Twin Prime Numbers. Bounded gaps between primes. Author: I.F.M.M. Nelen. Supervisor: Prof. Dr. F. Beukers

Transcription

1 Twin Prime Numbers Bounded gaps between primes Author: I.F.M.M. Nelen Supervisor: Prof. Dr. F. Beuers 28 januari 215

2 Table of Contents Twin Prime Numbers 1 Abstract Introduction 2 Prime Numbers Twin Prime Numbers Heuristic Approach To Twin Prime Numbers Bounding The Gaps 4 Admissible Tuples and Sets Outline of Maynard s Proof 6 Estimating The Sums 9 Technical Lemmas Changing the Variables Rewriting the second sum Relating the Variables Choosing Suitable y Comparing the sums 29 Obtaining a lower bound of M for small Length of Bounded Gaps

3 Twin Prime Numbers Abstract In this thesis an in-depth explanation will be given of the proof Maynard gave in his article. In which the steps of the proofs will be expanded. He used a refinement on the GPY sieve to study -tuples and small gaps between primes. This will show that lim infp n+1 p n ) 6, and, by assuming the Elliott-Halberstam conjecture, that lim infp n+1 p n ) 12 and lim infp n+2 p n ) 6. Introduction Prime Numbers For thousands of years people have been fascinated by numbers. First the positive integers where used to describe quantities. Then these numbers were extended by the negative integers and fractions. But the fascination with positive integers or natural numbers stayed. The building blocs of natural numbers are prime numbers, every natural number greater than two is a prime number or a unique product of prime numbers. The grees started with examining prime numbers and found there are infinitely many. The proof of this theorem is based on the divisibility of numbers p and p + 1. Because there are infinitely many prime numbers one may wonder about the distribution of those prime numbers. For x > let πx) denote the number of prime numbers not exceeding x. Because there are infinitely many prime numbers πx) as x. Gauss 1792) and Legendre1798) proposed the distribution of prime numbers around a number x is asymptotic to x/ log x. This means heuristically there is a 1 in log x chance of a number close to x to be a prime number[9]. It is nown as the prime number theorem. 2

4 Twin Prime Numbers When one loos at the first prime numbers one can see some patterns emerge. For example 17, 19 or 41, 43 and 17, 19. These are all prime numbers which differ by exactly 2. One may wonder if there are infintely many of such pairs. This can be phrased as there are infinitely many pairs of natural numbers p, p + 2) with p and p + 2 both prime. This conjecture is nown as the twin prime conjecture. One may wonder if there are more patterns in prime numbers instead of 2 one may tae a difference of 4 or 6 between pairs of primes. These prime pairs are called respectively cousin and sexy primes. Heuristic Approach To Twin Prime Numbers To estimate the number of twin primes up to a natural number x one can use the distribution of the prime numbers and the prime number theory. This states that a number smaller than x has at least probability 1/ log x of being a prime number. This means if we pic two numbers smaller than x the probability of both of them being a prime numbers is at least 1/log x) 2 but only when the event of p is prime is independent of the event p+2 is prime. This isn t true if p 1 mod 3 and prime then p + 2 mod 3 thus p + 2 isn t prime. One needs to correct for this dependence by some correction factor. The probability for an arbitrary number to be divisible by a number q is 1/q. So the probability for two arbitrary numbers not to be divisible by a number q is 1 1/q) 2. For two numbers p and p + 2 this is different because we need p mod q and p 2 mod q which is in 2/q of the cases. The ratio between these factors is the correction factor. Thus the correction factor for any number q > 2 becomes: 1 2 q ) 1 1 q )2 For q = 2 one has 1 modulo class which is restricted for p. This correction factor becomes: ) )2 = 2 Being divisible by a small prime number is independent of the other small primes. Thus one may multiply the correction factors of the small prime numbers. In fact one may multiply over all prime numbers because when q is large the correction factor converges to 1. This suggests a definition of a 3

5 twin prime constant of: C = 2 q prime q q ) 1 1 1, q )2 This is the total correction factor over all primes q. One may guess the estimate of the number of twin prime pairs smaller than an integer x is: x C log x) 2 This would mean by a heuristic approach there would be infinitely many pairs of primes which differ 2. Because the formula for the estimate of the number of pairs under x goes to infinity when x goes to infinity. 4

6 Bounding The Gaps There are two main ways to attempt to prove the twin-prime conjecture. One can try to find the difference between the n th prime p n and the next prime p n+1. And prove for infinitely many n the difference between them is 2. Or one can use GPY method which taes sets of numbers of the same length and proves at least two of them have to be prime. Admissible Tuples and Sets We are interested in sets which when you add n all of them could, in theory, be prime. If you tae the tuple, 2, 4 then it is nown when you add a arbitrary integer n. One of the n, n + 2, n + 4 is always divisible by 3. This is a restriction on the tuple. Thus when one needs to find prime numbers this tuple is not used. Definition 1. A -tuple H = h 1,..., h is called an admissible set when there is no integer q such that q n + h i) for all n Z The tuples used here are all tuples of natural numbers but this is not needed. One may tae tuples of lineair forms such as m, m+n, m+4n for two natural numbers m, n. The next conjecture isn t proven for any > 1. It is proven for any admissible set of linear forms provided no two satisfy a linear equation over the integers. Unfortunatly most of the questions mathematicians are interested in do not satisfy these conditions. Such as the twin prime conjecture which satisfies the linear equation q p = 2. Conjecture 1. Prime -tuples conjecture). Let H = {h 1,..., h } be admissible. Then there are infinitely many integers n such that all of n + h 1,..., n + h are prime. If one wants to prove the twin prime conjecture the prime -tuples conjecture has to be proven for = 2 and h 2 = h When a bound needs to be proven a less strict conjecture is needed. 5

7 Conjecture 2. Let H = {h 1,..., h } be admissilbe. Then there are infinitely many integers n such that at least two of n + h 1,..., n + h are prime. When this conjecture was proven for some H we have lim inf n p n+1 p n max i,j h i h j. The breatrough of finding a bounded gaps between primes is made by Zhang. Who in his paper proved lim inf n p n+1 p n = B for a B < and even gave an upper bound of B 7. As well in his paper he stated This result is not optimal... to replace the [upper bound of B] by a value as small as possible is an open problem that will not be discussed in this paper [1]. The mathematical community wored together in the Polymath8 project to show this bound could indeed be improved. This project features different mathematicians all over the world who with modern technology try to mae a breatrough in a certain mathematical problem. In Polymath8 new ways to calculate the bounds were found. The results followed each other in a span of wees. 1 In less than a year the bound has been lowered to less than 6. This step was a major breatrough in the project because a different method was used to find this bound. All other improvements where made by small improvements in the proof by Zhang. This improvement uses a different method. In my thesis I will give the proof of Maynard [6] in a extended and simplified way. In the next chapter an outline of the proof will be given. 1 For the complete timeline loo at title=timeline of prime gap bounds 6

