A talk given at Univ. of Illinois at Urbana-Champaign (August 28, 2012) and the 6th National Number Theory Confer. (October 20, 2012) and the China-Korea Number Theory Seminar (October 27, 2012) Various new observations about primes Zhi-Wei Sun Nanjing University Nanjing 210093, P. R. China [email protected] http://math.nju.edu.cn/∼zwsun October 27, 2012 Abstract This talk focuses on the speaker's recent discoveries about primes. We will talk about the speaker's new way to generate all primes or primes in certain arithmetic progressions, and his various conjectures for products of primes, sums of primes, recurrence for primes, representations of integers as alternating sums of consecutive primes. We will also mention his new observations (mainly made in August 2012 at UIUC) about twin primes, squarefree numbers, and primitive roots modulo primes. 2 / 42 Part I. On functions taking only prime values 3 / 42 Representing primes by polynomials Euler: x2 − x + 41 is prime for every x = 1;:::; 40. Theorem (Rabinowitz, 1913). Let p > 1 be an integer and let K p be the imaginary quadratic field Q( 1 − 4p). Let OK be the ring of algebraic integers in K. Then x2 − x + p is a prime for all x = 1;:::; p − 1 if and only if K has class number one, i.e., OK is a principal ideal domain. Theorem (conjectured by Gauss and proved by H. Stark). The onlyp imaginary quadratic field having class number one are those Q( −d) with d 2 f1; 2; 3; 7; 11; 19; 43; 67; 163g. Thus, for any p > 41, x2 − x + p cannot take prime values for all x = 1;:::; p − 1. In general, any non-constant polynomial P(x1;:::; xn) with integer coefficients cannot take primes for all x1;:::; xn 2 Z. 4 / 42 Mills' Theorem Theorem (Mills, 1947). There is a real number A such that M(n) = bA3n c takes only prime values. 5=8 Sketch of the Proof. Since pn+1 − pn = O(pn ) (A. E. Ingham, 1937), one can construct infinitely many primes P0; P1; P2;::: with 3 3 Pn < Pn+1 < (Pn + 1) − 1: 3−n Then the sequence un = Pn is increasing while the sequence 3−n vn = (Pn + 1) is decreasing. As un < vn, we see that A = limn!1 un 6 B = limn!1 vn, hence 3n 3n 3n Pn = un < A < Pn + 1 = vn : 3n So bA c = Pn is a prime for all n = 1; 2; 3;:::. 5 / 42 A problem on central binomial coefficients Let p be an odd prime. For k = 0;:::; (p − 1)=2 we have 2k (2k)! = 6≡ 0 (mod p); k k!2 but for k = (p + 1)=2;:::; p − 1 we have 2k (2k)! = ≡ 0 (mod p): k k!2 Conjecture (Sun, Feb. 20, 2012). Let p > 5 be a prime. Then 2k p − 1 ± : k = 1;:::; k 2 cannot be a reduced system of residues modulo p. Moreover, if 2k p > 11 then k (k = 1;:::; (p − 3)=2) cannot be pairwise distinct modulo p. Remark. Later I realized that for any positive integer m the largest n such that 2k (k = 1;:::; n) are pairwise distinct modulo m pk p should be O( m) and probably less than 4:53 m. 6 / 42 A function taking only prime values Conjecture (Z. W. Sun, Feb. 21, 2012). For n = 1; 2; 3;::: define 2k s(n) as the least integer m > 1 such that k (k = 1;:::; n) are pairwise distinct modulo m. Then s(n) is always a prime! I also guessed that s(n) < n2 for n = 2; 3; 4;:::. I calculated s(n) for n = 1;:::; 2065. For example, s(1) = 2; s(2) = 3; s(3) = 5; s(4) = s(5) = s(6) = 11; s(7) = s(8) = s(9) = 23; s(10) = 31; s(11) = ··· = s(14) = 43; s(15) = s(16) = s(17) = s(18) = 59; s(19) = 107; s(20) = 149: After I made the conjecture public via a message to Number Theory List, Laurent Bartholdi computed s(n) for n = 2001;:::; 5000, and his computational result supports my conjecture. 7 / 42 Artin's conjecture Let p be an odd prime. If a 2 Z is not divisible by p and ak 6≡ 1 (mod p) for k = 1;:::; p − 2, then we say that a is a primitive root mod p (or the order of a mod p is p − 1). Artin's Conjecture (1927). If a 2 Z is neither −1 nor a square, then there are infinitely many primes p having a as a primitive root modulo p (moreover, the set of such primes has a positive asymptotic density inside the set of all primes). Progress: (1) C. Hooley (1967): The conjecture follows from the Generalized Riemann Hypothesis. (2) R. Gupta & M. R. Murty (1984): The conjecture holds for infinitely many a (with help of sieve methods). (3) R. Heath-Brown (1986): There are at most two exceptional primes a for which Artin's conjecture fails. 