QUOTIENTS OF FIBONACCI NUMBERS

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1 QUOTIENTS OF FIBONACCI NUMBERS STEPHAN RAMON GARCIA AND FLORIAN LUCA Abstract. There have been many articles in the Monthly on quotient sets over the years. We take a first step here into the p-adic setting, which we hope will spur further research. We show that the set of quotients of nonzero Fibonacci numbers is dense in the p-adic numbers for every prime p. The nth Fibonacci number F n is defined by the recurrence relation F n+ = F n+1 +F n with initial conditions F 0 = 0 and F 1 = 1. Let F = {1,, 3, 5, 8, 13, 1,...} denote the set of nonzero Fibonacci numbers and let R(F) = {F m /F n : m, n N} be the set of all quotients of elements of F. Binet s Formula F n = 1 5 (ϕ n ϕ n ), (1) in which ϕ = = and ϕ = 1 5 = , implies that F n ϕ n / 5 tends to zero exponentially fast as n [13, p.57]. So as a subset of R, R(F) accumulates only at the points ϕ k for k Z [4, Ex.17]. On the other hand, R(F) is a subset of the rational number system Q, which can be endowed with metrics other than the one inherited from R. For each prime p, there is a p-adic metric on Q, with respect to which Q can be completed to form the set Q p of p-adic numbers. Our aim here is to prove the following theorem. Theorem. R(F) is dense in Q p for every prime p. By this, we mean that the closure of R(F) in Q p with respect to the p-adic metric is Q p. Although there have been many papers in the Monthly on quotient sets over the years [1 4,6,7,11,1], we are unaware of similar work being undertaken in the p-adic setting. We hope that this result will spur further investigations. Definition 3. If p is a prime number, then each r Q\{0} has a unique representation r = ±p k a b, (4) in which k Z, a, b N, and a, b, p are pairwise relatively prime. The p-adic valuation ν p : Q Z is defined by ν p (0) = + and, for r as in (4), by ν p (r) = k. It satisfies ν p (xy) = ν p (x) + ν p (y) for x, y Q. The p-adic absolute value on Q is r p = p νp(r) and the p-adic metric on Q is d p (x, y) = x y p. The p-adic number system Q p is the completion of Q with respect to the p-adic metric. Ostrowski s theorem asserts that Q p, for p prime, and R are essentially the only completions of Q that arise from an absolute value [8]. Thus, it is natural to seek results like Theorem, since similar questions in R have long since been explored. 1

2 S.R.GARCIA AND F.LUCA In fact, Q p is a field that contains Q as a subfield. Its elements can be represented as infinite series, convergent with respect to the p-adic metric, of the form n=k a np n, in which k Z and a n {0, 1,..., p 1}. The p-adic valuation, norm, and metric extend uniquely to Q p in a manner that respects these series representations. Example 5. In Q, we have = 1 since N 1 n ( 1) = 1 N = 1 (1 N ) = N = N n=0 tends to zero as N. Manipulating p-adic numbers is analogous to handling decimal expansions of real numbers. Instead of powers 10 n with n running from some nonnegative N to, we have powers p n with n running from N (potentially negative) to +. Further details on Q p can be found in [5, 8]. To prove our theorem, we require a few preliminary results. In what follows, we write a b to denote that the integer b is divisible by the integer a. Proofs of the following assertions can be found in [13, p.8, p.73, & p.81]. Lemma 6. (a) If j k, then F j F k. (b) For each m N, there is a smallest index z(m) so that m F kz(m) for all positive integers k. (c) ν p (F z(p)p j ) = ν p (F z(p) ) + j for any odd prime p and j N. We also need a convenient dense subset of Q p. We freely make use of the fact that d p (x, y) 1/p k if and only if ν p (x y) k. Lemma 7. For each prime p, the set {p r n : n, r N} is dense in Q p. Proof. Since Q p is the closure of Q with respect to the p-adic metric, it suffices to show that each rational number can be arbitrarily well approximated, in the p-adic metric, by rational numbers of the form p r n with n, r N. If x = p k j=0 a jp j Q p, then let N k + 1 and n = N k 1 j=0 a j p j so that ( ν p (x p k n) = ν p p k a j p j) ( = k + ν p a j p j) k + (N k) = N. j=n k j=n k An algebraic integer is a root of a monic polynomial with integer coefficients. For instance,, 5, and ϕ are algebraic integers. They are roots of x, x 5, and x x 1, respectively. The following is [10, Cor.1, p.15]. Lemma 8. A rational algebraic integer is an integer. We also require a few facts about arithmetic in the field K = Q( 5). Let O K denote the set of algebraic integers in K. It is a ring; in fact, O K = {a+bϕ : a, b Z} [10, Cor., p.15]. In particular, ϕ, ϕ, and 5 each belong to O K. An ideal in O K is a subring i of O K so that αi i for all α O K. A prime in O K refers to a prime ideal in the ring O K. A prime ideal is an ideal p O K with the property that, for any α, β O K, the condition αβ p implies that α p or β p. To avoid confusion, we refer to the primes, 3, 5, 7,... as rational primes.

