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Journal of Theory 180 (2017) 643–659

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Journal of Number Theory

www.elsevier.com/locate/jnt

Twin primes and the parity problem

M. Ram Murty a,∗,1, Akshaa Vatwani b a Department of , Queen’s University, Kingston, Ontario K7L 3N6, Canada b Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, N2L 3G1, Canada a r t i c l e i n f o a b s t r a c t

Article history: We formulate a conjecture regarding the equidistribution Received 21 April 2016 of the Möbius function over shifted primes in arithmetic Received in revised form 8 May 2017 progressions. Our main result is that such a conjecture for Accepted 9 May 2017 a fixed even h, in conjunction with the Elliott– Available online 13 July 2017 Halberstam conjecture, can resolve the parity barrier and Communicated by the Principal Editors produce infinitely many primes p such that p +h is also prime. © 2017 Elsevier Inc. All rights reserved. MSC: 11N35 11N05

Keywords: Twin primes The parity problem Equidistribution in arithmetic progressions

1. Introduction

Let Λ(n)denote the von Mangoldt function, log p if n = p, Λ(n)= 0otherwise.

* Corresponding author. E-mail addresses: [email protected] (M. Ram Murty), [email protected] (A. Vatwani). 1 Research partially supported by an NSERC Discovery grant. http://dx.doi.org/10.1016/j.jnt.2017.05.011 0022-314X/© 2017 Elsevier Inc. All rights reserved. 644 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659

Let h be a fixed even integer. It is conjectured that Λ(n)Λ(n + h) ∼ S(h)x, (1.1) n≤x where S(h)is the singular series defined as 1 1 S(h)= 1+ 1 − . p − 1 (p − 1)2 p|h ph

If h =2, this gives an asymptotic formula for the number of twin primes. It is believed that sieve methods cannot resolve this conjecture because of the parity problem. However, recently some breakthroughs have been made in proving the infinitude of bounded gaps between primes beginning with the work of Yitang Zhang [27] and later by Maynard [17]. The Polymath project [21] highlights the limitations of these new methods and amplifies the parity problem in resolving the twin prime conjecture. The term “parity principle” was first coined by Atle Selberg [23], who came across this phenomenon in 1946, in his work on the ingenious sieve that bears his name today. He described this principle as follows: “Sets of tend to be very evenly distributed with respect to the parity of their number of prime factors unless they have been particularly produced, constructed or selected in a way that has a built in bias.” The discussion by Selberg in [22] indicates that sieve methods are unable to distinguish whether an integer has an odd or an even number of prime factors. This is commonly referred to as the parity problem. The parity problem can also be explained in the context of Bombieri’s asymptotic sieve [1], which highlights that classical sieve methods are unable to sift out having exactly r prime factors, irrespective of the choice of r. There have been a number of attempts (cf. [13,5,8,10]) in various settings, to break the parity barrier by postulating additional analytic data into the sieve machinery. In this article, we follow this line of thought. A related result is due to J. Friedlander and H. Iwaniec [9] who assumed an estimate for certain bilinear forms, relying upon cancellations arising from sign changes of the Möbius function, in order to circumvent the parity problem and show the infinitude of primes of the form a2 + b4. A problem closely related to the parity principle is that of showing significant cancel- lation in the summatory function of the Möbius function: M(x):= μ(n). n≤x

1 + The assertions M(x) = o(x)and M(x)  x 2 are equivalent to the theorem and the Riemann hypothesis respectively. Ahigher rank version of this was conjectured by S. Chowla [4], in terms of the related Liouville function λ(n), defined by (−1)Ω(n) where Ω(n)is the total number of prime factors of n, counted with multiplicity. M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 645

Conjecture (Chowla). Let h1 <...

