Riddles of Representations of Integers

A talk given at Southchina Normal Univ. (Dec. 6, 2018) and Nanjing Normal Univ. (Nov. 25, 2019) and Qinghai Normal Univ. (Nov. 24, 2020) Riddles of Representations of Integers Zhi-Wei Sun Nanjing University Nanjing 210093, P. R. China [email protected] http://math.nju.edu.cn/∼zwsun November 24, 2020 Part I. Representations of positive integers involving primes 2 / 52 Prime numbers An integer p > 1 is called a prime if it cannot be written as ab with 1 < a 6 b < p. Primes below 100: 2; 3; 5; 7; 11; 13; 17; 19; 23; 29; 31; 37; 41; 43; 47; 53; 59; 61; 67; 71; 73; 79; 83; 89; 97: Fundamental Theorem of Arithmetic. Any integer n > 1 can be written as a product of finitely many primes. If we ignore the order of the factors, such a decomposition is unique. + For n 2 Z = f1; 2; 3;:::g we denote the nth prime by pn. Thus p1 = 2; p2 = 3; p3 = 5; p4 = 7; p5 = 11; p6 = 13;:::: Prime Number Theorem. For x > 1 let π(x) be the number of primes not exceeding x. Then x π(x) ∼ as x ! +1: log x 3 / 52 Goldbach's conjecture Goldbach's Conjecture (1742): Every even number n > 2 can be written in the form p + q with p and q both prime. Goldbach's weak Conjecture [proved by I. M. Vinorgradov (1937) and H. Helfgott (2013)]. Each odd number n > 6 can be written as a sum of three primes. Goldbach's conjecture implies that for any n > 2 there is a prime p 2 [n; 2n] since 2n 6= p + q if p and q are smaller than n. Jing-run Chen's Theorem (1973). If an even number is large enough, then it can be written as p + P2, where p is a prime, and P2 is either a prime or a product of two primes. Lemoine's Conjecture (1894). Any odd integer n > 6 can be written as p + 2q, where p and q are primes. 4 / 52 Conjectures for twin primes, cousin primes and sexy primes Conjecture (Sun, 2012-12-22) Any integer n > 12 can be written as p + q with p; p + 6; 6q ± 1 all prime. Remark. I have verified this for n up to 109. Conjecture (Sun, 2013-01-03) Let + A =fx 2 Z : 6x − 1 and 6x + 1 are both primeg; + B =fx 2 Z : 6x + 1 and 6x + 5 are both primeg; + C =fx 2 Z : 2x − 3 and 2x + 3 are both primeg: Then A+B = f2; 3;:::g; B+C = f5; 6;:::g; A+C = f5; 6;:::gnf161g: Also, if we set 2X := X + X then 2A ⊇ f702; 703;:::g; 2B ⊇ f492; 493;:::g; 2C ⊇ f4006; 4007;:::g: 5 / 52 A conjecture refining Bertrand's Postulate Bertrand's Postulate (proved by Chebyshev in 1850). For any positive integer n, the interval [n; 2n] contains at least a prime. Conjecture (Sun, 2012-12-18) For each positive integer n, there is an integer k 2 f0;:::; ng such that n + k and n + k2 are both prime. Conjecture (Sun, 2013-04-15)p For any positive integer n there is a 2 2 positive integer k 6 4 n + 1 such that n + k is prime. 6 / 52 My conjecture on Heath-Brown primes Heath-Brown's Theorem (2001). There are infinitely many primes of the form x3 + 2y 3 where x and y are positive integers. Conjecture (Sun, 2012-12-14). Any positive integer n can be 3 3 written as x + y (x; y 2 N = f0; 1;:::g) with x + 2y prime. In general, for each positive odd integer m, any sufficiently large m m integer can be written as x + y (x; y 2 N) with x + 2y prime. Conjecture (Sun, 2013-04-15) For any integer n > 4 there is a positive integer k < n such that 2n + k and 2n3 + k3 are both prime. 7 / 52 A general hypothesis on representations Let's recall Schinzel's Hypothesis H. If f1(x);:::; fk (x) are irreducible polynomials with integer coefficients and positive leading coefficients such that there is no prime dividing the product f1(q)f2(q):::fk (q) for all q 2 Z, then there are infinitely many + n 2 Z such that f1(n); f2(n);:::; fk (n) are all primes. The following general hypothesis is somewhat similar to Schinzel's Hypothesis. General Conjecture on Representations (Sun, 2012-12-28) Let f1(x; y);:::; fm(x; y) be non-constant polynomials with integer + coefficients. Suppose that for large n 2 Z , those f1(x; n − x);:::; fm(x; n − x) are irreducible, and there is no prime Qm + dividing all the products k=1 fk (x; n − x) with x 2 Z. If n 2 Z + is large enough, then we can write n = x + y (x; y 2 Z ) such that jf1(x; y)j;:::; jfm(x; y)j are all prime. 