PERIOD LENGTHS OF CONTINUED FRACTIONS INVOLVING FIBONACCI NUMBERS R.A. Mollin Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada, TN 1N e-mail ramollin@math.ucalgary.ca K. Cheng Department of Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada, TN 1N (Submitted October 001-Final Revision July 00) 1. INTRODUCTION For given nonsquare positive integers C 5(mod 8), we investigate families {D k (X)} k N of integral polynomials of the form D k (X) = A k X +B k X+C where (B k /) (A k /) C =, and show that the period length of the simple continued fraction expansion of (1+ D k (X))/ is a multiple of k, and independent of X. For each member of the families involved, we show how to easily determine the fundamental unit of the underlying quadratic order Z[(1+ D k (X))/]. We also demonstrate how the simple continued fraction expansion of (1+ D k (X))/ is related to that of (1 + C)/. As applications, we present infinite families of continued fractions related to the Fibonacci numbers. This continues work in [11]-[1] and corrects errors in [15] (see Theorem 3.1). In 199, Nyberg [1] found the first example of a parametric family in which a fundamental unit can be easily produced even though the period length of the continued fraction gets arbitrarily large. Since this discovery, there have been a number of generalizations: Dan Shanks [16]-[17] (see also Yamamoto []) in 1969-1971, Hendy [5] in 197, Bernstein []-[3] in 1976, Williams [0] in 1985, Levesque and Rhin [7] in 1986, Azuhata [1] in 1987, Levesque [6] in 1988, Halter-Koch [] in 1989, Mollin and Williams [13] in 199, Williams [19] in 1995, and van der Poorten and Williams in [15] in 1999. In [1], we found infinite families of quadratic Pellian polynomials D k (X) such that the continued fraction expansions D k (X) have unbounded period length for arbitrary k, while for fixed k and arbitrary X > 0, have constant period length. In this paper, we continue the investigation for simple continued fraction expansions of (1 + D k (x))/, where D k (X) are quadratic polynomials of Eisenstein type (see Section 3 below). We are also able to find explicit fundamental units of the order Z[(1 + D k (X))/]. A consequence of the result is an infinite family of continued fraction expansions whose fundamental units and discriminant are related to Fibonacci numbers. Ostensibly the first such finding of these types of continued fractions was given in [1], and the findings herein appear to be the second, albeit distinctly different from those given in [1]. The relatively small fundamental units for the underlying quadratic order which we explicitly determine means that we have large class numbers h D(X) for Z[(1 + D(X))/]. The reason behind this fact is Siegel s class number result [18] which tells us that for positive discriminants, lim log(h R)/ log( ) = 1 161
where h is the class number of Q( ), ε is its fundamental unit, and R = log(ε ) is the regulator of Q( ). Hence, a small fundamental unit will necessarily mean a large class number.. NOTATION AND PRELIMINARIES The background for the following together with proofs and details may be found in most standard introductory number theory texts (see for example [9], and for a more advanced exposition with detailed background on quadratic orders, see [8]). Let = d D 0 (d N, D 0 > 1 squarefree) be the discriminant of a real quadratic order O = Z + Z[ ] = [1, ] in Q( ), U the group of units of O, and ε the fundamental unit of O. Now we introduce the notation for continued