8 Outline of Maynard s Proof The proof of Maynard is based on the GPY method named after Goldston, Pintz and Yildrim[2]. This method uses the distribution of primes to study prime tuples and small gaps between primes. Given θ > we say the primes have level of distribution θ if, for any A > max a,q)=1 q x θ πx) πx; q, a) ϕx) A x log x) A 1) where πx; q, a) is the number of primes up to x of the form a mod q. We have the following results. Theorem 1. Bombieri-Vinogradov Theorem) The primes have a level of distribution θ for any θ < 1/2. Conjecture 3. Elliott-Halberstam Conjecture) The primes have a level of distribution θ for any θ < 1. Now the basic idea behind the approach of Maynard is when H = {h 1,..., h } is a fixed admissible set one considers the sum SN, ρ) = N n 2N χ P n + h i ) ρ)w n 2) where χ P is the characteristic function of the primes. Thus χ P p) = 1 when p is prime and otherwise. ρ > and w n are non negative weights. If one can show SN, ρ) >. Then at least one term in 2) has a positive contribution. This means there exists some integer n [N, 2N] such that at least ρ + 1 of the n + h i are prime. Thus if SN, ρ) > for all large N, then there are infinitely many integers n for which at least ρ + 1 of the n + h i are prime. Thus there are infinitely many bounded intervals containing ρ + 1 primes. The hard part in this is choosing the weights such that this happens. The 7

9 difference between these sieves and the GPY sieve is in the λ s. standard Sielberg sieve the following weights are chosen. In the w n = d n+h i) d<r λ d ) 2, λ d = µd)log R/D) 3) But Goldston, Pintz and Yildrim used different weights of the form λ d = µd)f log R/D) 4) for a suitable smooth function F and µ the Möbius function. They chose F x) = x +l for suitable l N, which has been shown to be essentially optimal when is large. This will prove the existence of bounded gaps between primes when the level of distribution of primes θ > 1/2 but not when θ < 1/2. Thus the existence of bounded gaps between primes then relies on the the Elliott-Halberstam Conjecture conjecture 3) which has not been proven. Thus a weaer condition is needed to prove the existence of bounded gaps. Zhang used in his paper a modified form of the distribution of primes. Maynard uses in his paper a different weight. Maynard s weights aren t only based on a d i+1 n + h i) but uses different d i n + h i i. Which gives sieve weights of the form w n = λ d1,...,d ) 2 5) d i n+h i i In Maynard s proof he choses his weights to be zero unless n lies in a fixed residue class v mod W ), where W = p D p the product of all primes numbers smaller than D. Thus will remove some minor complications when dealing with small prime factors. He choses D to be D = log log log N to be sure that W log log N) 2 by the prime number theorem. To estimate the sum 2) it is split in two parts. S 2 = S 1 = N n 2N n v mod W ) N n 2N n v mod W ) d i n+h i i λ d1,...,d ) χ P n + h i ) d i n+h i i 2 λ d1,...,d So S = S 2 ρs 1 with a few minor resrictions added. In the same way as Zhang s proof the first part is rewriting both sums untill they are more 8 2 6) 7)

10 or less smooth functions. In the sum itself you have some functions which can t be specifically calculated. By rewriting both sums, and changing the variables the sum will be easier to handle. This will be the first part of his proof. The second part will be proving the desired result will be achieved when the sum is calculated. The next part is finding the length of the tuple in which two prime numbers will be found. To finish the proof a tuple is given and the bound will be set at 6. 9

11 Estimating The Sums We will use a proposition to evaluate the sums. Proposition 1. Let the primes have exponent of distribution θ >, and let R = N θ/2 δ for some small fixed δ >. Let λ d1,...,d be defined in terms of a fixed piecewise differentiable function F by λ d1,...,d = µd i )d i ) r 1,...,r d i r i i r i,w )=1 i µ r i) 2 ϕr i) F log r 1 log R,..., log r log R ) whenever d i, W ) = 1 and let λ d1,...,d = otherwise. Let F be supported on R = {x 1,..., x ) [, 1] : x i 1}. This means if λ d1,...,d then d 1...d R and d i, d j ) = 1 when i j. Then we have S 1 = 1 + o1))ϕw ) NLogR) W +1 I F ), S 2 = 1 + o1))ϕw ) NLogR) +1 W +1 log N provided I F ) and J m) J m) F ) = 1 I F ) = for each m where... 1 j=1 F t 1,..., t )dt 1...dt J m) F ), F t 1,..., t )dt m ) 2 dt 1...dt m 1 dt m 2...dt 1

12 Technical Lemmas The following Lemmas are used troughout the proofs Lemma 1. τ d) < Rlog R) 1 8) d<r Proof. τ d) = d<r d 1,...,d d 1...d <R 1 = R d 1,...,d i d i<r 1 R < R d 1,...,d d i <R 1 d i = R d<r ) 1 < Rlog R) 1 9) d Lemma 2. Generalized Möbius inversion) If A d1,...,d with support in d i N, i d i < R. Define B r1,...,r = r i d i A d1,...,d 1) then A d1,...,d = r i d i µr i )B r1,...,r 11) Lemma 3. R a=1 µa) 2 ϕa) log R 12) log Proof. p R p prime 1+ 1 p 1 ) = log1+ 1 p 1 ) p R p R R a=1 1 < c+log log R 13) p 1 µa) 2 ϕa) < ) log R 14) p 1 p R p prime 11

