Sieves in Number Theory

Sieves in Number Theory Lecture Notes Taught Course Centre 007 Tim Browning, Roger Heath-Brown Tyeset by Sanro Bettin All errors are the resonsibility of the tyesetter. In articular there are some arguments which, as an exercise for the tyesetter, have been fleshe out or re-interrete, ossibly incor- rectly. Tim s lectures were neater an more concise. Corrections woul be gratefully receive at sanro.bettin@bristol.ac.uk

Contents Introuction Sieve of Eratosthenes 5 3 Large Sieve 0 4 Selberg sieve 9 5 Sieve limitations 3 6 Small gas between rimes 37

Chater Introuction Sieves can be use to tackle the following questions: i Are there infinitely many rimes such that + is also rime? ii Are there infinitely many rimes such that = n + for some n N? iii Are there infinitely many rimes such that 4 + is also rime? iv Is every sufficiently large n a sum of two rimes? v Is it true that the interval n, n + contains at least one rime for every n N? These roblems are still oen, but, using Sieves methos, some stes towars their solutions have been one. For examle, in 966 Chen rove a weaker version of iv stating that every sufficiently large n is a sum of a rime an a P where P r enotes the numbers that have at most r rime factors. These roblems are also relate to imortant roblems in other Mathematics branches, such as Artin s rimitive root conjecture, which says that, for all a Z with a 0, ±, there exists infinitely many rimes such that a is a rimitive root moulo. Proosition. If iii is true, then Artin s conjecture is true for a =, i.e. there exists infinitely many rimes such that is a rimitive root moulo.

Proof. Let = k +, with k N, an q = 4 + = 8k + 5 be rimes. Recall that for all rime r where a { if r ± mo 8, = r if r ±3 mo 8, is the Legenre symbol. Therefore q = an so there oesn t exist any x such that x mo 8. Furthermore, by Fermat s little theorem, 4 = q mo q an so the orer of moulo q must be,, 4,, or 4. It s easily checke that the orer can t be, or 4, an it can t be either because otherwise = q 4 mo q an so k+ mo q. It remains to show that mo q. If it weren t so, we woul have 4k mo q an so there woul be two ossibilities: k mo q or k mo q. The first is imossible for the same reason as before, the secon is imossible because it woul imly that = = = =. q q q q The funamental goal of sieve theory is to rouce uer an lower boun for sets of the tye SA,, z = #{n A n > z }, where A is a finite subset of N, is a subset of the set of rimes P an z > 0. Examles.. Let A = {n N n x} an = { P 3 mo 4}, then SA,, x =#{n x n, P 3 mo 4} #{n x n = a + b for some corime a, b N}, so through this function we can etect sums of two squares.. Let A = {n N n x} an x < z x. Then SA, P, z = #{n x n > z} where πx = #{ P x}. = πx πz, 3

3. Let A = {nn n n N, n N }. Then SA, P, x = #{, N P N < < N N} an this is relate to Golbach conjecture. 4

Chater Sieve of Eratosthenes The Möbius function is the function µ : N {0, ±} efine by if n =, µn = 0 if P such that n, r if n = r with,..., r istinct rimes. Lemma. For all n N we have µ = n { if n =, 0 otherwise. Proof. Suose n = e er r with,..., r rimes an e,..., e r N. Then µ = r r µ = + + + r = r = 0. r n r Lemma Abel s artial summation formula. Let λ, λ,... be an increasing sequence of real numbers that goes to an c, c,... a sequence of comlex numbers. Let Cx = λ c n x n an φ : [λ, [ R be of class C. Then λ n X c n φλ n = X λ Cxφ x x + CXφX,. for all X λ. Moreover, if CXΦX 0 as X, then c n φλ n = Cxφ x x,. λ n= rovie that either sie is convergent. 5

Proof. One has CXφX c n φλ n = c n φx φλ n = λ n X = λ n X X c n φ x x = λ λ λ n x X λ λ n X n X This roves.. To rove. it s enough to let X go to infinity. Cxφ x x. c n φ x x Let Π = Π, z :=,, z A := {n N n A}, for all N. Alying lemma, we can write SA,, z = µ = µ#a. n A, n,π= = n A n,π Π,z.3 Now, suose that there exist X, R an a comletely multilicative function ω, with ω 0, such that Then we can rove the following ω = 0 P \, #A = ω X + R N..4 Theorem A Sieve of Eratosthenes. Let X, R, ω as above an assume furthermore that. R = Oω. k 0 such that Π,z ω log k log z + O 3. y > 0 such that #A = 0 for > y. 6

Then we have where SA,, z = XW z + O W z = x + y log z k+ ex log y, log z log z, z ω. Proof. Assume all the hyothesis in the theorem. For all δ > 0, we have F t, z := ω δ t ω, t, Π Π using Rankin s trick. Since + x e x for all x R, using multilicativity of ω we euce that F t, z t δ + ω t ω δ ex = ex δ log t + δ δ Π Π Π ω. δ Now, writing δ = η an using the inequality e x + xe x for x > 0, we see that η = exη log + η log η + ηz η log, since every rime Π, z is less then z. Therefore F t, z t ex η log t ex ω η Π t ex η log t + ω + ηz η Π Π ω log. Now, alying lemma to c = ω log an φx =, we have log x Π,z ω log z log = by hyothesis. Hence Π,x k log log z + O, ω log x log x x + Π,z F t, z t ex η log t + k log log z + kz η log z. 7 ω log log z

