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It is easy to predict the main term for S(x) by using the formula Λ(n)= µ(d) log d . − Xd n | Inserting this in (1.2) we arrange S(x) as
S(x)= λd Ad(x) Xd6x where λ = µ(d) log d and A (x) are the congruence sums d − d (1.5) Ad(x)= an . nX6x n 0 (mod d) ≡ Hence the problem reduces to that of the asymptotic evaluation of the sums Ad(x). In practice there is no problem to obtain a bound of the correct order of magnitude with considerable uniformity. We assume that 1 8 (1.6) A (x) d− τ(d) A(x) d ≪ uniformly in d 6 x1/3. Of course this bound is crude and we need more precise formulas if we are insisting on asymptotic results. We assume that we can approximate Ad(x) by writing
(1.7) Ad(x)= g(d)A(x)+ rd(x) , so g may be regarded as a density function and rd(x) may be regarded as a rather small remainder term. In the leading term we assume that g is multi- plicative and satisfies 1 (1.8) 0 6 g(p) < 1, g(p) p− , ≪ for all primes. Moreover, some regularity in the distribution of g(p) is required. We express this by assuming that for y > 2 10 (1.9) g(p) = log log y + c + O (log y)− Xp6y with some constant c. Now we can write by (1.7) S(x)= H(x)A(x)+ R(x) where (1.10) H(x)= µ(d)g(d) log d − Xd6x and R(x) is the remainder term
(1.11) R(x)= λd rd(x) . Xd6x ASYMPTOTIC SIEVE FOR PRIMES 1043
The hypothesis (1.9) allows us (see (2.4)) to extend (1.10) to the complete series getting by partial summation, 1 (1.12) H(x)= H + O (log x)− where H is the positive constant (1.13) H = µ(d)g(d) log d . − Xd That this is positive follows since the series is also given by the infinite product 1 (1.14) H = (1 g(p)) 1 1 − . − − p Yp Hence we conclude that 1 (1.15) S(x)= HA(x) 1+ O (log x)− + R(x) . This would finish the job if only one could ignore the remainder term R(x), but in general this is nothing but a wishful dream. A primary reason is that it is impossible in practice to control the individual error terms rd(x) uniformly for d 6 x. We assume that the error terms satisfy 2 222 (R) µ (d) r (t) 6 A(x)(log x)− | d | dX6D for all t 6 x with some D = D(x) such that
2 (R1) x 3 < D(x) < x . The assumption (R) is quite realistic since D(x) is not necessarily required to be near x. But such a choice of D(x) necessitates some additional hypothesis by means of which one would be able to avoid an encounter with the extremely large moduli. We assume an estimate for bilinear forms of the following type: 222 (B) γ(n)µ(mn)amn 6 A(x)(log x)− Xm N
This is required for every C with 1 (B3) 1 6 C 6 xD− . Remarks. Since C is relatively small the coefficients γ(n,C) behave more or less as constant; indeed we have γ(n, 1) = 1. A thorough examination of γ(n,C) can be found in [DIT]. Finally, for simplicity we assume that the sequence is supported on A squarefree numbers, that is
(1.16) an =0 if µ(n) = 0 . In Section 9 we remove this assumption at the expense of strengthening slightly the hypothesis (R) and introducing a few extra minor conditions. Our main result is Theorem 1. Assuming the above hypotheses, we have log δ(x) (1.17) a log p = HA(x) 1+ O , p log ∆(x) Xp6x where the implied constant depends only on the function g. In practice (B) can be established in the range (B1) for δ = (log x)α and ∆ = xη with some positive constants α, η. For these choices the error term in (1.17) becomes O(log log x/ log x) and this cannot be improved by refining our arguments. If log δ(x) log ∆(x) then the error term exceeds