13 Lemma 4. u,w )=1 u>d µu) 2 ϕu) 2 1 D 15) Proof. u,w )=1 u>d µu) 2 ϕu) 2 = p>d ) p 1) 2 16) If one loos at the logarithm of the right hand side we get ) p 1) 2 < p 1) 2 < n 1) 2 < 2 17) D p>d log p>d 2 n>d 2 Changing the Variables To prove this proposition a change of variables will be introduced. This approach is based on the elementary combinatorial ideas of Selberg. [7] We assume that the primes have a fixed level of distribution θ, and R = N θ/2 ɛ. The weight λ d1,...,d is restricted to tuples with d = d i and d < R, d, W ) = 1 and µd) 2 = 1. This implies all the d i are pairwise coprime and square-free. Proposition 2. Let y r1,...,r = µr i )ϕr i ) d 1,..,d r i d i i λ d1,...,d d i 18) Let y max = sup y r1,...,r then r 1,...,r S 1 = N W r 1,...r y2 r 1,...,r ϕr i) + maxϕw ) NLogR) Oy2 W ) 19) D 12

14 Proof. Expand out the square in S 1 and then swap the order of summation S 1 = 2 λ d1,...,d = λ λ d1,...,d e1,...,e 1 d i n+h i i d 1,...,d e 1,...,e N n<2n n v modw N n<2n n v modw [d i,e i ] n+h i i 2) The inner sum runs over all n such that N n < 2N with n v mod W and n+h i mod [d i < e i ] for all i. The W, [d i, e i ] are pairwise coprime. If not it would mean [d i, e i ], [d j, e j ]) = d > 1 thus d d i, d j but the product over the d i is square free. This is a contradiction so they are pairwise coprime. Because they are coprime there exists, by the chinese remainder theorem, a residue class a mod q with q = W [d i, e i ] such that the summation runs over n a mod q and N n < 2N There is N/q times a n modulo this residue class in an interval with lenght N. This calculation has an error which approaches 1 when N. The inner sum becomes N/q + O1) when the integers W, [d i, e i ] are pairwise coprime, when they are not the sum is zero. This restriction on the integers W, [d i, e i ] will be denoted by. This gives S 1 = N W d 1,...,d e 1,...,e λ d1,...,d λ e1,...,e [d i, e i ] + O λ λ d1,...,d e1,...,e ) 21) d 1,...,d e 1,...,e To ease notation we put y max = sup r1,...,r y r1,...,r. The assumptions on λ state it is only non zero when d i < R, thus the errorterm contributes λ λ d1,...,d e1,...,e d 1,...,d e 1,...,e λ max) d 1,...,d e 1,...,e 2 λ 2 max d<r τ d)) 2 22) Where τ d) means in how many ways d can be written as a product of integers.. This is bigger than the number of ways d can be written with divisors which are square free. This can be estimated in order of magnitude by: λ 2 max d<r τ d)) 2 λ 2 maxr 2 LogR) 2 23) 13

15 In this estimation Lemma 1 is used to estimate the sum. In the main sum we remove the dependencies between the d i and the e j variables. We use the identity: 1 [d i, e i ] = 1 e i, d i ) = 1 d i e i d i e i u i d i,e i ϕu i ) This means the least common multiple can be rewritten to the product divided by the greatest common divisor. This is a combination of all u i dividing the greatest common divisor and the equality n d ϕn) = d. Thus the sum is the greatest common divisor. The main term becomes S 1 = N W u 1,...,u ϕu i ) ) d 1,...,d e 1,...,e u i d i,e i i λ d1,...,d λ e1,...,e d i e i Now recall the requirement on the summation which is that W, [d 1, e 1 ],..., [d, e ] are alle pairwise coprime. The weight λ is only supported on integers d 1,..., d which are coprime with W. Thus the requirement of W being coprime with the least common multiples can be dropped. Similarly the requirements of d i, d j ) = 1 for all i j and e i, e j ) = 1 for all i j may be dropped. The only restriction left from the pairwise coprimality of W, [d 1, e 1 ],..., [d, e ] is d i, e j ) = 1 for all i j. To be certain the requirement will be held without a requirement in the sum, multiply the main term with s i,j d i,e j µs i,j ). This wors by the equality d n µd) is 1 when n = 1 and otherwise. So the sum for a pair i, j) will not be empty only when d i, e j ) = 1. This applies to all i, j with i j. The main term becomes S 1 = N W u 1,...,u ) ϕu i ) s 1,2,...,s, 1 1 i,j i j µs i,j ) d 1,...,d e 1,...,e u i d i,e i i s i,j d i,e j i j 24) λ d1,...,d λ e1,...,e d i e i 25) Assume u i, s i,j ) 1 this means s i,j > 1 and s i,j d i, e i, e j. Thus e i, e j ) s i,j > 1 and λ e1,...,e = unless e i, e j ) = 1. So there may be a restriction to s i,j by only summing over the ones which are coprime to u i and u j. Because 14

16 when this is not true λ e1,...,e does not support the give tuple. In a similar way the restriction may be expanded by demanding s i,j to be coprime with s i,a and s b,j for all a j and b i. The summation over these will be further noted by. To mae our sum S 1 more straightforward we will introduce a change of variables. Let y r1,...,r = µr i )ϕr i )) d 1,...,d r i d i i λ d1,...,d d i 26) This change of variables has to be invertible. When it is not invertible the sum will not run over all possible tuples, for d 1,..., d square free. We will prove our change of variables is invertible. This follows by using Lemma 2 r 1,...,r d i r i i y r1,...,r ϕr i) = λ d 1,...,d µd i)d i 27) This means any choice of y r1,...,r supported on r 1,..., r when the product r i = r is square-free, r < R and r, W ) = 1 will give a suitable choise of λ d1],...,d. By changing our variable we need to re-estimate the λ max. Let y max = sup r1,...,r y r1,...,r. Since d/ϕd) = e d 1/ϕd) for d square-free. Tae r = r i/d i then by the change of variables and the definition of λ max λ max sup d 1,...,d d i square-free y max d i ) r 1,...,r d i r i i r i<r r i square-free µr 1 ) 2 ϕr i ) ) 28) by changing the variable from d i to r the limits of the sum change r i < R r < R/ d i and because d i r i the other limit changes r i square-free r, d i) = 1. When r is used instead of a sum over r 1,.., r there need to be corrected by the number of ways the r 1,..., r i can form a certain product. This correction is smaller than the number of ways the product can be written by all it s divisors. So we can estimate this term true multiplying by τ r ). 15

