 1 2 )ϱ ) g 0 1 )g 0 2 ) k 0 + 2l 0 )! 1 k 0 + 2l 0 + 1)! µ 1 2 )ϱ ) ϕ ) 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 ol k 0+2l 0 +1 ). 3.2) l 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 ) l OL l0+ε ), 3.4) ϑ 1 ) an ϑ 2 ) are the multiplicative functions supporte on squarefree integers such that ϑ 1 p) = 1 ν ) 1 p, ϑ 2 p) = 1 ν ) 1 p 1. p p 1 8
9 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 wellknown result on the zerofree 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 πi ϑ D/) s s 1)S L l0 1+ε. s =1/L s l 0+1 9
10 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 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 ) 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
11 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 ) 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 squarefree 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
12 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 HeathBrown[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
13 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) 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
14 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 t 2 n t 1 n 2 mo ) l t 1 t 2 l 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 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 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 ) e 0 a1 n + a 2 n + l 0 ) + mn ),
15 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 ) This estimate is wellknown 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 squarefree. 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 BirchBombieri 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
16 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 squarefree, 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 squarefree 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 ) g 0 1 )g 0 2 ). 4.2) T 1 = Σ 1 + Σ 31, µ 1 2 )ϱ ) g 0 1 )g 0 2 ), µ 1 2 )ϱ ) g 0 1 )g 0 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
17 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 ) g 0 1 )g 0 2 ), µ 1 2 )ϱ ) g 0 1 )g 0 2 ) T 1 T 1 Σ 1 + Σ 2 + Σ 3, 4.3) 1 µ 1 2 )ϱ ) g 0 1 )g 0 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) 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) ) 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
18 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 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
19 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 OL A ). 4.14) q P q<d ϱ 1 q) q 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
20 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
21 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 δ2 2 k0 + 2l 0 + log 293)k 0 ). Together with 3.1), this implies that T 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 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 + δ )k0 ) log 293)k 0 < 2 < 1 292! 292π )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 < log log185100) log26500) < κ 1 < exp{ 1200}. 4.21) 21
22 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
23 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 ) g 0 1 )g 0 2 ). ϕ ) 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 ol k 0+2l 0 +1 ), 5.5) k 0 + 2l 0 + 1)! l ) κ 2 = δ ϖ)1 + δ2 2 k0 + 2l 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 ) Slog D) k 0+2l 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 oxl k 0+2l 0 +1 ). 5.6) k 0 + 2l 0 + 1)! l 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 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 ) 22l0 + 1) l κ 2 = k 0k 0 + 2l 0 + 1)1 + 4ϖ) κ 1 2l 0 + 1)2l 0 + 2) κ 1 ) 1 + 4ϖ)k 0 + 2l 0 )! 2l0 l 0 ), k 0 1 κ 2 ) k 0 + 2l κ ) 1) ϖ < 10 8.
24 Thus, by 4.21), both of the constants κ 1 an κ 2 are extremely small. It follows by simple computation that ω > ) 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 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 BombieriVinograov 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
25 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 ϖ. 25
26 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 ϖ < ν + ν 1 < 5 8 8ϖ, then γ is of Type I or II. For example, if ϖ < 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 ϖ or ν + ν ϖ. 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 ϖ. 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 ϖ ν 4 ν 3 ν 2 ν 1, ν 4 + ν 3 + ν 2 + ν 1 < 1 + log 2 L together imply that ϖ ν 3 + ν 4 < 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 SiegelWalfisz theorem, that for all the choices of β above the SiegelWalfisz 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
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