Conjectures and Results on Super Congruences

Transcription

1 Conjectures and Results on Suer Congruences Zhi-Wei Sun Nanjing University Nanjing , P. R. China htt://math.nju.edu.cn/ zwsun Feb. 8, 2010

2 Part A. Previous Wor by Others

3 What are suer congruences? A suer congruence is a -adic congruence which haens to hold not just modulo a rime but a higher ower of. Examle. (Wolstenholme For any rime > 3 we have =1 1 0 (mod 2 and ( 2 1 = 1 ( 2 1 (mod Remar. It is easy to see that (/2 1 = ( = =1 and ( 2 1 = 1 =1 =1 + = =1 (/2 =1 ( (mod ( 1 + =1 1 (mod 2.

4 Suer Congruences for Aéry Numbers In 1978 Aéry roved that ζ(3 = n=1 1/n3 is irrational! During his roof he used the sequence {B(n/A(n} n=1 of rational numbers to aroximate ζ(3, where A(0 = 1, A(1 = 5, B(0 = 0, B(1 = 6, and both {A(n} n 0 and {B(n} n 0 satisfy the recurrence (n u n+1 = (2n + 1(17n n + 5u n n 3 u n 1 (n = 1, 2,.... In fact, A(n = n =0 ( n 2 ( n + and these numbers are called Aéry numbers. Dedeind eta function in the theory of modular forms: 2 η(τ = q 1/24 (1 q n with q = e 2πiτ n=1 Note that q < 1 if τ H = {z C : Im(z > 0}.

5 Beuers Conjecture (1985. For any rime > 3 we have the suer congruence ( 1 A a( (mod 2, 2 where a(n (n = 1, 2, 3,... are given by η 4 (2τη 4 (4τ = q (1 q 2n 4 (1 q 4n 4 = n=1 a(nq n. n=1 S. Ahlgren and Ken Ono [J. Reine Angew. Math. 518(2000]: The Beuers conjecture is true! Outline of their roof. First show that a( can be exressed as a secial value of the Gauss hyergeometric function 4 F 3 (λ defined in terms of Jacobi sums. Then rewrite Jacobi sums in terms of Gauss sums and aly the Gross-Koblitz formula to exress Gauss sums in terms of the -adic Gamma function Γ (x. Finally use combinatorial roerties of Γ (x and some sohisticated combinatorial identities involving harmonic numbers H n = 0< n 1/.

6 Gaussian hyergeometric series The rising factorial (or Pochhammer symbol: Note that (1 n = n!. (a n = a(a + 1 (a + n 1 = Classical Gaussian hyergeometric series: Γ(a + n. Γ(a r+1f r (α 0,..., α r ; β 1,..., β r x = n=0 (α 0 n (α 1 n (α r n x n (β 1 n (β r n n!, where 0 α 0 α 1 α r < 1 and 0 β 1 β r < 1.

7 Greene s character sum analogue of hyergeometric series Let be an odd rime. For two Dirichlet characters A and B mod, define the normalized Jacobi sum ( A := B( 1 J(A, B = B( 1 A(x B(1 x. B x=0 Given Dirichlet characters A 0, A 1,..., A r and B 1,..., B r mod, Greene [Trans. AMS, 1987] defined r+1f r (A 0, A 1,..., A r ; B 1,..., B r x = ( r ( A0 χ Ai χ χ(x. 1 χ B i χ χ i=1 r+1f r (A 0, A 1,..., A r ; B 1,..., B r x = A r B r ( 1 r F r 1 (A 0, A 1,..., A r 1 ; B 1,..., B r 1 xy y=0 A r (yār B r (1 y (Greene, 1987.

8 Connections to ellitic curves over F Let ε be the trivial character with ε (x = 1 for all x 0 (mod. Let φ be the Legendre character given by φ (x = ( x. Set r+1f r (x := r+1 F r (φ, φ,..., φ ; ε,..., ε x. Consider the curve over Q defined by 2E 1 (λ : y 2 = x(x 1(x λ. If λ Q \ {0, 1}, then 2 E 1 (λ is an ellitic curve with ( 2 E 1 (λ = 16λ 2 (λ 1 2 (discriminant j( 2 E 1 (λ = 256(λ2 λ λ 2 (λ 1 2 (j invariant.

