PATTERNS OF PRIMES IN ARITHMETIC PROGRESSIONS

PATTERNS OF PRIMES IN ARITHMETIC PROGRESSIONS JÁNOS PINTZ Rényi Institute of the Hungarian Academy of Sciences CIRM, Dec. 13, 2016

2 1. Patterns of primes Notation: p n the n th prime, P = {p i } i=1, d n = p n+1 p n. Abbreviation: i.o. = infinitely often, Z + = {1, 2,... } Twin Prime Conjecture {n, n + 2} P 2 i.o. d ν = 2 i.o. Polignac Conjecture (1849) Definition 2 h d n = h i.o. H = H k = {h i } k i=1, 0 h 1 < h 2 < h k is admissible if the number of residue classes covered by H mod p, ν p (H) < p for every prime p. Dickson Conjecture (1904). H k admissible = {n + h i } k i=1 Pk i.o. 2 / 35

3 Hardy Littlewood Conjecture (1923). H = H k admissible = x 1 n x log k x σ(h), {n+h i } P k σ(h) = p ( 1 ν ) ( p(h) 1 1 ) k p p 3 / 35

4 Remark 1. HL conjecture implies: Strong HL conjecture: H = H k admissible = x 1 (log x) σ(h) k n x {n+h i } P k n+h / P if h [0,h k ], h h i Proof. By Selberg s upper bound sieve 1 n x {n+h i } P k n+h P x (log x) k+1. Remark 2. Dickson Conjecture Strong Dickson Conjecture. 4 / 35

5 2. Primes in Arithmetic Progressions (AP) Conjecture (Lagrange, Waring, Erdős Turán (1936)). The primes contain k-term AP s for every k. Conjecture (Erdős Turán (1936)). If A Z + has positive upper density then A contains k-term AP s for every k. Roth (1953): This is true for k = 3. Szemerédi (1975) This is true for every k. 5 / 35

6 3. History before 2000 (2004) Erdős (1940) Bombieri Davenport (1966) d n < (1 c 0 ) log n i.o. c 0 > 0 fix. H. Maier (1988) d n < (log n)/4 i.o. Van der Corput (1939) AP s. d n < (log n)/2 i.o. P contains infinitely many 3-term Heath-Brown (1984) There are infinitely many pairs n, d such that n, n + d, n + 2d P, n + 3d P 2. Definition n = P k -number if it has at most k prime factors (ALMOST PRIMES). 6 / 35

7 4. Results after 2000 Green Tao Theorem (2004 2008): P contains k-term AP s for every k. Goldston Pintz Yıldırım (2005 2009): lim inf n d n/ log n = 0. GPY (2006 2010): lim inf n d n/(log n) c = 0 if c > 1 2. Zhang (2013 2014): H k admissible, k > 3.5 10 6 = n + H k contains at least 2 primes i.o. 7 / 35

8 Maynard (2013 2015): This is true for k 105. Polymath (2014): This is true for k 50. Maynard (2013 2015), Tao unpublished: H k admissible = n + H k contains at least ( 1 4 + o(1)) log k primes i.o. 8 / 35

9 5. Patterns of primes in arithmetic progressions A common generalization of the Green Tao and Maynard Tao theorem is Theorem 1 (J. P. 2017) Let m > 0 and A = {a 1,..., a r } be a set of r distinct integers with r sufficiently large depending on m. Let N(A) denote the number of integer m-tuples {h 1,..., h m } A such that there exist for every l infinitely many l-term arithmetic progressions of integers {n i } l i=1 where n i + h j is the j th prime following n i prime for each pair i, j. Then (5.1) N(A) m # { (h 1,..., h m ) A } m A m = r m. 9 / 35

10 Theorem 1 will follow by the application of Maynard s method from the weaker Theorem 2 (J. P. 2017) Let m be a positive integer, H = {h 1,..., h k } be an admissible set of k distinct non-negative integers h i H, k = Cm 2 e 4m with a sufficiently large absolute constant C. Then there exists an m-element subset (5.2) {h 1, h 2,..., h m} H such that for every positive integer l we have infinitely many l-element non-trivial arithmetic progressions of integers n i such that n i + h j P for 1 i l, 1 j m, further n i + h j is always the j th prime following n i. 10 / 35

