When do Fibonacci invertible classes modulo M form a subgroup?

Transcription

1 Calhoun: The NPS Institutional Archive DSace Reository Faculty and Researchers Faculty and Researchers Collection 2013 When do Fibonacci invertible classes modulo M form a subgrou? Luca, Florian Annales Mathematicae et Informaticae 41 (2013), (Proc. 15th International Conference on Fibonacci Numbers and Their Alications). htt://hdl.handle.net/10945/38822 Downloaded from NPS Archive: Calhoun

2 Annales Mathematicae et Informaticae 41 (2013) Proceedings of the 15 th International Conference on Fibonacci Numbers and Their Alications Institute of Mathematics and Informatics, Eszterházy Károly College Eger, Hungary, June 25 30, 2012 When do the Fibonacci invertible classes modulo M form a subgrou? Florian Luca a, Pantelimon Stănică b, Aynur Yalçiner c a Instituto de Matemáticas, Universidad Nacional Autonoma de México C.P , Morelia, Michoacán, México; fluca@matmor.unam.mx b Naval Postgraduate School, Alied Mathematics Deartment Monterey, CA 93943; stanica@ns.edu c Deartment of Mathematics, Faculty of Science, Selçuk University, Camus Konya, Turkey; aynuryalciner@gmail.com Abstract In this aer, we look at the invertible classes modulo M reresentable as Fibonacci numbers and we ask when these classes, say F M, form a multilicative grou. We show that if M itself is a Fibonacci number, then M 8; if M is a Lucas number, then M 7. We also show that if x 3, the number of M x such that F M is a multilicative subgrou is O(x/(log x) 1/8 ). Keywords: Fibonacci and Lucas numbers, congruences, multilicative grou MSC: 11B39 1. Introduction Let {F k } k 0 be the Fibonacci sequence given by F 0 = 0, F 1 = 1 and F k+2 = F k+1 + F k for all k 0, with the corresonding Lucas comanion sequence {L k } k 0 satisfying the same recurrence with initial conditions L 0 = 2, L 1 =1. The distribution of the Fibonacci numbers modulo some ositive integer M has been extensively studied. Here, we ut F M = {F n (mod M) : gcd(f n, M) = 1} and ask when is F M a multilicative grou. We resent the following conjecture. 265

3 266 F. Luca, P. Stănică, A. Yalçiner Conjecture 1.1. There are only finitely many M such that F M is a multilicative grou. Shah [5] and Bruckner [1] roved that if is rime and F is the entire multilicative grou modulo, then {2, 3, 5, 7}. We do not know of many results in the literature addressing the multilicative order of a Fibonacci number with resect to another Fibonacci number, although in [3] it was shown that if F n F n+1 is corime to F m and F n+1 /F n has order s {1, 2, 4} modulo F m, then m < 500s 2. Moreover, Burr [2] showed that F n (mod m) contains a comlete set of residues modulo m if and only if m is of the forms: {1, 2, 4, 6, 7, 14, 3 j } 5 k, where k 0, j 1. In this aer, we rove that if M = F m is a Fibonacci number itself, or M = L m, then Conjecture 1.1 holds in the following strong form. Theorem 1.2. If M = F m and F M is a multilicative grou, then m 6. If M = L m and F M is a multilicative grou, then m 4. We also show that for most ositive integers M, F M is not a multilicative grou. Theorem 1.3. For x 3, the number of M x such that F M is a multilicative subgrou is O(x/(log x) 1/8 ). In articular, the set of M such that F M is a multilicative subgrou is of asymtotic density Proof of Theorem 1.2 We first deal with the case of the Fibonacci numbers. It is well-known that the Fibonacci sequence is urely eriodic modulo every ositive integer M. When M = F m, then the eriod is at most 4m. Thus, #F M 4m, Let ω(m) be the number of distinct rime factors of m. Assume that X is some ositive integer such that π(x) ω(m) + 4. (2.1) Here, π(x) is the number of rimes X. Then there exist three odd rimes < q < r X none of them dividing m. For a trile (a, b, c) {0, 1,..., (4m) 1/3 }, we look at the congruence class F a F b q F c r (mod M). There are ( (4m) 1/3 + 1) 3 > 4m #F M such elements modulo M, so they cannot be all distinct. Thus, there are (a 1, b 1, c 1 ) (a 2, b 2, c 2 ) such that F a1 Fq b1 Fr c1 F a2 Fq b2 Fr c2 (mod M). Hence, F a1 a2 Fq b1 b2 Fr c1 c2 1 (mod M). Observe that the rational number x = F a1 a2 Fq b1 b2 Fr c1 c2 1 cannot be zero because F, F q, F r are all larger than 1 and corime any two. Thus, M divides the numerator of the nonzero rational number x, and so we get F m = M F a1 a2 F q b1 b2 F r c1 c2. (2.2)

