Some sophisticated congruences involving Fibonacci numbers

Transcription

1 A tal given at the National Center for Theoretical Sciences (Hsinchu, Taiwan; July 20, 2011 and Shanghai Jiaotong University (Nov. 4, 2011 Some sohisticated congruences involving Fibonacci numbers Zhi-Wei Sun Nanjing University Nanjing , P. R. China htt://math.nju.edu.cn/ zwsun Nov. 4, 2011

2 Abstract The well-nown Fibonacci numbers lay imortant roles in many areas of mathematics and they have very nice number-theoretic roerties. We will focus on some sohisticated congruences on Fibonacci numbers including the recent determination of F ( 5 modulo 3, where is an odd rime. We will also mention some conjectures related to 5-adic valuations for further research. 2 / 42

3 Part A. On Fibonacci quotients and Wall-Sun-Sun rimes 3 / 42

4 Fibonacci numbers and Lucas numbers The Fibonacci sequence {F n } n 0 is defined by F 0 = 0, F 1 = 1, F n+1 = F n + F n 1 (n = 1, 2, 3,.... Thus F 0 = 0, F 1 = F 2 = 1, F 3 = 2, F 4 = 3, F 5 = 5, F 6 = 8, F 7 = 13, F 8 = 21, F 9 = 34, F 10 = 55, F 11 = 89, F 12 = 144. The Lucas numbers L 0, L 1, L 2,... are given by L 0 = 2, L 1 = 1, L n+1 = L n + L n 1 (n = 1, 2, 3,.... Relations between Fibonacci numbers and Lucas numbers: L n = F n 1 + F n+1 and 5F n = L n 1 + L n+1. 4 / 42

5 More nown results on Fibonacci numbers An exlicit exression. F n = (n 1/2 =0 ( n 1 Y. Bugeaud, M. Mignotte and S. Sise [Ann. of Math. 163(2006]: The only owers in the Fibonacci sequence are. F 0 = 0, F 1 = F 2 = 1, F 6 = 2 3, and F 12 = Ke-Jian Wu and Z. W. Sun [Math. Com. 78(2009]: Let and a = M = Then, for any x a (mod M, the number x 2 cannot be written in the form F n /2 ± m with a rime and m, n N = {0, 1,...}. 5 / 42

6 Lucas sequences Let N = {0, 1, 2,...} and Z + = {1, 2, 3,...}. Fix A, B Z. The Lucas sequence u n = u n (A, B (n N and its comanion v n = v n (A, B (n N are defined as follows: u 0 = 0, u 1 = 1, and u n+1 = Au n Bu n 1 (n = 1, 2, 3,...; v 0 = 2, v 1 = A, and v n+1 = Av n Bv n 1 (n = 1, 2, 3,.... Examle : F n = u n (1, 1, L n = v n (1, 1; u n (2, 1 = n, v n (2, 1 = 2. Universal Formulae: Let = A 2 4B, and let α = A + 2 and β = A 2 be the two roots of the equation x 2 Ax + B = 0. Then u n = { α n β n α β n 1 α β if 0, = n(a/2 n 1 if = 0. 0 <n And v n = α n + β n. 6 / 42

7 Basic roerties of Lucas sequences As (α n β n (β n + β n = α 2n β 2n, we have u 2n = u n v n. Note also that vn 2 un 2 =(α n + β n 2 ((α βu n 2 =(α n + β n 2 (α n β n 2 = 4(αβ n = 4B n. Lucas Theorem. Let A, B Z with (A, B = 1. Then (u m, u n = u (m,n. In articular, m n u m u n. Examle. F 2n = F n L n, L 2 n 5Fn 2 = 4( 1 n, (F m, F n = F (m,n. 7 / 42

8 Legendre symbols Let be an odd rime and a Z. The Legendre symbol ( a is given by ( a 0 if a, = 1 if a and x 2 a (mod for some x Z, 1 if a and x 2 a (mod for no x Z. It is well nown that ( ab = ( a ( b for any a, b Z. Also, ( { 1 = ( 1 ( 1/2 1 if 1 (mod 4, = 1 if 1 (mod 4; ( 2 = ( 1 (2 1/8 = { 1 if ±1 (mod 8, 1 if ±3 (mod 8; ( { 5 ( 1 if ±1 (mod 5, = = 5 1 if ±2 (mod 5. 8 / 42