the main ≫ term and (1.17) follows from the upper bound sieve. Therefore for the proof we may assume that x > ∆(x)A, ∆(x) > δ(x)A, and δ(x) > A for any fixed positive constant A. Note also that it suffices to prove the Theorem with ∆(x) in the inequality (B1) replaced by ∆(x)A (and similarly for δ(x)). Most of the assumptions made above are standard ones from sieve theory. We have tried to choose them so that they are not difficult to verify in practice and have made no attempt to choose a minimal set. For example, some of the information about the multiplicative function g is stronger than necessary. The assumption (1.9) is as strong as the prime number theorem; it could be weakened considerably but it is not worth the effort. What needs to be said is that (1.9) is not crucial for breaking the parity problem. In the classical sieve machinery all of the above assumptions but (B) are present in one form or another and, of these, (R) is the fundamental one. As already mentioned, with (R) alone one cannot capture the primes, even 1 ε for D(x) = x − , which is the case treated by Bombieri [B]. It is clear that D(x)= x cannot happen in practice. Actually, for thin sequences one cannot A expect (R) to hold with D(x) > A(x) and for such sequences the classical sieve does correspondingly worse. ASYMPTOTIC SIEVE FOR PRIMES 1045
By introducing into the sieve axioms the additional hypothesis (B) we shall be able to resolve the “parity problem” and detect primes. Our theorem gives conditions under which one can guarantee the expected number of primes in the sequence . Note that in case D(x)= x1 ε very little is required in the A − range (B1) of the bilinear forms, namely 1 ε 1 ε x 2 − ′ Thus µ(d)rd(x) has constant sign in long intervals, namely for all d satisfying x 6 x L+1 < d L with any integer L< √x. 1046 JOHN FRIEDLANDER AND HENRYK IWANIEC Of course such a sequence is rather artificial and we certainly expect that the sieve given here is quite capable of settling, for instance, the twin prime problem. The stumbling block is that in this case we have no idea how to prove that the relevant sequence satifies the condition (B). Even proving (R) to such a high level is currently beyond reach. Fortunately there are other interesting sequences for which these hypotheses can be verified. There are certainly other options for the introduction of bilinear form estimates to obtain primes with the sieve, see for example [DFI], but it is not yet clear what is the optimal general strategy if indeed there is one. Our choice was motivated by the desire to treat thin sequences, in particular an being the number of representations n = a2 + b4. In this case (R) was established in 3/4 ε [FI] with D(x) = x − . The bilinear form bound (B) valid in the range 1/4+ε 1/2 A x Acknowledgment. We are happy to thank the Institute for Advanced Study for pleasant working conditions and Enrico Bombieri for helpful con- versations. We thank Andrew Granville, Mark Watkins, and the referee for pointing out inaccuracies in the previous draft and making suggestions which led us to improvements. J.F. also wishes to express thanks to Macquarie Uni- versity for their hospitality during part of this work. 2. Technical reductions What we actually need for the proof of Theorem 1 are the following hy- potheses: 2 3 (R ) µ (d)τ (d) r (t) A(x)(log x)− ′ 5 | d |≪ dX6D 3 (B ) τ (m) γ(n)µ(mn)a A(x)(log x)− ′ 5 mn ≪ Xm N In this section we derive (R′) and (B′) from (R) and (B), respectively, using the crude bound (1.6). As soon as (R′), (B′) are established the hypotheses (R), (B) and (1.6) are no longer required. We begin with a lemma which is reminiscent of a result due to Wolke [W]. Lemma 1. Fix k > 2. Any squarefree integer n > 1 has a divisor d 6 n1/k such that τ(n) 6 (2τ(d))k . Proof. Write n = p ...p with p < < p and put d = p ...p ; 1 r 1 · · · r 1 [r/k] thus d 6 n1/k and 2τ(d) = 21+[r/k] > 2r/k = τ(n)1/k completing the proof. We use Lemma 1 for k = 3. First we estimate the following sums: 4 τ5(d) 3 Ad(x) 6 τ10(d)Ad(x)= τ11(n)an Xd6x Xd6x nX6x 4 12 6 τ(n) an 6 (2τ(d)) Ad(x) nX6x d6Xx1/3 by the lemma. Then by (1.6) this is bounded by O (A(x)V (x)) where ♭ 1 20 220 V (x)= d− τ(d) (log x) . ≪ Xd6x Here, and hereafter, ♭ indicates that the summation is restricted to squarefree integers. Therefore P 4 220 (2.1) τ (d) 3 A (x) A(x)(log x) . 5 d ≪ Xd6x Similarly, but using (1.9), ♭ 4 11 (2.2) τ (d) 3 g(d) 6 (1 + 11g(p)) (log x) . 5 ≪ Xd6x pY6x Combining (2.1) and (2.2) we infer that ♭ 4 220 (2.3) τ (d) 3 r (t) A(x)(log x) . 5 | d |≪ Xd6x Finally (R′) follows from (R) and (2.3) by H¨older’s inequality. The derivation of (B′) from (B) is similar. 1048 JOHN FRIEDLANDER AND HENRYK IWANIEC We shall also need, and conclude this section by proving, the following consequence of (1.9): 6 (2.4) µ(d)g(d) σ (log y)− ≪ ν Xd6y (d,ν)=1 uniformly in ν > 1, y > 2, where 1 (2.5) σν = 1+ √p . Yp ν | Let 2 6 z 6 √y and P (z) be the product of the primes p < z. We split the sum (2.4), say Gν (y), as follows: Gν (y)= µ(m)µ(n)g(m)g(n) mn6XXy,(mn,ν)=1 m P (z),(n,P (z))=1 | = µ(m)g(m) µ(n)g(n) mX6√y nX6y/m m P (z),(m,ν)=1 (n,νP (z))=1 | + µ(n)g(n) µ(m)g(m). nX6√y √y ♭ g(n) 6 (1 + g(p)) log y ≪ nX6√y pY6√y we get G (y) ( G (w, z) + H(y, z)) log y ν ≪ | ν | for some w with √y 6 w 6 y, where Gν(w, z)= µ(n)g(n) Xn6w (n,νP (z))=1 and H(y, z)= g(m). m>X√y m P (z) | We estimate H(y, z) by using Rankin’s trick as follows: ε 2ε ε 2ε H(y, z) 6 y− g(m)m = y− 1+ g(p)p mXP (z) p 1 by choosing ε = (log z)− . To estimate Gν(w, z) we consider F (w, z)= µ(n)f(n) Xn6w (n,P (z))=1 where f is another multiplicative function which satisfies the hypothesis (1.9). By (1.9) we deduce that f(p) g(p) 6 h(z) − u 1 10 10 h(z) p− + (log z)− σ (log z)− . ≪ ≪ ν p>Xz,p ν | We also have the inequality F (w, z) G (w, z) 6 h(z) f(m)g(n) | − ν | mnX6w which can be seen from the identity f(p1...pr) g(p1...pr)= f(p1...pj 1) (f(pj) g(pj)) g(pj+1...pr). − − − 16Xj6r Therefore 2 Gν (w, z)= F (w, z)+ O h(z)(log w) . It remains to estimate F (w, z). To this end we take the simple function f(n)= 1 n− for which we know that c log w 6 F (w, z) z− + exp (log z) . ≪ − log z This can be proven by a standard contour integration using the bound for the corresponding zeta function s 1 1 s 1 3 µ(n)f(n)n− = ζ(s + 1)− (1 p− − )− (log( s + 1))(log z) − ≪ | | (n,PX(z))=1 p 10 3 log y 9 G (y) σ (log z)− (log y) + exp (log y) . ν ≪ ν −2 log z 2 We choose log z = (log y)(log log y)− obtaining (2.4). 