4 When do the Fibonacci invertible classes modulo M form a subgrou? 267 We now use the fact that α k 2 F k α k 1 for all k = 1, 2..., where α = (1 + 5)/2, to deduce from (2.2) that so that therefore α m 2 F m (F F q F r ) (4m)1/3 < (α X 1 ) 3(4m)1/3, m < 3(4m) 1/3 X + 2 3(4m) 1/3 < 3(4m) 1/3 X, m < 6 3X 3/2. (2.3) Let us now get some bounds on m. We take X = m 1/2. Assuming X > 17 (so, m > 17 2 ), we have, by Theorem 2 in [4], that Since 2 ω(m) m, we have that π(x) > X log X = 2m1/2 log m. ω(m) log m log 2. Thus, inequality (2.1) holds for our instance rovided that 2m 1/2 log m > log m log 2 + 4, which holds for all m > Now inequality (2.3) tells us that m < 6 3m 3/4, therefore m < (6 3) 4 < (2.4) Let us reduce the above bound on m. Since = > m, it then follows that ω(m) 5, therefore it is enough to choose X = 23 to be the 9th rime and then inequality (2.1) holds. Thus, (2.3) tells us that m /2 < We covered the rest of the range with Mathematica. That is, for each m [10, 1200], we took the first two odd rimes and q which do not divide m and checked whether for some ositive integer n 4m both congruences F n 1 (mod F m ) and Fq n 1 (mod F m ). The only m s that assed this test were m = 10, 11. We covered the rest by hand. The only values m that satisfy the hyothesis of the theorem are m = 1, 2, 3, 4, 5, 6. If M = L m, then, the argument is similar to the one above u and we oint out the differences only. The eriod of the Fibonacci numbers modulo a Lucas number

5 268 F. Luca, P. Stănică, A. Yalçiner L m is at most 8m, and so #F M 8m. As before, one takes X as in (2.1), and the trile (a, b, c) {0, 1,..., 2m 1/3 }, imlying an inequality as in (2.2), namely L m = M F a1 a2 Since for all k 1, α k 1 L k α k+1, then F q b1 b2 F r c1 c2. (2.5) α m 1 L m (F F q F r ) 2m1/3 α 6(X+1)m1/3, and so, m < 6m 1/3 X m 1/3 < 13m 1/3 X, which imlies m < 13 3/2 X 3/2. (2.6) The argument we used before with X = m 1/2 works here, as well, rendering the bound m < 13 6 = 4, 826, 809. We can decrease the bound by using the fact that the roduct of all rimes u to 19 is 9,699,690 > 4,826,809, and so, ω(m) 7, therefore, it is enough to choose X = 31 (the 11th rime) for the inequality (2.1) to hold. We use X = 31 in the formula before (2.6) to get m 192 m 1/3 1 < 0, which imlies m < 14 3 = 2744 (to see that, label y := m 1/3 and look at the sign of the olynomial y 3 192y 1). To cover the range from 10 to 2744, we used the same trick as before (which works, since by F 2m = L m F m, then gcd(f, L m ) = gcd(f, F 2m /F m ) gcd(f, F 2m ) = F gcd(,2m) ). To seed u the comutation we used the fact that one can choose one of the rimes, q to be 5, since a Lucas number is never divisible by 5. The only m s that assed the test were 10, 12, 15, 21, which are easily shown (by dislaying the corresonding residues) not to generate a multilicative grou structure. The only values of m, for which we do have a multilicative grous structure for F M when M = L m are m {1, 2, 3, 4}. 3. Proof of Theorem 1.3 Consider the following set of rimes { ( ) 5 P = > 5 : = 1, ( ) 11 = ( ) } 46 = 1. Here, for an integer a and an odd rime, we use ( a ) for the Legendre symbol of a with resect to. Let M be the set of M such that F M is a multilicative subgrou. We show that M is free of rimes from P. Since P is a set of rimes of relative density 1/8 (as a subset of all rimes), the conclusion will follow from the Brun sieve (see [6, Chater I.4, Theorem 3]). To see that M is free of rimes from, observe that since F 3 = 2, F 4 = 3, and F M is a multilicative subgrou, it follows that there exists n such that F n 6 (mod M). If M for some P, it follows that F n 6 0 (mod ). (3.1)

6 When do the Fibonacci invertible classes modulo M form a subgrou? 269 Since ( 5 ) = 1, it follows that both 5 and α are elements of F. With the Binet formula, we have F n = αn β n. 5 Put t n = α n, ε n = ( 1) n. Thus, β n = ( α 1 ) n = ε n t 1 n, so congruence (3.1) becomes t n ε n t 1 n 6 0 (mod ) 5 giving t 2 n 6 5 t n ε n 0 (mod ). Thus, one of the quadratic equations t t ± 1 = 0 must have a solution t modulo. Since the discriminants of the above quadratic equations are 176 = and 184 = 4 46, resectively, and since neither 11 nor 46 is a quadratic residue modulo, we get the desired conclusion. 4. Comments The bound O(x/(log x) 1/8 ) of Theorem 1.3 is too weak to allow one to decide via the Abel summation formula whether M M is finite or not. Of course Conjecture 1.1 would imly that the above sum is finite. We leave it as a roblem to the reader to imrove the bound on the counting function of M [1, x] from Theorem 1.3 enough to decide that indeed the sum of the above series is convergent. Acknowledgment. F. L. was suorted in art by Project PAPIIT IN and a Marcos Moshinsky Fellowshi. P. S. acknowledges a research sabbatical leave from his institution. References [1] G. Bruckner, Fibonacci Sequence Modulo a Prime 3 (mod 4), Fibonacci Quart. 8 (1970), [2] S. A. Burr, On Moduli for Which the Fibonacci Sequence Contains a Comlete System of Residues, Fibonacci Quart. 9 (1971), [3] T. Komatsu, F. Luca, On the multilicative order of F n+1/f n modulo F m, Prerint, [4] J. B. Rosser, L. Schoenfeld, Aroximate formulas for some functions of rime numbers, Illinois J. Math. 6 (1962), M

7 270 F. Luca, P. Stănică, A. Yalçiner [5] A. P. Shah, Fibonacci Sequence Modulo m, Fibonacci Quart. 6 (1968), [6] G. Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, Cambridge University Press, 1995.