ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS


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1 ON FIBONACCI NUMBERS WITH FEW PRIME DIVISORS YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK Abstract If n is a positive integer, write F n for the nth Fibonacci number, and ω(n) for the number of distinct prime divisors of n We give a description of Fibonacci numbers satisfying ω(f n) 2 Moreover, we prove that the inequality ω(f n) (log n) log 2+o() holds for almost all n We conjecture that ω(f n) log n for composite n, and give a heuristic argument in support of this conjecture Introduction Let F n be the nth Fibonacci number and L n be the nth Lucas number In a previous paper [3], the following result was proved Theorem The only solutions to the equation F n = y m, m 2 are given by n = 0,, 2, 6 and 2 which correspond respectively to F n = 0,,, 8 and 44 Moreover, the only solutions to the equation L n = y p, m 2 are given by n = and 3 which correspond respectively to L n = and 4 The proof involves an intricate combination of Baker s method and the modular method; it also needs about one week of computer verification using the systems pari [] and magma [2] In this paper we use the above theorem, together with wellknown results of Carmichael and Cohn to give a description of Fibonacci numbers with at most two distinct prime divisors If n is a positive integer, write ω(n) for the number of distinct prime divisors of n We prove that the inequality ω(f n ) (log n) log 2+o() holds for almost all n We conjecture that ω(f n ) log n for composite n, and give a heuristic argument in support of this conjecture 2 The Theorems of Carmichael and Cohn We need the following two celebrated results on Fibonacci numbers The first is due to Carmichael [4] and the second to Cohn [5] Theorem 2 Let n > 2 and n 6, 2 then F n has a prime divisor which does not divide any F m for 0 < m < n; such a prime is called a primitive divisor of F n Date: February 3, Mathematics Subject Classification Primary B39, Secondary K65 Key words and phrases Fibonacci numbers, arithmetic functions, prime divisors F Luca s work is funded in part by grants SEPCONACyT E and E S Siksek s work is funded by a grant from Sultan Qaboos University (IG/SCI/DOMS/02/06)
2 2 YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK Theorem 3 Let 0 < m < n and suppose that the product F m F n is a square, then (m, n) = (, 2), (, 2), (2, 2) or (3, 6) 3 Fibonacci numbers with ω(f n ) 2 Write ω(m) for the number of distinct prime factors of m Notice that ω(f n ) > 0 for n > 2 and that for integers m and n, with < m < n and mn 6, 2, we have () ω(f mn ) > ω(f m ) + ω(f n ), if (m, n) =, because first F m and F n both divide F mn and are coprime when the indices are coprime, and secondly Theorem 2 implies that there is a prime number dividing F mn which does not divide the product F m F n Lemma 3 Suppose n is a positive integer and and ω(f n ) 2 n =, 2, 4, 8, 2 or n = l, 2l, l 2 for some odd prime number l Then either Proof The Lemma follows straightforwardly from the fact that ω(f n ) 3 in the following cases: (a) if 6 n, or 24 n, or (b) if pq n where p, q are distinct odd primes, or (c) if l 3 n or 2l 2 n or 4l n for some odd prime l, unless n = 2 These may be verified using a combination of Theorem 2 and inequality () For example, if 4l n where l 3 is an odd prime then ω(f n ) ω(f 4l ) > ω(f 2l ) > ω(f l ), where the strict inequalities follow from Theorem 2 Thus ω(f n ) 3 If n = 2l with l > 3 then F n = F l L l and ω(f n ) = 2 implies that F l and L l are prime numbers (because of Theorem ) Consider now the case n = l 2 with ω(f n ) = 2 Then ω(f l ) = and, by Theorem, F l = p = prime If l = 5 then F l = 5 and F 25 = If l 5 then p does not divide F n /F l so that F n = pq t, say By Theorem 3, the product F l F n = p 2 q t is not a square, thus t is odd We have proved the following result Theorem 4 The only solutions to the equation ω(f n ) = are given by n = 4, n = 6 or n is an odd prime number for which F n is prime The only solutions to the equation ω(f n ) = 2 are given for even n by n = 8 or n = 2 or n = 2l where l is some odd prime number for which F n = F l L l where F l and L l are both prime numbers For odd n, the only possible cases are n = l or n = l 2 Moreover if ω(f n ) = 2 and n = l 2 then F l must be prime, say F l = p; and if l 5, then F n = pq t where q is a prime number distinct from p, and the exponent t is odd A short computer search reveals examples of all of the above possibilities Thus