17 λ max y max sup d 1,...,d d 1 square-free y max sup d 1,...,d d d i d i ϕd i ) ) µd) ϕd) ) r <R/ d i r, d i)=1 r <R/ d i r, d i)=1 µr ) 2 τ r ) ϕr ) µr ) 2 τ r ) ϕr ) 29) 3) The product we can rewrite by using the equality stated and rewriting 1 = µd) 2 to remove the requirement of needing d i to be square-free. If we let u = dr and by using the fact that τ dr ) τ r ) both sums can be combined an the estimate becomes y max u<r µu) 2 τ u) ϕu) y max logr) 31) This is by combining Lemma 1) and Lemma 3. By substituting the change of variables and with the estimation of the error term, we obtain S 1 = N W u 1,...,u ϕu i )) s 1,2,...,s, 1 1 i,j i j µs i,j )) µa i µb i )) ϕa i )ϕb i ) )y a 1,...,a y b1,...,b +OymaxR 2 2 LogR) 4 ) 32) where a i = u i i j s i,j and b j = u j i j s i,j The functions ϕ and µ are multiplicative and all factors are pairwise coprime thus µa i ) = µu i ) i j µs i,j) the same for µb i ), ϕa i ) and ϕb i ) but only when all s i,j are coprime with all other terms in the a i and b i. All terms with these not square free do not contribute to the sum and µs i,j ) 3 = µs i,j ) hence the sum can be rewritten to as S 1 = N W u 1,...,u µu i ) 2 ϕu i ) ) s 1,2,...,s, 1 1 i,j i j µs i,j ) 2 ϕs i,j ) 2 y a 1,...,a y b1,...,b +Oy 2 maxr 2 LogR) 4 ) 33) 16

18 Again there is no contribution from s i,j with s i,j, W ) 1 because of the restricted support of y. Thus only s i,j = 1 and s i,j > D need to be considered. When s i,j > D the contribution to the sum is y2 maxn W u<r u,w )=1 µu) 2 ϕu) µs i,j)2 ϕs i,j ) 2 s i,j >D µs) 2 ϕs) 2 s ) The first sum can be rewritten in much the same way as in Lemma 3 to ϕw ) LogR) /W. Because one needs to correct for the terms u which are not coprime with W this correction term is ϕw )/W. Thus the ratio between W and the numbers which are coprime to it. The last sum is convergent thus can be rewritten as a contstant. For the middle sum we use Lemma 4. If we combine the Lemma with the other estimations then the equation of 34) is estimated by y2 maxϕw ) NLogR) W +1 D 35) When s i,j = 1 and all other terms are in our error term we get S 1 = N W u 1,...,u y2 u 1,...,u ϕu i) + maxϕw ) NLogR) Oy2 W +1 + y D maxr 2 2 LogR) 4 ) 36) To prove the lemma it now suffices to show the first error term dominates the second. Recall R 2 = N θ 2δ N 1 2δ, W N δ and R LogR) 3. Thus N/W D R 2 LogR) 3 hence N/W D N 1 2δ R 2 and the first term dominates. This ends the proof and S 1 = N W u 1,...u y2 u 1,...,u ϕu i) + maxϕw ) NLogR) Oy2 W ) 37) D 17

19 Rewriting the second sum In a similar way we estimate S 2 = m=1 Sm) 2 where S m) 2 = χ P n + h m ) λ d1,...,d ) 2 38) N n<2n n v modw ) d1,...,d d i n+h i i In the next lemma we estimate S m) 2 in a similar way to S 1 Proposition 3. Let y m) r 1,...,r = µr i )gr i ) d 1,..,d r i d i i d m=1 λ d1,...,d ϕd i) 39) where g is the totally multiplicative function defined on primes by gp) = p 2. Let y max m) = sup y r m) 1,...,r then r 1,...,r S m) 2 = N ϕw )LogN r 1,...r y m) r 1,...,r ) 2 max) 2 ϕw )NLogR) 2 gr i) +Oym) W 1 D )+O y2 maxn LogN) A ) 4) Proof.By interchanging the order of summation and expanding out the squares we get a resemblence with S 1 in equation 2). S m) 2 = λ λ d1,...d e1,...e d 1,...,d e 1,...,e N n<2n n v mod W [e i,d i ] n+h i χ p n + h m ) 41) The inner sum is over + 1 residue classes which can be written as a sum over a single residue. This residue class is q = W [d i, e i ] with W, [d 1, e 1 ],..., [d, e ] are pairwise coprime. Because of the χ-prime function the inner sum for S2 m is only non zero when d m = e m = 1. By the same reason the inner sum will contribute X N /ϕq) and an error term. Where X N = N n<2n χ P n) 42) 18

20 It means we have at least the sum of prime numbers between N and 2N divide by the quantity of numbers which are coprime by q. Let EN, q) = sup a,q)=1 N n<2n n amodq) χ P n) 1 ϕq) N n<2n Thus the size of the contribution to the error term becomes χ P n) 43) OEN, q)) 44) It s easy to see by summing the two we get the desired result. Thus by the same notation of the restriction the sum becomes S m) 2 = X N ϕw ) d 1,...,d e 1,...,e e m=d m=1 λ d1,...,d λ e1,...,e ϕ[d i, e i ]) + O λ λ d1,...,d e1,...,e d 1,...,d e 1,...,e EN, q)) 45) In a similar way to the first sum we estimate the error term. The support of λ d1,...,d restricts q to be only square free and smaller than R 2 W. Given a square free integer r there are only τ 3 r) possibilities of d 1,..., d, e 1,..., e for which W [d i, e i ] = r. Recall λ max y max LogR). Hence the error term becomes Rewrite the sum y 2 maxlogr) 2 = y 2 maxlogr) 2 r<r 2 W r<r 2 W µr) 2 EN, r)τ 3 r) 46) τ 3 r)µr)en, r) 1/2) µr)en, r) 1/2) 47) Then you will get by using Cauchy Schwartz on both parts between bracets and EN, r) N/ϕr) ymaxlogr) 2 2 1/2 µr) 2 τ3 2 r) N 1/2 µr) 2 EN, r) ϕr) r<r 2 W r<r 2 W 48) Remember the primes have level of distribution θ thus the last sum N/LogN A ) 1/2 for any A >. In the first sum they are smaller than N 1/2 LogN) B for a B. Hence this can be rewritten to 19