Choosing η = log z, we obtain F t, z t ex log t log z..5 log z Moreover, by artial summation lemma with c = ω, φx =, we can conclue x that Π, >y ω F t, z F y, z = t = y t log y log z ex k + log z log y log z k+ ex. log z F y, z y y + t ex y log t log z F t, z t t t.6 Finally, by hyothesis 3 an.3-.4 we have SA,, z = Π, y µ#a = = XW z + O X Π, y Π, >y = XW z + O X + y log z where we use hyothesis an.5-.6. µxω µω + log z k+ ex + O Π, y Π, y ω log y log z We aly the revious theorem to the roblem of twine rimes. Corollary A. Brun s theorem. We have, + P < Proof. It follows from a slight moifie version of Theorem A. Corollary A.. For z x 4 log log x, we have φx, z = #{n x n z} <z x µ R, 8

Proof. Exercise. Note that Lemma 3 Merten s formula. We have e γ log z, where γ is Euler s constant. z Proof. See Hary Wright, theorem 49. 9

Chater 3 Large Sieve Lemma 4. Let F : [0, ] C be a ifferentiable function with continuous erivative. Then, if we exten F by erioicity to all R with erio, we have for all z N z Proof. We have that Therefore a, a,= a F z a F = F α + F 0 F α α + α a 0 F t t. F α α, a α F α + F t t. 3. Now, let δ = z, so that the intervals I = I a := I a δ, a + δ, for z, a an a, =, are all isjoints an containe in [0, ]. Integrating 3. over I, we obtain a α δ F F α α + F t t α I = I I a F α α + I F α α + δ I a I I F t t α F t t, 0

since, if α I, then [ a, α] I. Summing over a an an multilying by z we obtain a F z F α α + F t t z a, a,= z a, a,= z F α α + 0 I 0 I F α α. Theorem B Analytic large sieve inequality. Let {a n } n N be a sequence in C, x N an Sα = n x a n e nα, where e β = exπiβ. Then z a, a,= S a z + 4πx a n. Proof. Alying lemma 4 with F α = Sα, we obtain z a, a,= a S z By Parseval s ientity we have that 0 0 Sα α + Sα α = n x a n n x 0 S αsα α. an, since S α = π n x na ne nα, by Cauchy s inequality an Parseval s equality, we get S αsα α a n 4π n a n 4π x a n, 0 n x n x n x that comletes the roof. Remark. Montgomery-Vaughan 974 an Selberg rove ineenently that 4π can be remove from the analytic large sieve inequality. Moreover, is the best ossible coefficient of x.

Next we euce a sieve metho from Theorem B. We nee the following lemma about Ramanujan sums. Lemma 5. For all, n N, let c n = a, a,= e na. Then., = c n = c nc n;. c n = D,n µ D D; 3., n = c n = µ. Proof.. By Bézout s ientity we have c n =. By lemma, we have c n = a = D,n a, a, = = s, s,= e e e na = ns na r, r, = r, s, a,= µ = D a, D µd D = µ D D,n e nr + s e nr = c nc n µd D, a D nad e since a e { na 0 if n, = if n. 3. It s a secial case of the revious oint.

Theorem C Arithmetic large sieve inequality. Let P an A = {n N n x}. For each, let Ω = {w,,..., w ω, } be a set of ω resiue classes moulo an ut ω = 0 if /. Finally, let SA,, z = {n A n w i, mo i ω Π, z} an SA,, z = #SA,, z. Then SA,, z z + 4πx, Lz where Lz = z µ ω ω. Proof. Let =, t a square-free integer iviing Π, z. By Chinese remainer theorem, for every i = i,..., i t with i j ω j there exists a unique W i, such that 0 W i, < an W i, w ij, j mo j for j t. Let s call ω = t j= ω j the total numbers of the ossible W i, as we vary i. Now let n SA,, z. Then n W i,, = for all an i. Hence, by lemma 5 item 3, we have µ = c n W i, = a, a,= Summing over i an n SA,, z, we euce that µsa,, zω = c n W i, = a, a,= an Wi, e e i awi,. an e n an therefore, by Cauchy-Schwartz inequality, µsa,, zω a, a,= e i awi, a, a,= an e n. 3

The first term on the right han sie is awi, e = e a, i a, W i,,w i, a,= a,= = µ W i,,w i, D,W i, W i, = Dµ D D = ω Wi, W ωω D ω i, a = W i,,w i, D = Dµ D D D = ω E = ω ω, c Wi, W i, µeωe E W i, W i, D W i, W i, where we use lemma 5 item. Hence we have µ SA,, z ω ω a, a,= an e n. an this equality is obviously true also if is not square-free or if it oesn t ivie Π, z. Summing over z an alying Theorem B with a n = if n SA,, z, 0 otherwise, we obtain LzSA,, z z + 4πxSA,, z. Given a rime let s efine q to be the smallest ositive integer such that q is not a square moulo or, i.e. the Legenre symbol is equal to. Note that, q being the Legenre symbol comletely multilicative, q P. Moreover, q = if ±3 mo 8, since { if ± mo 8, = 0 if ±3 mo 8. The best result known is q θ+ε for all ε > 0 unconitionally, where θ = 4 e = 0, 56..., while, assuming the Riemann hyothesis, it is q log. This roblem is linke to Artin s conjecture on rimitive roots. Using Theorem C, we can now rove the following corollary. 4

