Some of my number-theoretic conjectures and related progress

Transcription

1 A tal given at the 6th China-Japan Conference on Number Theory (Shanghai; August 15-17, 2011) Some of my number-theoretic conjectures and related progress Zhi-Wei Sun Nanjing University Nanjing , P. R. China zwsun August 16, 2011

2 Abstract and References In recent years the author raised hundreds of number-theoretic conjectures some of which involve representations of integers, binary quadratic forms, congruences for sums of binomial coefficients, and series for powers of π. In this tal we will introduce my main conjectures and related progress, and announce prizes for proofs of some conjectures. References 1. Z. W. Sun, Open Conjectures on Congruences, arxiv: Z. W. Sun, List of conjectural series for powers of π and other constants, arxiv: Z. W. Sun, Conjectures and results on x 2 mod p 2 with 4p = x 2 + dy 2, arxiv: / 34

3 Part A. On Representations of Integers 3 / 34

4 Mixed Sums of Triangular Numbers and Other Terms Triangular numbers: T x = x(x + 1)/2 (x Z). Conjecture (Sun [J. Combin. Number Theory 1(2009)]). Any positive integer n 216 can be written in the form p + T x with p zero or a prime. Remar. This has been verified for n Prize $1000 for the first rigorous proof. In 2005 Z. W. Sun [Acta Arith. 2007] investigated what ind of mixed sums ax 2 + by 2 + ct z or ax 2 + bt y + ct z (with a, b, c Z + ) are universal (i.e., all natural numbers can be so represented). This project was completed via three papers: Z. W. Sun [Acta Arith. 2007], S. Guo, H. Pan & Z. W. Sun [Integers, 2007], and B. K. Oh & Sun [JNT, 2009]. Theorem (B. K. Oh and Sun [JNT 2009]). Any positive integer n can be written as the sum of a square, an odd square and a triangular number. 4 / 34

5 Mixed Sums of Squares and Triangular Numbers Ken Ono and K. Soundararajan [Invent. Math. 130(1997)]: Under GRH (the generalized Riemann hypothesis), we have Ramanujan s assertion that any positive odd integer greater than 2719 can be represented by the form x 2 + y z 2. 2n + 1 = (2x) 2 + (2y + 1) z 2 n = 2x 2 + 4T y + 5z 2. Another Version of the Ono-Soundararajan Theorem. Under GRH, 2x 2 + 5y 2 + 4T z represents all integers greater than B. Kane and Z. W. Sun [Trans. AMS 362(2010), ] determined completely when the general form ax 2 + by 2 + ct z (a, b, c Z + ) represents sufficiently large integers and established similar results for the forms ax 2 + bt y + ct z and at x + bt y + ct z. In particular, the form ax 2 + y 2 + T z represents sufficiently large integers if and only if each odd prime divisor of a is congruent to 1 or 3 modulo 8. Tools: The theory of ternary quadratic forms and modular forms. 5 / 34

6 Polygonal Numbers Polygonal numbers are nonnegative integers constructed geometrically from the regular polygons. For m = 3, 4, 5,..., the m-gonal numbers are given by ( ) n p m (n) = (m 2) + n (n = 0, 1, 2,...). 2 Note that p 3 (n) = T n and p 4 (n) = n 2. Those p 5 (n) = n(3n 1)/2 (n N) are called pentagonal numbers. Those p 6 (n) = n(2n 1) = T 2n 1 (n N) are called hexagonal numbers. Fermat s Assertion (proved by Cauchy in 1813). Any natural number n can be written as the sum of m m-gonal numbers. 6 / 34

7 Diagonal Representations by Polygonal Numbers n =p 3 (x 1 ) + p 3 (x 2 ) + p 3 (x 3 ) n =p 4 (x 1 ) + p 4 (x 2 ) + p 4 (x 3 ) + p 4 (x 4 ) n =p 5 (x 1 ) + p 5 (x 2 ) + p 5 (x 3 ) + p 5 (x 4 ) + p 5 (x 5 ) n =p 6 (x 1 ) + p 6 (x 2 ) + p 6 (x 3 ) + p 6 (x 4 ) + p 6 (x 5 ) + p 6 (x 6 ) Diagonal Representation: n = p 4 (x 1 ) + p 5 (x 2 ) + p 6 (x 3 ) Conjecture (Z. W. Sun, 2009). Let m 3 be an integer. Then any n N can be written in the form p m+1 (x 1 )+p m+2 (x 2 )+p m+3 (x 3 )+r (x 1, x 2, x 3 N, r {0,..., m 3}). Moreover, {r(n) : n = 1, 2, 3,...} = {1, 2, 3,...}, where r(n) = {(x, y, z) N 3 : n = p 4 (x) + p 5 (y) + p 6 (z)}. Remar. $500 for the first rigorous proof. 7 / 34