8 / 42 A conjecture implying Artin's conjecture Conjecture (Sun, Feb. 22-23, 2012). Let a 2 Z with jaj > 1. For + n 2 Z define fa(n) as the least integer m > 1 such that those ak (k = 1;:::; n) are pairwise incongruent modulo m. (i) fa(n) is a prime for all sufficiently large n. (ii) If a is not a square, then for any sufficiently large n, fa(n) is the least prime p > n having a as a primitive root mod p; (iii) If a is a square, then for any sufficiently large n, fa(n) is just the least prime p > 2n such that a; a2;:::; a(p−1)=2 are pairwise distinct modulo p. + Example. f−3(n) with n 2 Z is the least prime p > n such that −3 is a primitive root mod p. p 2k k Motivation. By Stirling's formula, k ∼ 4 = kπ. 9 / 42 Conjectures on Lucas sequences Let A be an integer with jAj > 2. Define u0(A) = 0; u1(A) = 1; and un+1(A) = Aun(A) − un−1(A); v0(A) = 2; v1(A) = A; and vn+1(A) = Avn(A) − vn−1(A): Conjecture (Sun, Feb. 26, 2012). (i) If 2 + A is not a square, then there are infinitely many odd primes p such that those vk (A) mod p with k = 1; :::; (p − "p)=2 are pairwise distinct, A2−4 where "p = ( p ). (ii) If 2 − A is not a square, then there are infinitely many odd primes p such that those uk (A) mod p with 1 6 k 6 (p − "p)=2 are pairwise distinct. + Conjecture (Sun, Feb. 26, 2012). Let n 2 Z be sufficiently large (n > 2jAj or n > 100 may suffice). Then tA(n) (the smallest integer m > 1 such that vk (A)(k = 1; :::; (p − "p)=2) are pairwise distinct modulo m) is prime. Moreover, if A + 2 is not a square, then tA(n) is the smallest odd prime p such that p − "p > 2n and those vk (A) mod p (k = 1; :::; (p − "p)=2) are pairwise distinct. 10 / 42 Generate all primes in a combinatorial manner + Theorem 1 (Sun, Feb. 29, 2012) For n 2 Z let S(n) denote the smallest integer m > 1 such that those 2k(k − 1) mod m for k = 1;:::; n are pairwise distinct. Then S(n) is the least prime greater than 2n − 2. Remark. (a) The range of S(n) is exactly the set of all primes! + (b) I also showed that for any d 2 Z whenever n > d + 2 the + least prime p > 2n + d is just the smallest m 2 Z such that 2k(k + d)(k = 1;:::; n) are pairwise distinct modulo m. (c) I proved that the least positive integer m such that those k 2 = k(k − 1)=2 (k = 1;:::; n) are pairwise distinct modulo m, is just the least power of two not smaller than n. 11 / 42 Another theorem + Theorem 2 (Sun, March 2012) (i) Let d 2 f2; 3g and n 2 Z . Then the smallest positive integer m such that those k(dk − 1) (k = 1;:::; n) are pairwise distinct modulo m, is the least power of d not smaller than n. (ii) Let n 2 f4; 5;:::g. Then the least positive integer m such that 18k(3k − 1) (k = 1;:::; n) are pairwise distinct modulo m, is just the least prime p > 3n with p ≡ 1 (mod 3). Remark. We are also able to prove some other similar results including the following one: + For n > 5 the least m 2 Z such that those 18k(3k + 1) (k = 1;:::; n) are pairwise distinct modulo m, is just the first prime p ≡ −1 (mod 3) after 3n. 12 / 42 One more theorem + Theorem (Sun, March 2012). (i) For d; n 2 Z let λd (n) be the smallest integer m > 1 such that those (2k − 1)d (k = 1;:::; n) are pairwise incongruent modulo m. Then λd (n) with d 2 f4; 6; 12g and n > 2 is the least prime p > 2n − 1 with p ≡ −1 (mod d). (ii) Let q be an odd prime. Then the smallest integer m > 1 such that those kq(k − 1)q with k = 1;:::; n are pairwise incongruent mod m, is just the least prime p > 2n − 1 with p 6≡ 1 (mod q): Remark. In the proof I used the Brun-Titchmarsh theorem which + asserts that if a; q 2 Z , gcd(a; q) = 1 and x > q then 2x jfp x : p ≡ a (mod q)gj : 6 6 '(q) log(x=q) 13 / 42 Previous results by others Theorem 1 (L.