fractions. Let α O. We denote the simple continued fraction expansion of α (in terms of its partial quotients) by: If α is periodic, we use the notation: α = q 0 ; q 1,, q n,. α = q 0 ; q 1 q, q k 1, q k, q k+1,..., p l+k 1, to denote the fact that q n = q n+l for all n k. The smallest such l = l(α) N is called the period length of α. If k = 0 is the least such nonnegative value, then α is purely periodic, namely α = q 0 ; q 1,..., q l 1. The convergents (for n 0) of α are denoted by We will need the following facts. x n y n = q 0 ; q 1,..., q n = q nx n 1 + x n q n y n 1 + y n. (.1) x j = q j x j 1 + x j (for j 0 with x = 0, and x 1 = 1), (.) y j = q j y j 1 + y j (for j 0 with y = 1, and y 1 = 0), (.3) and x j y j 1 x j 1 y j = ( 1) j 1 (j N). (.) In particular, we will be dealing with α = (1 + D)/ where D is a radicand. In this case, the complete quotients are given by (P j + D)/Q j where the P j and Q j are given by the recursive formulae as follows for any j 0 (with P 0 = 1 and Q 0 = ): P j + D q j =, (.5) Q j P j+1 = q j Q j P j, (.6) 16
and D = P j+1 + Q j Q j+1. (.7) Thus, we may write: (1 + D)/ = q o ; q 1,..., q n, (P n+1 + D)/Q n+1. (.8) We will also need the following facts for α = (1 + D)/. For any integer j 1, let g j 1 = x j 1 y j 1, (.9) then for any nonnegative integer j, g j 1 = P j y j 1 + Q j y j, (.10) and if l = l((1 + D)/), then Dy j 1 = P j y j 1 + Q j y j, (.11) P 1 = P jl = q 0 1 and Q 0 = Q jl =, (.1) and for any j N, Also, for any 1 j l 1, g jl 1 y jl 1D = ( 1) jl. (.13) q j < q 0. (.1) We close this section with a result on Eisenstein s equation, which will be quite useful in establishing results in the next section. In fact, this gives more detail to the fact exhibited in (.13) for this case. It is know that the Diophantine Equation: x Dy = with gcd(x, y) = 1 (.15) was studied by Eisenstein in search of a criterion for its solvability, a question already asked by Gauss. Several criteria are known (for instance, see [8, pp. 59-61]). In particular, when D 5(mod 8), it is known that (.15) has a solution if and only if the fundamental unit of Z[(1 + D)/] is not in the order Z[ D]. Moreover, it can be shown that (.15) is solvable if and only if the ideal I = [, 1 + D] is principal in Z[ D]. For instance, if D = 37, there is no solution to (.15) since ε D = 6 + 37 Z[ D], whereas for D = 13, there is a solution since ε 13 = (3 + 13)/ Z[ 13]. Theorem.1: Let D 5(mod 8) be a nonsquare positive integer such that ε D Z[ D]. For j 0, let x j and y j be as above in the simple continued fraction expansion of (1 + D)/ and let l = l((1 + D)/). If l is even, then all solutions of x Dy = with gcd(x, y) = 1, (.16) are given by (x, y) = (g jl 1, y jl 1 ) (.17) 163
for j 1, whereas there are no integer solutions of x Dy =, with gcd(x, y) = 1. (.18) If l is odd, then all positive solutions of (.16) are given by (x, y) = (g jl 1, y jl 1 ) for j 1, whereas all positive solutions of (.18) are given by (x, y) = (g (j 1)l 1, y (j 1)l 1 ). Proof: See [9, Theorem 5.3., p. 6]. For an algorithm, relying only on continued fraction expansions, which finds all primitive solutions of x Dy = a for any integer a and any nonsquare D > 0, see [10]. 3. RESULTS We make the following assumptions throughout. We let A, B, C, k, X N and C 5(mod 8) not a perfect square. Suppose that (x, y) is the smallest positive solution of x Cy = with gcd(x, y) = 1. Set A = y and B = x and define, for each k N, B k + A k C = (B + A C) k / k 1. Also, set and for m N define 1 + C = c 0 ; c 1,..., c n, c 0 1, w m = c 1,..., c n, c 0 1, c 1,..., c n, c 0 1,..., c 1,..., c n, which is m iterations of c 1,..., c n, c 0 1 followed by one iteration of c 1,..., c n. Theorem 3.1: Let D k (X) = A k X + B k X + C. Then the fundamental solutions of u D k (X)v = is (u, v) = ((A k X + B k)/, A k /). In other words, If q 0 = A k X/ + c 0, then: (a) if n 0 is even, ε Dk (X) = A k X + B k + A k Dk (X). 1 + D k (X) = q 0 ; w k 1, q 0 1 with l ( 1 + ) D k (X) = k(n + 1); 16