11 In fact we prove a stronger result, namely Theorem 3 (J. P. 2017) There is some C, such that for all k 0 and all k > Ck0 2 e 4k 0 there is some c > 0, such that for all admissible tuples {h 1,..., h k } the number N(x) of integers n x, such that n + h i is n c -pseudo prime, and among these k integers there are at least k 0 primes, satisfies N(x) x. These N(x) log k x integers n x contain an m-term AP if x > C 0 (m). 11 / 35

12 6. Structure of the proof of the Maynard Tao theorem (i) Key parameters: H = {h 1,..., h k } given (0 h 1 < h 2 <... < h k ) N large, we look for primes of the form n + h i with n [N, 2N) R = N θ/2 ε, where θ is a level of distribution of primes: (6.1) max a q x θ (a,q)=1 π(x) π(x, q, a) ϕ(q) A x (log x) A holds for any A > 0 where the symbol of Vinogradov means that f (x) = O(g(x)) is abbreviated by f (x) g(x). Remark. θ = 1/2 admissible: Bombieri Vinogradov theorem (1965). θ > 0 Rényi (1947) (6.2) W = D 0 = C (k) suitably large p D 0 p (6.3) n ν 0 (mod W ) (ν 0 + h i, W ) = 1 for i = 1,..., k. 12 / 35

13 (ii) We weight the numbers n ν 0 (mod W ), n [N, 2N) by w n, so that w n 0 and on average w n would be large if we have many primes among {n + h i } k i=1, ( ) 2 (6.4) w n = λ d1,...,d k. d i n+h i i ( k ) (6.5) λ d1,...,d k = µ(d i )d i ( k whenever i=1 i=1 r 1,...,r k d i r i i (r i,w )=1 ( k ) 2 µ r i i=1 k i=1 ϕ(r i ) ) d i, W = 1 and λ d1,...,d r = 0 otherwise. y r1,...,r k 13 / 35

14 (6.6) y r1,...,r k = F ( log r1 log R,..., log r ) k log R where F is piecewise differentiable, real, F and F bounded, supported on { } k (6.7) R k = (x 1,..., x k ) [0, 1] k : x i 1. (iii) Let χ P (n) denote the characteristic function of P, (6.8) ( k ) S 1 := w n, S 2 := w n χ P (n+h i ), n N n<2n n ν 0 (mod W ) n N n<2n n ν 0 (mod W ) i=1 i=1 14 / 35

15 If we succeed to choose F, thereby λ d and w n in such a way that ( r k Z +) (6.9) S 2 > S 1, or S 2 > (r k 1)S 1 resp. we obtain at lest two, or k 0 primes, resp. among n + h 1,..., n + h k = bounded gaps between primes or even k 0 primes in bounded intervals i.o. 15 / 35

16 (iv) First step towards this: evaluation of S 1 and S 2. Proposition 1. We have as N ( ( )) 1 1 + O D 0 ϕ(w ) k N(log R) k (6.10) S 1 = I W k+1 k (F ), ( ( )) 1 1 + O D 0 ϕ(w ) k N(log R) k+1 k (6.11) S 2 = J (j) k (F ), W k+1 j=1 (6.12) I k (F ) = (6.13) 1 J (j) k (F ) =... 0 1 0 ( 1 0 1 0 1... 0 F (t 1,..., t k ) 2 dt 1... dt k, F (t 1,..., t k )dt j ) 2 dt 1... dt j 1 dt j+1... dt k. 16 / 35

17 After this we immediately obtain Corollary. If the sup is taken with F k as before and (6.14) M k = sup k J (j) k (F ) θmk, r k = I k (F ) 2 j=1 and let H be a fixed admissible sequence H = {h 1,..., h k } of size k. Then there are infinitely many integers n such that at least r k of the n + h i (1 i k) are simultaneously primes. Proposition 2. M 105 > 4 and M k > log k 2 log log k 2 for k > k 0. 17 / 35

