PAijpam.eu THE PERIOD MODULO PRODUCT OF CONSECUTIVE FIBONACCI NUMBERS

International Journal of Pure and Applied Mathematics Volume 90 No. 014, 5-44 ISSN: 111-8080 (printed version); ISSN: 114-95 (on-line version) url: http://www.ipam.eu doi: http://dx.doi.org/10.17/ipam.v90i.7 PAipam.eu THE PERIOD MODULO PRODUCT OF CONSECUTIVE FIBONACCI NUMBERS Narissara Khaochim 1, Prapanpong Pongsriiam 1, Department of Mathematics, Faculty of Science Silpakorn University Ratchamankanai Rd, Nakornpathom, 7000, THAILAND Abstract: Let F n be the nth Fibonacci number. The period modulo m, denoted by s(m), is the smallest positive integer k for which F n+k F n (mod m) for all n 0. In this paper, we find the period modulo product of consecutive Fibonacci numbers. For instance, we prove that, for n 1, n(n+1)(n+)(n+), if n 0 (mod ), s(f n F n+1 F n+ F n+ ) = n(n+1)(n+)(n+), if n 0,9 (mod 1), n(n+1)(n+)(n+), if n,6 (mod 1). AMS Subect Classification: 11B9 Key Words: Fibonacci sequence, divisibility, Fibonacci entry point, the period modulo m 1. Introduction Let (F n ) n 0 be the Fibonacci sequence given by F n+ = F n+1 +F n, for n 0, where F 0 = 0 and F 1 = 1. These number are famous for possessing wonderful properties, see [,, 5, 6, 15] for additional references and history. In particular, we will be concerned with the divisibility properties and the periodic nature of the Fibonacci sequence. Received: September 5, 01 Correspondence author c 014 Academic Publications, Ltd. url: www.acadpubl.eu

6 N. Khaochim, P. Pongsriiam Let m be a positive integer. The order of appearance of m in the Fibonacci sequence, denoted by z(m), is defined as the smallest positive integer k such that m F k (some authors also call it the order of apparition, the rank of apparition, or Fibonacci entry point). The period modulo m of the Fibonacci sequence, denoted by s(m), is defined as the smallest positive integer k such that F n+k F n (mod m) for all n 0. There are several results about z(m) and s(m) in the literature. For instance, Stanley [14] shows that if m is an integer greater than then s(f m ) = m if m is even and s(f m ) = 4m if m is odd. It is also well known that z(f m ) = m for m (see for example [4, p. 1]). For other classical results on z(m) and s(m), we refer the reader to [1, 4, 6, 1, 1, 14, 15, 16, 17]. Recently, D. Marques [7, 8, 9, 10, 11] has obtained a formula of z(m) for various special numbers m. Particularly, he obtains the following [10] in 01. Theorem 1. (Marques [10]) (i) For n, (ii) For n, z(f n F n+1 F n+ ) = (iii) For n 1, z(f n F n+1 F n+ F n+ ) = z(f n F n+1 ) = n(n+1). { n(n+1)(n+), if n 1 (mod ), n(n+1)(n+), if n 0 (mod ). n(n+1)(n+)(n+), if n 0 (mod ), n(n+1)(n+)(n+), if n 0,9 (mod 1), n(n+1)(n+)(n+) 6, if n,6 (mod 1). The above theorem motivates us to study the period s(n) when n is the product of consecutive Fibonacci numbers. Our main results are the formulas for s(f n F n+1 ), s(f n F n+1 F n+ ), and s(f n F n+1 F n+ F n+ ). This task is a bit more difficult than the calculation of, for example, z(f n F n+1 F n+ F n+ ) in Theorem 1. This is because the latter only requires the smallest k such that F k 0 (mod F n F n+1 F n+ F n+ ) while our task is to find the smallest k such that F k 0 (mod F n F n+1 F n+ F n+ ) and F k+1 1 (mod F n F n+1 F n+ F n+ ) (see Lemma ). But with the aid of Lemma, our task become easy. We will give some lemmas in the next section. Then we will give the main results in the last section.