1050 JOHN FRIEDLANDER AND HENRYK IWANIEC 3. Combinatorial identities For f : N C, an arithmetic function, we split → f(n) if n 6 z f(n 6 z)= 0 if n > z 0 if n 6 z f(n > z)= f(n) if n>z. We have f(n > z)= µ(b)f(c > z). Xbc n | Split the M¨obius function µ(b)= µ(b 6 y)+ µ(b>y) getting f(n > z)= µ(b 6 y)f(c > z)+ µ(b>y)f(c > z). Xbc n Xbc n | | In the first sum insert f(c > z)= f(c) f(c 6 z) getting the identity (a variant − of that of Vaughan), f(n > z)= µ(b 6 y)F n µ(b 6 y)f(c 6 z)+ µ(b>y)f(c > z), b − Xb n Xbc n Xbc n | | | where F = f 1. Split the last sum, say f(n; y, z), into three subsums ∗ f1(n; y, z)= µ(b>sy)f(c>sz), Xbc n | f2(n; y, z)= µ(sy > b>y)f(c > z), Xbc n | f3(n; y, z)= µ(b>sy)f(sz > c > z). Xbc n | Here s can be any number > 1. We choose s to be the power of 2 in the interval ∆√x ∆√x 6 n F (n; y)= µ(b y)F b , Xb n | F (n; y, z)= µ(b 6 y)f(c 6 z). Xbc n | ASYMPTOTIC SIEVE FOR PRIMES 1051 We have 3 (3.1) f(n > z)= F (n; y) F (n; y, z)+ f (n; y, z). − j Xj=1 We apply the integral operator eY eZ 1 Ih(Y, Z)= h(y, z)(yz)− dy dz ZY ZZ to obtain an identity with smoothed parts. Note that if h is independent of one or both variables the integration over the relevant variable(s) does nothing. When we apply this operator to (3.1) we denote the results by changing y, z to Y, Z as appropriate. For example for the truncated function f(n > z) this operator gives f(n) if n > eZ n 6 f(n > Z)= f(n) log Z if Z 3 (3.2) Λ(n > Z)= L(n; Y ) L(n; Y, Z)+ Λ (n; Y, Z). − j Xj=1 We shall take n 6 x and make the choices 1 (3.3) Y = Z =∆− √D . Note that when y changes over the segment Y ρn = λν > 0 . Xν n | 1052 JOHN FRIEDLANDER AND HENRYK IWANIEC Clearly Λ(n > Z)= ρnΛ(n > Z), for squarefree n since Z > ∆. Therefore, (3.4) S(x, Z)= anΛ(n > Z)= anρnΛ(n > Z) nX6x nX6x 3 = T (x; Y ) T (x; Y, Z)+ S (x; Y, Z) − j Xj=1 where T (x; Y )= anρnL(n; Y ) nX6x T (x; Y, Z)= anρnL(n; Y, Z) nX6x Sj(x; Y, Z)= anρnΛj(n; Y, Z), j = 1, 2, 3 . nX6x The sum in (1.2) S(x)= ap log p Xp6x is closely approximated by S(x; Z); precisely we have 1 (3.5) 0 6 S(x) S(x; Z) 6 a log p A(x)(log x)− − p ≪ pX6eZ by (1.4). It remains to treat the various sums on the right side of (3.4). 4. Evaluation of T (x; Y ) We are able to treat the sum T (x; y) for individual x,y. We split this by means of log n = log n log b, getting b − T (x; y)= T1(x; y)+ T2(x; y). We have (4.1) T1(x; y)= anρn log n µ(b) nX6x Xb n b6| y = µ(b) anρn log n. Xb6y nX6x n 0 (mod b) ≡ Here the inner sum is, by partial summation, x dt (4.2) Vb(x) log x Vb(t) − Z1 t ASYMPTOTIC SIEVE FOR PRIMES 1053 where (4.3) Vb(x)= an λν nX6x Xν n n 0 (mod b) | ≡ = λνA[ν,b](x)= λν g([ν, b])A(x)+ r[ν,b](x) . Xν Xν b We have g([ν, b]) = g(ν)g (ν,b) and b µ(b)g = µ(d) µ(b)g(b). (ν, b) Xb6y Xd ν bX6 y | d (b,ν)=1 6 This last inner sum is, by (2.4), bounded by O σν(log x)− . By this and (4.2) we deduce that the contribution to (4.1) coming from the main term in (4.3) is 5 3 A(x)(log x)− λ g(ν)τ(ν)σ A(x)(log x)− . ≪ | ν | ν ≪ Xν The remainder terms in (4.3) make a contribution to (4.1) which is (log x) λ µ(b)r (x) ≪ | ν [ν,b] | Xν Xb ♭ 2 (log x) τ (d) r (x) A(x)(log x)− ≪ 3 | d |≪ d We insert the approximation (1.7) for A[ν,b](x). The remainder terms give a contribution which may be estimated in the same way as those for T1(x; y) giv- ing the same bound A(x)(log x) 2. Here, however, the main term provides ≪ − also the main term for the total sum S(x). This is (4.4) A(x) λ g(ν) µ(b)g b log b . − ν (ν,b) Xν Xb6y We extend the summation to all b making an error A(x)(log x) 3, by (2.4). ≪ − Then, we evaluate the complete inner sum as µ(η) µ(b)g(b) log ηb = µ(η) µ(b)g(b) log b Xη ν (b,νX)=1 Xη ν (b,νX)=1 | | 1054 JOHN FRIEDLANDER AND HENRYK IWANIEC and this is equal to H if ν = 1 and equal to zero otherwise. Thus the sum − 2 (4.4), and also T2(x,y) are given by HA(x)+ O A(x)(log x)− . Combining this with the bound for T1(x,y) we obtain 2 (4.5) T (x,y)= HA(x)+ O A(x)(log x)− . 