8 Part B. On Combinatorial Congruences 8 / 34

9 On Fibonacci quotient mod p Z. W. Sun [Sci. China Math. 53(2010)]) determined p 1 ( 2 ) /m mod p 2, where p is an odd prime and m 0 (mod p). In particular, ) p 1 ( 2 p 1 ( ) 2 ( 1) 2 ( 1) (p 1)/2 (mod p 2 ), ( p ) ( ) 1 2F 5 p ( p 5 ) (mod p 2 ) (p 5), where {F n } n 0 is the Fibonacci sequence with F 0 = 0, F 1 = 1 and F n+1 = F n + F n 1 (n = 1, 2, 3,...). Theorem (Sun and Tauraso [Adv. in Appl. Math. 2010]). For any odd prime p 5 we have p 1 ( 1) ( ) 2 5 F p ( p 5 ) (mod p). p =1 9 / 34

10 Determination of F p ( p 5 ) mod p 3 Conjecture (Sun and Tauraso [Adv. in Appl. Math. 2010]). Let p 2, 5 be a prime and let a Z +. Then p a 1 ( 1) ( 2 ) ( p a ) ( ) 1 2F 5 p a p ( a 5 ) (mod p 3 ). Recently Hao Pan and Z. W. Sun confirmed the conjecture. The proof is very sophisticated and we use an important technique of Andrew Granville [2004]. In the same paper Pan and Sun also proved the following conjecture of Roberto Tauraso: p 1 =1 L 0 (mod p) for any prime p > 3, 2 where L 0 = 2, L 1 = 1 and L n+1 = L n + L n 1 (n = 1, 2, 3,...). 10 / 34

11 Another congruence for F p ( p 5 ) mod p 3 Theorem (Z. W. Sun, 2010) Let p 2, 5 be a prime. Then (p 1)/2 ( 2 ) ( p ) ( ( 16) 1 + F ) p ( p 5 ) (mod p 3 ). 5 2 The proof is also very sophisticated and quite difficult. One of the lemmas is as follows. Lemma (Conjectured by Sun and proved by Tauraso) Let p > 5 be a prime and let H (2) = 0<j 1/j2 for = 0, 1, 2,.... Then p 1 ( ) ( ) 2 ( ( 1) H (2) p ) 2 5 Fp ( p 5 ) (mod p), 5 2 p p 1 ( ) 2 ( 2) H (2) 2 3 q p(2) 2 (mod p) where q p (2) denotes the Fermat quotient (2 p 1 1)/p. 11 / 34

12 A further conjecture Conjecture (Sun, 2010) Let p be an odd prime and let a Z +. (i) If p a 1, 2 (mod 5), or a > 1 and p 3 (mod 5), 4p a /5 ( ) ( ) 2 5 ( 1) p a (mod p 2 ). If p a 1, 3 (mod 5), or a > 1 and p 2 (mod 5), then 3p a /5 ( ) ( ) 2 5 ( 1) p a (mod p 2 ). (ii) If p 1, 7 (mod 10) or a > 2, then 7p a /10 ( 2 ) ( ) 5 ( 16) p a (mod p 2 ). If p 1, 3 (mod 10) or a > 2, then ) 9p a /10 ( 2 ( 16) ( 5 p a ) (mod p 2 ). 12 / 34

13 A conjecture on 5-adic valuations For a prime p the p-adic valuation ν p (m) of an integer m refers to sup{a N : p a m}. Conjecture (Z. W. Sun). For each n = 1, 2, 3,..., we have ( n 1 ( )) ( ( )) 2 2n ν 5 F 2+1 ν 5 (2n + 1)n 2. n Also, if a, b Z + and a b then the sum 5 1 a 1 ( ) 2 5 2a F 2+1 modulo 5 b only depends on b. 13 / 34

14 Central binomial coefficients and 3-adic valuations Theorem (Strauss, Shallit and Zagier, 1992) ( n 1 ( )) (( )) 2 2n ν 3 = 2ν 3 (n) + ν 3 n for n = 1, 2, 3,.... Theorem (Sun and Tauraso [Int. J. Number Theory 7(2011)]) Let p > 3 be a prime and let a be a positive integer. Then p a 1 ( ) 2 + d ( p a d Remar. Note that ( 2 are given by ( ) n 3 ) +2p a 0<<d ( 1) 1 ( ) d 3 ) = ( 2,), where multi-nomial coefficients 1,..., n = ( n )!. 1! n! (mod p 2 ). 14 / 34