This conjecture is known only in the case k =1, where again, the statement is equivalent to the prime number theorem. In general, one expects what is called the Möbius randomness law (cf. [11]), that is, for any “reasonable” of complex numbers an, the sum of the Möbius function twisted by this sequence is relatively small. A fundamental result of this type is Halász’s mean value theorem (see [6], Theo- rem 6.2), which completely determines the asymptotic behaviour of the sum −1 | | ≤ ∈ N n≤x x g(n)for any multiplicative function g, with g(n) 1for all n . Amore general conjecture along these lines was formulated by P.D.T.A Elliott [7]. Recently, K. Matomäki, M. Radziwiłł and T. Tao [16,24] formulated a slightly modified, “cor- rected” form of Elliott’s conjecture. They also succeeded in proving the averaged and logarithmically averaged (for k =2) versions of the Chowla and Elliott conjectures. The

Elliott conjecture essentially states that if g1, ..., gk : N → C are multiplicative functions absolutely bounded by 1, and are “reasonably distributed” (in a certain explicit sense, see Conjecture 1.5 of [16]), then for any a1, ..., ak, b1, ..., bk ∈ N, we have g1(a1n + b1) ...gk(akn + bk)=o(x), n≤x provided that aibj − ajbi =0for any 1 ≤ i

It will be convenient to use the notation μh to denote μ(n + h). One would expect this conjecture to be of the same level of difficulty as (1.1). Such estimates are inherently related to the parity problem in sieve theory, as we will see. It seems that Hildebrand [14] was the first to study such problems and derive a partial analogue of Halász’s theorem for the subsequence of shifted primes. Later, these results have been improved by Timofeev [25] and Khripunova [15]. Some results in this direction have also been obtained by J. Pintz in [20]. 646 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659

In finding small gaps between primes, one of the main tools is the behaviour of primes in arithmetic progressions.

θ Elliott–Halberstam Conjecture EHΛ(x ). For any A > 0, we have y x max max Λ(n) − A . (1.3) y≤x a φ(q) (log x)A q≤xθ (a,q)=1 n≤y n≡a (mod q)

This conjecture is true and is called the Bombieri–Vinogradov theorem when θ<1/2. This conjecture alone cannot resolve the parity problem and prove either (1.1) or (1.2), for instance cf. [21]. Let h be a fixed non-zero integer. We postulate a conjecture regarding the equidistribution of the Möbius function on shifted primes {p + h}, in arithmetic progressions. We show that this more powerful conjecture does break the parity barrier.

η Shifted Möbius Elliott–Halberstam Conjecture EHμh (x ). For any A > 0, we have 1 x max max Λ(n)μ(n + h) − Λ(n)μ(n + h) A . (1.4) y≤x a φ(q) (log x)A q≤xη (a,q)=1 n≤y n≤y n≡a (mod q)

The conjecture (1.2) with a stronger error term would simplify the statement of this conjecture, but our main theorem below shows why we have not done this.

Theorem 1.1. Let h =0 be a fixed even integer. Suppose that the conjectures θ C 1−θ EHΛ(x (log x) ) and EHμh (x ) are true for some fixed θ<1 and a suitably large fixed C. We then obtain the following:

(a) The assertions (1.1) and (1.2) are equivalent. (b) We have Λ(n)Λ(n + h) ≥ (1 − o(1))S(h)(1 −A(h))x, n≤x

where 1 A = 1 − . (1.5) h p(p − 1) ph p>2

In particular, part (b) of the theorem shows that the twin prime conjecture follows θ C 1−θ from EHΛ(x (log x) )and EHμ2 (x ). Note that if the Elliott–Halberstam conjecture 1− holds, that is, we have EHΛ(x )for any small  > 0, then we only need EHμ2 (x )for any small positive  in order to show the infinitude of twin primes! M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 647

We also remark that our proof goes through if we frame the equidistribution con- θ η ≡− jectures EHΛ(x )and EHμh (x )for the fixed residue class n h (mod q)instead of taking the maximum over all residue classes co-prime to q. This may be of interest in the light of extensions of the Bombieri–Vinogradov theorem proved in [2,3].