21 y2 maxn LogN) A 49) As in the treatment of the first sum the conditions of all d i, e j ) = 1 can be rewritten by multiplying the expression by s i,j d i,e j µs i,j ). Again the same requirements may restrict s i,j and will be denoted by. In the same way the ϕ[d i, e i ]) term can be split, because they are square-free, by the equation 1 ϕ[d i, e i ]) = 1 ϕd i )ϕe i ) u i d i,e i gu i ) Where g is the totally multiplicative function defined on the primes by gp) = p 2. This transforms the main term to X N ϕw ) u 1,...,u gu i )) s 1,2,...,s, 1 1 i.j µs i,j )) d 1,...,d e 1,...,e u i d i,e i i s i,j d i,e j i j d m=e m=1 λ d1,...,d λ e1,...,e ϕd i) ϕe i) 5) A similar substition may be used in this situation we have only one extra demand r m = 1. When r m then y r m) 1,...,r is. Thus let y m) r 1,...,r = µr i )gr i ) d 1,..,d r i d i i d m=1 λ d1,...,d ϕd i) 51) Hence the main term becomes in a similar way as equations??) to 32) X N ϕw ) u 1,...,u µu i) 2 gu i ) ) s 1,2,...,s, 1 1 i.j i j µs i,j) gs i,j ) )ym) a 1,...,a y m) b 1,...,b 52) Again there are two different cases for s i,j is 1 or > D. When s i,j isn t 1 the contribution is by the same calculation as S 1 ym) max) 2 N ϕw )LogN u<r u,w )=1 µu) 2 gu) ) 1 s µs) 2 gs) 2 ) 1) 1 µs i,j)2 gs i,j ) 2 s i,j >D 2

22 ym) max) 2 ϕw ) 2 NlogR) 1 W 1 D LogN 53) And by the prime number theorem X N = N/LogN + ON/LogN) 2 ) which contributes to the error term by ym) max) 2 N ϕw )LogN) 2 u<r u,w )=1 µu) 2 gu) ) 1 ym) max) 2 ϕw ) 2 NlogR) 3 W 1 54) Which will be absorbed by the first error term. With this we can rewrite the sum S m) 2 = N ϕw )LogN r 1,...r y Which ends the proof. m) r 1,...,r ) 2 Relating the Variables max) 2 ϕw )NLogR) 2 gr i) +Oym) W 1 D )+O ym) maxn LogN) A ) 55) Next the new variables of S 1 will be related to S m 2 by the following lemma. Lemma 5. if r m = 1 then y m) r 1,...,r = a m y r1,...,r m 1,a m,r m+1,...,r ϕa m ) + O y maxϕw )LogR W D ) Proof. Substitue the expression from??) to 27) in definition 51) we get y m) r 1,...,r = µr i )gr i )) d 1,...,d r i d i i d m=1 µd i )d i ϕd i ) ) a 1,...,a d i a i i y a1,...,a ϕa i) 56) By switching the summation of d and a variables the last sum is over d. This sum we can calculate explicitly. Because all the d i, a i and r i are square-free 21

23 and the functions are multiplicative d 1,...,d d i a i,r i d i i d m=1 Hence we get µd i )d i ϕd i ) = i m d i d i a i /r i µd i )d i ϕd i ) µr i )r i ϕr i ) = i m µa i /r i ) µr i )r i ϕa i /r i ) ϕr i ) 57) y m) r 1,...,r = µr i )gr i )) a 1,...,a r i a i i y a1,...,a ϕa i) i m µa i )r i ϕa i ) 58) Looing at the support of y r m) 1,...,r we can restrict the summation over a i to a i, W ) = 1. Thus either a i = r i or a i > D r i. First loo at the contribution of a i r i for j m. Split the last sum in three parts. The part of a j > D r j. The part of a m and the rest. The first and last sums converge thus restricted to a constant. The middle sum is estimated by 14). y max µa j gr i )r i ) )2 ϕa j ) 2 ) a j >D r j a m<r a m,w )=1 µa m ) 2 ϕa m ) ) 1 i i j,m µa i ) 2 ϕa i ) 2 ) r i a i gr i )r i ϕr i ) 2 )y maxϕw )LogR y maxϕw )LogR 59) W D W D Thus estimate the main contribution when a j = r j for all j m y m) r 1,...,r = r i gr i ) ϕr i ) 2 ) a m y r1,...,r m 1,a m,r m+1,...,r ϕa m ) + O y maxϕw )LogR W D ) 6) Note that gp)p/ϕp) 2 = 1 + Op 2 ). Since the contribution is zero unless the product of the r i is coprime to W every r i > D thus the product in the expression may be replaced by 1 + OD 1 ). This will be dominated by the error term we already have. Thus this gives the result y m) r 1,...,r = a m y r1,...,r m 1,a m,r m+1,...,r ϕa m ) + O y maxϕw )LogR W D ) 61) This will end the proof and gives a way te relate both new variables. 22

24 Choosing Suitable y To complete the proof of proposition 1 we need a suitable choice of y. The choice of y is such that the ratio between the main terms of S 1 and S 2 is maximized. A second demand is that the y are smooth. Such that it has no dependence on the prime factorisation of the r i. Remember r = r i satisfies r, W ) = 1 and µr) 2 = 1. Choose y r1,...,r = F log r i log R,..., log r log R ) 62) for some piecewise differentiable function F : R R supported on R = {x 1,..., x ) [, 1] : x i 1}. When r is nog coprime to W or not square free set y r1,...,r = 1. With this choice suitable asymptotic estimates for S 1 and S 2 can be made. Lemma 6. Let κ, A 1, A 2, L >, Let γ be a multiplicative function satisfying and L w p z γp) p γp) log p p 1 A 1 κ log z/w A 2 for any 2 w z.let g be the totally multiplicative function defined on primes by gp) = γp)/p γp)). Finally, let G : [, 1] R be a piecewise differentiable function and let G max = sup t [,1] Gt) + G t) ). Then d<z µd) 2 gd)g log d log 1 z)κ ) = G Gx)x κ 1 dx+o A1,A log z Γκ) 2,κG LG max log z) κ 1 ) where G = p q γp) p ) 1 q 1 p )κ The constant implied by O is independent of L and G 23