15 On p-adic valuations Conjecture (Z. W. Sun [Acta Arith. 148(2011)]). For any odd prime p and positive integer n we have ( n 1 ( )) ( ( )) (p 1) 2n ν p ν p n.,..., n Remar. I d lie to offer $200 for the first rigorous proof. Theorem (Z. W. Sun [Acta Arith. 148(2011)]). For any prime p we have p 1 ( ) (p 1) pb p 1 + ( 1) p 1 2p (mod p 2 ),,..., where B 0, B 1, B 2,... are Bernoulli numbers. Also, an integer n > 1 is a prime if and only if n 1 ( ) (n 1) 0 (mod n).,..., 15 / 34

16 A conjecture on p 1 =1 ( 2 ) / 3 mod p 4 Conjecture (Z. W. Sun, Sci. China Math., in press). For any prime p > 7 we have p 1 =1 ( 2 ) 3 2 p 2 H p 1 13 p 1 27 =1 1 3 (mod p4 ), where H n denotes the harmonic number n =1 1/. Remar. It is nown that H p 1 /p 2 B p 3 /3 (mod p) for any prime p > 3 and p 1 =1 1/3 6 5 p2 B p 5 (mod p 3 ) for each prime p > 5. Motivation. =1 1 4( ) = ζ(4). Progress. (K. Hessami Pilehrood and T. Hessami Pilehrood, 2011) The conjectural congruence holds mod p. 16 / 34

17 On (p 1)/2 =1 ( 1) /( 3( ) 2 ) mod p Conjecture (Z. W. Sun, Sci. China Math., in press). For any prime p > 5 we have Motivation. (p 1)/2 =1 ( 1) 3( ) 2B 2 p 3 (mod p). ( 1) 3( ) = ζ(3). =1 Related result (Sun [Sci. China Math. 53(2010)]): For any prime p > 3 we have (p 1)/2 =1 1 2( ) ( 1) (p 1)/ E p 3 (mod p), where E 0, E 1, E 2,... are Euler numbers. (Compare this with the identity =1 1/(2( ) 2 ) = π 2 / / 34

18 On divisibility of binomial coefficients Catalan numbers: C n = 1 ( ) 2n = n + 1 n ( ) ( ) 2n 2n (n = 0, 1, 2,...). n n + 1 An Observation of Sun. Let and l be positive integers. If all prime factors of divide l, then ln + 1 ( ) n+ln ln for every n = 0, 1, 2,.... Conjecture (Sun, 2010). Let and l be positive integers. If ( ) ln + 1 n + ln ln for all sufficiently large positive integers n, then each prime factor of divides l. 18 / 34

19 Part C. Series for powers of π and L(2, ( 3 )) 19 / 34

20 Fast convergent series for L(2, ( 3 )) With a Dirichlet character χ, the series L(s, χ) = converges very slow. n=1 χ(n) n s (Re(s) > 1). Conjecture (Z. W. Sun, 2010). Set ( ( )) ( 3 K = L 2, = ) 3 2 = Then =1 (15 4)( 27) 1 =1 3( ) 2 2 ( 3 ) = K, (1) (5 1)( 144) =1 3( ) 2 2 ( 4 ) = 45 K. 2 2 (2) Progress. (1) was recently confirmed by K. Hessami Pilehrood and T. Hessami Pilehrood (arxiv: ) via the Hurwitz ζ(s, x). 20 / 34

21 Fast covergent series for π 2 Conjecture (Z. W. Sun, 2010). We have (10 3)8 =1 3( ) 2 2 ( 3 ) = π2 2, (1) (11 3)64 =1 =1 ) 2 ( 3 3( 2 (35 8)81 3( 2 ) 2 ( 4 2 ) =8π 2, (2) ) =12π 2. (3) Progress. (2) has been confirmed by J. Guillera via applying a Barnes-integrals strategy of the WZ-method. Theorem (Sun, 2010). We have 2 n H (2) n 1 n ( ) = π3 2n 48 n n=1 n 1, where H(2) n 1 = = / 34