5. Estimation of T (x; Y, Z) It again suffices to estimate the contribution for given y, z and here this is given by the sum T (x; y, z)= µ(b) Λ(c) λνA[ν,bc](x). Xb6y Xc6z Xν We insert the approximation (1.7) for A[ν,bc](x). For the contribution coming from the main terms, arguing as with T1(x; y), we are now led to estimate the sum ♭ bc b ♭ c µ(b) Λ(c) g (ν,bc) = µ(b) g (ν,b) Λ(c) g (ν,c) Xb6y Xc6z Xb6y Xc6z c∤b c∤b and by (2.4), (1.8) this is τ(ν)σ (log y) 6(log νz). Summing over ν, as with ≪ ν − T (x; y) we get a contribution A(x)(log x) 3. 1 ≪ − The contribution coming from the remainder terms is bounded by ♭ ♭ ♭ 2 λ Λ(c) r (x) 6 (log yz) r (x) A(x)(log x)− | ν| | [ν,bc] | [ν,d] ≪ Xν Xb6y Xc6z Xν dX6yz c∤b 2 2 1 by (R′) since yz 6 e ∆− D < ∆− D. From the two estimates we conclude that 2 (5.1) T (x; y, z) A(x)(log x)− . ≪ 6. Estimation of S1(x; Y, Z) As before this may be estimated for individual y and z. We have S1(x; y, z)= µ(b)Λ(c)ρbcdabcd . XXXbcd6x b>sy,c>sz Note that the conditions of summation imply that c is located in a short interval (in the logarithmic scale), namely 1 (6.1) δ− √x = Λ(c) λνA[cd,ν](x) . Xc Xd Xν Of course, we have lost the possibility of detecting cancellation due to the sign changes of µ(b), but we have gained the congruence sums which can be 3 evaluated asymptotically due to (R′). Note that [cd, ν] 6 cdν < δ ∆√x < D. We insert (1.7) and estimate the resulting remainder by (R′) getting 2 (6.3) S (x; y, z) 6 A(x)L(x)M(x)+ O A(x)(log x)− | 1 | where L(x), M(x) come from the main terms, namely ♭ L(x)= Λ(c)g(c), Xc ♭ M(x)= λν g([d, ν]) . Xd Xν Here c, d run over the segments (6.1) and (6.2) respectively. By (1.8) we infer that (6.4) L(x) log δ . ≪ For the estimation of M(x) we write g([d, ν]) = g(d)g(ν/(ν, d)) and note that λνg(ν/(ν, d)) > 0 . Xν The latter is a property of any upper-bound sieve, and it holds for any multi- plicative function f(ν) with 0 6 f 6 1 in place of g(ν/(ν, d)). This positivity property allows us to introduce the factor (δ2/d)ε into M(x) and extend the summation to all d (Rankin’s trick) getting 2ε ♭ ε M(x) 6 δ λν g(ν) g(d/(ν, d))d− Xν Xd 2ε ε = δ (1 + g(p)p− ) λν g(ν)h(ν) Yp Xν where h(ν) is the multiplicative function with ε ε 1 1 h(p)=(1+ p− )(1 + g(p)p− )− < 2(1 + g(p))− . 1056 JOHN FRIEDLANDER AND HENRYK IWANIEC Note that g(p)h(p) < 1, so the sieve theory applies giving ε 1 λ g(ν)h(ν) (1 g(p)h(p)) = (1 g(p))(1 + g(p)p− )− . ν ≪ − − Xν p 1 (6.5) M(x) (1 g(p)) (log ∆)− . ≪ − ≪ p We remark that it is for this bound alone that we introduced the λν. Without doing so we would lose a logarithm, and this very tight bound cannot afford such a loss. The optimal choice of λν belongs to the theory of the 2-dimensional sieve. 7. Estimation of S2(x; Y, Z) In this section and the next we finally encounter the sums where (due to contamination of the variables coming from the support of ρ) it is necessary to perform the integrated estimate. Whenever an occurs we assume n 6 x without writing this condition every time. We have S2(x; y, z)= µ(b)Λ(c)ρebcaebc, Xe y 6 (log x) λν1ν2 µ(b)aν1ν2bk . | | y sy Xν1 Xν2 Xk In order to remove ν2 from the range of the inner summation we take advantage of the integration over y (still not integrating over z). eY dy 6 (7.1) S2(x; Y, z) (log x) λν1ν2 µ(b)aν1ν2bk | | Z Y y Xν1 Xν2 Xk ∆ y eY dy 6 (log x) τ3(ℓ) µ(b)abℓ . Z Y y ∆ Xℓ y ASYMPTOTIC SIEVE FOR PRIMES 1057 By (B′) we conclude that 1 (7.2) S (x; Y, z) A(x)(log x)− . 