9 Connections to ellitic curves over F Suose λ(λ 1 0 (mod. Then is a rime of good reduction for 2 E 1 (λ, and we define 2a 1 (; λ := E 1 (λ where 2 E 1 (λ denotes the number of F -oints of 2 E 1 (λ including the oint at infinity. It is nown that and 2a 1 (; λ = φ (x(x 1(x λ x=0 2F 1 (λ = φ ( 1 2a 1 (; λ = ( 1 ( x(x 1(x λ. x=0

10 Conjectures of Rodriguez-Villegas In 2001 Rodriguez-Villegas conjectured 22 congruences which relate truncated hyergeometric series to the number of F -oints of some family of Calabi-Yau manifolds. Here we list some of them. ( 2 2 ( 1 16 = ( 1 (/2 (mod 2, =0 ( 2 ( 3 ( 27 (mod 2, 3 =0 ( 2 ( 4 ( 2 2 (mod 2, =0 =0 ( 6 3 =0 64 ( ( 2 3 ( 1 64 [q ]q (mod 2, (1 q 4n 6 (mod 2. n=1

11 A theorem of Stienstra and Beuers J. Stienstra and F. Beuers [Math. Ann. 27(1985]: [q ]q (1 q 4n 6 = = n=1 { 4x 2 2 if = 1 (mod 4 & = x 2 + y 2 with 2 x & 2 y, 0 if 3 (mod 4; [q ]q = (1 q n 2 (1 q 2n (1 q 4n (1 q 8n 2 n=1 { 4x 2 2 if = 1, 3 (mod 8 & = x 2 + 2y 2 with x, y Z, 0 if 5, 7 (mod 8; [q ]q (1 q 2n 3 (1 q 6n 3 n=1 { 4x 2 2 if = 1 (mod 3 & = x 2 + 3y 2 with x, y Z, 0 if 2 (mod 3.

12 Progress on Rodriguez-Villegas conjectures The congruences we list have been confirmed, see, E. Motenson, J. Number Theory 99(2003; Trans. AMS 355(2003; Proc. AMS 133(2005. Many of the 22 conjectures remain oen.

13 Ramanujan s series for 1/π Here are 5 of the 17 Ramanujan series recorded by him in 1914: =0 ( 2 3 ( 1 (4 + 1 (1/23 (1 3 = (4 + 1 ( 64 = 2 π, = ( 4 =0 =0 = (1/23 ( ( 8 (1/23 ( (1/23 (1 3 (1/2 (1/4 (3/4 (1 3 =0 ( 2 3 = ( = 4 π, =0 ( 2 3 = (6 + 1 ( 512 = 2 2 π, =0 ( 2 3 = ( = 16 π, =0 ( 4,,, = ( ( 1024 = 8 π. =0 Remar. The first one was actually roved by G. Bauer in 1859.

14 Hamme s Conjectures L. Van Hamme [1997] conjectured the -adic analogues of the above first 4 identities and W. Zudilin [JNT, 2009] obtained the -adic analogue of the last identity. (/2 =0 (/2 =0 (/2 =0 ( 2 3 (4 + 1 ( 64 ( 1 (mod 3, ( 2 3 ( 1 ( (mod 4, ( 2 3 ( 2 (6 + 1 ( 512 (mod 3, (/2 ( 2 3 ( =0 ( 4,,, ( ( 1024 =0 ( 1 5 (mod 4, ( 1 3 (mod 3.

15 Progress on Hamme s conjectures The first of the above congruence was roved by E. Mortenson [Proc. AMS 136(2008] and the second one was recently shown by Ling Long, while the last was confirmed by Zudilin via the WZ method. The third and the fourth remain oen. The -adic Gamma function lays an imortant role in Hamme s formulation of those conjectures. It is defined in the following way: Γ (n := ( 1 n (n = 1, 2, 3,... 1<<n n and Γ (x = lim n x Γ (n for any adic integer x.