3 QUOTIENTS OF FIBONACCI NUMBERS 3 The product of two ideals i, j in O K is defined by ij = {αβ : α i, β j}; it is also an ideal in O K. Positive powers of ideals are defined inductively by i 1 = i and i n = i(i n 1 ) for n =, 3,.... We say that i divides j if i j. For an ideal i in O K, we write α β (mod i) to mean that β α i. The familiar properties of congruences hold when working modulo an ideal. For instance, α β (mod i) implies that α j β j (mod i) for j N. Similarly, if p and q are distinct prime ideals, α β (mod p), and α β (mod q), then α β (mod pq). Lemma 9. Let a, b Z and j N. If a b (mod p j ), then p j divides b a in Z. Proof. Suppose that a, b Z and a b (mod p j ). Then, b a = p j α for some α O K. Since α = (b a)/p j is a rational algebraic integer, it is an integer by Lemma 8. Consequently, p j divides b a in Z. Each rational prime p gives rise to an ideal p = po K in O K This ideal factors uniquely as a product of prime ideals [10, Th.16, p.59]. In fact, since Q( 5) is a quadratic extension of Q we have [10, p.74]: Lemma 10. Let K = Q( 5) and let p be a rational prime. Then p = po K is the product of at most two prime ideals (not necessarily distinct). We can be a little more specific. The only rational prime p for which p is a square of a prime ideal is p = 5; this reflects the factorization 5 = ( 5) [10, Thm.4, p.7]. Proof of Theorem. There are two cases, according to whether p is odd or p =. Let p be an odd rational prime and let p = po K. For each j N, Lemma 6 ensures that p j F z(p)p j. Since 5 O K, (1) reveals that and hence ϕ z(p)pj ϕ z(p)pj = 5F z(p)p j 0 (mod p j ), ϕ z(p)pj ϕ z(p)pj (mod q j ) for each prime q in O K that divides p. Since ϕ = 1/ϕ, we have so that ϕ z(p)pj ϕ z(p)pj (mod q j ), ϕ 4z(p)pj 1 (mod q j ) and ϕ 4z(p)pj 1 (mod q j ). (11) Setting x = ϕ 4z(p)pj and y = ϕ 4z(p)pj in the identity implies that F 4z(p)p j m F 4z(p)p j x m y m = (x y)(x m 1 + x m y + + y m 1 ), m N, = ϕ4z(p)pjm ϕ 4z(p)pj m ϕ 4z(p)pj ϕ 4z(p)pj = (ϕ 4z(p)pj ) m 1 + (ϕ 4z(p)pj ) m ( ϕ 4z(p)pj ) + + ( ϕ 4z(p)pj ) m 1 m (mod q j ) by (11). Now Lemma 10 ensures that F 4z(p)p j m F 4z(p)p j m (mod p j )