(b) and if n is odd, 1 + D k (X) = q 0 ; w k 1, q 0 1 with l ( 1 + ) D k (X) = k(n + 1). Proof: We observe that (A k /) D k (X) = ((A kx + B k ))/), which is not a perfect square. Thus, by [8, Theorem 3..1, p. 78], ε (Ak /) D k (X) = A k X + B k + 1 A k D k(x) = A k X + B k + A k Dk (X). Let X j /Y j be the j th convergent of (1 + D k (x))/, so from (.9), G j = X j Y j. Since ( ) A k X + B k ( Ak ) D k (X) =, then by Theorem.1, there is a j N such that A k X + B k A k = G jl 1 Y jl 1. We now show that j = 1, namely, via Theorem.1, that ε (Ak /) D k (X) = ε Dk (X), since l will be shown to be even, and by so doing that the continued fraction expansions in (a)-(b) hold. First, we deal with part (a). If x j /y j is the j th convergent of (1 + C)/, then by (.1), A k X/ + c 0 ; w k 1 = A k X/ + c 0, w k 1, c 0 1 + y k(n+1) /y k(n+1) 1 = A k X + (c 0 1 + y k(n+1) /y k(n+1) 1 )x k(n+1) 1 + x k(n+1) (c 0 1 + y k(n+1) /y k(n+1) 1 )y k(n+1) 1 + y k(n+1) = 165
A k X + ((c 0 1)y k(n+1) 1 + y k(n+1) )x k(n+1) 1 + x k(n+1) y k(n+1) 1 ((c 0 1)y k(n+1) 1 + y k(n+1) )y k(n+1) 1 + y k(n+1) y k(n+1) 1, and using the facts: g k(n+1) 1 = (c 0 1)y k(n+1) 1 + y k(n+1) (see (.10) and (.1)) and x k(n+1) 1 y k(n+1) x k(n+1) y k(n+1) 1 = ( 1) k(n+1) = ( 1) k (see (.)), this equals: A k X + x k(n+1) 1(g k(n+1) 1 y k(n+1) ) + x k(n+1) y k(n+1) 1 (g k(n+1) 1 y k(n+1) )y k(n+1) 1 + x k(n+1) 1 y k(n+1) 1 = A k X + x k(n+1) 1g k(n+1) 1 + ( 1) k+1 g k(n+1) 1 y k(n+1) 1. (3.19) However, by Theorem.1, since n is even and (B k /) (A k /) C =, then B k / = g k(n+1) 1 and A k / = y k(n+1) 1. Moreover, g k(n+1) 1 + y k(n+1) 1 C = (gn + y n C) k / k 1 = (g k(n+1) 1 + y k(n+1) 1 C) / = (g k(n+1) 1 + y k(n+1) 1 C + g k(n+1) 1y k(n+1) 1 C)/ = g k(n+1) 1 + ( 1)k+1 + g k(n+1) 1 y k(n+1) 1 C where the last equality follows from Equation (.13) given that l((1 + C)/) = n + 1. Hence, B k = g k(n+1) 1 + ( 1)k+1 and A k = g k(n+1) 1 y k(n+1) 1. Thus A k + B k = g k(n+1) 1 y k(n+1) 1 + g k(n+1) 1 ) + ( 1)k+1 = g k(n+1) 1 (y k(n+1) 1 + g k(n+1) 1 + ( 1) k+1 = by (.9) so g k(n+1) 1 x k(n+1) 1 + ( 1) k+1 A k + B k = g k(n+1) 1 x k(n+1) 1 + ( 1) k+1. 166
Plugging this into (3.19), we get, A k X/ + c 0 ; w k 1 = A kx + A k + B k = (A k X + A k + B k )/. A k A k / By (.1), q 0 1 = A k X + c 0 1 c j for any 0 < j < l. Thus, since a convergent X jl 1 /Y jl 1 can only occur at the end of the j th multiple of a complete period, then j = 1. In other words, A k X/ + c 0, w k 1 = X l 1 Y l 1. Since we have achieved the (l 1) th convergent of D k (X), then D is as given in (a), and l = l((1 + D k (X))/) = k(n + 1) for all k N and all even n N. Moreover, since X l 1 = (A k X + A k + B k )/ and Y l 1 = A k /, then by (.9), G l 1 = X l 1 Y l 1 = (A kx + B k )/, which completes the proof of part (a). To prove (b), we assume that n is odd. Then by Theorem.1, B k / = g k(n+1) 1, A k / = y k(n+1) 1, and A k X/ + c 0, w k 1 = A k X/ + x k(n+1) 1 /y k(n+1) 1 = A k X/ + (g k(n+1) 1 + y k(n+1) 1 )/y k(n+1) 1 = A k X + A k + B k A k, where the penultimate