22 Conjectural series for ζ(3) and 1/π 2 Conjecture (Z. W. Sun). We have ( )( 64) =1 5( ) 2 4 ( 3 ) = 14ζ(3), ( ) 18n 2 ( ) + 7n + 1 2n 2 n ( ) 1/4 2 ( ) 3/4 2 ( 128) n = 4 2 n n π 2, n=0 n=0 40n n + 5 ( 256) n ( 2n n ) 2 n ( n ) 2 ( 2 )( 2(n ) n ) = 24 π 2. Related p-adic congruences of ( ): For any odd prime p we have p ( ) ( ) 3 ( 64) 3p p5 B p 3 (mod p 6 ), (p 1)/ ( 64) ( 2 ) 4 ( ) 3 3p 2 +6 ( ) 1 p 4 E p 3 (mod p 5 ). p 22 / 34

23 Conjectural series for 1/π I have made over 100 conjectural series for 1/π of the form (b + c) a = C m π, where a, b, c, m Z, bm is nonzero, and C 2 is rational. Such series are closely related to the theory of modular forms, and the full list of my conjectural series is available from arxiv: Examples of my conjectural series: ( )( ) a = , 2 π where a denotes the coefficient of x in (x x + 1) ; ( ) P (8) = 520 π, ( ) P (18) = π, where P n (x) = n ( n ) 2 ( 2 n ) ( 1) x 2 n. $520 for a proof of ( ). 23 / 34 ( )

24 Part D. On x 2 mod p 2 with p = x 2 + dy 2 24 / 34

25 On p = x 2 + 7y 2 and 4p = x y 2 Conjecture (Z. W. Sun [JNT 131(2011)]). Let p be an odd prime. Then p 1 ( ) 2 3 { 4x 2 2p (mod p 2 ) if ( p 7 ) = 1 & p = x 2 + 7y 2 (x, y Z), 0 (mod p 2 ) if ( p 7 ) = 1, i.e., p 3, 5, 6 (mod 7). Remar. Recently Z. H. Sun confirmed the congruence in the case ( p 7 ) = 1 via Legendre polynomials. Conjecture (Z. W. Sun [JNT 131(2011)]). Let p be an odd prime. Then p 1 ( 2 ) 2 ( 3 ) 64 { x 2 2p (mod p 2 ) if ( p 11 ) = 1 & 4p = x y 2 (x, y Z), 0 (mod p 2 ) if ( p 11 ) = 1, i.e., p 2, 6, 7, 8, 10 (mod 11). 25 / 34

26 Apéry numbers and polynomials Define Apéry polynomials by n ( ) n 2 ( ) n + 2 A n (x) := x (n = 0, 1, 2,...). Those Apéry numbers A n = A n (1) play important roles in Apéry s proof of the irrationality of ζ(3). Theorem (Z. W. Sun, 2011) Let p be an odd prime. Then p 1 p 1 ( ) 1 ( 1) A ( 2) ( 1) A 4 { 4x 2 2p (mod p 2 ) if p = x 2 + y 2 (2 x), 0 (mod p 2 ) if p 3 (mod 4). Conjecture (Z. W. Sun, 2010). For any odd prime p, we have { p 1 4x 2 2p (mod p 2 ) if p = x 2 + 2y 2, A 0 (mod p 2 ) if p 5, 7 (mod 8). 26 / 34

27 On congruences involving D (x) 3 D n := n ( )( ) n n + = n ( )( ) n In combinatorics, D n is the number of lattice paths from (0, 0) to (n, n) with steps (1, 0), (0, 1) and (1, 1). We define D n (x) := n ( n )( n + ) x. P n (x) = D n ((x 1)/2) is the Legendre polynomial of degree n. Conjecture (Sun [JNT, in press]). Let p be an odd prime. Then ( ) p 1 { 1 D ( 3) 3 4x 2 2p (mod p 2 ) if p = x 2 + 3y 2, p 0 (mod p 2 ) if p 2 (mod 3); p 1 4x 2 2p (mod p 2 ) if p = x y 2, D (3) 3 12x 2 2p (mod p 2 ) if p = 3x 2 + 5y 2, 0 (mod p 2 ) if ( p 15 ) = / 34

28 On S n (x) = n ( n ) x Define polynomials S n (x) := n ( ) n 4 x (n = 0, 1, 2,...). Conjecture (Sun). Let p be an odd prime. Then p 1 S (12) ( 1) [3 x] (4x 2 2p) (mod p 2 ) if p = x 2 + y 2 (2 x, 3 xy), ( xy 3 )4xy (mod p2 ) if p = x 2 + y 2 (2 x, 3 xy), 0 (mod p 2 ) if p 3 (mod 4). p 1 (4 + 3)S (12) p ( ( )) 3 p (mod p 2 ). 1 n 1 (4 + 3)S (12) Z for all n = 1, 2, 3,.... n 28 / 34