Corollary C.. Let ε > 0 an E ε N = #{rimes N q > N ε }. Then E ε N ε. Proof. Since E ε N E ε N if ε < ε, we can suose ε N. Let A = {,..., N }, = { P = n N ε } an Ω = {v mo = }. Thus ω = #Ω = n for all an h := ω ω = + 3 v if. Theorem C imlies that But an so SA,, N N + 4πN N µ h = + 4πN µ E ε N = N, q>n ε N, 3h N, q>n ε + 4πN h N, h. q>n ε E ε NSA,, N 3 + 4πN. 3. Moreover, we have SA,, N = #{n N N, n m = m N } ε #{n = m k N N ε ε / < j < N ε for j k = ε }. 3.3 Inee if n = m k N with N ε ε < j < N ε for j k = ε, then for all k j m we have = for all j k an =, since N m N / ε ε = mn ε an so m N ε. Thus = =. Using the fact that B log log B, the equation 3.3 gives SA,, N,... k n N ε ε < j <N ε N log ε ε N, m k [ ] N > N k ε N ε ε log N,... k N ε ε < j <N ε ε k,... k j <N ε = N log ε ε ε ε log N 3.4 5

since log ε > ε log N for N large enough eening on ε. To comlete the roof it is enough to ut together 3. an 3.4. We now woul like to tackle the following questions: for a, b N how likely is it that the conic C a,b := {ax + by = z, x, y, z 0, 0, 0} P Q has a rational oint? If MH is efine as MH = #{a, b N a, b H, C a,b Q }, what is the ratio MH H as H goes to infinity? We are now going to euce by Theorem C a artial answer to this roblem, but first, we nee to state some efinitions an results. by Let K = R or Q for some rime. The Hilbert symbol for K is the function efine a, b K = for all a, b K. Write { x, y, z K 3 \ {0} s.t. ax + by = z, otherwise a, b K = We ll nee the following roerties: { a, b K = Q a, b K = R. Proosition 3. Let K = Q for some rime or R an let a, a, b K. Then:. a, b = b, a. aa, b K = a, b K a, b K bimultilicativity, { a or b > 0, 3. a, b = a, b < 0, 6

β 4. If > an a = α u, b = β v for uv, then a, b = αβ u v α, where the last two factors are Legenre symbols. Proof. See 3 of Serre s A course in arithmetic. It s worthwhile to know the following theorem that roves the C a,b satisfy the Hasse rincile. Theorem Hasse-Minkowski. There exists x, y, z Q 3 \ {0} such that ax + by = z iff a, b = an a, b = for all rimes. Proof. See Serre s A course in arithmetic. Now we are reay to rove the following Corollary C.. We have MH H log H ε. Proof. Let M H, H = #{a, b N a H, µa =, b H, C a,b Q }. Clearly, we have M H, H a H µa M a H, where M a H = #{b H a, b = > }. If we efine = { P > }, A = {b H} an Ω = {v mo v, a, v = }, then M a H SA,, z z > 0. Let s now fix a square-free a H an assume H H. Since a is square-free we can write a = α u for u an α {0, }. Thus, by roosition 3 item 4, we have that if >, 7

Ω = { v = theorem C, we therefore obtain where an g = +. L a z = v α} an so ω = if α =, 0 otherwise. Alying M a H z + 4πH, L a z µ z, a Now, let ε > 0 an note that + +ε efine ν :=, we have L a z = z, D a, ε + + ε νa a, a, ε + = iff +ε ε + ε = z, a g ε. If we take z = a an we z, a νa. + ε ν + ε Moreover, we have that z = a H H, thus M H, H µa M a H H νa µa H νa + ε. + ε a H a H Hary an Ramanujan rove that a H β νa M H, H H log H β HH log H ε. a H an so we obtain Finally, note that C uv,bq imlies C u,b Q, so, writing a = uv for u square-free, we get MH v H H M v, H H log H ε. Remark C... The result rove in the revious corollary can be imrove. Hooley an Serre rove that H log H MH H log H. In fact, 8

Chater 4 Selberg sieve Eratosthenes sieve investigates the function SA,, z = n N,, n,π=, via the equality <z SA,, z = µ = µ#a. n A Π n, Π, where Π = Π, z = The basic sieve roblem is to fin some arithmetic functions µ ± : N R such that { µ if n, Π =, 4. 0 if n, Π > ; so that n, Π µ + n, Π { if n, Π =, 0 if n, Π >, µ #A = µ SA,, z µ + = µ + #A. Π n A n A Π n, Π Writing #A as #A = ωx n, Π + R with ω comletely multilicative, this gives Sa,, z X Π µ + ω + Π 4. µ + R. 4.3 Selberg sieve arose out of an effort to minimize 4.3 subject to 4.. The key iea is to relace µ + by a quaratic form, otimally chosen. We ll nee the following lemmas 9

Lemma 6. Let ζ > 0 an {λ i } i N R. Then l Π, l, l<ζ hols for all Π with < ζ if an only if for all l < ζ, l Π. µly l = ωλ µ y l = δ Π, l δ, δ<ζ ωδλ δ δ 4.4 Proof. If y l = δ Π, l δ, δ<ζ l Π, l, l<ζ ωδλ δ δ µly l = l Π, l, l<ζ = δ Π, δ, δ<ζ for all l < ζ, l Π, we have that µl ωδλ δ δ = µ ωλ. δ Π, l δ, δ<ζ ωδλ δ δ = δ Π, δ, δ<ζ ωδλ δ δ µm = µ m δ δ Π, δ, δ<ζ l, l δ ωδλ δ δ µl µm Vice versa, if 4.4 hel for another {y l } l<ζ with {y l } l<ζ {y l } l<ζ, then there woul exist a maximal l < ζ, l Π such that y l y l, an this is a contraiction since 0 = µly l y l = µ ly l y l 0. l Π, l l, l<ζ m δ 4.5 Lemma 7. Let Π an z, ζ > 0. For all a Π, let G a ζ, z = am Π,z, m<ζ gm, with gm the multilicative arithmetic function efine by gm = ωm m Then, if 0 ω <, we have G ζ, z G ζ, z ω. m ω. 0

Proof. We have that G ζ, z = m Π,z, m<ζ = l l gl gm = l gl lm Π,m, l =, m Π, m < ζ m < ζ l m Π, m,=l, m<ζ gm = l gm = l gm = G ζ, z l gl m Π, m < ζ l gl, lm Π,m l =, lm <ζ gm glm since gm 0. To conclue the roof it s enough to observe that gl = + g = + ω ω = ω. l We are now reay to rove the following Theorem D Funamental theorem for Selberg sieve. Let z > 0, y > an ω a comletely multilicative arithmetic function such that 0 ω < Ω an #A = ωx + R. Then SA,, z X G y, z + Π,z, <y 3 ν R, where ν =, G y, z = gl = ωl l l Π,z, l< y l gl, ω.