2. Decomposition of Λ

Recall (cf. Ex 1.1.6, [18]) that Λ(n)= μ(d)log(1/d). d|n

We can write Λ(n)= μ(d)log(1/d)+ μ(d)log(1/d):=Λy(n)+Λy(n). d|n d|n d≤y d>y

We want to apply this decomposition only when the argument n is square-free. Since Λ(n) = μ2(n)Λ(n) except when n = pm with m ≥ 2, and there are only O(x1/2 log x) such n ≤ x, we have Λ(n)Λ(n + h)= Λ(n)μ2(n + h)Λ(n + h)+O(x1/2 log x) n≤x n≤x 2 2 = Λ(n)μ (n + h)Λy(n + h)+ Λ(n)μ (n + h)Λy(n + h) n≤x n≤x + O(x1/2 log x)

1/2 := S1(y)+S2(y)+O(x log x). (2.1)

The condition that n + h is square-free will be needed to evaluate S2. Although this condition complicates the evaluation of S1, we show in the next section that we get the same asymptotic value as we get for S1 without this condition.

θ C 3. Evaluation of S1 using EHΛ(x (log x) )

We define the singular series by − 2C p 1 if h is even, h =0 , S(h)= 2 p>2,p|h p−2 (3.1) 0ifh is odd, where 1 C = 1 − . 2 (p − 1)2 p>2 648 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659

In order to evaluate S1, we will need the following special case of the Wiener–Ikehara Tauberian theorem due to D.J. Newman [19]. For a more general treatment of this, we refer the reader to the article [26].

Theorem 3.1 (Newman). Let |an| ≤ 1. We consider the series

∞ a F (s):= n , ns n=1 which is absolutely convergent for Re(s) > 1. If F (s) can be analytically continued to ≥ ∞ s ≥ Re(s) 1, then the series n=1 an/n converges for Re(s) 1.

In particular, the proof of this theorem shows that for Re(s) ≥ 1, we have

∞ a n = F (s). ns n=1

We need the following preliminary propositions.

Proposition 3.2. For a fixed even integer h, let (p − 1)2 g (d):= . (3.2) h p(p − 1) − 1 p|d ph

The series

∞ μ(d)g (d) f(s):= h , (Re(s) > 0), φ(d)ds d=1 (d,h)=1 can be analytically continued to Re(s) ≥ 0.

Proof. We have the following Euler product for f(s)for Re(s) > 0, g (p) (p − 1) f(s)= 1 − h = 1 − . (p − 1)ps (p(p − 1) − 1)ps ph ph

Multiplying and dividing by the factor (1 − 1/ps+1), for each prime, we obtain

− f(s)=ζ(s +1) 1G(s), (3.3) where ζ(s)is the Riemann-zeta function and G(s)is the Euler product M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 649

− − 1 1 1 1 (p − 1) G(s)= 1 − 1 − 1 − . ps+1 ps+1 (p(p − 1) − 1)ps p|h ph

If we show that G(s)is absolutely convergent for Re(s) ≥ 0, then the expression (3.3) gives the analytic continuation of f(s)to Re(s) ≥ 0. We can write G(s)as ⎛ ⎞ −1 −1 − 1 − 1 ⎝ − 1 ⎠ G(s)= 1 s+1 1 s+1 1 p p s+1 − 1 p|h ph p 1 p(p−1) ⎛ ⎞ −1 −1 1 1 1 1+O p(p−1) = 1 − 1 − ⎝1 − ⎠ , ps+1 ps+1 ps+1 p|h ph

−1 since (1 − xp) =1 + O(xp), where xp =1/p(p − 1). Continuing with this notation, we can simplify the above expression further as