3 FIBONACCI NUMBERS WITH FEW PRIME DIVISORS 3 examples of ω(f n ) = with n prime are: F 5 = 5, F 7 = 3, F = 89, F 3 = 233, examples of ω(f n ) = 2 with n = 2l are: F 0 = 55, F 4 = 3 29, F 22 = 89 99, F 26 = , examples of ω(f n ) = 2 with n = l, l 2 are: F 9 = 34, F 9 = 37 3, F 25, F 3 = ,, F 2 = , Remark Lemma 3 can also be proved using some results from [7] For example, one of the results from [7] is that, if τ(m) denotes the number of divisors of the positive integer m, then we have τ(f n ) F τ(n) for any positive integer n, with equality only for n =, 2, 4 Assuming that ω(f n ) 2, one gets immediately that (up to a few exceptions) τ(n) 4, giving that n is the product of two distinct prime numbers or the square of a prime number, which together with Theorem 2 yields the conclusion of Lemma 3 4 Fibonacci numbers rarely have few prime factors Theorem 4 in effect tells us that ω(f n ) 3 for almost all n We mean by this that the set of integers for which this inequality holds has density Indeed, much more is true For example, by the following theorem, ω(f n ) C holds for almost all n, whatever the value of C Theorem 5 The inequality ω(f n ) (log n) log 2+o() holds for almost all n Proof By Theorem 2, we know that if a divisor d of n is not, 2, 6 or 2 then F d has a primitive prime factor This translates immediately in saying that ω(f n ) τ(n) 4, where τ(n) is the total number of divisors of n Certainly, τ(n) 2 ω(n) Since ω(n) = ( + o()) log log n for almost all n, the desired inequality follows 5 Heuristic results Given the present state of knowledge in analytic number theory, it seems unrealistic to obtain precise results as to whether each of the possibilities from Theorem 4 occurs finitely many times or infinitely often There is, however, a standard heuristic argument which suggests that there are infinitely many primes l with F l prime, but only finitely many primes l with ω(f 2l ) = 2 or ω(f l 2) = 2 The heuristic argument goes as follows (compare with [6, page 5]): from the Prime Number Theorem, the probability that a positive integer m is prime is / log m Thus the expected number of primes l such that F l is also prime is log F l A for some positive constant A Since this last series diverges (albeit very slowly), it is reasonable to guess that there are infinitely prime F l Applying the same heuristic argument suggests that there are only finitely many primes l with ω(f 2l ) = 2 or ω(f l 2) = 2 For example, the expected number of primes l with ω(f 2l ) = 2 is log F l log L l B l l 2 <
4 4 YANN BUGEAUD, FLORIAN LUCA, MAURICE MIGNOTTE, SAMIR SIKSEK where B is some positive constant In fact we can go even further We conjecture the following Conjecture 5 ω(f n ) log n holds for all composite positive integers n In order to justify this conjecture, let us make the following heuristic principle Heuristic 52 Let k : N N be any function Let A be a subset of positive integers such that there is no algebraic reason for a A to have more than k(a) prime factors Assume that for every a A there exists a proper divisor of a, let us call it ã >, such that the greatest common divisor of ã and a/ã has at most one prime factor and the series (log log ã) k (log log(a/ã)) k2 (k )! (k 2 )! log ã log(a/ã) a A k +k 2 k(a)+ is convergent Then ω(a) k(a) should hold only for finitely many a A Heuristic 52, is based on the fact that the probability