Proof. Let {λ } N R with λ = an efine where [a, b] = the inequality 4., inee µ + =,, =[, ] λ λ, ab is the least common multile of a an b. This choice of a,b µ+ satisfies µ + = n, Π [, ] n,π λ λ =, n,π λ λ = n,π an if n, Π = then n, µ + = µ + = λ =. Thus 4.3 hols, that is Π λ 0 Sa,, z X Π = XM + E, µ + ω + Π µ + R 4.3 say. Now, assume that λ = 0 for y. As a consequence we have that µ + = 0 for y. Thus M = Π µ + ω = [, ] Π λ λ ω[, ] [, ] By conition Ω, we can efine gk = ωk k µk 0, we have gk = k ωk = l k k k µ l ω l =, Π,, < y, ω 0 k ω = k µlωl ωk l l k l ωl = µk µl ωl. l k ω λ ω λ, ω,. 0 an, if ωk 0 an = l k Therefore, by Möbius inversion formula, if ω 0 an µ 0, we have k/l µl ωk/l ω = k. gk

Thus M =, Π,, < y, ω 0 = l Π, l< y, ωl 0 gl ω λ Π, l, < y ω λ, ω, = ωλ = yl gl, l Π, l< y, ωl 0, Π,, < y, ω 0 ω λ ω λ k, gk 4.6 say. Alying lemma 6 with = an ζ = y, we get = l Π, l< y µly l = l Π, l< y, ωl 0 µly l = l Π, l< y, ωl 0 So, by Cauchy s inequality, we obtain µl gl l Π, l< y = G y, zm, l Π, l< y, ωl 0 y l µl gl. gl y l gl since Π is square-free an by 4.6. Therefore we have M G y,z an the equality hols if an only if the equality hols in Cauchy s inequality, or, in equivalence, if there exists a constant c such that y l gl = cµl gl l Π, s.t. l < y, ωl 0. So, to obtain the best estimate, we have to choose y l = cµlgl an if that hols, alying again lemma 6 with = an η = y, we get = l Π, l< y µly l = l Π, l< y µl gl = cg y, z. Thus to obtain the otimal estimate we have to fin if there exist some λ such that y l = µlgl G y,z for all l < ζ, l Π. So, alying lemma 6 with ζ = y, we fin that the sought 3

λ exist an have to be λ = µ ω l Π, l, l< z = µ G g y, z ω µly l = j Π, j< z µ ωg y, z gj = µ G y, z l Π, l, l< z µl gl ω y G, z, 4.7 using the notation of lemma 7. With this choice of λ we have M = G y,z becomes SA, w, z X G y, z + Π, <y µ + R. an 4.3 Therefore, to conclue it s enough to observe that by 4.7 an lemma 7 we have λ since G ζ, z = Gζ, z an so µ + = λ λ for all square-free. =[, ] =[, ] ν ν = a = 3 ν, a a=0 Theorem D can be use to obtain an uer boun for the function φx, z = #{n x n z}. To rove it we ll nee the following lemmas. Lemma 8. Let H k z = l,k=, l<z µl ϕl, where ϕl is the Euler s φ function. Then H k z ϕk k log z. 4

Proof. Firstly we rove the statement for k =. We have that H z = l<z µl ϕl = l= h <z, < < h h i = i= h <z, < < h, α i α α h h = κn<z n, where κn = n is the square-free kernel of n. Thus, On the other han, we have an so H z = l<z = l k l k H z = κn<z µn ϕn = l k µl ϕl n,k=, n <z/l n n<z n<z, l=n,k µl ϕl H kz = k n log z. µn ϕn = l k µn ϕn = l k H k z ϕk k n <z/l, n,z/l= µl ϕl H k z l H k z = log z. µln ϕln k ϕk H kz Lemma 9. For all h N we have S = x µ h ν x + log x h, S = x µ hν + log x h, where ν =. Proof. We have that S x µ x hν = xs. 5

Moreover, S = x = µ,..., h x,..., h, = h = µ µ h h = = h, i x µ µ h h µ + log x h. x Remark 9.. Using Perron s formula, one can rove that S x log x h, anyway this imrovement oesn t have any effect on our final result about φx, z, in fact that just forces us to use an asymtotic inequality instea of a simle inequality. Now we are reay to rove the following Corollary D.. We have i φx, z x log z + z + log z 3, ii πx x. log x Proof. If we efine A = {n N n x}, we have that φx, z = SA, P, z. Moreover, we have #A = xω + R, with ω = for all an R <. Alying Theorem D with y = z we have x φx, z Gz, z + 3 ν R, where Gz, z = l Π, l<z ωl l Thus, alying lemma 8 with k =, we fin ϕx, z l l ΠP,z, <z ω = µl ϕl. l<z x log z + <z 3 ν µ 6

