QUOTIENTS OF FIBONACCI NUMBERS

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 = 1 + 5 = 1.618... and ϕ = 1 5 = 0.618..., 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

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 + + + 3 + 4 + = 1 since N 1 n ( 1) = 1 N 1 + 1 = 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.

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 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-165973. 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) 590 599. [] J. Bukor and J. T. Tóth, On accumulation points of ratio sets of positive integers, Amer. Math. Monthly 103 (1996) 50 504. [3] S. R. Garcia, Quotients of Gaussian Primes, Amer. Math. Monthly 10 (013) 851 853. [4] S. R. Garcia, V. Selhorst-Jones, D. E. Poore, and N. Simon, Quotient sets and Diophantine equations, Amer. Math. Monthly 118 (011) 704 711. [5] F. Q. Gouvêa, p-adic numbers: an introduction, second ed., Universitext, Springer-Verlag, Berlin, 1997.

QUOTIENTS OF FIBONACCI NUMBERS 5 [6] S. Hedman and D. Rose, Light subsets of N with dense quotient sets, Amer. Math. Monthly 116 (009) 635 641. [7] D. Hobby and D. M. Silberger, Quotients of primes, Amer. Math. Monthly 100 (1993) 50 5. [8] N. Koblitz, p-adic numbers, p-adic analysis, and zeta-functions, second ed., Graduate Texts in Mathematics, vol. 58, Springer-Verlag, New York, 1984. [9] T. Lengyel, The order of the Fibonacci and Lucas numbers, Fibonacci Quart. 33 (1995) 34 39. [10] D. A. Marcus, Number fields, Springer-Verlag, New York-Heidelberg, 1977, Universitext. [11] A. Nowicki, Editor s endnotes, Amer. Math. Monthly 117 (010) 755 756. [1] P. Starni, Answers to two questions concerning quotients of primes, Amer. Math. Monthly 10 (1995) 347 349. [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 91711 E-mail address: stephan.garcia@pomona.edu URL: http://pages.pomona.edu/~sg064747 School of Mathematics, University of the Witwatersrand, Private Bag X3, Wits 050, Johannesburg, South Africa E-mail address: Florian.Luca@wits.ac.za