2 ≪ 8. Estimation of S3(x; Y, Z) Recall that S3(x; y, z)= µ(b) Λ(c)ρebaebc . Xe b>syX z Here we have, for the first time, taken advantage of the fact that ρebc = ρeb since c is prime. In essence this sum is of the same type as that in the previous section but with b and c interchanged. Thus by analogy to the previous argument, we put c into the inner summation because its range is appropriate for (B′). However, the fact that c is weighted by Λ rather than µ necessitates splitting it further in order to obtain a factor with sign changes from which we may hope to detect cancellation. We write µ(c)Λ(c)= µ(m) log m = λ+(c) λ (c) C − − Xm c | 1 (recall c is squarefree) with C = xD− where + + m λ (c)= µ(m) log C Xm c | + C λ−(c)= µ(m) log m . Xm c | Note that the definition of λ−(c) is close to γ(c, C) in (B2); precisely we have C 1 λ−(c)= γ(c, t)t− dt . Z1 This gives a contribution to S3(x; y, z) which, by virtue of (B′) applied log2 s times, satisfies 1 (8.1) S−(x; y, z) 6 τ (ℓ) µ(c)λ−(c)a A(x)(log x)− . | | 4 cℓ ≪ Xℓ z + The contribution to S3(x; y, z) from λ (c) is arranged as follows + + m S (x; y, z)= µ(b)ρeb µ(ℓ) log C aebℓm Xe b>syX z<ℓmXX6sz x dt = µ(b) ρeb µ(ℓ) aebℓm . Z t C b>syX ebz Put d = ebℓ, so d < D and d is squarefree because of (1.16). Here the inner sum is equal to A (min x, ebsz ) A (max dt, ebz ). Applying (1.7) this inner d { } − d { } sum becomes (8.2) g(d) a + r (min x, ebsz ) r (max dt, ebz ) . n d { } − d { } dt µ(b)ρebg(ebℓ)= g(eℓ) µ(ν1)g(ν1) λν1ν2 µ(b)g(b) (b,eℓX)=1 Xν1 Xν2 e (b,eℓνX1)=1 | bmin Now we can apply (R′) obtaining 1 (8.4) S′(x; y, Z) A(x)(log x)− . ≪ Collecting (8.1), (8.3) and (8.4) we conclude that 1 (8.5) S (x; y, Z) A(x)(log x)− . 3 ≪ Combining (3.4), (3.5), (4.5), (5.1), (6.6), (7.2) and (8.5) we complete the proof of Theorem 1. 9. A generalization Our condition that an are supported on squarefree numbers may not be convenient in some applications. Therefore, in this section we establish a result ASYMPTOTIC SIEVE FOR PRIMES 1059 which does not require this condition. Throughout we assume that = (a ) A n satisfies all the previous hypotheses except possibly for (1.16). We replace (1.16) by the following condition: For all primes p we have 2 2 2 (9.1) 0 6 g(p ) 6 g(p), g(p ) p− . ≪ We may remark that, although this condition was not needed before it would 2 have been natural to include it in the form g(p ) = 0 because Ap2 (x)=0 due to (1.16). Now we need to have some control on the size of the coefficients an. We assume that 2 2 2 (9.2) an 6 x− 3 A(x) . nX6x Remarks. The condition (9.2) holds for sequences = (an) satisfying ε 2 +2ε A an 6 n and A(x) > x 3 . Actually our argument requires a slightly weaker condition, for example the bound 2 1 2 1 (9.3) an 6 x− A(x) D(x) 2 nX6x would be sufficient. We furthermore need a slightly stronger version of (R), namely for any t 6 x we assume 3 2 (R ) r (t) 6 A(x)L− 3 | d | d 24 where L = (log x)2 and 3 restricts to cubefree moduli. P Theorem 2. Assume the following hypotheses: (1.4), (1.6), (1.8), (1.9), (B), (B1), (B2), (B3), (9.1), (9.2), (R3) and (R1). Then (1.17) holds true for = (a ). A n By analogy with the squarefree case, before proving Theorem 2 we show that (R3) implies 3 1 217 (R ) τ(d) r (t) A(x)L− (log x) . 3′ | d |≪ d Lemma 2. Fix k > 2. Any n > 1 has a divisor d 6 n1/k such that k log k τ(n) 6 (2τ(d)) log 2 . 