an so to obtain item i it s enough to aly lemma 9. To euce item ii, we have just to observe that by item i we have an choose z = x log x. πx φx, z + πz x log z + O z log z 3 + z Remark D... In the revious corollary we obtaine a better estimate than the one we coul obtain from corollary A.. This is ue to the fact that the main terms of theorems A an B are basically the same, but the error term of the Selberg sieve is much better than the one of the sieve of Eratosthenes. We can also use Theorem D to estimate πx; k, a = #{rimes x a mo k} for given corime a an k. Corollary D.. Let = {rimes x k} an let { if k, ω = 0 otherwise. Then Proof. Exercise. SA,, z k x ϕk log z + Π,z,,k=, <z 3 ν R. Dirichlet theorem of rimes in a rogression assures that πx; k, a goes to infinity as x if k, a = otherwise it s clearly 0 or. k, a = rimes a mo k have analytic ensity, that is ϕk lim s that coincie with arithmetic ensity a mo k s log s πx; k, a lim x πx 7 = ϕk = ϕk In fact, Dirichlet showe that if

but be aware that the two statements aren t equivalent. More recisely, we have πx; k, a x ϕk log x with an error term that s not uniform in k. Siegel an Walfisz rove the following result uniform in k. Theorem Siegel-Walfisz. Let a, k =. For all N > 0 there exists a c = cn > 0 such that for any k logx N we have uniformly in k an where li x := x πx; k, a = ϕk li x + O x ex c log x, u log u is the logarithmic integral function. Moreover, if the generalize Riemann hyothesis hols, we have that, for any k uniformly in k. πx; k, a = ϕk li x + O x logkx, x logx, As a consequence of theorem D, we can rove the following corollary, that gives an estimate for πx; k, a that is worse than the revious ones, but that hols for a bigger range of k. Corollary D.3 Brun-Titchmarsh. Let a, k = an k x. Then 4 x x log log x k πx; k, a ϕk log x + O ϕk k log x, k uniformly in k. Proof. Let A = {n x n a mo k} an = { P k}. Then Moreover #A = x k ω πx; k, a SA,, z + z k +. + R, where ω = { if k, 0 if k 8

an R <. Hence, by Corollary D. an Lemma 9, we have SA,, z k x ϕk k log z + Π,z,,k=, <z Taking z = x x 5 k k we comlete the roof. 3 ν µ = x ϕk log z + O z log z 3. Remark D.3.. If we coul relace by δ for some δ > 0 in Corollary D.3, we woul have as a consequence that the Lanau-Siegel zeros on t exist. We now state the following theorem. Theorem E Bombieri-Vinograov Theorem, 965. For all A > 0, there exist c = ca > 0 an B = BA > 0 such that for K = x log x B. max a Z/kZ k K li x πx; k, a ϕk C x log x A Proof. See Davenort, Multilicative number Theory it s rove using the large sieve. Combining Theorems D an E, we can stuy Titchmarsh ivisor roblem, that is to comute the orer of the function Sx = x + a, for a N fixe an where n := n. In 930 Titchmarsh was able to rove that Sx = Ox. The following corollary goes beyon that estimate roviing the asymtotic behaviour of Sx. Corollary E.. For all a N, there exists c > 0 such that x log log x Sx = cx + O. log x 9

Proof. For all n N we have that where Thus Sx = x = x, a,= +a, +a n = n, n δn = x δn, { if n is a square, 0 otherwise. δ + a = πx;, a + O x, since x δ + a n x δn + a = O x. x Now, let A > 0 an let B = BA > 0 as in Theorem E. Write πx;, a = πx;, a + x, a,= say. Theorem E imlies that S x = xlog x B, a,= = S x + S x, xlog x B, a,= = li x Moreover we have that xlog x B, a,= <t, a,= li x ϕ + for some c > 0. Hence S x = cx + O by 4.8. S x xlog x B, a,= ϕ + O xlog x B x, a,= x log x A πx;, a + O x xlog x B x, a,= πx;, a πx;, a li x ϕ. ϕ = c log t + O, 4.8 x log log x. Finally, Corollary D.3 imlies log x x ϕ log x x log log x, log x 30

Chater 5 Sieve limitations The otimization roblem for the uer boun sieve requires minimising the functional Lµ + := X Π,z µ + ω + Π µ + R, subject to µ + n, Π { if n, Π =, 0 if n, Π >. This is almost a roblem of linear rogramming. To obtain a linear rogramming roblem in stanar form, we nee to write µ + = µ + µ + with µ + i 0 an try to minimize the linear functional Lµ + := X Π,z µ + µ + ω + Π µ + + µ + R, subject to µ + µ + n, Π { if n, Π =, 0 if n, Π > an µ + 0, µ + 0. 3

Now, efine where δ i,j is Kronecker s elta an c = k X ω + R Π, k=, x = µ +k Π, k=, b = δ,n,π A n;,k = n A { k if n, 0 otherwise, with n A, Π an k =,. Then, what we are trying to minimize is c T x, uner the conitions Ax b an x 0. The ual roblem is to maximize y T b, subject to y 0 an y T A c T. Note that, if the conitions Ax b, x 0, y T A c T an y 0 hol, we have that c T x y T Ax y T b. 5. Moreover, the strong uality theorem assures that there exist x an y such that the equality hols in 5. an clearly those vectors are solutions for the linear rogramming roblem an its ual. Thus, tackling the ual roblem, we can obtain informations about the best uer boun it s ossible to obtain through sieve methos. Now, in this case the ual roblem is maximizing the function uner the conitions X ω y n 0. Jy = R n A, n y n, n A, n,π= y m X ω + R, Note that, taking y n = for any n, we obtain Jy = SA,, z. Moreover, for any subset à A such that X ω à R, 5. 3

taking y n = if n Ã, 0 otherwise, we fin Jy = SÃ,, z. Thus, for any à A satisfying 5., we have Lµ + SÃ,, z an it s easy to show that we can ro the conition à A. We now give an examle where the uer boun given by Selberg sieve is otimal. Let Ωn the number of factor of n counte with multilicity an let λn = Ωn be the Liouville function. Set A ± = {n N n x, λn = }. Now, SA +, P, z = #{n A + n z} = #{n x λn =, n z}. Clearly, if z > x 3 we have that SA +, P, z = πx πz = x x log x + O log + Oz. x We now want to fin an uer boun for SA +, P, z using Selberg sieve. We nee the following lemma. Lemma 0. Let Λx = n x λn. Then there exists c > 0 such that Λx E cx, where E c x = x ex c log x. Proof. Let s consier Mertens function Mx = n x µn. It s well known that µn n= = n s. Moreover, using Perron formula, if x isn t an integer we have that ζs Mx = k+i x s πi k i ζs s s, for any k >. Using Cauchy theorem an the zero free region for ζs, one can rove that Mx = OE c x. Moreover, note that if n/l is square-free, then we have λn = 33