25 Proof. This proof is divided in two lemmas in [3]. Lemmas two and three prove this result with a slight notation difference. Which again is divided in multiple lemmas and based on explicit estimates of selbergs upper bounds. Lemmas 5.3 and 5.4 [5] The next two lemmas will finish the estimation of S 1 and S2 m this will conclude the proof of Proposition 1. The next lemma will estimate S 1 Lemma 7. Let y r1,...,r be given in terms of a piecewise differentiable function F supported on R = {x 1,.., x ) [, 1] : sum x i 1} by 62). Let F max = sup F t 1,..., t ) + δf t 1,..., t ) 63) t 1,...,t ) [,1] δt i Then S 1 = ϕw ) Nlog R) W +1 I F ) + O F maxϕw 2 ) Nlog N) W +1 ) 64) D where I F ) = F t 1,..., t ) 2 dt 1...dt 65) Proof. We substitute the choice of y 62) into the expression of S 1 which was rewritten in lemma 2). There are no restrictions given for the -tuples which are supported on y so we will add the restrictions to the sum. This gives S 1 = N W u 1,...u u i,u j )=1 i j u i,w )=1 i µu i ) 2 ϕu i ) )F log u i log R,..., log u log R )2 +O y2 maxnlogr) ϕw ) W D ) 66) Note that the requirement of u i, u j ) = 1 can be dropped at the cost of an reasonably sized error. This is because every prime divider they have in common has to be bigger than D for they both are coprime with W. This error is of size F 2 maxn W p>d u 1,...u p u i,u j u i,w )=1 i µr i ) 2 ϕu i ) 24

26 F 2 maxn W p>d 1 p 1) 2 u<r u,w )=1 µu) 2 ϕu) ) F 2 maxnlog R) ϕw ) W +1 D 67) The second sum is rewritten in the same way as in the estimations of 2 to ϕ log R) /W while in the first sum hier moet nog een afschatting and rewrite it to 1/D. Which will give the desired result. Thus the sum that needs to be evaluated is u i,...,u u i,w )=1 i µu 1 ) 2 ϕu i ) )F log u i log R,..., log u log R )2 68) By -fold application of Lemma.6 the sum can be estimated. Tae for each application κ = 1 and { 1, p W γp) = 69), otherwise L 1 + p W log p p and A 1 and A 2 constants of suitable size. log D 7) Lemma 8. If γp), κ stated as above then the lower limit L stated in Lemma 6 will be log D Proof. One chooses κ = 1 and no terms are counted when p < D and use the maximum of log z/w. The prime counting function πx) will be used. But not x/ log x a stricter approximation is needed. A stricter approxomation is πx) = x 2 dt/ log t + Ox/log x)a [1] πx) = [ ] t x log t 2 x log x + x 2 t d dt 1 log t Loo at D <p<r log p/p over primes. D <p<r log p/p = R D log x x dπx) = ) dt + Ox/log x) A ) x log x) 2 + Ox/log x)3 ) 71) [ πx) log x ] R R πx) d x D D dx ) log x x 25

27 By the prime number theorem [4, p. 352] the first part is O1) for the second part the above estimation of πx) is used. This gives. = O1) + R D ) x log x + x log x log x) 2 x 2 1 ) x 2 dx R 1 = O1) + D x 1 xlog x) 2 dx = C + log R/D This constant C > thus our estimation becomes log D C + log R/D log R C 72) Thus L = log D It will be shown how this wors when = 2. When κ = 1 and because W is square-free G = 1 1 ) = ϕw ) 73) p W p W = = Thus we get u 1 <R µu 1 ) 2 ϕu 1 ) ϕw )log R) W u 1 <R µu 1 ) 2 ϕu 1 ) ϕw )log R) W 1 = ϕw )2 log R) 2 W 2 I 2 F )+ u 2 <R 1 ϕw )log R) W 1 µu 2 ) 2 ϕu 2 ) F log u 1 log R, log u 2 log R )2 F x, log u 2 log R )2 dx + O ϕw )F maxlog 2 D ) ) W 1 F x, y) 2 dx + O ϕw )F 2 maxlog D ) W O ϕw )2 Fmaxlog 2 D ) log R W 2 )dy+o ϕw )F maxlog 2 D ) ) W = ϕw )2 log R) 2 W 2 I 2 F ) + O ϕw )2 Fmaxlog 2 D ) log R W 2 ) 74) If this is applied times to the sum in equation 68) we get u 1,...,u u i,w )=1 i µu 1 ) 2 ϕu i ) )F log r i log R,..., log r log R )2 26 )dy +O ϕw )F maxlog 2 D ) ) W

28 = ϕw ) log R) W I F ) + O ϕw ) Fmaxlog D )log R) 1 W ) 75) And by combining 75), 67) and 66) this ends the proof of lemma 7) To end the proof of the proposition the following lemma needs to be proved. Lemma 9. Let y r1,...,r, F and F max be as defined in Lemma 7. Then we have S m) where 2 = ϕw ) Nlog R) +1 W +1 J m) log N F ) + O F maxϕw 2 ) Nlog N) W +1 ) 76) D J m) F ) = F t 1,..., t )dt m ) 2 dt 1...dt m 1 dt m+1...dt 77) Proof. The proof is similar to the proof of Lemma 7. First estimate y r m) 1,...,r. Recall y r m) 1,...,r = unless r m = 1 and r = r i satisfies r, W ) = 1 and µr) 2 = 1. Then y r m) 1,...,r = is given by Lemma 5. First the case when y r m) 1,...,r is checed. Substitute 62) in the expression of Lemma 5 and this gives y m) r 1,...,r = u,w )r i)=1 µu) 2 ϕu) F log r i log R,..., log r m 1 log R, log u log R, log r m+1 log R,..., log r log R ) +O F maxϕw ) log R W D ) 78) Which maes y max m) ϕw )F max log R)/W. Now estimate the sum over u. Lemma 6 is used with κ = 1, γp) = { 1, p W r i, otherwise 79) L 1 + p W r i log p p p<log R log p p + p W r i p>log R log log R log R log log N 8) and with A 1, A 2 suitable fixed constants. In the same way as in lemma 7 it is easy to see G = ϕw ) ϕr i)/w r i which gives 27

29 y m) where r 1,...,r = log R) ϕw ) W F m) r 1,...,r = 1 F ϕr i ) )F r m) r 1,...,r + O F maxϕw ) log R ) 81) i W D log r1 log R,..., log r m 1 log R, t m, log r m+1 log R,..., log r ) dt m 82) log R If this is substituted in the expression 4). The term of y max is to the power of two. Thus the error term becomes F 2 maxϕw ) Nlog R) )/W +1 D ). The sum obtained is S m) 2 = ϕw )Nlog R)2 F 2 max W 2 log N r 1,...,r r i,w )=1 i r i,r j )=1 i j r m=1 F 2 O max ϕw ) Nlog R) ) W +1 D ) µr i ) 2 ϕr i ) F r m) gr i )r 1,...,r ) 2 + i 83) In the same way as in the first sum the condition r i, r j ) = 1 can be removed. This will introduce an error of size 1 ϕw )Nlog R)2 Fmax 2 ϕp)2 µr) 2 ϕr) W 2 log N gp) 2 p 2 gr)r p>d r<r r,w )=1 The first sum will converge to a constant of 1/D. The second sum can be estimated by ϕw ) 1 log R) 1 /W 1 ϕw ) 1 log N) 1 /W 1 which give an error term of r 1,...,rm 1,r m+1,...,r r i,w )=1 i F 2 maxϕw ) Nlog N) W +1 D 84) Now the last part of the sum needs to be evaluated. In the same way as the first sum we will use Lemma 6. Thus the sum we evaluate is µr i ) 2 ϕr i ) F r m) gr i )r 1,...,r ) 2 85) i 1 i i j 28