16 Part B. My Results and Conjectures

17 Some Joint Wor H. Pan and Z. W. Sun [Discrete Math. 2006]. ( ( 2 d (mod (d = 0,...,, + d 3 =0 =1 ( 2 0 (mod for > 3. Sun & R. Tauraso [arxiv: , Adv. in Al. Math.]. a 1 ( ( 2 a (mod 2, 3 =0 =1 ( B 3 (mod 3 for > 3, L. L. Zhao, H. Pan and Z. W. Sun [Proc. AMS, 2010] 2 ( 3 0 (mod.

18 My own results Observe that if /2 < < then ( 2 = (2! 0 (mod (! 2 and so =0 ( 2 (/2 m =0 where m is an integer with m. ( 2 m (mod, In 2009 I [arxiv: , arxiv: , ] determined =0 ( 2 m mod 2, (/2 =0 in terms of linear recurrences. ( 2 m mod 2, =0 ( 3 m mod

19 Some articular congruences due to me =0 =0 ( 2 2 ( 1 ( 2 ( 3 3 ( 2 ( 1 =0 (/2 =0 (mod 2, (mod 2, (/2 =0 (/2 =0 ( 2 ( ( 2 ( 3 ( (1 2F 5 ( 5 (mod 2, ( 2 ( ( ( F ( 5 where {F n } n 0 is the Fibonacci sequence defined by (mod 2, (mod 2. (mod 2, F 0 = 0, F 1 = 1, F n+1 = F n + F n 1 (n = 1, 2, 3,....

20 Some articular congruences due to me =0 =0 =0 ( (( 4 5 ( 3 1 (mod, { 2 (mod if ±2 (mod 7, 7 1 (mod otherwise. ( 4 1 (mod if 1 (mod 5 & 11, 5 1/11 (mod if 2, 3 (mod 5, 9/11 (mod if 4 (mod 5. If 1 (mod 3 then =0 ( (/3 (mod.

21 My recent results on suer congruences Recall that Euler numbers E 0, E 1,... are given by E 0 = 1, ( n E n = 0 (n = 1, 2, 3, It is nown that E 1 = E 3 = E 5 = = 0 and sec x = ( 1 n x 2n ( E 2n x < π. (2n! 2 n=0 Z. W. Sun [arxiv: ]. =0 (/2 =0 ( 2 2 ( 1 (/2 2 E 3 (mod 3, ( 2 8 ( 2 + ( E 3 (mod 3..

22 My recent results on suer congruences (/2 =1 (/2 =0 (/2 =0 ( 2 (/2 2 ( ( 1 =1 1 2( 2 ( 1 (mod 2, (/2 3 8 =1 ( ( 1(+1/ (2 1 =0 + ( 1 (/2 3 (/2 32 =1 =1 ( 2 ( 2 (mod 4, (mod 4. Remar. Via some advanced tools, R. Osburn and C. Schneider [Math. Com. 78(2009] roved that (/2 ( 2 2 ( 1 (/ (/2 ( 8 2 (mod 3.

23 Conjecture 1 I have formulated over 50 conjectures on suer congruences, see Z. W. Sun, Oen Conjectures on Congruences, arxiv: Conjecture 1. For any rime > 3 we have (/2 =1 (/2 =1 (/2 =0 /2<< /2<< ( 2 ( 1 (+1/2 8 3 E 3 (mod 2, 1 2( ( 1 (/ E 3 (mod, ( ( 1 (/2 + 2 E 3 (mod 3, ( E 3 (mod 3, 2 ( E 3 (mod 3.

24 A Remar to Conjecture 1 Remar. I have roved that the first, the second and the third congruences are equivalent. Note that lim + ( = 1 π and =0 1 2( = π

25 Conjecture 2 Let > 3 be a rime. Then =0 ( { 2 3 4x 2 2 (mod 2 if ( 7 = 1 & = x 2 + 7y 2, 0 (mod 2 if ( 7 = 1. Moreover, and (/2 =0 ( 2 3 ( (mod 4 =0 ( 2 3 ( ( E 3 (mod 4. Remar. Quite recently M. Jameson and K. Ono claimed to have a roof of the first congruence via the theory of modular forms and K 3 surfaces related to Calabi-Yau manifolds.