µn/l = n µn/. Hence Λn = µn/ = µm E c x. n x m x n x Remark 0.. Note that Riemann hyothesis is true if an only if Mx = Ox +ε for any ε > 0 an if an only if Λx = Ox +ε for any ε > 0. Now, let s go back to our sets A ±. We have that #A ± = #{n x λn =, n} = #{m x λm = λ}. Observe that if λm = λ, then λλm = ant it s 0 otherwise. Thus #A ± = λλm m x = [ x ] λ x Λ = x x E + O c, by lemma 0 an if x. Therefore, we have to take X = x Alying Theorem D to this roblem we obtain the remainer term 3 ν R x µ 3 ν E c <y Π, <y <y µ 9ν <y µ E c x an ω = for all., where we use Cauchy s inequality an we are assuming y < x. Now, by lemma 9, we have <y µ 9ν Moreover, we have x µ E c x <y <y x log y ex log 9 y. ex c log x/y c log x/y. 34

Thus Π, <y 3 ν R x log 5 y ex c log x/y an taking y E x, we fin 3 ν R x log 5 x ex Π, <y c 4 log x Therefore, taking y E x an z y, theorem D gives us SA + X x, P, z G y, z + O log, x where G y, z G y, y = l Π, l< y by lemma 8. Thus Taking y = E x, we fin ωl l SA +, P, z SA +, P, z an so, since we alreay knew that SA +, P, z = l x log x. ω = µl l< ϕl log y, y x x log y + O log. x x log x + O x log x 3 x x log x + O log + Oz x for z > x 3, we have that with Selberg sieve we are able to rove an otimal uer boun for x z x log x. Therefore Selberg s coefficients µ+ are otimal solutions to the minimization roblem for Lµ + an, corresonly, A + is otimal for the ual roblem. We now turn to the lower boun sieve roblem. It s clear that also this roblem can be exresse as a linear rogramming roblem with the new conition { µ if n, Π =, 0 otherwise. n, Π 35

Obviously, the choice µ = 0 for all satisfies this conition, an the corresoning inequality is SA,, z X Π µ ω Π µ R = 0. Now, for A = A an z > x, we have that SA, P, z =, so the coefficients µ = 0 are essentially otimal for our linear rogramming roblem an thus so is A for the ual roblem. In articular, since A an A + have the same inuts, ω, X an OR, it is not ossible for Sieve machinery to istinguish them. Therefore, for x < z < x, log x we can t rove that SA +, P, z x log x through sieve methos an thus that πx x. log x This roblem is ue to the fact that integers n with Ωn are seen by sieves as same as integers n with Ωn. This henomenon is known as Parity roblem an it s a big limitation for sieve methos. To tackle this kin of roblems is therefore necessary to insert some other machinery that oesn t come from sieve methos. 36

Chater 6 Small gas between rimes As a consequence of the rime number theorem stating that πx = #{ x} one can rove that an thus N n= n+ n log n N x, log x lim inf n n+ n log n lim su n n+ n log n. The twin rimes conjecture, saying that there are infinitely many rimes such that + is also rime, leas to think that the much weaker statement lim inf n n+ n log n = 0 is true. Hary an Littlewoo were the first to obtain some results in this irection. In 96, they rove that lim inf n uner the generalize Riemann hyothesis. n+ n log n 3 5, years, one of the last being Mayer s roof 986 of lim inf n Other rogresses have been one over the n+ n log n < 4. Finally, in 005 Golston, Pintz an Yilirim manage to rove that this lim inf is 0 an other results towars the twin rime conjecture. The following are results they were able to obtain. 37

Theorem. We have lim inf n n+ n log n = 0. Theorem. We have lim inf n n+ n log n log log n < Moreover, enote with BVθ the following statement max max a,q= y x Λn y q x ϕq x θ n y, log A x n a mo for any A > 0. BVθ Note that Bombieri-Vinograov Theorem Theorem E states that BVθ hols for any θ < an that the Elliott-Halberstam conjecture imlies that BVθ hols for any θ <. The following are conitional results rove by Golston, Pintz an Yilirim. Theorem 3. If BVθ hols for some θ >, then lim inf n n+ n <. Theorem 4. If BVθ hols for all θ <, then lim inf n n+ n 0. Theorem 5. If BVθ hols for all θ <, then lim inf n n+ n 6. We are now going to rove the first 4 theorems. The fifth can be obtaine in a similar way with some refinements. Let H > 0 an H [0, H] Z. H is sai amissible if for any rime there exists n such that n + h h H. For examle, {0, } is amissible, but {0,, 4} isn t, since the conition fails for = 3. Clearly if a set H isn t amissible, there aren t infinitely many n such that n + h is rime for all h H. 38