1060 JOHN FRIEDLANDER AND HENRYK IWANIEC Proof. Write n = ℓm where ℓ collects those primes occuring with exponent > k and m is made up of those having exponent < k. Thus (ℓ,m)=1. Put [α/k] dℓ = p pYα ℓ k 1/k so that dℓ 6 ℓ and k α k > τ(dℓ) = k + 1 (α +1) = τ(ℓ) . pYα ℓ pYα ℓ k k α1 αr Write m = p1 ...pr with p1 < < pr. Setting dm = p1 ...p[r/k] we have 1/k · · · dm log 2 log 2 k r (2τ(dm)) > 2 > (αj + 1) log k = τ(m) log k . Yj Take d = dℓdm completing the proof. We proceed to the proof of (R3′ ). By Lemma 2 with k = 3, the left-hand side is bounded by 1 1 3 3 2 3 2 27 τ(d)5 r (t) 6 27 τ(d)10 r (t) r (t) | d | | d | | d | X1 X1 X1 d6x 3 d6x 3 d6x 3 by Cauchy’s inequality. In the first sum we replace r (t) by A (x)+ g(d)A(x) | d | d and estimate as follows. By (1.6) 3 10 1 18 218 τ(d) A (x) A(x) d− τ(d) A(x)(log x) , d ≪ ≪ X1 X1 d6x 3 d6x 3 while, by (9.1) and (1.9), 3 10 τ(d)10g(d) 6 1 + 210g(p) + 310g(p2) (log x)2 . ≪ X1 pY6x d6x 3 For the second sum we apply (R3). These give (R3′ ). Now we are ready to prove Theorem 2. We shall apply Theorem 1 for the sequence ˜ = (˜a ) with A n 2 (9.4)a ˜n = µ (n)an . Letg ˜(d) be the multiplicative function given by g(p) g(p2) (9.5)g ˜(p)= − . 1 g(p2) − ASYMPTOTIC SIEVE FOR PRIMES 1061 Notice thatg ˜(p) satisfies (1.8) and (1.9) due to (9.1). For any d squarefree we write (9.6) A˜d(x)= a˜n =g ˜(d)A˜(x) +r ˜d(x) nX6x n 0 (mod d) ≡ where A˜(x)= A˜1(x). Although (B) for and ˜ coincide, it takes some effort to verify (R) for A A r˜d, that is ♭ 222 (9.7) R(t, D)= r˜ (t) 6 A˜(x)(log x)− . | d | dX6D We shall derive this from (R3). Detecting squarefree numbers by the M¨obius 2 formula µ (n)= ν2 n µ(ν) we get, for d squarefree P | ˜ Ad(x)= µ(ν)A[ν2,d](x) =g ˜(d)GA(x)+ µ(ν)r[ν2,d](x) Xν Xν where (9.8) G = 1 g(p2) , − Yp so that (9.9)g ˜(d)G = µ(ν)g [ν2, d] . Xν In particular, ˜ (9.10) A(x)= GA(x)+ µ(ν)rν2 (x) . Xν Hence r˜ (x)= µ(ν) r 2 (x) g˜(d)r 2 (x) . d [ν ,d] − ν Xν Therefore we have R(t, D) E log x, where ≪ ♭ ♭ E = r 2 (t) . | [ν ,d] | Xν dX6D Note that every k has at most τ(k) representations as k = [ν2, d] and k is cubefree if it has any. By (R3′ ) it follows that the error terms r[ν2,d](t) with ν 6 L and d 6 D contribute 1 217 (9.11) E A(x)L− (log x) . 1 ≪ 1062 JOHN FRIEDLANDER AND HENRYK IWANIEC From the remaining terms we get ♭ ♭ 2 E2 6 A[ν2,d](x)+ g [ν , d] A(x) ν>LX Xd6x 1 B = E21 + E22 + O A(x)(log x)L− (log L) where E21 + E22 is the splitting of the first term according to ν >L√D or L<ν 6 L√D and the second term is estimated by using (1.8), (1.9), and (9.1). Here B depends only on g. We estimate as follows: 1 1 2 2 E 6 a τ(n) 1 6 a2 τ(n)3 1 21 n n nX6x νX2 n nX6x nX6x νX2 n | | ν>L√D ν>L√D 1 1 1 1 17 1 17 x− 3 A(x)D− 4 L− 2 x 2 (log x) 6 A(x)L− 2 (log x) ≪ by (9.2) and (R1). For the second sum we apply Cauchy’s inequality getting 2 ♭ 3 E 6 A 2 (x) a τ(n) = S S , 22 ν n 1 2 L<νX6L√D nX6x say. Here ♭ 2 1 B S = g(ν )A(x)+ r 2 (x) A(x)L− (log L) 1 ν ≪ L<νX6L√D by (1.8), (9.1), and (R3). To estimate S2 we apply Lemma 2 with k = 3 getting by (1.6) 9 log 3 3 S2 = anτ(n) 6 Ad(x) (2τ(d)) log 2 nX6x d6Xx1/3 9 log 3 1 +8 39+8 A(x) d− τ(d) log 2 A(x)(log x) . ≪ ≪ Xd6x Gathering the above estimates we conclude the proof of (9.7), but with A(x) in place of A˜(x) (actually we saved at least a factor log x which is needed for clearing the implied constants below). Note, however, that along the above lines we have, by (9.10), also proved 1 (9.12) A˜(x)= GA(x) 1+ O (log x)− . This formula gives (9.7) for the sequence ˜ because x is large and it gives Ad (1.6) because 1 8 1 8 A˜ (x) 6 A (x) d− τ(d) A(x) d− τ(d) A˜(x). d d ≪ ≪ This completes the