− − 1 1 1 1 1 O (x ) G(s)= 1 − 1 − 1 − + p ps+1 ps+1 ps+1 ps+1 p|h ph   − − 1 1 O (x ) 1 1 = 1 − 1+ p 1 − . ps+1 ps+1 ps+1 p|h ph

Since (1 − p−s−1)−1 =1 + O(1/ps+1), we obtain

− 1 1 O (x ) O(x ) G(s)= 1 − 1+ p + p . ps+1 ps+1 p2s+2 p|h ph

Keeping in mind that xp =1/p(p − 1), it is easy to see that G(s)converges absolutely for Re(s) ≥−1. 2

Proposition 3.3. We have

∞ μ(d)g (d)log(1/d) A h = S(h), h φ(d) d=1 (d,h)=1 where Ah and S(h) are as defined by (1.5) and (3.1) respectively.

Proof. From the definition of f(s)above, we have

∞ μ(d)g(d)log(1/d) f (s)= , (3.4) φ(d)ds d=1 2d 650 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 which is absolutely convergent for Re(s) > 0. From the previous lemma we see that f (s) can be analytically continued to Re(s) ≥ 0. Indeed, we have    ζ (s +1) f (s)= ζ(s +1)−1G(s) = G(s)ζ(s +1)−1 + G(s) − ζ(s +1)−1 . ζ(s +1)

It can be checked that G(s)is absolutely convergent for Re(s) ≥ 0. Since ζ(s +1)−1 and − ζ ζ (s +1)have a simple zero and a simple pole respectively at s =0, with residue 1, at s =0we obtain:

− − 1 1 1 1 (p − 1) f (0) = G(0) = 1 − 1 − 1 − . p p (p(p − 1) − 1) p|h ph

We have obtained that the series

∞ μ(d)g(d)log(1/d) , (Re(s) > 0), φ(d)ds d=1 (d,h)=1 can be analytically continued to Re(s) ≥ 0. Using the Tauberian Theorem 3.1 obtained by Newman, we see that the above series in fact converges for Re(s) ≥ 0and is equal to f (0) at s =0. This gives

∞ μ(d)g(d)log(1/d) A = A f (0) = A G(0) = S(h), h φ(d) h h d=1 (d,h)=1 where the last equality is easy to check. This completes the proof. 2

We now prove the following lemma evaluating S1.

Lemma 3.4. Take h to be a fixed positive even integer, y = xθ for a fixed θ<1, and let θ C C be a fixed suitably chosen constant. Then assuming EH Λ(x (log x) ), we have

S1(y) ∼ S(h)x.

It is instructive to observe how this lemma would be proved if we drop the factor 2 μ (n + h)from S1. Then, S¯1 := Λ(n)Λy(n + h) n≤x = Λ(n) μ(d)log(1/d) n≤x d|n+h d≤y M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 651

= μ(d)log(1/d) Λ(n). d≤y n≤x n≡−h (mod d)  θ If the gcd (d, h) > 1, then this sum is bounded by y log y n|h Λ(n) x log x log h. We may thus assume (d, h) =1. Then μ(d)log(1/d) Λ(n) d≤y n≤x (d,h)=1 n≡−h (mod d)   x = μ(d)log(1/d) Λ(n) − φ(d) d≤y n≤x (d,h)=1 n≡−h (mod d) μ(d)log(1/d) + x . φ(d) d≤y (d,h)=1

By Lemma 2.1 of Goldston and Yıldırım [12], we have

μ(d)log(1/d) √ = S(h)+O(e−c log y). (3.5) φ(d) d≤y (d,h)=1

θ θ A For the first term, we take y = x and apply EHΛ(x )to see that it is  x/(log x) for any A > 0. This proves the lemma for S¯1. 2 Proof of Lemma 3.4. Recall (cf. Ex 1.1.7, [18]) that μ (n) = d2|n μ(d). This gives S1 = Λ(n) μ(d)log(1/d) μ(e) n≤x d|n+h e2|n+h d≤y = μ(d)μ(e)log(1/d) Λ(n) d√≤y n≤x e≤ x+h n≡−h (mod [d,e2]) [d,e2]≤x+h