18 7. A stronger version of the Maynard Tao theorem Theorem 3 gives a stronger form in 3 aspects: (i) all numbers n + h i are almost primes, having all prime factors greater than n c(k) ; (ii) the number of such n s is at least c (k)n, the true order log k N of magnitude of n [N, 2N) with all n + h i being n c -almost primes; (iii) for every l we have (infinitely) many l-term AP s with the same prime pattern. Remark: properties (i) and (ii) are interesting in themselves, but crucial in many applications, in particular in showing (iii). 18 / 35

19 Let P (n) denote the smallest prime factor of n. The following Lemma shows that the contribution of n s to S 1 with at least one prime p (n + h i ), p < n c 1(k) is negligible if c 1 (k) is suitably small ( R = N θ/2 ε). Lemma 1. We have (7.1) S 1 = N n<2n ( n ν 0 (mod W ) k ) P (n+h i ) <n c 1 (k) i=1 c 1 (k) log N w n k,h S 1. log R 19 / 35

20 Corollary. We get immediately property (i). Further (7.2) w n λ 2 max y 2 max(log R) 2k (log R) 2k since (n + h i ) has in this case just a bounded number of prime factors, so the sum over the divisors can be substituted by the largest term (apart from a factor depending on k). So we get (7.3) S 1 := n N n<2n n ν 0 (mod W ) ( k ) P (n+h i ) i=1 #{i;n+h i P} r k >n c 1 (k) by which we obtained property (ii). 1 S 1(1 + O(c 1 (k)) (log R) 2k, 20 / 35

21 8. Green Tao theorem: structure of proof Original Szemerédi theorem. If A Z N has a positive density then A contains m-term AP for every m, if N > C(m). Relative Szemerédi theorem (Green Tao). If A Z N is a pseudorandom set, B A has a positive relative density within A, i.e. with a measure ν(n) obeying the linear forms condition ν(n) (8.1) lim N n N,n B n N,n A ν(n) = δ > 0 then B contains m-term AP for every m, if N > C(δ, m). 21 / 35

22 Definition. A set A Z N is a pseudorandom set if there is a measure ν : Z N R + which satisfies the linear forms condition if the following holds. Let (L ij ), 1 i l, 1 j t rational numbers with all numerators and denominators at most L 0, b i Z N, l l 0, m m 0. Let ψ i (x) = t L i,j x j + b i, where the t-tuples j=1 (L ij ) 1 j t Q t are non-zero and no t-tuple is a rational multiple of another. Then ( (8.2) E ν(ψ 1 (x) ) )... ν(ψ m (x)) x Z t N = 1 + o L0,l 0,m 0 (1). 22 / 35

23 Remark 1. The primes up to N form a set of density 1/ log N 0 as N. Therefore we cannot use the Szemerédi theorem. Step 1. To formulate and show a generalization of the Szemerédi theorem where the set Z N = [1, 2,..., N] can be substituted by some sparse set satisfying some regularity condition like (8.2). This result is called Relative Szemerédi Theorem. 23 / 35

24 Remark 2. Another condition, the correlation condition in the original work of Green and Tao could be avoided by a different proof of Conlon Fox Zhao (2015). Step 2. To find a suitable pseudo-random set A where the set P of the primes can be embedded as a subset of positive density. This was proved in an unpublished manuscript of Goldston and Yıldırım (2003). This set A is the set of almost primes; the measure (µ is the Möbius function, c > 0 small) ( (8.3) ν(n) = d n,d R ( µ(d) 1 log d )) 2 R = N c. log R 24 / 35

25 9. Combination of the methods of Green Tao and Maynard Tao Difficulty: the original Maynard Tao method produces directly (without using any further ideas) only at least c/ log log N (9.1) N integers n [N, 2N) with at least k 0 = 1 (1 + o(1)) log k 4 primes among {n + h i } k i=s. The expected number of n s with this property is N (9.2) c 2 (k 0 ) (log N), k 0 which is much more. 25 / 35