THE PERIOD MODULO PRODUCT OF... 7. Preliminaries We recall some facts on Fibonacci numbers for the convenience of the reader. Let n and m be positive integers. The following results are well known and will be used throughout this article: For n, F n F m if and only if n m. (1) For n 1, F n 1 F n+1 Fn = ( 1) n (The Cassini s formula). () k ( ) k For k,n 1 and r 0, F kn+r = F nf k n 1 F +r () For n 1, F n+1 F n 1 ( 1)n (mod F n ). (4) Identity (1) and () can be found, for example, in [, 5, 6, 15]. Identity () and (4) might be less well known, so we will give a proof here for completeness. Let α = 1+ 5 and β = 1 5. It is well known that F n = αn β n α β. By solving the equation, α n β n = (α β)f n, and α α n β β n = (α β)f n+1, for α n and β n, we obtain α n = αf n +F n 1, β n = βf n +F n 1. Let k,n 1 and r 0. Then F kn+r = αkn+r β kn+r α β = 1 [(αf n +F n 1 ) k α r (βf n +F n 1 ) k β r] α β = 1 α β = 1 α β = k k k ( ) k (αf n ) F k n 1 αr k [( ) ] k F nf k n 1 (α+r β +r ) ( ) k F nf k n 1 F +r. ( ) k (βf n ) F k n 1 βr This proves (). Next we prove (4). Let n 1. By the Cassini s formula, we have F n F n+ F n+1 = ( 1)n+1. So F n+1 = F nf n+ ( 1) n+1 ( 1) n+ ( 1) n (mod F n ) and F n 1 = (F n+1 F n ) F n+1 ( 1)n (mod F n ). This proves (4). The next result is actually an equivalent definition for the period of the Fibonacci sequence modulo m. It appeared for example in [1, 16, 17].

8 N. Khaochim, P. Pongsriiam Lemma. Let m beapositive integer. Then s(m) is equal to the smallest positive integer k such that F k 0 (mod m) and F k+1 1 (mod m). Proof. Let m be a positive integer and let k be the smallest positive integer such that F k 0 (mod m) and F k+1 1 (mod m). We will prove that F n+k F n (mod m) for all n 0 by strong induction on n. It is clearly true for n = 0 and n = 1. Assume that it is valid for any integer such that 0 n. Thus F (n+1)+k = F n+k+1 = F n+k +F n+k 1 F n +F n 1 F n+1 (mod m). Therefore F n+k F n (mod m) for all n 0. Suppose that r is a positive integer such that F n+r F n (mod m) for all n 0. Then F r F 0 0 (mod m) and F r+1 F 1 1 (mod m). Thus k r. Therefore k is the smallest positive integer such that F n+k F n (mod m) for all n 0. We conclude that s(m) = k. It is shown in [4] that z(n) s(n) for every n 1. So it is natural to define the quantity t(n) = s(n) z(n). So t(n) is an integer for all n 1. A useful result on t(n) is obtained by Vinson[16] as follows. Lemma. (Vinson [16]) The following statement holds. (i) t(m) = 4 if m > and z(m) is odd, (ii) t(m) = 1 if 8 m and z(p) but 4 z(p) for every odd prime p dividing m, and (iii) t(m) = for other m.. Main Results In this section, we give the proof of the main theorems. As mentioned before, Lemmagives anequivalent definitionofs(n)andwewill uselemmawithout further referring. Theorem 4. s(f 1 F ) = s(1) = 1,s(F F ) = s() =, and s(f n F n+1 ) = n(n+1) for every n. Proof. It is straightforward to verify that s(f 1 F ) = s(1) = 1,s(F F ) = s() =. So we let n. By Theorem 1, we have z(f n F n+1 ) = n(n + 1), which is even. Then by Lemma, we see that t(f n F n+1 ) = 1 or. Therefore s(f n F n+1 ) = n(n+1) or s(f n F n+1 ) = n(n+1). (5)

THE PERIOD MODULO PRODUCT OF... 9 First we consider F n(n+1)+1 (mod F n ). By (), we have n+1 ( ) n+1 F n(n+1)+1 = Fn Fn+1 n 1 F +1 Fn 1 n+1 (mod F n ). We consider the following two cases: Case 1: If n is even, then by (4), we have F n+1 n 1 (F n 1 )n F n 1 ( 1) n(n ) F n 1 F n 1 1 (mod F n ). This is because F n 1 < F n. Case : Assume that n is odd. Then by (4), Fn 1 n+1 (F n 1 )n+1 ( 1) n(n+1 ) ( 1) n+1 (mod F n ). Since ( 1) n+1 = 1 if and only if 4 n+1, we see that Fn 1 n+1 1 (mod F n) if and only if n (mod 4). We conclude that F n(n+1)+1 1 (mod F n ) if and only if n (mod 4). (6) Next we consider F n(n+1)+1 (mod F n+1 ). By (), we have F n(n+1)+1 = n ( ) n F n+1 Fn n F +1 Fn n (mod F n+1 ). Similar to the proof of (6), we apply (4) to obtain the following: Ifnisodd,thenFn n (F n )n 1 F n ( 1) (n+1)(n 1 ) F n F n 1 (mod F n+1 ). If n is even, Fn n (F n )n ( 1) (n+1)(n ) ( 1) n (mod F n+1 ), which is congruent to 1 (mod F n+1 ) if and only if 4 n. Therefore we obtain F n(n+1)+1 1 (mod F n+1 ) if and only if n 0 (mod 4). (7) Since gcd(f n,f n+1 ) = 1, we have F k 1 (mod F n F n+1 ) if and only if F k 1 (mod F n ) and F k 1 (mod F n+1 ). (8) From (6), (7) and (8), we see that there is no n N satisfying F n(n+1)+1 1 (mod F n F n+1 ). Hence s(f n F n+1 ) n(n + 1). Now by (5), we can conclude that s(f n F n+1 ) = n(n+1). This completes the proof. Remark 1. We can also directly calculate F n(n+1)+1 (mod F n F n+1 ) by applying Identity () and (4) as follows. F n(n+1)+1 = (n+1) ( (n+1) ) F n F(n+1) n 1 F +1