30 By applying Lemma 6 1 times with κ = 1 and γp) = { p 2 p 1, p W, otherwise 86) L 1 + p W log p p log D 87) and A 1, A 2 suitable fixed constants. It is easy to see that error term 84) dominates the dominant error term we get when using this lemma 1 times. Thus this gives S m) where 2 = ϕw ) Nlog R) +1 W +1 J m) log N F ) + O F maxϕw 2 ) Nlog N) W +1 ) 88) D J m) F ) = as required F t 1,..., t )dt m ) 2 dt 1...dt m 1 dt m+1...dt 89) 29

31 Comparing the sums In the previous chapter we estimated the sums. Now it needs to be proven that by taing these estimations and a suitable -tuple that there are infinitely many integers n such that several, but at least two, of the n + h i are prime. In the next proposition we will prove this and we will introduce a way to calculate the bound. Proposition 4. Let the primes have level of distribution θ >. Let δ > and H = h 1,..., h an admissible set. Let I F ) and J m) F ) be given as in Proposition 1, let S denote the set of piecewise differentiable functions F : [, 1] R supported on R = {x 1,..., x ) [, 1] : x i 1} with I F ) and J m) for each m. Let m=1 M = sup J m) I F ), r = θm 2 9) F S Then there are infinitely many integers n such that at least r of the n + h i 1 i ) are prime. In particular, lim inf n p n+r 1 p n ) max 1 i,j j h i h j ). This proposition will tell when we have an upper bound θm > 2 for a then we will have infinitely many n where n + h i with h i in a admissible set of order. Where at least two will be prime. Proof.Let S = S 2 ρs 1. If S > for all large N and ρ > 1 then there are infinitely many integers n such that at least two of the n + h i are prime. Put R = N θ/2 ɛ for a small ɛ >. By the definition of M, a F S can be chosen such that m=1 J m) F ) > M ɛ)i F ) = m=1 J m) F ) ɛi F ). By using Proposition 1, we can rewrite both sums with given such that λ d1,...,d S = ϕw ) Nlog R) W +1 log R log N m=1 3 J m) F ) ρi F ) + o1) )

32 ϕw ) Nlog R) I F ) W +1 θ ) 2 ɛ)m ɛ) ρ + o1) 91) When ρ = θm /2 δ by choosing ɛ sufficiently small then S > for all large N. Thus there are infinitely many integers n for which at least ρ + 1 of the n + h i are prime. Since ρ + 1 = θm /2 when δ is sufficiently small the result is proven. To get the result of bounded gaps between primes an esimation of M and θ is needed. We need θ to be as large as possible. Remember if the primes have level of distribution θ if for any A >, we have max a,q)=1 q x θ πx) πx; q, a) ϕq) A x log x) A 92) By the Bombieri-Vinogradov theorem it is proven when θ < 1/2. If we find M > 4 for a and an admissible set H with elements there are infinitely many n N such that two of the n + h i are prime. Obtaining a lower bound of M for small K Let S denote the set of piecewise differentiable function F : [, 1] R supported on R = {x 1,..., x ) : x i 1} such that I F ) and J m) F ) for each m. If a lower bound is obtained, with these requirement, this will be M = sup F S m=1 J m) F ) I F ) 93) To obtain this lower bound we will consider approximations to the optimal function F of the form F t 1,..., t ) = { P t1,..., t ), ift 1,..., t ) R, otherwise 94) for polynomials P. Lemma 1. Let P j = sum tj i denote the jth symmetric power sum polynomial. Then we have 1 P 1 ) a P b a! j dt 1...dt = + jb + a)! G b,j) 95) R 31

33 where G b,j x) = b! b r=1 ) r b 1,...,b r 1 r r b i=b jb 1 )! b i! 96) Proof. First by induction on it follows R 1 t i ) a t a i i dt 1,..., dt = a! a i! + a + a i)! 97) First consider the integration with respect to t 1. The limits of the integration are and 1 i=2 t i for t 2,..., t ) R 1. By substituing v = t 1 /1 i=2 t i) we find 1 i=2 t i 1 t i ) a t a i i )dt 1 = i=2 i=2 t a i i ))1 t i ) a+a i+1 i=2 a!a 1! = a + a 1 + 1)! t a i i )1 t i ) a+a i+1 i=2 1 1 v) a v a 1 dv 98) Where the beta function identity is used 1 ta 1 t) b dt = a!b!/a + b + 1)!. The next step will give a fraction before the main term of a+a 1 +1)!a 2!/a+ a 1 + a 2 + 2)! thus the identity follows by induction. By the multinomial theorem, P b j = t j i )b = b 1,...,b b i=b b! b i! Thus by applying this the following result will be obtained R 1 P 1 ) a P b j dt 1...dt = b 1,...,b b i=b b! b 1 i! R t jb i i 99) t i ) a t b idt 1,..., dt = b!a! + a + jb)! b 1,...,b b i=b jb i )! b i! 1) 32