29 On p = x 2 + 5y 2 (Q( 5) has class number 2) Conjecture (Sun). Let p be an odd prime. Then p 1 S ( 4) S ( 64) p 1 4x 2 2p (mod p 2 ) if p 1, 9 (mod 20) & p = x 2 + 5y 2, 2x 2 2p (mod p 2 ) if p 3, 7 (mod 20) & 2p = x 2 + 5y 2, 0 (mod p 2 ) if ( 5 p ) = 1. And p 1 ( p ) ( ( )) 1 (8 + 7)S ( 64) p p Moreover, (mod p 2 ). 1 n 1 (8 + 7)S ( 64) Z for all n = 1, 2, 3,.... n 29 / 34

30 On p = x y 2 (Q( 30) has class number 4) Conjecture (Sun). Let p be an odd prime. Then p 1 S (36) 4x 2 2p (mod p 2 ) if ( 2 p ) = ( p 3 ) = ( p 5 ) = 1, p = x y 2, 12x 2 2p (mod p 2 ) if ( p 3 ) = 1, ( 2 p ) = ( p 5 ) = 1, p = 3x y 2, 8x 2 2p (mod p 2 ) if ( 2 p ) = 1, ( p 3 ) = ( p 5 ) = 1, p = 2x y 2, 2p 20x 2 (mod p 2 ) if ( p 5 ) = 1, ( 2 p ) = ( p 3 ) = 1, p = 5x 2 + 6y 2, 0 (mod p 2 ) if ( 30 p ) = 1. p 1 ( p ) ( ( )) 6 (8 + 7)S (36) p (mod p 2 ). 15 p 1 n 1 (8 + 7)S (36) Z for all n = 1, 2, 3,.... n Remar. I would lie to offer $300 for the first correct proof. 30 / 34

31 On p = x y 2 (Q( 190) has class number 4) Conjecture (Sun) Let p be an odd prime. Then p 1 S (5776) 4x 2 2p (mod p 2 ) if, ( 2 p ) = ( p 5 ) = ( p 19 ) = 1, p = x y 2, 8x 2 2p (mod p 2 ) if ( 2 p ) = 1, ( p 5 ) = ( p 19 ) = 1, p = 2x y 2, 2p 20x 2 (mod p 2 ) if ( p 19 ) = 1, ( 2 p ) = ( p 5 ) = 1, p = 5x y 2, p 1 2p 40x 2 (mod p 2 ) if ( p 5 ) = 1, ( 2 p ) = ( p 19 ) = 1, p = 10x y 2 0 (mod p 2 ) if ( 190 p ) = 1. ( )S (5776) p Moreover, ( p ( ( p ) 19)) 1 n 1 ( )S (5776) Z for all n = 1, 2, 3,.... n (mod p 2 ). 31 / 34

32 On p = x y 2 Conjecture (Sun, 2011). p 1 ( 2n ) n ( )( )( ) n n n n ( 324) n 2 n=0 4x 2 2p (mod p 2 ) if p = x y 2 (x, y Z), 2x 2 2p (mod p 2 ) 2p 12x 2 (mod p 2 ) 2p 6x 2 (mod p 2 ) 20x 2 2p (mod p 2 ) 10x 2 2p (mod p 2 ) 28x 2 2p (mod p 2 ) if 2p = x y 2 (x, y Z), if p = 3x y 2 (x, y Z), if 2p = 3x y 2 (x, y Z), if p = 5x y 2 (x, y Z), if 2p = 5x y 2 (x, y Z), if p = 7x y 2 (x, y Z), 14x 2 2p (mod p 2 ) if 2p = 7x y 2 (x, y Z), 0 (mod p 2 ) if ( 105 p ) = 1. Remar. The quadratic field Q( 105) has class number eight. I d lie to offer $1050 for the first correct proof of the conjecture. 32 / 34

33 Related series for 1/π Conjecture (Z. W. Sun, 2011). We have n=0 357n n ( 2n n ) n ( n For any prime p > 5, we have )( n )( ) 2 ( 324) n = 90 π. p 1 ( ) n ( )( )( ) 357n n n n n ( 324) n n 2 n=0 ( ) 1 ( ( p p (mod p p 15)) 2 ). For my general philosophy about connections between series for 1/π and x 2 mod p 2 with p = x 2 + dy 2, see the survey Z. W. Sun, Conjectures and results on x 2 mod p 2 with 4p = x 2 + dy 2, arxiv: / 34

34 Than you! You are welcome to solve my conjectures! 34 / 34