40 N. Khaochim, P. Pongsriiam F (n+1) n 1 (Fn 1 )n+1 ( 1) n(n+1) 1 (mod F n ) n ( ) n F n(n+1)+1 = F n+1 Fn n F +1 Fn n (F n )n ( 1) (n+1)n 1 (mod F n+1 ). Since gcd(f n,f n+1 ) = 1, we have F n(n+1)+1 1 (mod F n F n+1 ), as required. Note. also that, from this point on, we will apply (4) without referring to it.. Theorem 5. s(f 1 F F ) = s() = and for n, we have s(f n F n+1 F n+ ) = { n(n+1)(n+), if n 1 (mod ), n(n+1)(n+), if n 0 (mod ). Proof. It easy to verify that s(f 1 F F ) = s() =. So we let n. We split the proof into two cases: Case 1: Assume that n is even. Then n + is even and n(n+1)(n+). Therefore, by Theorem 1, z(f n F n+1 F n+ ) is even. So, by Lemma, t(f n F n+1 F n+ ) = 1 or. We will show that t(f n F n+1 F n+ ) =. By Lemma, it suffices to find a prime p F n F n+1 F n+ such that 4 z(p). Since n is even, 4 n or 4 n+. Then, by (1), F 4 F n or F 4 F n+ and therefore F 4 F n F n+1 F n+. That is F n F n+1 F n+. Since z() = 4, we have t(f n F n+1 F n+ ) =, as desired. In conclusion, if n is even s(f n F n+1 F n+ ) = z(f n F n+1 F n+ ) = n(n+1)(n+). Case : Assumethatnisodd. ByTheorem1, z(f n F n+1 F n+ ) = n(n+1)(n+) which is an even number. So, by Lemma, we obtain t(f n F n+1 F n+ ) = 1 or. Thus s(f n F n+1 F n+ ) = n(n+1)(n+) or s(f n F n+1 F n+ ) = n(n+1)(n+). (9) We consider F n(n+1)(n+)+1 (mod F n+1 ). By (), we have F n(n+1)(n+)+1 = n(n+) ( ) n(n+) F n+1 Fn(n+) n F +1 Fn n(n+) (mod F n+1 ).

THE PERIOD MODULO PRODUCT OF... 41 Since n is odd, n(n+) is odd. So, by (4), we have F n(n+) n (F n) n(n+) 1 F n ( 1) (n+1)n(n+) 1 F n F n 1 (mod F n+1 ). Thus F n(n+1)(n+)+1 1 (mod F n F n+1 F n+ ). Therefore s(f n F n+1 F n+ ) n(n+1)(n+). Nowby(9), weconcludethats(f n F n+1 F n+ ) = n(n+1)(n+). Thiscompletes the proof. Remark. We can also calculate F n(n+1)(n+)+1 (mod F n F n+1 F n+ ) by applying () as follows F n(n+1)(n+)+1 = (n+1)(n+) ( (n+1)(n+) ) F n F(n+1)(n+) n 1 F +1 F (n+1)(n+) n 1 (Fn 1 )(n+1)(n+) ( 1) n(n+1)(n+) 1 (mod F n ). n(n+) ( ) n(n+) F n(n+1)(n+)+1 = F n+1 Fn(n+) n F +1 Fn n(n+) (Fn) n(n+) ( 1) (n+1)(n)(n+) 1 (mod F n+1 ). n(n+1) ( ) n(n+1) F n(n+1)(n+)+1 = F n+ Fn(n+1) n+1 F +1 F n(n+1) n+1 (F n+1) n(n+1) ( 1) (n+)(n)(n+1) 1 (mod F n+ ). Since the number F n,f n+1,f n+ are pairwise relatively prime, we have as required. Theorem 6. For n 1, F n(n+1)(n+)+1 1 (mod F n F n+1 F n+ ), n(n+1)(n+)(n+), if n 0 (mod ), n(n+1)(n+)(n+) s(f n F n+1 F n+ F n+ ) =, if n 0,9 (mod 1), n(n+1)(n+)(n+), if n,6 (mod 1).