9 Congruence roerties of Lucas sequences Let be an odd rime. Then ( u (mod and v A In fact, (mod. v = α + β (α + β = A A (mod ; also u = (α β(α β (α β +1 = (+1/2 (mod and hence u ( (mod if. When = A2 4B 0 (mod, we have ( u u A, A2 = 4 ( A 1 2 ( (mod. 9 / 42

10 Congruence roerties of Lucas sequences Theorem. If is an odd rime not dividing B, then u ( 0 (mod. ( Proof. If, then u ( = u ( = 0 (mod. If ( = 1, then u ( = u +1 = Au + v A ( + A = 0 (mod. If ( = 1, then u and 2Bu 1 = Au v A ( = u 1 ( A = 0 (mod. 10 / 42

11 Congruence roerties of Lucas sequences The congruence u ( is actually an extension of Fermat s little theorem. If a 0, 1 (mod, then a 1 1 a 1 = u 1 (a+1, a = u ( (a+1 2 4a (a+1, a 0 (mod. Examle. If is an odd rime, then ( 5 ( F = (mod, L 1 5 (mod, and F ( 5 0 (mod. For an odd rime, we call the integer F ( 5 / a Fibonacci quotient. 11 / 42

12 D. D. Wall s question In 1960 D. D. Wall [Amer. Math. Monthly] studied Fibonacci numbers modulo a ositive integer systematically. Let be an odd rime and let n( be the smallest ositive integer n with F n. Can F n( be a multile of 2? Wall found no such rimes. One can show that any ositive integer d divides some ositive Fibonacci numbers. Let n(d be the smallest ositive integer such that d F n(d. Then Wall s question is whether n( n( 2 for any odd rime. 12 / 42

13 Zhi-Hong Sun and Zhi-Wei Sun s contribution Theorem (Zhi-Hong Sun and Zhi-Wei Sun [Acta Arith. 60(1992]. Let 2, 5 be a rime. (i We have n( = n( 2 2 F ( 5. (ii We have F ( (mod (mod 5 1 (mod. (iii If 2 F ( 5, then the first case of Fermat s last theorem holds for the exonent, i.e., there are no ositive integers x, y, z with xyz such that x + y = z. (iv If 1, 9 (mod 20 and = x 2 + 5y 2 with x, y Z, then F ( 1/4 4 xy. 13 / 42

14 Zhi-Hong Sun and Zhi-Wei Sun s contribution (v We can determine F (±1/2 and L (±1/2 mod in the following way: { 0 (mod if 1 (mod 4, F ( ( 5 /2 F (+( 5 /2 L ( ( 5 /2 L (+( 5 /2 2( 1 (+5/10 ( 5 5( 3/4 (mod if 3 (mod 4; { ( 1 (+5/10 ( 5 5( 1/4 (mod if 1 (mod 4, ( 1 (+5/10 ( 5 5( 3/4 (mod if 3 (mod 4; { 2( 1 (+5/10 ( 5 5( 1/4 (mod if 1 (mod 4, 0 (mod if 3 (mod 4; { ( 1 (+5/10 5 ( 1/4 (mod if 1 (mod 4, ( 1 (+5/10 ( 5 5(+1/4 (mod if 3 (mod 4. The theorem was obtained via exressing the sum r (mod 10 ( n in terms of Fibonacci and Lucas numbers. 14 / 42

15 Wall-Sun-Sun rimes R. Crandall, K. Dilcher and C. Pomerance [Math. Com. 66(1997] called those rimes satisfying F (/5 0 (mod 2 Wall-Sun-Sun rimes. Heuristically there should be infinitely many Wall-Sun-Sun rimes though they are very rare. Wall-Sun-Sun rimes were also introduced in many aers and boos including the famous boo R. E. Cradall and C. Pomerance, Prime Numbers: A Comutational Persective, Sringer, U to now no Wall-Sun-Sun rimes have been found. The current search record is due to F. G. Dorais and D. W. Klyve (2010: There are no Wall-Sun-Sun rimes below / 42

16 Connection to cyclotomic fields S. Jaubec [Math. Com. 67(1998]: Let = 2l (mod 8 and q be odd rimes with l a rime and 5 (mod q. Suose that the order of q modulo l is (l 1/2. If q divides the class number of the real cyclotomic field Q(ζ + ζ 1, then q must be a Wall-Sun-Sun rime. 16 / 42