26 Conjecture 3 Let be an odd rime. Then =0 ( 2 2 ( 3 64 { x 2 2 (mod 2 if ( 11 = 1 & 4 = x y 2 (x, y Z, 0 (mod 2 if ( 11 = 1, i.e., 2, 6, 7, 8, 10 (mod 11. Furthermore, ( =0 ( 2 2 ( (mod 4. Remar. It is well-nown that the quadratic field Q( 11 has class number one and hence for any odd rime with ( 11 = 1 we can write 4 = x y 2 with x, y Z.

27 Conjecture 4 Let > 3 be a rime. If 1 (mod 3 and = x 2 + 3y 2 with x, y Z, then =0 ( x 2 2 (mod 2 and If 2 (mod 3, then =0 ( 2 Furthermore, and (/2 = (mod 2 and =0 =0 =0 ( 2 3 ( (mod 4 (3 + 1 ( x 2 3 (mod 2. ( (mod 2. ( 2 3 ( E 3 (mod 4.

28 Conjecture 5 Let > 3 be a rime. Then =0 ( 3,, 24 =0 =0 ( 2 ( 3,, ( 27 ( 3 {( 2(/3,, ( 216 (/3 (mod 2 if 1 (mod 6, 0 (mod if 5 (mod 6. 4x 2 2 (mod 2 if 1, 4 (mod 15 & = x y 2, 20x 2 2 (mod 2 if 2, 8 (mod 15 & = 5x 2 + 3y 2, 0 (mod 2 if ( 15 = 1. = ( 27 ( 2 2 ( 3 ( 4 3 (mod 3.

29 Conjecture 6 Let > 5 be a rime. Then ( 4,,, ( 2 10 =0 4x 2 2 (mod 2 if 1, 9 (mod 20 & = x 2 + 5y 2, 2( x 2 (mod 2 if 3, 7 (mod 20 & 2 = x 2 + 5y 2, 0 (mod 2 if 11, 13, 17, 19 (mod 20. =0 ( 2 2 ( x 2 2 (mod 2 if 1, 7 (mod 24 & 4 = x 2 + 6y 2, 2x 2 2 (mod 2 if 5, 11 (mod 24 & 2 = x 2 + 6y 2, 0 (mod 2 if 13, 17, 19, 23 (mod 24. ( 2 2 ( 3 ( ( (mod 3. 3 =0

30 Conjecture 7 Let > 3 be a rime. If ( 7 = 1 and = x 2 + 7y 2 with ( x 7 = 1, then ( 2 ( 4 ( 2 ( 63 2x (mod 2 3 2x and =0 =0 ( 2 If ( 7 = 1, then =0 ( 2 ( 4 ( ( (mod and ( 2x x =0 ( 2 (mod 2. ( (mod.

31 Conjecture 8 Let be an odd rime. Then (3 + 1 ( 8 =0 ( 2 If 1 (mod 4, then =0 If 3 (mod 4, then and =0 3 ( E 3 (mod 4. ( 2 3 ( ( ( 8 0 (mod 3. =0 ( 2 3 ( 8 0 (mod 2 ( 2 2 ( ( 2 0 (mod 3.

32 Conjecture 9 Let > 5 be a rime. If 1 (mod 4, then =0 If 3 (mod 4. Then and 3( (mod 2. ( ( 2 1 ( 8 0 (mod 2, =0 2 ( ( 2 1 ( 64 0 (mod 2. =0 Remar. The last congruence is equivalent to =0 ( H 0 (mod with H = 0<j 1 j.

33 Conjecture 10 Let > 3 be a rime. If 7 (mod 12 and = x 2 + 3y 2 with y 1 (mod 4, then and =0 ( ( 2 2 ( 3 ( 16 ( 1( 3/4 4y 3y =0 If 1 (mod 12, then (mod 2 ( ( ( 16 ( 1(+1/4 y (mod 2. ( 1 =0 ( 3 ( (mod 2.

34 Recall that the Pell sequence {P n } n 0 is given by P 0 = 0, P 1 = 1, and P n+1 = 2P n + P n 1 (n = 1, 2, 3,.... Conjecture 11 Let be an odd rime. If 1, 3 (mod 8 and = x 2 + 2y 2 with x, y Z and x 1, 3 (mod 8, then =0 P ( 8 =0 ( { (mod 2 if 8 1, ( 1 ( 3/8 ( 2x 2x (mod 2 if 8 3. P ( 8 We also have =0 P ( 8 ( 1 =0 ( 2 2 ( 1(x+1/2 ( x + 2 2x P (mod 2. ( (mod if 5 (mod 8, 8 ( (mod 2 if 7 (mod 8.