If we set k = #H, to verify that H is amissible, it s enough to check the conition for all rimes k. Thus the set H = { i k < < < k } is amissible for any k an so we can fin amissible sets of any carinality. Now, let N N an let H be an amissible set. Define S 0 := logn + h log 3N N<n N h H, n+h rime Clearly, if we were able to rove that S 0 > 0 for infinitely many N, we woul have that there exist infinitely many n such that h H, n+h rime logn + h > log 3N an so that for infinitely many n there exist at least two h such that n + h is rime. Unfortunately this is not the case, since S 0 as N. Thus, we try to sum just on the n that are more likely to give more than one h such that n + h is rime. To o that, we try with Selberg s iea, multilying the summans by n λ an let this being essentially suorte on almost rimes. Therfore, we consier the sum S := logn + h log 3N N<n N h H, n+h rime. h H n+h λ The otimal values of the λ are still not known in this context an so we try to use the Selberg s sieve ones, that are essentially log + κ ζ/ λ µ, log ζ. where log + x := { log x if x, 0 if x 0, an κ is the imension of the sieve. We take κ k = #H an we write κ = k + l, with l 0. As before, if we are able to rove that S > 0 for infinitely many N we ll obtain 39

that there exist at least two h such that n + h is rime for infinitely many n. Now, assume that k, l an H [0, H] are fixe an that N 0 on [0, ξ], we can write S in the form S =, ξ λ λ = D ξ Λ D N<n N,, h H n+h N<n N, D h H n+h h H, n+h rime h H, n+h rime where Λ D := [, ]=D λ λ. Since λ, we have that ξ N. Since the λ are suorte logn + h log 3N logn + h log 3N, Λ D #{, [, ] = D} D. Moreover, λ = 0 unless is squarefree an so the same hols for Λ. Let s fix an h 0 H an let H 0 = H \ {h 0 }. We have Sh 0 := Λ D logn + h = Λ D D ξ D ξ N<n N, n+h 0 rime, D h H n+h = D ξ Λ D N+h 0 < N+h 0, D h H 0 h 0 +h log, N+h 0 < N+h 0, D h H h 0+h log since D, = being > N ξ an D ξ. Now, let a,..., a ν0 D be the classes in the set {x mo D h H 0 x h 0 + h 0 mo D} that are corime to D. If D is rime we have ν 0 D #H 0 k k = D k an ν 0 D is clearly multilicative, so ν 0 D D k hols for all square-free D. Moreover, we have Sh 0 = ν 0 D Λ D D ξ j= N+h 0 < N+h 0, a j mo D log. 40

If we efine y, q, a = y, a mo q log y ϕq, we have that Sh 0 = ν 0 D N Λ D ϕq + O N + h 0, D, a j + O N + h 0, D, a j D ξ j= = N ϕd Λ Dν 0 D + O Λ D ν 0 D max max y, D, a a,d= y 3N D ξ D ξ = N Λ D ν 0 D + O D +k max ϕd max y, D, a. 6. a,d= y 3N D ξ D ξ We now nee the following lemma. Lemma. If BVθ hols, then for all t N an A > 0 we have q t max max y, q, a x a,q= y x log x. A q x θ Remark.. The revious lemma is in some sense surrising, since q can be greater than any ositive ower of q. Now, assume that BVθ hols. Alying lemma, with x = 3N an ξ = 3N θ, from 6. we have that Sh 0 = N D ξ Λ D ν 0 D ϕd For any rime, it is easy to show that + O Nlog N A. ν 0 = #{n mo h H h n mo } = ν, say. Thus ν 0 is ineenent of h 0 an so Sh 0 = Λ D D ξ h 0 H N<n N D h H n+h = kn D ξ Λ D ν 0 D ϕd 4 h 0 H, n+h 0 rime + O Nlog N A. logn + h 0

With a similar but easier argument, one can rove that Λ D νd Λ D log 3N = Nlog 3N + O Nlog N A, ϕd D ξ N<n N, D h H n+h D ξ with the main ifference that this time we on t nee to assume that BVθ hols an we can take ξ = 3N θ for all θ <. We nee the following two lemmas. Lemma. We have Λ D ν 0 D k +! l +! = SH ϕd l +! k + l +! log ξ k+ + O log ξ k+, D ξ where SH = ν k. Remark.. Note that ν = #{n mo h H, n h mo } = k if > H. Thus >H ν k = >H k + = >H + O <. Moreover we have that SH > 0 iff ν < iff H is amissible. Lemma 3. We have Λ D νd = SH ϕd D ξ k + l! l! l! k + l! log ξ k + O log ξ k. Alying the revius lemmas we obtain k + l! l! S =NSH l +! k + l +! log ξ k kl + l + log ξ+ l + k + l + log 3N + O Nlog N k+ an, since we chose ξ = 3N θ, we have k + l! l! S =NSH l +! k + l +! l + k + l + + O k θ log 3N k+ Nlog N k+. kl + + 6. We are now reay to rove theorems 3 an 4. 4

Proofs of theorems 3 an 4. Suose that BVθ hols for some θ >. k = l, from 6. we have that S > 0 for N large enough if Then, taking an since θ > comletes the roof of theorem 3. θ > l + l + l + l + l it is certainly ossible to fin an l that satisfies this inequality. This Now, suose that BVθ hols for some 0 < θ <. The set H = {, 3, 7, 9, 3, 9, 3} is amissible with k = 7 an, taking l =, 6. imlies that S > 0 for N large enough. To rove theorems an we nee to moify slightly our arguments. Let S := N<n N H h=, n+h rime logn + h log 3N h H n+h λ. As before, if S > 0, we can fin two rime <, > N such that < H. Now, suose that BVθ hols an take again ξ = 3N θ. Proceing in a similar way as for Sh 0 with h 0 H, one can rove that the contribution to S of h 0 / H is where N D ξ Λ D ν 0D ϕd + O Nlog N A, ν 0D = #{x mo D h Hx h 0 + h 0 mo D}. We now nee the analogous of lemmas an 3 for ν 0D. Lemma 4. We have 0D k + l! l! = S H {h 0 } ϕd l! k + l! log ξ k + O log ξ k. D ξ Λ D ν 43