17 Bernoulli olynomials and Fibonacci quotients Bernoulli numbers B 0, B 1, B 2,... are given by n ( n + 1 B 0 = 1, B = 0 (n = 1, 2, 3,.... =0 Bernoulli olynomials are those n ( n B n (x = B x n (n = 0, 1, 2,.... =0 Theorem (A. Granville and Z. W. Sun [Pacific J. Math. 1996]. Let 2, 5 be a rime. Then ( a ( a 5 B 1 B F ( q (5 (mod for a = 1, 2, 3, 4, and for a = 1, 3, 7, 9 we have ( a ( a 15 B 1 B 1 F ( q (5+2q (2 (mod, where q (m denotes the Fermat quotient (m 1 1/. 17 / 42

18 Part B. On F ( 5 mod 3 and some suer congruences involving Fibonacci numbers 18 / 42

19 A curious identity and its consequence H. Pan and Z. W. Sun [Discrete Math. 2006]. If l, m, n {0, 1, 2,...} then = l ( ( ( l m ( 1 m 2 n 2l + m =0 l =0 ( l ( ( 2 n n l m + n 3 l On the basis of this identity, for d, r {0, 1, 2,...} the authors constructed exlicit F (d, r and G(d, r such that for any rime > max{d, r} we have 1 r C +d. { F (d, r (mod if 1 (mod 3, G(d, r (mod if 2 (mod 3, where C n denotes the Catalan number 1 n+1( 2n n. 19 / 42

20 Some congruences involving central binomial coefficients Let be an odd rime. H. Pan and Z. W. Sun [Discrete Math. 2006]. 1 ( ( 2 d (mod (d = 0,...,, + d 3 =0 1 ( 2 0 (mod for > 3. Sun & R. Tauraso [AAM 45(2010; IJNT 7(2011]. a 1 ( ( 2 a (mod 2, 3 =0 1 1 ( 1 ( 2 ( B 3 (mod 3 for > 3, 5 F ( 5 (mod ( / 42

21 Some auxiliary identities Sun and Tauraso [Adv. in Al. Math. 2010]: Let n Z + and d N. Then ( ( 2 d + = ( ( 2n n + d, + d <n 0 <n+d 0 <n and d ( 2 ( 1 +d + F 2d = + d 0<<n 0 <n+d ( 2n ( 1 F 2(n+d, ( 1 +d ( 2 + ( 2n ( 1 L + d 2(n+d 0 <n+d ( 2n 1 = L 2d ( 1 n+d 2 δ d,0. n + d 1 21 / 42

22 On 1 =0 ( 2 /m mod 2 Z. W. Sun [Sci. China Math. 53(2010]: Let be an odd rime and let m Z with m. Then 1 =0 ( 2 m In articular, 1 =0 1 ( 2 ( 1 =0 ( m 2 4m + u ( m 2 4m (m 2, 1 (mod 2. ( 2 2 ( 1 ( 1/2 (mod 2, ( ( 1 2F 5 ( 5 (mod 2 ( 5. (Note that ( 1 n 1 u n ( 3, 1 = u n (3, 1 = F 2n = F n L n. 22 / 42

23 Two conjectures on Fibonacci and Lucas numbers Conjecture (Sun and Tauraso [Adv. in Al. Math. 2010] Let 2, 5 be a rime and let a Z +. Then a 1 =0 ( 1 ( 2 ( a ( 1 2F 5 a ( a 5 (mod 3. Conjecture (Roberto Tauraso, Jan For any rime > 5 we have 1 L 2 0 (mod. These two conjectures are very sohisticated and difficult to rove. Theorem (Hao Pan and Z. W. Sun, arxiv: The two conjectures are true! 23 / 42

24 Another congruence for F ( 5 mod 3 Theorem (Z. W. Sun, arxiv: Let 2, 5 be a rime and let a Z +. Then ( a 1/2 ( 2 ( ( a F ( a ( a 5 (mod =0 The roof is also very sohisticated and quite difficult. Below is a ey lemma. Lemma. Let 2, 5 be an odd rime. For any a Z + we have ( L a 1 a F a F 2 (mod 4. a ( a 5 If we set H (2 1 ( 2 ( 1 =0 = 0<j 1/j2 for = 0, 1, 2,..., then H (2 ( ( 2 F ( 5 (mod. 24 / 42