35 Define the sequence {S n } n 0 is given by S 0 = 0, S 1 = 1, and S n+1 = 4S n S n 1 (n = 1, 2, 3,.... Conjecture 12. Let > 3 be a rime. If 7 (mod 12 and = x 2 + 3y 2 with y 1 (mod 4, then ( S 2 2 ( 4 ( 1 (+1/4 4y (mod 2 3y and =0 =0 S 4 We also have =0 S 4 ( 2 2 ( ( 1 ( 3/4 6y 7 3y (mod 2. ( { (mod 2 if 1 (mod 12, 0 (mod if 2 (mod 3. If ±1 (mod 12, then ( ( 1 2 ( 1 S ( 1 (/2 S (mod 3.

36 Conjecture 13. For any n Z + we have 6 (mod 25 if n 0 (mod 5, ( 1 n/5 1 n 1 ( 2 4 (mod 25 if n 1 (mod 5, (2n + 1n 2( F 2n 2+1 n 1 (mod 25 if n 2, 4 (mod 5, =0 9 (mod 25 if n 3 (mod 5. Also, if a, b Z + and a b then the sum 5 1 a 1 ( 2 5 2a F 2+1 =0 modulo 5 b only deends on b. Remar. I have roved that if 2, 5 is a rime then ( 2 ( ( F 2 ( 1 /5 1 (mod 2, 5 =0 =0 F 2+1 ( 2 ( ( 1 /5 5 (mod 2.

37 Conjecture 14. For any rime and ositive integer n we have and ( n 1 ( ( 1 ν ν (n,..., =0 ( n 1 ( ( n 1 ( 1 ν ( 1 ν (n,,..., =0 where ν (n = max{a N : a n} is the -adic order of n. Remar. I have roved that an integer > 1 is a rime if and only if ( ( 1 0 (mod.,..., =0 He also showed that if n Z + is a multile of a rime then n 1 ( ( 1 0 (mod.,..., =0

38 Conjectures 15 and 16 Conjecture 15(Sun and Tauraso [Adv. in Al. Math., in ress] Let 2, 5 be a rime and let a Z +. Then a 1 =0 ( 1 ( 2 ( a ( 1 2F 5 a ( a 5 where {F n } n 0 is the Fibonacci sequence. Remar. I have roved the congruence mod 2. (mod 3, Conjecture 16 Let be an odd rime. If 1, 2 (mod 5, then 4 5 =0 ( 1 ( 2 If 1, 3 (mod 5, then 3 5 =0 ( 1 ( 2 ( 5 ( 5 (mod 2. (mod 2.

39 Conjectures 17 and 18 Conjecture 17 Let be an odd rime and let a Z +. If 1 (mod 3 or a > 1, then 5 6 a =0 ( 2 ( 3 16 a (mod 2. Remar. The author [S09f] roved that /16 ( 3 (mod 2 for odd rime and a Z +. a a /2 =0 ( 2 Conjecture 18. For any nonnegative integer n we have ( 1 n 2 { (2n + 1 2( 2n 16 1 (mod 9 if 3 n, n 4 (mod 9 if 3 n. =0. Also, (3 1 a 1/2 ( 2 3 2a 16 ( 1a 10 (mod 27 =0 for every a = 1, 2, 3,....

40 Conjectures 19 and 20 Conjecture 19. Let be an odd rime. Then we have =1 2 ( 3 3 q 2 (2 (mod 2, where q (2 denotes the Fermat quotient (2 1/. Remar. Zhao, Pan and Sun [Proc. AMS, 2010] roved that 0 (mod for any odd rime. =1 2 ( 3 Conjecture 20. Let > 3 be a rime. Then ( 1 =0 ( 2 ( (( 1 ( 3 (3 1 (mod 3. 3 Remar. I have roved the congruence mod 2.

41 More Conjectures on Congruences For more conjectures of mine on congruences, see Z. W. Sun, Oen Conjectures on Congruences, arxiv: You are welcome to solve my conjectures!

42 Than you!