Let 1 ≤ z ≤ x be chosen later. Then, when e >z, the terms in this sum are x  (log x)2 μ2(d)μ2(e) +1 . [d, e2] [d,e2]≤2x e>z

Letting δ =(d, e), we can write d = dδ, e = eδ, so that [d, e2] = de 2δ2. Hence, when e >z, the contribution to the sum is 652 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 x  (log x)2 +1 de 2δ2 de 2δ2≤2x eδ>z ⎛ ⎞ ⎛ ⎞ 1 1 1  x(log x)2 ⎝ ⎠ +(logx)2 ⎝ 1⎠ d e 2 δ2 ≤ ≤  ≤  z ≤  2 2 d 2x e 2x δ>z/e δ 2x e > δ d 2x/e δ x(log x)4 x(log x)3 x(log x)4  +  . z z z

Choosing z =(logx)B, B = A +4, with A sufficiently large, we obtain x S = μ(d)μ(e)log(1/d) Λ(n)+O . 1 (log x)A d≤y n≤x e≤z n≡−h (mod [d,e2])  Again if (de, h) > 1, we can show that the sum is bounded by yz log y n|h Λ(n) xθ(log x)B+2. We may thus assume (de, h) =1, to get x S = μ(d)μ(e)log(1/d) Λ(n)+O 1 (log x)A d≤y n≤x e≤z n≡−h (mod [d,e2]) (de,h)=1 μ(d)μ(e)log(1/d) = x φ([d, e2]) d≤y e≤z (de,h)=1 ⎛ ⎞ ⎜ ⎟ ⎜ x ⎟ + μ(d)μ(e)log(1/d) ⎝ Λ(n) − ⎠ φ([d, e2]) d≤y n≤x e≤z n≡−h (mod [d,e2]) (de,h)=1 x + O . (log x)A

For the second term above, letting r denote [d, e2], it is clear that r ≤ xθ(log x)2B in the given range of d, e. Since the number of pairs of natural numbers (n, m)such that [n, m] = r is a multiplicative function of r, it is easy to see that this number can be bounded above by τ3(r), where τ3(r)is the number of ways of writing r as a product of three positive integers. Then, by the Cauchy–Schwarz inequality and the trivial bound x x Λ(n) −  +1, φ(r) r n≤x n≡−h (mod r) the term under consideration is M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 653 ⎛ ⎞ ⎜ ⎟ ⎜ x ⎟  log x τ3(r) ⎝ Λ(n) − ⎠ φ(r) r≤xθ (log x)2B n≤x n≡−h (mod r) ⎛ ⎞ 1/2 x  log x ⎝ τ (r)2 +1 ⎠ 3 r r≤xθ (log x)2B ⎛ ⎞ 1/2 ⎜ ⎟ ⎜ x ⎟ × ⎝ Λ(n) − ⎠ . φ(r) r≤xθ (log x)2B n≤x n≡−h (mod r)

θ 2B Using elementary estimates for the first term in the above product and EHΛ(x (log x) ) for the second, we finally obtain μ(d)μ(e)log(1/d) x S (y)=x + O . 1 φ([d, e2]) (log x)A d≤y e≤z (de,h)=1

To finish the proof of the lemma, we will show that as y →∞,

μ(d)μ(e)log(1/d) S := ∼ S(h). 11 φ([d, e2]) d≤y e≤z (de,h)=1

Since d, e are square-free, we can write [d, e2]as the product of the co-prime integers e2 and d/(d, e), so that φ([d, e2]) = eφ(e)φ(d)/φ((d, e)). Hence,