25 A further conjecture Conjecture (Sun, Let be an odd rime and let a Z +. (i If a 1, 2 (mod 5, or a > 1 and 3 (mod 5, 4 a /5 ( ( 2 5 ( 1 a (mod 2. =0 If a 1, 3 (mod 5, or a > 1 and 2 (mod 5, then 3 a /5 ( ( 2 5 ( 1 a (mod 2. =0 (ii If 1, 7 (mod 10 or a > 2, then 7 a /10 ( 2 ( 5 ( 16 a (mod 2. =0 If 1, 3 (mod 10 or a > 2, then 9 a /10 =0 ( 2 ( 16 ( 5 a (mod / 42

26 Some other congruences involving Fibonacci numbers Theorem (Sun. For any rime > 5, we have 1 ( F 2 0 (mod if ±1 (mod 5, 12 1 (mod if ±13 (mod 30, =0 1 (mod if ±7 (mod 30. Theorem (Sun. Let 2, 5 be a rime. Then 1 ( 2 ( ( F 2 ( 1 /5 1 (mod 2, 5 =0 1 =0 ( 1/2 =0 ( 1/2 =0 F 2+1 ( 2 F 2 16 F ( ( 1 /5 5 (mod 2, ( 2 ( 1 ( 1/2+ /5 (mod 2, ( 2 ( 1 ( 1/2+ /5 5 + ( 5 4 (mod / 42

27 A conjecture on 5-adic valuations Conjecture (Sun. For any n Z + the number s n := ( 1 n/5 1 (2n + 1n 2( 2n n n 1 =0 ( 2 F 2+1 is a 5-adic integer and furthermore 6 (mod 25 if n 0 (mod 5, 4 (mod 25 if n 1 (mod 5, s n 1 (mod 25 if n 2, 4 (mod 5, 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. 27 / 42

28 A conjecture on q-fibonacci numbers Recall that the usual q-analogue of n N is given by [n] q = 1 qn 1 q = 0 <n which tends to n as q 1. For any n, N with n, [ ] n 0<r n = [r] q ( 0<s [s] q( 0<t n [t] q q is a natural extension of the usual binomial coefficient ( n. A q-analogue of Fibonacci numbers introduced by I. Schur is defined as follows: F 0 (q = 0, F 1 (q = 1, and F n+1 (q = F n (q + q n F n 1 (q (n > 0. Conjecture (Sun Let a and m be ositive integers. Then we have the following congruence in the ring Z[q]: 5 a m 1 [ ] q 2(+1 2 F 2+1 (q 0 (mod [5 a ] 2 q. q =0 q 28 / 42

29 Four series involving Fibonacci and Lucas numbers In Oct Sun observed the following identities: =0 F 2 2( 2 ( 2 F2+1 = 4π2 25 5, ( = 2π 5 5, =0 L 2 2( 2 ( 2 = π2 5, L2+1 ( = 2π 5. In fact, they can be obtained by utting x = ( 5 ± 1/2 in the nown identities arcsin x ( 2 = 2 ( x 2+1 x 2 and ( ( = 2 arcsin 2 x 2 2. =0 Note that sin π 10 = and sin 3π = / 42

30 Corresonding conjectural congruences Conjecture (Sun, Let 2, 5 be a rime and set q := F ( 5 /. Then ( 3/2 =0 ( 3/2 =0 1 1 F 2 2( 2 L 2 2( 2 ( 2 F2+1 ( ( q q2 5 2 q 15 4 q2 (mod 2, ( ( ( 1(+1/2 5 (mod 2, ( 1 2 q q2 ( 2 ( L2+1 5 ( ( 1(+1/2 2 q q2 (mod 2, (mod 2. In Oct K. Hessami Pilehrood and T. Hessami Pilehrood [arxiv: ] roved the last two congruences. 30 / 42

31 A conjecture related to = x y 2 and = 3x 2 + 5y 2 Conjecture (Z. W. Sun, Set. 18, Let > 5 be a rime. (i If 1, 4 (mod 15 and = x y 2 (x, y Z with x 1 (mod 3, then 1 =0 1 =0 ( 2 ( 3 27 F 2 15 ( 2 ( 3 ( x 2x 27 L 4x x (mod 2 (mod 2, and 1 (3 + 2 =0 ( 2 ( 3 27 L 4x (mod / 42

32 A conjecture related to = x y 2 and = 3x 2 + 5y 2 (ii If 2, 8 (mod 15 and = 3x 2 + 5y 2 (x, y Z with y 1 (mod 3, then 1 =0 ( 2 ( 3 27 F 5y 4y (mod 2 and 1 =0 ( 2 ( 3 27 F 1 =0 ( 2 ( 3 27 L 4 3 y (mod 2. Remar. Sun has many other similar conjectures. 32 / 42

33 Part C. Proof of 1 L 0 (mod for any rime > / 42

34 Granville s wor Let be an odd rime. Glaisher roved that q ( (mod. A. Granville [Integers 4(2004] confirmed the following conjecture of L. Sula: 1 q ( (mod. Define q(x = x + (1 x 1 Granville showed that if > 3 then and G(x = 1 G(x G(1 x + x G(1 x 1 (mod, x 2. q(x 2 2x G(x 2(1 x G(1 x (mod. 34 / 42

35 A lemma and a roosition Combining the last two congruences we obtain Lemma. Let > 3 be a rime. Then ( x + (1 x (1 x 2x (1 x 1 2 (mod Proosition. Let A and B be nonzero integers, and let α and β be the two roots of the equation x 2 Ax + B = 0. Let be an odd rime not dividing AB. Then ( v (A, B A and ( v (A, B A 2 2A Aα 1 α 1 A 2 2β2 α A 2 1 2β2 α 2 ( B (mod, 2 A α B (mod / 42

36 Proof of the roosition By the lemma and Fermat s little theorem, ( 1 x + (A x A 2 A 2 ( (x/a + (1 x/a (1 x/a 2 2 ( x A 2 1 (1 A/x 2 (mod. Note that v (A, B = β + α = β + (A β and αβ = B. So we have ( v (A, B A 2 1 2A 2 (A β A 2 1 2β 2 (1 Aα/B 2 (mod and hence the first desired congruence holds since Aα B = α / 42

37 Proof of the roosition (continued On the other hand, α (A v (A, B = α (A α β = (B +α 2 +( α 2 B and hence 2 ( A α 2 v (A, B ( ( α 2 + (B ( α 2 B = 1 2B 2 1 = 2B 2 1 2(αβ 2 (1 ( α 2 /B 2 2 2( α 2 2 (Aα B 2 1 (Aα 2α4 α 2 2 (Aα B 2 1 2Aα3 1 (1 B/( α 2 2 α A (mod. ( 2 Therefore the second desired congruence follows. 37 / 42

38 Proof of 1 L / 2 0 (mod Let > 5 be a rime. We rove 1 L / 2 0 (mod. Let α and β be the two roots of the equation x 2 x 1 = 0. Alying the Proosition with A = 1 and B = 1, we get ( L α 2 1 α 2 2β2 (mod, (1 2 ( L α α 2 1 ( α 2β2 2 (mod. (2 1 2α (2 2 = (1 + ( 1 j αj j 2 j=1 1 ( α + ( α = 2 + α+ + ( α + ( (1 + α α 2 + (1 1 ( α α 2 (mod, 38 / 42

39 Proof of 1 L / 2 0 (mod so (1 can be rewritten as ( L (1 + 2(1 + α β 2 α 2 1 4(1 α β 2 ( α 2 (mod. (3 Multilying (2 by 2(1 α and then subtracting it from (3 we obtain ( (2α L ( 1 4α (1 α 2 4(1 + α β 2 1 =(4L 4L 2 2 α 2 (mod. α 2 39 / 42

40 Proof of 1 L / 2 0 (mod Now that L 1 (mod and L 2 = α 2 +β 2 (α 2 +β 2 = ( (α + β 2 2αβ = 3 3 (mod, we have ( (2α L 1 1 Similarly, 2 ( ( (2β L α 1 2 = 10 β (mod. 2 α (mod. 2 As 2α 1 + (2β 1 = 2L 2 0 (mod, we finally obtain 1 1 L 2 = α + β 2 0 (mod. 40 / 42

41 More conjectures on congruences For more conjectures of mine on congruences, see Z. W. Sun, Oen Conjectures on Congruences, arxiv: which contains 100 unsolved conjectures raised by me. You are welcome to solve my conjectures! 41 / 42

42 Than you! 42 / 42