4 N. Khaochim, P. Pongsriiam Proof. Let n. We split the proof into three cases: Case 1: Assumethatn 0 (mod ). ByTheorem1, z(f n F n+1 F n+ F n+ ) = n(n+1)(n+)(n+) n(n+1)(n+)(n+). Since 4 n(n + 1)(n + )(n + ),. So, by Lemma, we have t(f n F n+1 F n+ F n+ ) = 1 or. We will show that t(f n F n+1 F n+ F n+ ) =. ByLemma, itsufficestofindaprimep F n F n+1 F n+ F n+ such that 4 z(p), or to show that 8 F n F n+1 F n+ F n+. Since n 0 (mod ), we have n 1,,4,5 (mod 6). (i) Assume that n 4,5 (mod 6). Then 6 n + or 6 n+1. By (1), we have F 6 F n+ or F 6 F n+1 and therefore F 6 F n F n+1 F n+ F n+. That is 8 F n F n+1 F n+ F n+, as desired. (ii) Assume that n 1, (mod 6). Then n 1,,7,8 (mod 1). So 4 n + k for some k {0,1,,}. By (1), we have F 4 F n+k for some k {0,1,,}. That is F n+k for some k {0,1,,} and so F n F n+1 F n+ F n+. Thus is an odd prime dividing F n F n+1 F n+ F n+ and 4 z(), as required. Therefore t(f n F n+1 F n+ F n+ ) =. We conclude that s(f n F n+1 F n+ F n+ ) = z(f n F n+1 F n+ F n+ ) = n(n+1)(n+)(n+). Case : Assume that n 0,9 (mod 1). By Theorem 1, z(f n F n+1 F n+ F n+ ) = n(n+1)(n+)(n+). Sincen 0,9 (mod 1), wehave 6 nor6 n+. So n(n+1)(n+)(n+). Therefore z(f n F n+1 F n+ F n+ ) is even. By Lemma, we have t(f n F n+1 F n+ F n+ ) = 1 or. We will show that t(f n F n+1 F n+ F n+ ) =. By Lemma, it suffices to show that 8 F n F n+1 F n+ F n+. Since 6 n or 6 n +, we have F 6 F n or F 6 F n+. So F 6 F n F n+1 F n+ F n+. That is 8 F n F n+1 F n+ F n+, as desired. Therefore t(f n F n+1 F n+ F n+ ) = and hence s(f n F n+1 F n+ F n+ ) = z(f n F n+1 F n+ F n+ ) = n(n+1)(n+)(n+). Case : Assume that n,6 (mod 1). Then, by Theorem 1, we have z(f n F n+1 F n+ F n+ ) = n(n+1)(n+)(n+). 6 We will show that t(f n F n+1 F n+ F n+ ) =. By Lemma, it suffices to show that z(f n F n+1 F n+ F n+ ) is even and 8 F n F n+1 F n+ F n+.

THE PERIOD MODULO PRODUCT OF... 4 (i) Assume that n (mod 1). Then 6 n + and n + 1. So n(n+1)(n+)(n+) 6. Thus z(f n F n+1 F n+ F n+ ) is even. Since 6 n +, F 6 F n+. So F 6 F n F n+1 F n+ F n+. That is 8 F n F n+1 F n+ F n+. (ii) Assume that n 6 (mod 1). Then 6 n and n +. So n(n+1)(n+)(n+) 6. Thus z(f n F n+1 F n+ F n+ ) is even. Since 6 n, F 6 F n. So F 6 F n F n+1 F n+ F n+. That is 8 F n F n+1 F n+ F n+. In any case, z(f n F n+1 F n+ F n+ ) is even and 8 F n F n+1 F n+ F n+. Therefore We conclude that t(f n F n+1 F n+ F n+ ) =. s(f n F n+1 F n+ F n+ ) = z(f n F n+1 F n+ F n+ ) = n(n+1)(n+)(n+). This completes the proof. Acknowledgments The first author receives scholarship from DPST Thailand. The second author receives financial support from Faculty of Science, Silpakorn University, Thailand, contract number RGP 555-07. References [1] A. Andreassian, Fibonacci sequences modulo m, The Fibonacci Quarterly, 1, No.1 (1974), 51 65. [] A. Benamin and J. Quinn, The Fibonacci numbers-exposed more discretely, Math. Mag., 76, No. (01), 18 19. [] R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Second edition, Addison Wesley, USA (1994). [4] J. H. Halton, On the divisibility properties of Fibonacci numbers, The Fibonacci Quarterly, 4, No. (1996), 17 40. [5] D. Kalman, and R. Mena, The Fibonacci numbers exposed, Math. Mag., 76, No. (00), 167 181.

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