equality comes from (.9). As in the proof of (a), X l 1 = (A k X + A k+ B k )/ and Y l 1 = A k /, with the last equality as in the proof of (a). Hence, (1 + D k (X))/ has the continued fraction expansion as given in (b) and l((1 + D k (X))/) = k(n + 1). Moreover, by (.9), G l 1 = (A k X + B k)/. This completes the proof. Remark 3.1: We observe that in the proof of Theorem 3.1, A k D k(x) = (A k X + B k), is an RD-type, and we verified that ε (Ak /) D k (X) = A k X + B k + 1 A k D k(x) = A k X + B k + A k Dk (X). This is the underlying kernel of the proof and shows that, even though ERD types are avoided for the D k (X) since such types have period lengths no bigger than 1, we nevertheless exploit the order Z[(1 + (A k + B k) )/] in Z[(1 + D k (X))/] to achieve our goals. 167
Remark 3.: We see that for a fixed C (and so a fixed n N), we may let k in which case l((1 + D k (X))/) = l((1 + D k (1))/) for all X N. We also see that we have infinitely many distinct radicands D k (X) for a fixed k N with l((1 + D k (X))/) = l((1 + D k (X + 1))/) for all X N. Remark 3.3: It is well known that if a discriminant is of ERD-type, then log(ε ) e(log ) where e = 1 or (for example see [8]). Since ε (Ak /) D k (X) = ε Dk (X), (see Remark 3.1), then we necessarily have small regulator R = log(ε Dk (X)) compared to log( D k (X)), and so large class number h Dk (X). For example, h D1 (150) = h 9507773061 = 1656, R = log(ε D1 (150)) = 16.76068..., and log( 9507773061) = 11.8768.... We conclude with an application of Theorem 3.1 which motivated this paper s title. In [1], the authors found an infinite sequence of radicands D k = (F 6k + 1) + (8F 6k + ), where F j is the j th Fibonacci number, such that l((1 + D k )/) = 6k + 1. By employing Theorem 3.1(a), we get D k (X) = F kx + (0F k + 8( 1) k )X + 5 = L kf k X + (0F k + 8( 1) k )X + 5 with l((1 + D k (X))/) = k for any such k, where L k is the k th Lucas number. ACKNOWLEDGMENTS The first author s research is supported by NSERC Canada grant #A88, which also supports the second author, his graduate student. We also welcome the opportunity to thank the referee for comments that made the paper more concise and to the point. REFERENCES [1] T. Azuhata. On the Fundamental Units and the Class Numbers of Real Quadratic Fields II. Tykyo J. Math 10 (1987): 59-70. [] L. Bernstein. Fundamental Units and Cycles. J. Number Theory 8 (1976): 6-91. [3] L. Bernstein. Fundamental Units and Cycles in the Period of Real Quadratic Fields, Part II. Pacific J. Math. 63 (1976): 63-78. [] F. Halter-Koch. Einige Periodische Kettenbruchentwicklungen und Grundeinheiten Quadratischer Ordnung. Abh. Math. Sem. Univ. Hamburg 59 (1989): 157-169. [5] M.D. Hendy. Applications of a Continued Fraction Algorithm to Some Class Number Problems. Math. Comp. 8 (197): 67-77. [6] C. Levesque. Continued Fraction Expansions and Fundamental Units. J. Math. Phys. Sci. (1988): 11-1. [7] C. Levesque and G. Rhin. A Few Classes of Periodic Continued Fractions. Utilitas Math. 30 (1986): 79-107. [8] R.A. Mollin. Quadratics; CRC Press, Boca Raton, New York, London, Tokyo (1996). [9] R.A. Mollin. Fundamental Number Theory with Applications. CRC Press, Boca Raton, New York, London, Tokyo (1998). [10] R.A. Mollin. Simple Continued Fraction Solutions for Diophantine Equations. Expositiones Math. 19 (001): 55-73. [11] R.A. Mollin. Polynomial Solutions for Pell s Equation Revisited. Indian J. Pure Appl. Math. 8 (1997): 9-38. 168
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