4 4 S.R.GARCIA AND F.LUCA for any m N. However, F 4z(p)p j m/f 4z(p)p j is a rational integer by Lemma 6, so F 4z(p)p j m F 4z(p)p j by Lemma 9. Appealing to Lemma 6, we have so m (mod p j ) (1) ν p (F 4z(p)p (k+r)) = ν p (F 4z(p)p (k+r)) + r, (13) F 4z(p)p (k+r) F 4z(p)p (k+r) = p r l, l Z, gcd(l, p) = 1. (14) Set m = p r nl and j = k + r in (1) and use (14) to conclude that p r l F4z(p)p (k+r) p r nl F 4z(p)p (k+r) Since gcd(l, p) = 1, the preceding yields = F 4z(p)p (k+r) p r nl F 4z(p)p (k+r) p r nl (mod p k+r ). and hence p r F 4z(p)p k+3r nl F 4z(p)p k+4r ν p ( F4z(p)p k+3r nl F 4z(p)p k+4r n (mod p k ), ) p r n k. This holds for all n, r N and all k > r, so R(F) is dense in Q p by Lemma 7. If p =, we must replace the exponents 4z(p)p j in (11) with 3 j+ since z() = 3 and 4 =. The proof can proceed as before if we replace (13) with the corresponding statement for p =. It suffices to show that ν (F 3 j ) = j + for j N; see [9] for a more general result. We proceed by induction on j. The base case j = 1 is ν (F 6 ) = ν (8) = 3. Now suppose that ν (F 3 j ) = j + for some j. Let L n denote the nth Lucas number; these satisfy the recurrence L n+ = L n+1 + L n with initial conditions L 0 = and L 1 = 1. Since L n = ( 1) n + 5Fn [13, p.177] and since = F 3 divides F 3 j 1, we have ν (L 3 j ) = ν(l (3 j 1 )) = ν ( ( 1) 3 j 1 + 5F 3 j 1 ) = 1. Because F n = F n L n [13, p.177], we conclude that ν (F 3 j+1) = ν (F (3 j )) = ν (F 3 j ) + ν (L 3 j ) = (j + ) + 1 = (j + 1) +, which completes the induction. Acknowledgments: This work was partially supported by National Science Foundation Grant DMS We thank the anonymous referees for numerous helpful suggestions. References [1] B. Brown, M. Dairyko, S. R. Garcia, B. Lutz, and M. Someck, Four quotient set gems, Amer. Math. Monthly 11 (014) [] J. Bukor and J. T. Tóth, On accumulation points of ratio sets of positive integers, Amer. Math. Monthly 103 (1996) [3] S. R. Garcia, Quotients of Gaussian Primes, Amer. Math. Monthly 10 (013) [4] S. R. Garcia, V. Selhorst-Jones, D. E. Poore, and N. Simon, Quotient sets and Diophantine equations, Amer. Math. Monthly 118 (011) [5] F. Q. Gouvêa, p-adic numbers: an introduction, second ed., Universitext, Springer-Verlag, Berlin, 1997.

5 QUOTIENTS OF FIBONACCI NUMBERS 5 [6] S. Hedman and D. Rose, Light subsets of N with dense quotient sets, Amer. Math. Monthly 116 (009) [7] D. Hobby and D. M. Silberger, Quotients of primes, Amer. Math. Monthly 100 (1993) [8] N. Koblitz, p-adic numbers, p-adic analysis, and zeta-functions, second ed., Graduate Texts in Mathematics, vol. 58, Springer-Verlag, New York, [9] T. Lengyel, The order of the Fibonacci and Lucas numbers, Fibonacci Quart. 33 (1995) [10] D. A. Marcus, Number fields, Springer-Verlag, New York-Heidelberg, 1977, Universitext. [11] A. Nowicki, Editor s endnotes, Amer. Math. Monthly 117 (010) [1] P. Starni, Answers to two questions concerning quotients of primes, Amer. Math. Monthly 10 (1995) [13] S. Vajda, Fibonacci & Lucas numbers, and the golden section, Ellis Horwood Series: Mathematics and its Applications, Ellis Horwood Ltd., Chichester; Halsted Press [John Wiley & Sons, Inc.], New York, 1989, Theory and applications, With chapter XII by B. W. Conolly. Department of Mathematics, Pomona College, 610 N. College Ave., Claremont, CA address: stephan.garcia@pomona.edu URL: School of Mathematics, University of the Witwatersrand, Private Bag X3, Wits 050, Johannesburg, South Africa address: Florian.Luca@wits.ac.za

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