μ(d)log(1/d) μ(e)φ((d, e)) S = . 11 φ(d) eφ(e) d≤y e≤z (d,h)=1 (e,h)=1

We can extend the last sum to infinity with an error

1 φ((d, e)) 1 (d, e)  log y  log y . φ(d) eφ(e) φ(d) eφ(e) d≤y e>z d≤y e>z   On using (d, e) = r|(d,e) φ(r), and writing d = rd and e = re we find that this error is

1 1 1 (log y)3  log y  . rφ(r) φ(d) eφ(e) z r≤y d≤y/r e>z/r 654 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659

This is o(1) provided we take B>3 while choosing z =(logx)B. We have now obtained

∞ μ(d)log(1/d) μ(e)φ((d, e)) S ∼ 11 φ(d) eφ(e) d≤y e=1 (d,h)=1 (e,h)=1 μ(d)log(1/d) φ(p, d) = 1 − φ(d) p(p − 1) d≤y ph (d,h)=1 μ(d)log(1/d) = A g (d), h φ(d) h d≤y (d,h)=1 where Ah and gh(d)are as in (1.5) and Proposition 3.2 respectively. Invoking Proposi- tion 3.3 completes the proof. 2

η 4. Evaluation of S2 using EHμh (x )

Recall that 2 2 S2(y)= Λ(n)μ (n + h)Λy(n + h)= Λ(n)μ (n + h) μ(d)log(1/d) n≤x n≤x d|n+h d>y

2 n+h Writing n + h = de and noting that μ (n + h)μ( e ) = μ(n + h)μ(e), we have e S (y)= Λ(n)μ(n + h) μ(e)log 2 n + h n≤x e|n+h n+h e< y e = μ(e) Λ(n)μ(n + h)log . n + h e< x+h n≤x y n≡−h (mod e)

When (e, h) > 1, the contribution to the sum is

x(log x)2  log x μ2(e) Λ(n)  . y x+h n|e e< y

x ≤ x+h ≤ x Moreover, since there is at most one e with y

x(log x)2   y(log x)2. e M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 655

Hence, we have e x S (y)= μ(e) Λ(n)μ(n + h)log + O + y (log x)2 2 n + h y e< x n≤x y ≡− (e,h)=1 n h (mod e) x = S − S + O + y (log x)2 , 3 4 y where S3 := μ(e)loge Λ(n)μ(n + h), e< x n≤x y ≡− (e,h)=1 n h (mod e) S4 := μ(e) Λ(n)μ(n + h)log(n + h). e< x n≤x y ≡− (e,h)=1 n h (mod e)

We evaluate S3 and S4 in the following propositions.

x ≤ η η Proposition 4.1. Suppose y x , and EHμh (x ) holds. Then for any A > 0, we have   A S3 =(−S(h)+o(1)) Λ(n)μ(n + h) + O(x/(log x) ). n≤x

≤ η η Proof. Since x/y x , we have by EHμh (x ),   1 S = μ(e)loge Λ(n)μ(n + h) − Λ(n)μ(n + h) 3 φ(e) e< x n≤x n≤x y ≡− (e,h)=1 n h (mod e) μ(e)loge + Λ(n)μ(n + h) x φ(e) ≤ e< y n x (e,h)=1 μ(e)loge x = Λ(n)μ(n + h)+O A−1 . x φ(e) ≤ (log x) e< y n x (e,h)=1

We complete the proof by using (3.5) to obtain that the first term is   (−S(h)+o(1)) Λ(n)μ(n + h) . 2 n≤x

η Recall that conjecture EHμh (x ) predicts that the function Λ(n)μ(n + h)is equidis- tributed in arithmetic progressions, within a certain range. In order to evaluate S4, we 656 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 need equidistribution in arithmetic progression for the function Λ(n)μ(n + h) log(n +2). η Indeed, this is an easy consequence of partial summation on EHμh (x ), as illustrated below.