34 For computations b will be small, and so it is convenient to split ) the summation over the b i are non-zero. Given an integer r, there are ways of r choosing r of b 1,..., b to be non-zero. Thus b 1,...,b b i=b jb i )! b i! = b r=1 ) r b 1,...,b 1 b i=b jb i )! b i! 11) This gives the result. Now a lemma is used to write I F ) and J m) F ) in terms of manageable expression with this choice of P. Lemma 11. Let F be given in terms of a polynomial P by 94). Let P be given in terms of a polynomial expression in the symmetric power polynomials P 1 = t i and P 2 = t2 i by P = d a i1 P 1 ) b i P c i 2 for constants a i R and non-negative integers b i, c i. Then for each 1 m we have J m) F ) = where γ bi,b j,c i,c j,c 1,c 2 I F ) = 1 i,j d a i a j 1 i,j d c i c 1 = c 2 = b i + b j )!G ci +c a i a j,2) j + b i + b j + 2c i + 2c j )! c j ci c 1 ) ci c 1 ) γbi,b j,c i,c j,c 1 2G,c c 1) 1 +c 2 + b i + b j + 2c i + 2c j + 1)! = b i!b j!2c i 2c 1 )!2c j 2c 2 )!b i + b j + 2c i + 2c j 2c 1 2c 2 + 2)! b i + 2c i 2c 1 + 1)!b j + 2c j 2c 2 + 1)! and G is the polynomial given by Lemma 1 Proof. First consider I F ). Using Lemma 1 I F ) = P 2 dt 1...dt = a i a j 1 P 1 ) b i+b j P c i+c j 2 dt 1...dt R R = 1 i,j 1 i,j d b i + b j )!G ci +c a i a j,2) j + b i + b j + 2c i + 2c j )! 12) Thus the first part is proven. For the second sum remember that F is symmetric in t 1,..., t we see that J m) F ) is independent of m, thus is suffices to only consider J 1). The first integral becomes 33

35 1 i=2 1 P 1 ) b P c 2 dt 1 = c c = ) c c i=2 1 t 2 i ) c i=2 t i 1 t i ) b t 2c 2c 1 dt 1 Where P2 c is split such we have one part with t 1 and the rest which can be handled as a constant. c ) c 1 = c P 2) c 1 P 1 ) b+2c 2c +1 1 u) b u 2c c du = c = c c = ) c c P 2) c 1 P 1 ) b+2c 2c +1 b!2c 2c )! b + 2c 2c + 1)! Here P 1 = i=2 and P 2 = i=2 t2 i. Hence = 1 i,j d a i a j 1 c i F dt 1 ) 2 = c 1 = c 2 = c j ci c 1 d 1 a i i=2 t i 1 P 1 ) b i P c i 2 dt 1) 2 13) ) ) cj P 2) c 1 +c 2 1 P 1 ) b i+b j +2c i +2c j 2c 1 2c 2 +2 c 2 b i!b j!2c i 2c 1 )! b i + 2c i 2c 1 + 1)!b j + 2c j 2c 2 + 1)! 14) All terms are treated as constants in the other integrals except for the terms with P 1 and P 2 by Lemma 1 these terms are 1 P 1) b P 2) c b! = R 1 + b + c)! G c,2 1) 15) with b = c 1 + c 2 and c = b i + b j + 2c i + 2c j 2c 1 2c combining these, 15) and 14), will give the result. In Lemma 11 it has been proven that I F ) and J m) F ) can be written as quadratic forms of the a i. Thus both terms can be writen as matrices over coefficients a = a 1,..., a d ) of P. Moreover these will be positive definite real quadratic forms. Thus in particular M can be written as M = sup F S m=1 J m) F ) a T M 2 a = sup I F ) F S a T M 1 a 16) for two rational symmetric positive definite matrices M 1, M 2, which can be calculate explicitly in terms of for any choice of the exponents b i, c i. This can be maximized and has a solution 34

36 Lemma 12. Let M 1, M 2 be real, symmetric positive definite matrices. Then a T M 2 a a T M 1 a 17) is maximized when a is an eigenvector of M1 1 M 2 corresponding tot the largest eigenvalue of M1 1 M 2. The value of the ratio at its maximum is this largest eigenvalue. Proof. Multiplying a by a non zero scalar doesn t change the ratio, so we may assume without loss of generality that a T M 1 a = 1. By using Lagrangian multipliers, a T M 2 a is maximized subjest to a T M 1 a when La, λ) = a T M 2 a λa T M 1 a 1) 18) is stationary. This occurs when using the symmetricity of M 1, M 2 ) = δl δa i = 2M 2 2M 1 )a) i 19) for each i. This implies that recalling M 1 is positive definite so invertible) thus a T M 1 a = λ 1 a T M 2 a Length of Bounded Gaps M 1 1 M 2a = λa 11) By Lemma 12 a lower bound for M can be obtained. If F is chosen in the same way as in 94) the eigenvalues can be calculated. Theorem 2. i) lim infp n+1 p n ) 6 n Proof. When = 15 the largest eigenvalue of M 1 1 M 2 is λ > 4 111) Thus M 15 > 4. Because θ = 1/2 ɛ for every ɛ > by the Bombieri- Vinogradov Theorem. Proposition 4 can be used. Calculating r θm 2 = 1/2) /2 = 2. Hence if an admissible set with = 15 elements is given. The longest interval in the admissible set will give a upperbound for the interval length. The admissible set that can be chosen is of length 6 thus by Proposition 4, Theorem 2 is proven. 35

37 Theorem 3. Assume that the primes have level of distribution θ for all θ < 1. Then lim infp n+1 p n ) 12 n lim infp n+2 p n ) 6 n Proof. In the same way as in Theorem 2 it can be shown M 15 > 4 thus r 15 = 3 when θ < 1 in the same way r 5 = 2. By using the admissible set H = {, 2, 6, 8, 12} the first part is proven. Because r 15 = 3 with this distribution of primes by Proposition 4 this means there are infinitely many n such that there are 3 primes in the set n+h i with h i in the admissible set. This means if the same admissible set is used as in Theorem 2 the second part is proven. 36

38 Bibliografie [1] Tom M. Apostol. Introduction to Analytic Number Theory. Springer- Verlag, [2] C. Y. Yildrim D. A. Goldston, J. Pintz. Positive proportioh of small gaps in consecutive primes. arxiv: , 211. [3] J.Pintz C.Y.Yildrim D.A.Goldston, S.W.Graham. Small gaps between products of primes. Proc. London Math. Soc., 3: , 29. [4] G.H.Hardy and E.M.Wright. An Introduction to the Theory of Numbers. Oxford University Press, 28. [5] H.-E.Richert H. Halberstam. Sieve Methods. Academic Press, [6] James Maynard. Small gaps between primes. arxiv: v2, 213. [7] M.Ram Murty. Problems in Analytic Number Theory. Springer-Verslag New Yor, Inc, 21. [8] D.H.J. Polymath. The bounded gaps between primes polymath project - a retrospective. arxiv: , 214. [9] D. Zagier. Newman s short proof of the prime number theorem. The American Mathematical Monthly, 14:75 78, [1] Y. Zhang. Bounded gaps between primes. Annals of Mathematics, 179: