Introduction to Lucas Sequences

Transcription

1 A talk given at Liaoning Normal Univ. (Dec. 14, 017) Introduction to Lucas Sequences Zhi-Wei Sun Nanjing University Nanjing 10093, P. R. China zwsun Dec. 14, 017

2 Abstract Let A and B be integers. The Lucas sequences u n = u n (A, B) (n = 0, 1,,...) and v n = v 0 (A, B) (n = 0, 1,,...) are defined by and u 0 = 0, u 1 = 1, u n+1 = Au n Bu n 1 (n = 1,, 3,...) v 0 =, v 1 = A, v n+1 = Av n Bv n 1 (n = 1,, 3,...). They are natural extensions of the Fibonacci numbers and the Lucas numbers. Lucas sequences play important roles in number theory and combinatorics. In this talk we introduce various properties of Lucas sequences and their applications. In particular, we will mention their elegant application to Hilbert s Tenth Problem on Diophantine equations. / 41

3 Fibonacci numbers In a book in 10, Fibonacci considered the growth of an idealized (biologically unrealistic) rabbit population, assuming that: a newly born pair of rabbits, one male, one female, are put in a field; rabbits are able to mate at the age of one month so that at the end of its second month a female can produce another pair of rabbits; rabbits never die and a mating pair always produces one new pair (one male, one female) every month from the second month on. The puzzle that Fibonacci posed was: how many pairs will there be in one year? Suppose that in the n-th month there are totally F n pairs of rabbits. Then Note that F 0 = 0, F 1 = 1, F n+1 = F n + F n 1 (n = 1,, 3,...). F = 1, F 3 =, F 4 = 3, F 5 = 5, F 6 = 8, F 7 = 13, F 8 = 1, F 9 = 34, F 10 = 55, F 11 = 89, F 1 = / 41

4 A combinatorial interpretation of Fibonacci numbers Let f (n) denote the number of binary sequences not containing two consecutive zeroes. Clearly, f (1) = (both 0 and 1 meet the purpose), and f () = 3 (01, 10, 11 meet the purpose). Now let a 1,..., a n {0, 1}. Clearly a 1..., a n 1 meets the purpose if and only if the sequence a 1..., a n meets the purpose. Also, the sequence a 1..., a n 0 meets the purpose if and only if a n = 1 and the sequence a 1... a n 1 meets the purpose. Therefore f (n + 1) = f (n) + f (n 1). Since f (1) = F 3, f () = F 4 and also f (n + 1) = f (n) + f (n 1) for n = 1,, 3,..., we see that f (n) = F n+. 4 / 41

5 Edouard Lucas The name Fibonacci numbers or the Fibonacci sequence was first used by the French mathematician E. Lucas ( ). Lucas has several fundamental contributions to number theory. Lucas Congruence. Let p be a prime, and let a = k i=0 a ip i and b = k i=0 b ip i with a i, b i {0,..., p 1}. Then ( ) a k ( ) ai (mod p). b i=0 b i Lucas died in unusual circumstances. At the banquet of an annual congress, a waiter dropped some crockery and a piece of broken plate cut Lucas on the cheek. He died a few days later of a severe skin inflammation probably caused by septicemia. He was only / 41

6 Lucas sequences Let A and B be given numbers. For n N = {0, 1,,...} we define u n = u n (A, B) and v n = v n (A, B) as follows: u 0 = 0, u 1 = 1, and u n+1 = Au n Bu n 1 (n = 1,, 3,...); v 0 =, v 1 = A, and v n+1 = Av n Bv n 1 (n = 1,, 3,...). The sequence {u n } n N and its companion {v n } n N are called Lucas sequences. In 1876 E. Lucas introduced such sequences in the case A, B Z, and studied them systematically. Lucas sequences play important roles in number theory. 6 / 41

7 Special cases By induction on n N we find that u n (, 1) = n, v n (, 1) = ; u n (3, ) = n 1, v n (3, ) = n Those F n = u n (1, 1) and L n = v n (1, 1) are called Fibonacci numbers and Lucas numbers respectively. The Pell sequence {P n } n N is given by P n = u n (, 1), so P 0 = 0, P 1 = 1, P n+1 = P n + P n 1 (n = 1,, 3,...). The companion of the Pell sequence is {Q n } n N where Q n = v n (, 1). We also let S n = u n (4, 1) and T n = v n (4, 1); thus S 0 = 0, S 1 = 1, and S n+1 = 4S n S n 1 (n = 1,, 3,...); T 0 =, T 1 = 4, T n+1 = 4T n T n 1 (n = 1,, 3,...). 7 / 41

8 Binet Formulae The equation x = Ax B is called the characteristic equation of the Lucas sequences {u n (A, B)} n N and {v n (A, B)} n N. = A 4B is the discriminant, and the two roots are α = A + and β = A. Property 1 (Binet, 1843) For any n = 0, 1,,... we have u n = α k β n 1 k and v n = α n + β n. 0 k<n As A = α + β and B = αβ, we can prove the formulae by induction on n. If = 0, then α = β = A/, hence u n = ( ) A n 1 ( ) A n α n 1 = n and v n = α n =. 0 k<n 8 / 41

9 Explicit Formulae We can express u n and v n in terms of n, A and. In fact, ( un =(α β)u n = α n β n A + ) n ( = and thus = n 1 n k=0 k n 1 u n = ( ) n A n k (k 1)/ k (n 1)/ k=0 ( ) n A n 1 k k k + 1 (which also holds when = 0). Similarly, ( v n =α n + β n A + ) n ( A ) n = + = 1 n 1 n/ k=0 ( ) n A n k k. k A ) n 9 / 41

10 Special cases For the Fibonacci sequence {F n } n N and its companion {L n } n N, = 1 4( 1) = 5 and so 5Fn = ( 1 + ) n 5 ( 1 ) n ( 5, L n = 1 + ) n 5 +( For the Pell sequence {P n } n N and its companion {Q n } n N, = 4( 1) = 8 and so P n = (1 + ) n (1 ) n, Q n = (1 + ) n + (1 ) n. 1 ) n 5. For the sequence {S n } n N and its companion {T n } n N, = = 1 and so S n = 1 ( ( + 3) n ( 3) n), T n = (+ 3) n +( 3) n / 41

11 On ((A + A 4B)/) n Let A, B Z with = A 4B. For n N, as ( A + ) n ( A ) n un (A, B) =, ( A + ) n ( A ) n v n (A, B) = +, we have In particular, ( A ± ) n = v n(a, B) ± u n (A, B). ( 1 ± ) n 5 = L n ± F n 5, (1 ± ) n = Q n ± P n, ( ± 3) n = T n ± S n / 41

12 Relations between u n and v n Property. For n N we have v n = u n+1 Au n, u n = v n+1 Av n, vn un = 4B n. This can be easily proved since A = α + β, u n = α n β n, v n = α n + β n, where = A 4B, α = A + and β = A. In particular, L n = F n+1 F n, 5F n = L n+1 L n, L n 5F n = 4( 1) n. 1 / 41

13 Neighbouring formulae Property 3 (Neighbouring formulae). For any n N we have un+1 Au n+1 u n + Bun = B n, vn+1 Av n+1 v n + Bvn = B n. In other words, if n Z + = {1,, 3,...} then un u n 1 u n+1 = B n 1 and vn v n 1 v n+1 = B n 1. Proof. By Property, 4B n = vn un = (u n+1 Au n ) un and 4 B n = vn ( u n ) = vn (v n+1 4Av n ). So the desired results follow. Example. Fn+1 F n+1 F n Fn = ( 1) n, L n+1 L n+1 L n L n = 5( 1) n / 41

14 Addition formulae Property 4 (Addition Formulae). For any m, n N we have u m+n = u mv n + u n v m and v m+n = v mv n + u m u n. Property 5 (Double Formulae). For each n N we have u n = u n v n, v n = v n B n = v n + u n = u n + B n, u n+1 = u n+1 Bu n, v n+1 = v n v n+1 AB n. 14 / 41

15 Multiplication Formulae Property 6 (Multiplication Formulae). For any k, n N we have u kn = u k u n (v k, B k ) and v kn = v n (v k, B k ). Proof. Let A = v k and B = B k. Then = v k 4Bk = u k, α = v k + u k, β = v k u k. Hence ( (α ) n + (β ) n v k + ) n ( u k v k ) n u k = + ( A + ) kn ( A ) kn = + = v kn. That u kn = u k u n (v k, B k ) can be proved similarly. 15 / 41

16 Expansions in terms of A and B Property 7 (Expansions in terms of A and B). For any n Z + we have (n 1)/ ( ) n 1 k u n = A n 1 k ( B) k, k k=0 n/ n v n = n k k=0 Proof. Observe that u m x m 1 = m=1 = m 1 m=1 k=0 m 1 ( n k k α k β m 1 k x m 1 ) A n k ( B) k. (αx) k (βx) m 1 k = m=1 k=0 (αx) k k=0 j=0 (βx) j = 1 1 αx 1 1 βx = 1 1 Ax + Bx. 16 / 41

17 Expansions in terms of A and B (continued) Let [x m ]f (x) denote the coefficient of x m in the power series for f (x). Then u n 1 =[x n 1 ] u m x m 1 = [x n 1 1 ] 1 Ax + Bx m=1 =[x n 1 1 (x(a Bx))n ] 1 x(a Bx) n 1 = [x n 1 ] n 1 =[x n 1 ] x n 1 k (A Bx) n 1 k = = k=0 (n 1)/ k=0 (n 1)/ k=0 ( n 1 k k ( n 1 k k k=0 ) ( B) k A n 1 k k ) A n 1 k ( B) k. x k (A Bx) k 17 / 41

18 Expansions in terms of A and B (continued) v n =u n+1 Au n n/ ( ) n k = A n k ( B) k k = k=0 A n/ k=0 (n 1)/ ( k=0 n/ n = n k k=0 ( n k k ( n 1 k k ) ( n k k ) A n 1 k ( B) k ( n 1 k k ) A n k ( B) k. Corollary. For any n Z + we have F n = ( ) n 1 k and L n = k k N 0 k n/ )) A n k ( B) k n n k ( n k k ). 18 / 41

19 On linear index Property 8 (On linear index) (Z.-W. Sun [Sci. China Ser. A 35(199)]). Suppose that w n+1 = Aw n Bw n 1 for n = 1,, 3,.... Then, for any k Z + and l, n N we have n ( ) n w kn+l = ( Bu k 1 ) n j u j j k w l+j, in particular w n+l = j=0 j=0 n j=0 ( ) n ( B) n j A j w l+j. j Corollary. For any n N we have n ( ) n F n = F j and F n+1 = j n j=0 ( ) n F j+1. j 19 / 41

20 On linear index (continued) We prove the identity by induction on n. The case n = 0 is trivial. Now assume that the identity for n is valid. Then n ( ) n w k(n+1)+l =w kn+(k+l) = ( Bu k 1 ) n j u j j k w k+l+j j=0 n ( ) n = ( Bu k 1 ) n j u j j k (u kw l+j+1 Bu k 1 w l+j ) j=0 =u n+1 k + + j=0 n j=1 n j=1 ( ) n ( Bu k 1 ) n+1 j u j j 1 k w l+j ( ) n ( Bu k 1 ) n+1 j u j j k w l+j + ( Bu k 1 ) n+1 w l n+1 ( ) n + 1 = ( Bu k 1 ) n+1 j u j j k w l+j. 0 / 41

21 Lucas Theorem Property 9 (E. Lucas) Let A, B Z with (A, B) = 1. Then (u m, u n ) = u (m,n) for all m, n N. Proof. By induction, u n+1 A n (mod B). As (A, B) = 1, we have (u n+1, B) = 1. Since u n+1 Au n+1u n + Bu n = B n by Property 3, (u n, u n+1 ) B n and hence (u n, u n+1 ) (u n+1, B n ) = 1. By Property 8, for any n Z + and q, r N we have u nq+r = q j=0 ( ) q ( Bu n 1 ) q j u j j nu r+j ( Bu n 1 ) q u r = (u n+1 Au n ) q u r u q n+1 u r (mod u n ) and hence (u nq+r, u n ) = (u n, u r ). 1 / 41

22 Lucas Theorem (continued) Clearly, (u m, u 0 ) = (u m, 0) = u m = u (m,0). Now let r 0 = m and r 1 = n Z +. Write r i 1 = r i q i + r i+1 for i = 1,..., k with r 1 > r >... > r k > r k+1 = 0. For each i = 1,..., k, Therefore, (u ri 1, u ri ) = (u qi r i +r i+1, u ri ) = (u ri, u ri+1 ). (u m, u n ) =(u r0, u r1 ) = (u r1, u r ) =... =(u rk, u rk+1 ) = (u (m,n), u 0 ) = u (m,n). Remark. For m, n Z +, we can prove that { v (v m, v n ) = (m,n) if ord (m) = ord (n), (, v (m,n) ) otherwise. / 41

23 On positivity of u n and v n Property 10 (Z.-W. Sun [PhD thesis, 199]). u n 0 for all n N v n 0 for all n N A 0 and = A 4B 0. Proof. If A 0 and 0, then u n 0 and v n 0 by the explicit formulae in terms of A and. Suppose that u n 0 for all n N. Then A = u 0. If < 0, then u n+1 u nu n+ = B n > 0 and the decreasing sequence (u n+1 /u n ) n 1 has a real number limit θ. Since u n+ u n+1 = Au n+1 Bu n u n+1 = A B u n+1 /u n (n = 1,, 3,...), we see that θ Aθ + B = 0 and thus 0. Similarly, if v n 0 for all n N then A = v 0 0 and 0. 3 / 41

24 Periodicity of u n and v n modulo m Property 11 (Periodicity of u n and v n modulo m). Let m Z + with (B, m) = 1. Then there is a positive integer λ such that and also u n+λ u n (mod m) for all n N v n+λ v n (mod m) for all n N. Proof. Consider the m + 1 ordered pairs u i, u i+1 (i = 0,..., m ). By the Pigeon-hole Principle, two of them are congruent modulo m. Choose the least λ m such that for some 0 j < λ. Since u λ, u λ+1 u j, u j+1 (mod m) Bu j 1 = u j+1 Au j u λ+1 Au λ = Bu λ 1 (mod m) and (B, m) = 1, we get u j 1 u λ 1 (mod m). Continuing this process, we finally obtain u λ j, u λ j+1 u 0, u 1 (mod m). 4 / 41

25 Periodicity of u n and v n modulo m (continued) By the choice of λ, we must have j = 0 and thus u λ, u λ+1 u 0, u 1 (mod m). It follows that u n+λ u n (mod m) for all n = 0, 1,,.... For any n N, we also have v n+λ = u n+λ+1 Au n+λ u n+1 Au n = v n (mod m). This completes the proof. 5 / 41

26 Legendre symbols Let p be an odd prime and a Z. The Legendre symbol ( a p ) is given by ( ) a 0 if p a, = 1 if p a and x p a (mod p) for some x Z, 1 if p a and x a (mod p) for no x Z. It is well known that ( ab p ) = ( a p )( b p ) for any a, b Z. Also, ( ) { 1 = ( 1) (p 1)/ 1 if p 1 (mod 4), = p 1 if p 1 (mod 4); ( ) { = ( 1) (p 1)/8 1 if p ±1 (mod 8), = p 1 if p ±3 (mod 8). The Law of Quadratic Reciprocity: If p and q are distinct odd primes, then ( ) ( ) p q = ( 1) p 1 q 1. q p 6 / 41

27 Lucas sequences modulo primes Property 1. Let A, B Z with = A 4B, and let p be an odd prime. (i) u p ( p ) (mod p) and v p A (mod p). (ii) If p B, then p u p ( p ). Proof. Note that p ( ) p j for all j = 1,..., p 1. Thus (p 1)/ ( ) p u p p 1 u p = A p 1 k k k + 1 k=0 ( ) (p 1)/ (mod p) p and v p p 1 v p = (p 1)/ k=0 ( ) p A p k k k A p A (mod p). 7 / 41

28 Lucas sequences modulo primes Now we prove p u p ( p If p, then u p ( p ) = u p ( p If ( p ) = 1, then u p+1 = Au p + v p A ) under the condition p B. ) = 0 (mod p). and hence u p ( p ) = u p+1 0 (mod p). In the case ( p ) = 1, we have ( ) + A = 0 (mod p) p Bu p 1 = Au p u p+1 = Au p v p A (u p 1) 0 (mod p) and hence u p ( p ) = u p 1 0 (mod p) since p B. 8 / 41

29 Wall-Sun-Sun primes In 1960 D. D. Wall [Amer. Math. Monthly] investigated Fibonacci numbers modulo m. For d Z + let n(d) be the least n Z + such that d F n. Wall asked whether n(p) n(p ) for any odd prime p. Let p be an odd prime. Then F p ( p 5 ) = F p ( 5 ) 0 (mod p). p Z.-H. Sun and Z.-W. Sun [Acta Arith. 60(199)] proved that n(p) n(p ) p F p ( p 5 ) = x p + y p = z p for no x, y, z Z with p xyz. An odd prime p with p F p ( p 5 ) is called a Wall-Sun-Sun prime. There are no known Wall-Sun-Sun primes though heuristic arguments suggest that there should be infinitely many Wall-Sun-Sun primes. 9 / 41

30 Chebyshev polynomials The first kind of Chebyshev polynomials T n (x) (n N) and the second kind of Chebyshev polynomials T n (x) (n N) are given by Clearly, As cos nθ = T n (cos θ) and sin((n + 1)θ) = sin θ U n (cos θ). T 0 (x) = 1, T 1 (x) = x, T (x) = x 1, U 0 (x) = 1, U 1 (x) = x, U (x) = 4x 1. cos(nθ+θ) = cos θ cos nθ sin θ sin nθ = cos θ cos nθ cos(nθ θ) and sin(nθ +θ) = cos θ sin nθ +sin θ cos nθ = cos θ sin nθ sin(nθ θ), we have T n+1 (x) = xt n (x) T n 1 (x) and U n+1 (x) = xu n (x) U n 1 (x). 30 / 41

31 Chebyshev polynomials (continued) As T 0 (x) =, T 1 (x) = x, T n+1 (x) = xt n (x) T n 1 (x) (n = 1,,...), and U 0 (x) = 1, U 1 (x) = x, U n+1 (x) = xu n (x) U n 1 (x) (n = 1,,...), we see that T n (x) = v n (x, 1) and U n (x) = u n+1 (x, 1). Thus, for any n Z +, by Property 7 we have T n (x) = 1 n/ ( ) n n k (x) n k ( 1) k n k k and U n (x) = k=0 n/ k=0 ( n k k ) (x) n k ( 1) k. 31 / 41

32 Pell s equation Let d Z + \. It is well-known that the Pell equation y dx = 1 has infinitely many integral solutions. (Note that x = 0 and y = ±1 are trivial solutions.) Moreover, {y + dx : x, y Z and y dx = 1} is a multiplicative cyclic group. For any integer A, the solutions of the Pell equation y (A 1)x = 1 (x, y N) are given by x = u n (A, 1) and y = v n (A, 1) with n N. J. Robinson and his followers wrote u n (A, 1) and v n (A, 1) as ψ n (A) and χ n (A) respectively. To unify Matiyasevich s use of F n = u n (3, 1) and Robinson s use of ψ n (A) = u n (A, 1), we deal with Lucas sequences (u n (A, 1)) n 0. 3 / 41

33 On u n (A, 1) with n Z We extend the sequences u n = u n (A, 1) and v n = v n (A, 1) to integer indices by letting and u 0 = 0, u 1 = 1, and u n 1 + u n+1 = Au n for all n Z, v 0 =, v 1 = A, and v n 1 + v n+1 = Av n for all n Z. It is easy to see that u n (A, 1) = u n (A, 1) = ( 1) n u n ( A, 1) and v n (A, 1) = v n (A, 1) = ( 1) n v n ( A, 1) for all n Z. Lemma. Let A, X Z. Then (A 4)X + 4 X = u m (A, 1) for some m Z. Remark. For n N and A, it is easy to show that (A 1) n u n+1 (A, 1) A n. 33 / 41

34 Hilbert s Tenth Problem In 1900, at the Paris conference of ICM, D. Hilbert presented 3 famous mathematical problems. He formulated his tenth problem as follows: Given a Diophantine equation with any number of unknown quantities and with rational integral numerical coefficients: To devise a process according to which it can be determined in a finite number of operations whether the equation is solvable in rational integers. In modern language, Hilbert s Tenth Problem (HTP) asked for an effective algorithm to test whether an arbitrary polynomial equation P(z 1,..., z n ) = 0 (with integer coefficients) has solutions over the ring Z of the integers. However, at that time the exact meaning of algorithm was not known. 34 / 41

35 Two key steps to solve HTP Based on the above theorem, M. Davis, H. Putnam and J. Robinson [Ann. of Math. 1961] successfully showed that any r.e. set is exponential Diophantine, that is, any r.e. set A has the exponential Diophantine representation a A x x n 0[P(a, x 1,..., x n, x 1,..., xn ) = 0], where P is a polynomial with integer coefficients. Recall that the Fibonacci sequence (F n ) n 0 defined by F 0 = 0, F 1 = 1, and F n+1 = F n + F n 1 (n = 1,, 3,...) increases exponentially. In 1970 Yu. Matiyasevich took the last step to show ingeniously that the relation y = F x (with x, y N) is Diophantine! It follows that the exponential relation a = b c (with a, b, c N, b > 1 and c > 0) is Diophantine, i.e. there exists a polynomial P(a, b, c, x 1,..., x n ) with integer coefficients such that a = b c x x n 0[P(a, b, c, x 1,..., x n ) = 0]. 35 / 41

36 A key lemma The following lemma is an extension of a lemma of Yu. Matiyasevich (1970). Lemma (Sun [Sci. China Ser. A 35(199)]). Let A, B Z with (A, B) = 1, and let k, m N. (i) ku k m = u k u m. (ii) Suppose that A 0, A 4B, and A 1 or (k )B 0. Then u k u m ku k m. 36 / 41

37 Diophantine representation of C = u B (A, 1) with unknowns arbitrarily large Matiyasevi c and Robinson (1975) showed that for A > 1 and B, C > 0 there is a Diophantine representation of C = u B (A, 1) only involving three natural number variables. Lemma (Sun [Sci. China Ser. A 35(199)]). Let A, B, C Z with A > 1 and B 0. Then C = u B (A, 1) C B x > 0 y > 0(DFI ) x, y, z 0[DFI (C B + 1) = (z DFI (C B + 1)) ], where D = (A 4)C + 4, E = C Dx, F = 4(A 4)E + 1, G = 1 + CDF (A + )(A ) E, H = C + BF + (y 1)CF, I = (G 1)H + 1. Moreover, if C = u B (A, 1) with B > 0, then for any Z Z + there are integers x Z and y Z with DFI. 37 / 41

38 Diophantine representation of C = u B (A, 1) with integer unknowns Clearly C B x 0(C = B + x). However, if we use integer variables, we need three variables: C B x y z[c = B + x + y + z + z]. Thus, to save the number of integer variables involved, we should try to avoid inequalities. Note that u B (A, 1) u B (, 1) = B (mod A ). Lemma (Sun [Sci. China Ser. A 35(199)]). Let A, B, C Z with 1 < B < A / 1. Then C = u B (A, 1) (A C B) x 0 y(dfi ), where D, F, I are defined as before. 38 / 41

39 Diophantine representation of W = V B with integer unknowns J. Robinson showed that W = V B (with V > 1 and B, W > 0) if and only if there is an integer A > max{v 3B, W B } such that (V 1)W u B (A, 1) V (W 1) (mod AV V 1). Lemma (Sun [Sci. China Ser. A 35(199)]). Let B, V, W be integers with B > 0 and V > 1. Then W = V B if and only if there are A, C Z for which A max{v 4B, W 4 }, C = u B (A, 1) and (V 1)WC V (W 1) (mod AV V 1). Remark. A, V and W in this lemma are not necessarily positive, they might be negative. 39 / 41

40 Applications to Hilbert s Tenth Problem Matiyasevich (1975). There is no algorithm to determine for any P(x 1,..., x 9 ) Z[x 1,..., x 9 ] whether P(x 1,..., x 9 ) = 0 has solutions with x 1,..., x 9 N. Z.-W. Sun (199). There is no algorithm to determine for any P(z 1,..., z 11 ) Z[z 1,..., z 11 ] whether the equation has integral solutions. P(z 1,..., z 11 ) = 0 40 / 41

41 References For main sources of my work mentioned here, you may look at: 1. Z.-W. Sun, Reduction of unknowns in Diophantine representations, Sci. China Math. 35(199), Z.-W. Sun, Further results on Hilbert s tenth problem, arxiv: , Thank you! 41 / 41