Therefore, S =NSH k + l! l! log ξ k kl + l + log ξ+ l +! k + l +! H + l + k + l + h 0 =, h 0 / H S H h 0 S H Since S H {h} = S H if h H, we have that log 3N + O Nlog N k+. H h 0 =, h 0 / H S H h 0 S H = H h 0 = S H h 0 S H k 6.3 an the behaviour of the firs summan on the right is given by the following lemma Lemma 5. We have that L h= We are now reay to rove theorem. S H h S H L. Proof of theorem. Let s take H = c log 3N for c > 0 arbitrary. Since we chose ξ = 3n θ, by lemma 5 an 6.3 we have that S > 0 for sufficiently large N if kl + θ l + k + l + c > 0. 6.4 Bombieri-Vinograov Theorem Theorem E assures that BVθ hols for any θ < an so we can take θ = c 4. For k = l, 6.4 becomes l l + l + 3 > c c/ an so we have that S > 0 for l large enough. Thus for all N large enough there exist n n, n] an 0 h < h c log 3N such that = n + h an = n + h are rimes. Therefore log log 3N c log N c an this roves theorem since c > 0 was arbitrary. 44

We now give a sketch of the roof of lemma 3. Lemmas an 4 can be roven in a similar way. Proof of lemma 3. We nee to estimate M = D ξ Λ D νd ϕd, where Λ D = [, ]=D λ λ, log + k+l ζ/ λ = µ, log ζ N 0 ξ N an ν [0, k], ν = k if > H. Firstly, observe that for all δ > 0 πi δ s ξ s s = k+l+ log + ζ/ k+l, k + l! where δ inicates the line δ i, δ + i. Thus k + l! ξ s +s M = log ξ k+l πi s s F s, s k+l+ s s say, where = k + l! log ξ k+l M, F s, s = =, = δ δ µ µ ν[, ] [, ] + s s ν For > H ν = k an so F s, s is aroximately s s s +s. F s, s ζs + k ζs + k ζs + s + k. It is clear that the function Gs, s, efine by has an Euler rouct Gs, s = F s, s ζs + k ζs + k ζs + s + k, Gs, s = G s, s. Note that G0, 0 = SH. Moreover, if < H, then G is regular for Rs, Rs >, while, if > H, ν = k an so G = + O 3 for Rs >. Therefore 8 G is 45

uniformly convergent for Rs, Rs > 8. Thus on this region Gs, s is holomorhic in s an s an Gs, s ν k O + O 7 8 k. Now, take δ = log N. On Rs = Rs = δ we have that ξ s +s = O. Moreover ζ + s an ζ + s are Olog + t on Rs = δ, t. Therefore, for any T we have δ δ+it δ it ξ s +s s s k+l+ F s, s s s + t T log T O T k l an clearly the same hols taking t T instea of t T. Thus M = δ+it δ+it πi δ it δ it log + t + t O + t k+l+ + t k+l+ t t ξ s +s s s k+l+ F s, s s s + O T k l. It is known that there exists c > 0 such that for σ c, t T, one has that log T ζ + s 0 an that ζ + s an ζ + s are O log + t. Therefore, ζ + s 0 insie the rectangle Γ with vertices a ± T := δ ± it, b± T := c log T ± it with the usual orientation. Clearly, if s Γ an t T an suosing T < N c σ = δ we have that s + s 0. Alying Cauchy s formula, for t T we have a + T ξ s +s πi a s T s F s, s k+l+ s = b T b + T a + T ξ s +s + + πi a T b T b + s T s F s, s k+l+ s + ξ s +s + Res s s F s, s k+l+ ; s = s + ξ s +s + Res s s F s, s k+l+ ; s = 0. Now, we have that δ+it b T + δ it a T a + T b + T ξ s +s s s F s, s k+l+ s s T c log T T = O T k l log N δ δ + c log T Olog T O δ + t k+l+ T k+l+ t t 46

an δ+it δ it b + T b T ξ s +s T s s F s, s k+l+ s s = O T T T ξ δ ξ c log T log T O log N c log T log T O δ + t k+l+ c + t log T t k+l+ t. Thus, choosing T = ex log N we fin that M = δ+it ξ s +s Res πi δ it s s F s, s k+l+ ; s = s + ξ s +s + Res s s F s, s k+l+ ; s = 0 s + O e c log N for some c > 0. Let s comute the first resiue. Let C be the circle s = s + log N eiφ. We have ξ s +s Res s s F s, s k+l+ ; s = s since ζ = πi C ξ s +s s s k+l+ F s, s s log Nk log t k s k+l+ ζ + s, k s + logn eiϕ log t for σ = δ, t T an N big enough. Therefore, δ+it δ it ξ s +s Res s s F s, s k+l+ ; s = s s = O log N k. Now, let consier the secon resiue. The function Zs, s = Gs, s is regular an nonzero near s = s = 0. Let ξ s +s fs := Res s s F s, s k+l+ ; s = 0 = Res On the rectangle Γ with vertices δ ± it, c log T that k s + s ζ + s + s s ζs + s ζ + s fs ξrs s k+l+ log ξo., ξ s +s Zs + s s s l+ s + s ; s k = 0. ± it with the usual orientation, we have 47

Thus, alying Cauchy s formula, we have that the integral δ+it fs s πi δ it is the resiue of fs in s = 0 lus an error term that is O ex c log N. To conclue it is therefore sufficient to comute Res fs ; s = 0 an that is just a long caluclation. 48