26 Hope: By Theorem 1 (cf. 7 (i) (ii)) we obtain (9.3) c 3 (k)n log k N such numbers, which is still less than (9.2). Further idea: if we require additionally that all n + h i s should be almost primes, i.e. P (n + h i ) > n c 1(k), then we obtain also c(k)n/ log k N numbers n [N, 2N) which is already the true order of magnitude of such n s. 26 / 35

27 Solution: instead of embedding primes into the set of almost primes we embed the set of n s, n [N, 2N) with at least k 0 primes among {n + h i } k i=1 and ( k ) (9.4) P + h i ) > n i=1(n c 1(k) into the set of n s, n [N, 2N) with (9.4). Remark: in some sense we embed the set of almost prime k-tuples with at least k 0 primes into the set of all almost prime k-tuples. 27 / 35

28 Lemma 2. Let k be an arbitrary positive integer and H = {h 1,..., h k } be an admissible k-tuple. If the set N (H) satisfies with constants c 1 (k), c 2 (k) { ( k ) } (9.5) N (H) n; P + h i ) n i=1(n c 1(k) and (9.6) # { n X, n N (H) } c 2(k)X log k X for X > X 0, then N(H) contains l-term arithmetic progressions for every l. 28 / 35

29 Main idea of the proof: We use the measure ( ) k ϕ(w ) k Λ 2 R (Wn+ν 0+h i ), n [N, 2N) W log R (9.7) ν(n) := i=1 0 otherwise with R = N c 1(k) and (9.8) Λ R (u) = Remark. If P ( k i=1 d R,d u ) (u + h i ) µ(d) log R d. > N c 1(k) = R, then Λ R (u + h i ) = log R (the single term in the sum is that with d = 1) and ν(u) = (ϕ(w )/W ) k log k R does not depend on u. 29 / 35

30 The pseudorandomness of the measure ν can be proved by a generalization of the original Goldston Yıldırım method. The original GY method is exactly the case k = 1. The possible methods are either (i) analytic number theoretical (using the zeta-function) or (ii) Fourier series or (iii) real elementary. 30 / 35

31 Remark. The proof that we obtain consecutive primes by this procedure follows from the fact that the number of n [N, 2N) obtained is at least c(k)n/ log k N. If any of the numbers n + h, 0 h h k, h h i (i = 1, 2,..., k) were additionally prime then by Selberg s upper bound sieve we would find at most c (k)n/ log k+1 N such numbers (cf. the estimate in 1.) since all n + h i (i = 1, 2,..., k) are almost primes (similarly to the case of the Strong HL conjecture). So here we also need both properties 7 (i) and 7 (ii). 31 / 35

32 10. Sketch of the proof of Lemma 2 The proof is essentially the same for an arbitrary k as for the simplest case k = 1. So let k = 1. We choose a prime p < N c1(k) and try to evaluate (10.1) Sp = λ d λ e N n<2n,p n+h,n ν 0 (W ) [d,e] n+h 32 / 35

33 Distinguishing the cases (10.2) (10.3) (10.4) p [d, e] = N pw d = d p, p e = N pw d = d p, e = e p = N pw λ d λ e [d, e] λ d λ e [d, e] λ d λ e [d, e ] (or reversed) 33 / 35

34 we obtain in all cases an asymptotic of type (10.5) S p = N pw λd λ e [d, e, p]/p + O(R 2+ε ). Lemma (Selberg, Coll. Works 1991, Greaves 2000) (10.6) T p := λ d λ e [d, e, p]/p = r p r µ 2 (r) ϕ(r) (y r y rp ) 2. 34 / 35

35 However, by the definition of y r and F we have (10.7) ( ( ) ( )) 2 log r log r + log p (y r y rp ) 2 log p = F F F log R log R log R, (10.8) T p log p ϕ(w ) log R log R W (10.9) S = = S p log p p N W ϕ(w ) W, p<n c 1 (k) S p N W c 1(k) log N ϕ(w ) W k,θ c 1 (k)s 1 which is negligible if c 1 (k) is sufficiently small. 35 / 35