for a positive integer n to have k distinct prime factors is (k )! (log log n) k log n So, if a A and ω(a) = k k(a), then there exist nonnegative integers k and k 2 such that ω(ã) = k and ω(a/ã) = k 2 Furthermore, since the greatest common divisor of ã and a/ã has at most one prime factor, it follows that either k + k 2 = k k(a) or k + k 2 = k + k(a) + Thus, the above sum represents just the sum of the probabilities that ω(ã) = k and ω(a/ã) = k 2 assuming that such events are independent 5 Conjecture 5 follows from Heuristic 52 Let n be a composite integer Observe first that, by Theorem 2, if a divisor d of n is different from, 2, 6 or 2, then F d has a prime factor not dividing F d for any positive integer d < d Consequently, if n has at least 03 log n divisors, then we have (2) ω(f n ) max{, 03 log n 6} log n Let A be the set of Fibonacci numbers F n, where n runs through the composite integers having less than 03 log n divisors Let n be composite and write m for the largest proper divisor of n Clearly, m = n/p(n), where p(n) is the smallest prime factor of n Note that m n /2 It is known that gcd(f m, F n /F m ) p(n), therefore the two numbers F m and F n /F m share at most one prime factor Fix k and let k and k 2 be such that k + k 2 = k One checks immediately that both inequalities F m < e m and F n /F m < e n m hold Thus, we get (k )!(k 2 )! (log log F m) k (log log(f n /F m )) k2 k +k 2=k (k 2)! = k +k 2=k ( ) k 2 (log m) k (log(n m)) k2 k (k 2)! (log m + log(n m))k 2 <, k 2
5 FIBONACCI NUMBERS WITH FEW PRIME DIVISORS 5 where for the last inequality above we used Stirling s formula Using the inequalities log F m m n /2 and log(f n /F m ) (n m) n (because m divides n), it follows that k +k 2=k (log log F m ) k (log log(f n /F m )) k2 (k )!(k 2 )! log F m log(f n /F m ) n 3/2 k 2 For a fixed y, the function x (2ey/x) x is increasing for x < 2y Thus, taking k(f n ) = c 0 log n, where c 0 < 2 is some constant, we get that k +k 2 k(f n) k k(f n) (log log F m ) k (log log(f n /F m )) k2 (k )!(k 2 )! log F m log(f n /F m ) n 3/2 log n k 2 n 3/2 ( 2e c 0 ) c0 log n Choosing c 0 such that c 0 log(2e/c 0 ) = c < /2 (we can choose, say, c 0 = 03), we get that the right hand side of the above inequality is log n n 3/2 c and summing up over n we get a convergent series Hence, Heuristic 52 with a = F n and ã = F m for F n in A implies that ω(f n ) < 03 log n holds only for finitely many composite integers n, which, by (2) and the definition of A, implies that ω(f n ) log n holds for all composite integers n Actually, our choice of A takes into account an algebraic reason for which F n has more than 03 log n divisors References [] C Batut, K Belabas, D Bernardi, H Cohen and M Olivier, User s guide to PARIGP, version 2 (See also [2] W Bosma, J Cannon and C Playoust, The Magma Algebra System I: The User Language, J Symb Comp 24 (997), (See also [3] Y Bugeaud, M Mignotte and S Siksek, Classical and modular approaches to exponential Diophantine equations, I Fibonacci and Lucas Perfect Powers Submitted [4] R D Carmichael, On the numerical factors of arithmetical forms α n ± β n, Ann of Math (2) 5 (93), [5] J H E Cohn, Squares in some recurrence sequences, Pacific J Math 4 (972), [6] G H Hardy and E M Wright, Introduction to the theory of numbers, Oxford University Press, fourth edition, 975 [7] F Luca, Arithmetic functions of Fibonacci numbers, Fibonacci Quart 37 (999), Yann Bugeaud, Université Louis Pasteur, U F R de mathématiques, 7, rue René Descartes, Strasbourg Cedex, France address: Florian Luca, Instituto de Matemáticas, UNAM, Ap 5880, Morelia, Michoacán, Mexico address: Postal 63 (Xangari), CP Maurice Mignotte, Université Louis Pasteur, U F R de mathématiques, 7, rue René Descartes, Strasbourg Cedex, France address: Samir Siksek, Department of Mathematics and Statistics, College of Science, Sultan Qaboos University, PO Box 36, AlKhod 23, Oman address:
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