proof that ˜ satisfies all the hypotheses of Theorem 1. A ASYMPTOTIC SIEVE FOR PRIMES 1063 The constant (1.14) corresponding tog ˜ may be expressed as 1 1 − 1 (9.13) H˜ = (1 g˜(p)) 1 = HG− , − − p Yp with G as in (9.8), so that (9.12) yields 1 (9.14) H˜ A˜(x)= HA(x) 1+ O (log x)− . Therefore the formula (1.17) for the sequence ˜ = (˜a ) coincides with that for A n = (a ), as it has to. This completes the proof of Theorem 2. A n 10. A reduction of the bilinear form Our new axiom (B) in sieve theory can be made technically easier to verify in practice if the involved integer variables have no small prime divisors. Having this in mind, in this section we introduce an extra restriction into the inner sum in (B), namely (10.1) (n, Π) = 1 where Π is the product of all primes p < P and P is given subject to 35 (10.2) 2 6 P 6 ∆1/2 log log x . 35 Notice that this restriction is possible if ∆ > (log x)2 . We assume in place of (B) that * 226 (B ) γ(n)µ(mn)a A(x)(log x)− , ∗ mn ≪ Xm N For the proof it is enough to show that (B∗) implies (B) with δ, ∆ replaced by 2δ, 2∆2 respectively. Furthermore we only need to establish (B) in the following integrated form 2N C dt dw 222 ( B) γ(n,t)µ(mn)amn A(x)(log x)− Z Z t w ≪ R N 1 Xm w 1 1 for every N with ∆− √D To see this recall that (B) was used solely to estimate the sums S2 and S3 in Sections 7 and 8. In the case of S2 we note that (7.1) contains such an integration with y in place of w and γ(n,t) = 1. In the case of S3 we note that (8.1) may be integrated over z in place of w and there λ−(c) is the integral of γ(c, t) as required. We write n = n n where (n , Π) = 1 and n Π. The contribution to 0 1 0 1 | ( B) from terms with n1 > ∆ is estimated trivially as R 2 W1 6 (log C) τ(n0n1)µ (mn0n1)amn0n1 . XXXmn0n16x n1>∆ Every such n1 has a divisor d with ∆ < d 6 ∆P , so ♭ W1 6 (log C) anτ4(n) . Xd Π nX6x ∆ ♭ 1 32 232 G = ν− τ(ν) (log x) , 1 ≪ Xν6x 1 8 ε ε 1 8 ε 8 ε 1 G2 = d− τ(d) 6 ∆− d − τ(d) =∆− 1 + 2 p − , Xd Π Xd Π p In the case (10.3) we have γ(n0n1,t)= µ(d1)γ(n0, t/d1). dX1 n1 | Insert this and change variables t td , w wn getting → 1 → 1 2N C * dt dw W0 6 τ3(m) γ(n,t)µ(mn)amn . Z Z t w N/∆ 1 Xm w The same bound holds in the case γ(n,t) = 1, in fact in stronger form since then the integration in t is not needed and τ3(m) can be replaced by τ(m). By the argument that gave (B) (B ) it follows that (B ) implies ⇒ ′ ∗ * 223 τ (m) γ(n)µ(mn)a A(x)(log x)− 3 mn ≪ Xm N (the restriction (n, Π) = 1 only helps matters). Hence we conclude that 222 (10.5) W A(x)(log x)− . 0 ≪ Adding (10.4) to (10.5) we obtain ( B) completing the proof of Theorem 3. University of Toronto, Toronto, CanadaR E-mail address: [email protected] Rutgers University, New Brunswick, NJ E-mail address: [email protected] References [B] E. Bombieri, The asymptotic sieve, Mem. Acad. Naz. dei XL, 1/2 (1976), 243–269. [DFI] W. Duke, J.B. Friedlander, and H. Iwaniec, Equidistribution of roots of a quadratic congruence to prime moduli, Ann. of Math. 141 (1995), 423–441. [DIT] F. Dress, H. Iwaniec, and G. Tenenbaum, Sur une somme li´ee `ala fonction de M¨obius, J. Reine Angew. Math. 340 (1983), 51–58. [FI] E. Fouvry and H. Iwaniec, Gaussian primes, Acta Arith. 79 (1997), 249–287. [FI1] J. Friedlander and H. Iwaniec, Bombieri’s sieve, in Analytic Number Theory, Proc. Halberstam Conf., Allerton Park Illnois, June 1995, ed. B. C. Berndt et al., pp. 411–430, Birkh¨auser (Boston), 1996. 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