8 / 52 Two curious conjectures on primes Conjecture (Sun, 2013, Prize $200). Any integer n > 1 can be + 2 2 written as x + y (x; y 2 Z ) such that x + ny and x + ny are both prime. For example, 20 = 11 + 9 with 11 + 20 × 9 = 191 and 112 + 20 × 92 = 1741 both prime. Conjecture (Sun, 2013, Prize $1000). Any integer n > 1 can be + k written as k + m (k; m 2 Z ) with 2 + m prime. 7 Remark. I have verified this for n 6 10 . For example, 8 = 3 + 5 with 23 + 5 = 13 prime. Also, 9302003 = 311468 + 8990535 with 2311468 + 8990535 a prime of 93762 decimal digits. 9 / 52 Unification of Goldbach's conjecture and the twin prime conjecture Unification of Goldbach's Conjecture and the Twin Prime Conjecture (Sun, 2014-01-29). For any integer n > 2, there is a prime q with 2n − q and pq+2 + 2 both prime. We have verified the conjecture for n up to 108. Clearly, it is stronger than Goldbach's conjecture. Now we explain why it implies the twin prime conjecture. In fact, if all primes q with pq+2 + 2 prime are smaller than an even number N > 2, then for any such a prime q the number N! − q is composite since N! − q ≡ 0 (mod q) and N! − q > q(q + 1) − q > q: Example. 20 = 3 + 17 with 3, 17 and p3+2 + 2 = 11 + 2 = 13 all prime. 10 / 52 Super Twin Prime Conjecture If p; p + 2 and π(p) are all prime, then we call fp; p + 2g a super twin prime pair. Super Twin Prime Conjecture (Sun, 2014-02-05). Any integer n > 2 can be written as k + m with k and m positive integers such that pk + 2 and ppm + 2 are both prime. Example. 22 = 20 + 2 with p20 + 2 = 71 + 2 = 73 and pp2 + 2 = p3 + 2 = 5 + 2 = 7 both prime. Remark. If all those positive integer m with ppm + 2 prime are smaller than an integer N > 2, then by the conjecture, for each j = 1; 2; 3;:::, there are positive integers k(j) and m(j) with k(j) + m(j) = jN such that pk(j) + 2 and ppm(j) + 2 are both prime, and hence k(j) 2 ((j − 1)N; jN) since m(j) < N; thus 1 1 X 1 X 1 > ; p pjN j=1 k(j) j=1 which is impossible since the series on the right-hand side diverges while the series on the left-hand side converges by Brun's theorem. 11 / 52 Alternating sums of primes Let pn be the nth prime and define n−1 sn = pn − pn−1 + ··· + (−1) p1: For example, s5 = p5 − p4 + p3 − p2 + p1 = 11 − 7 + 5 − 3 + 2 = 8: Note that n n X X s2n = (p2k − p2k−1) > 0; s2n+1 = (p2k+1 − p2k ) + p1 > 0: k=1 k=1 Let 1 6 k < n. If n − k is even, then sn − sk = (pn − pn−1) + ··· + (pk+2 − pk+1) > 0: If n − k is odd, then n k X n−l X k−j sn − sk = (−1) pl − 2 (−1) pj ≡ n − k ≡ 1 (mod 2): l=k+1 j=1 So, s1; s2; s3;::: are pairwise distinct. 12 / 52 An amazing recurrence for primes We may compute the (n + 1)-th prime pn+1 in terms of p1;:::; pn. Conjecture (Z. W. Sun, J. Number Theory 2013). For any positive integer n 6= 1; 2; 4; 9, the (n + 1)-th prime pn+1 is the least positive integer m such that 2 2 2s1 ;:::; 2sn are pairwise distinct modulo m. 5 Remark. I have verified the conjecture for n 6 2 × 10 , and proved 2 2 that 2s1 ;:::; 2sn are indeed pairwise distinct modulo pn+1. Let 1 6 j < k 6 n. Then 0 < jsk − sj j 6 maxfsk ; sj g 6 maxfpk ; pj g 6 pn < pn+1: Also, sk + sj 6 pk + pj < 2pn+1. If 2 - k − j, then sk + sj = pk − pk−1 + ::: + pj+1 6 pk < pn+1: If 2 j k − j, then sk ≡ sj (mod 2) and hence sk + sj 6= pn+1. Thus 2 2 2sk − 2sj = 2(sk − sj )(sk + sj ) 6≡ 0 (mod pn+1). 13 / 52 Conjecture on alternating sums of consecutive primes Conjecture (Z. W. Sun, J. Number Theory, 2013). For any positive integer m, there are consecutive primes p ;:::; p (k < n) p k n not exceeding 2m + 2:2 m such that n−k m = pn − pn−1 + ··· + (−1) pk : p (Moreover, we may even require that m < p < m + 4:6 m if p n p 2 - m and 2m − 3:6 m + 1 < pn < 2m + 2:2 m if 2 j m.) Examples.

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