η ∈ Z  Proposition 4.2. If EHμh (x ) holds for some fixed h , h =0, then we have for any A > 0, max max Λ(n)μ(n + h)log(n + h) y≤x (a,q)=1 q

Proof. For any (a, q) =1, let us use the notation Λ(n)μ(n + h)=M(x)+Eμ(x, q, a), (4.1) n≤x n≡a (mod q) 1 where M(x) = φ(q) n≤x Λ(n)μ(n + h). Then by partial summation, Λ(n)μ(n + h)log(n + h) n≤y n≡a (mod q) =log(y + h) Λ(n)μ(n + h) n≤y n≡a (mod q) y 1 − Λ(n)μ(n + h)dt t + h ≤ 2 n t n≡a (mod q) y M(t) = M(y)log(y + h) − dt + O max |Eμ(t, q, a)| log y . t + h t≤y 2

Again using partial summation, we see that the first two terms give us the desired main term 1 Λ(n)μ(n + h)log(n + h). φ(q) n≤y

The term in parenthesis is the error term and can be checked to satisfy the required bound on average. 2

This allows us to bound S4 as follows. M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 657

x ≤ η η Proposition 4.3. If y x , and EHμh (x ) holds, then for any A > 0, we have

x S  . 4 A (log x)A

Proof. Applying Proposition 4.2, we obtain μ(e) x S4 = Λ(n)μ(n + h)log(n + h)+O A−1 . x φ(e) ≤ (log x) e< y n x (e,h)=1

By Lemma 2.1 of Goldston and Yıldırım [12], we have

μ(e) √  e−c log x, φ(e) e

Putting together Propositions 4.1 and 4.3, we have proved the following result regard- ing the evaluation of S2(y).

Lemma 4.4. For any A > 0, we have ⎛ ⎞ x S (y)=(−S(h)+o(1)) ⎝ Λ(n)μ(n + h)⎠ + O + O((xη + y)(log x)2), 2 (log x)A n≤x

x ≤ η η provided y x and the Shifted Möbius Elliott–Halberstam Conjecture EHμh (x ) holds.

5. Proof of Theorem 1.1

We choose y = xθ and η =1 − θ, so that (2.1) and Lemmas 3.4, 4.4 give us ⎛ ⎞ Λ(n)Λ(n + h) ∼ (S(h)+o(1)) ⎝x − Λ(n)μ(n + h)⎠ , n≤x n≤x

θ C 1−θ provided EHΛ(x (log x) )and EHμh (x )hold. Assuming these two conjectures, this shows that (1.1) and (1.2) are equivalent to each other. The second part of the theorem follows from the trivial bound ≤ 2 Λ(n)μ(n + h) Λ(n)μ (n + h) n≤x n≤x 658 M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 and the formula 2 Λ(n)μ (n + h) ∼Ahx. n≤x

We prove this formula as follows. We have Λ(n)μ2(n + h)= Λ(n) μ(d) n≤x n≤x d2|n+h = μ(d) Λ(n) √ d≤ x+h n≤x n≡−h (mod d2) = μ(d) Λ(n) d≤x1/4 n≤x n≡−h (mod d2) + μ(d) Λ(n). √ x1/4

We use the trivial bound on the inner sum to see that the second term is x  log x +1  x3/4 log x. √ d2 x1/4

For the first term, the contribution of d such that (d, h) > 1is  x1/4 log x. Using the Bombieri–Vinogradov theorem, we get μ(d) x μ(d) Λ(n)=x + O φ(d2) (log x)A d≤x1/4 n≤x d≤x1/4 (d,h)=1 n≡−h (mod d2) (d,h)=1 x = A x + O(x3/4)+O , h (log x)A which proves the required formula and completes the proof.

Acknowledgments

We would like to thank Professors John Friedlander, V. Kumar Murty and Yitang Zhang for helpful suggestions on a previous version of this paper. We are grateful to the referee for detailed comments that improved the quality of the paper. M. Ram Murty, A. Vatwani / Journal of Number Theory 180 (2017) 643–659 659

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