# PELL S EQUATION AND CONTINUED FRACTIONS. 1. Introduction

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1 PELL S EQUATION AND CONTINUED FRACTIONS I. GREGOR DICK Abstract. This paper first provides an exposition of the theory of simple continued fractions; it then discusses Pell s equation x 2 dy 2 = N and a method of Lagrange for obtaining its solutions; and finally it presents another method of solution relying on the continued fraction expansion of d which is shown to be equivalent to Lagrange s method. Pell s equation,. Introduction x 2 dy 2 = N, (.) is a Diophantine equation with d and N being nonzero integers. Its study is known to have been conducted as early as 400 B.C., in the context of a variety of geometric and combinatorial problems [4]. In more recent times the discovery of its solutions was proposed as a challenge by Fermat. John Pell contributed little to its analysis, but instead the equation acquired its eponym due to a misattribution by Euler. The objects of this paper are two-fold: firstly, to describe a method of Lagrange for solving (.); and then also to consider an alternative method relying on the theory of continued fractions, which provide a representation of real numbers, and whose properties we begin by exhibiting. 2. Continued Fractions 2.. Finite Simple Continued Fractions. Let n be a non-negative integer, and let {a i } 0 i n be a sequence of positive integers, except perhaps a 0 which may also be zero or negative. Define the finite simple continued fraction [a 0 ; a, a 2,..., a n ] thus: [a 0 ; a, a 2,..., a n ] = a 0 + a + a an. (2.) If the sequence {a n } is allowed to consist of any real numbers, then the corresponding continued fraction is not simple; however, in this paper we will restrict ourselves to the consideration of simple continued fractions, and so the appellation simple will often be omitted. It is clear that the value of a finite simple continued fraction is a rational number. The converse is also true: Theorem 2.. Let q Q. Then there exists a finite simple continued fraction [a 0 ; a, a 2,..., a n ] such that q = [a 0 ; a, a 2,..., a n ]. Such a continued fraction is said to be a continued fraction expansion of q. A proof is given in [] in which s expressed in the form a/b with (a, b) =, and the Euclidean algorithm is applied to a and b to yield the expansion of q. Date: October 6, This project was funded by a Vacation Scholarship from the Carnegie Trust. The author wishes to express his thanks to Chris Smyth of the University of Edinburgh for his kind supervision of this project.

2 2 I. GREGOR DICK Definition 2.2. Let x = [a 0 ; a, a 2,..., a n ] be a finite continued fraction. Then the a i are called partial quotients of the continued fraction, and, for 0 m n, [a 0 ; a, a 2,..., a m ] is called the m-th convergent to x. The continued fraction expansion of a rational number can also be shown to be unique up to a trivial pairing of expansions. Theorem 2.3. Let A = [a 0 ; a, a 2,..., a m ] and B = [b 0 ; b, b 2,..., b n ] be finite continued fractions with m n such that A = B. Then either m = n and a i = b i for all 0 i n, or m + = n with b m = a m and b n =. Again, a proof of this theorem is given in [] Infinite Simple Continued Fractions. The notation in the previous section can be expanded in the natural fashion to admit infinite simple continued fractions: [a 0 ; a, a 2,...] = a 0 + a +. (2.2) a It is not immediately clear that such a fraction is well-defined. The demonstration of its convergence is extraneous to the purposes of this paper, but is treated of fully in [], where it is furthermore shown to be irrational. For the present discussion, it is sufficient to consider one particular class of infinite continued fraction. Definition 2.4. Let ξ = [a 0 ; a, a 2,...] be an infinite continued fraction such that there exist n 0 and r > 0 with the property that a i = a i+r for all i n. Then ξ is said to be periodic with period r, and is written thus: ξ = [a 0 ; a,..., a n, a n+,... a n+r ]. If n = 0, ξ is said to be purely periodic, and can be written as follows: ξ = [a 0, a,..., a r ]. The following result is proved at length in []: Theorem 2.5. Let d be a positive integer not a perfect square. Then the continued fraction expansion of d is periodic and is of the form d = [a 0 ; a, a 2,..., a r, 2a 0 ], for some positive integer r. Furthermore, the following iterative scheme yields the partial quotients a i : a i = ξ i ξ i = m i + d where a 0 = m 0 = 0 and q 0 =. m i+ = a i m i + = d m2 i+, 3. Pell s Equation We return now to the study of equation (.).

3 PELL S EQUATION AND CONTINUED FRACTIONS Initial Observations. We begin by noting that when d < 0 and when d is a perfect square, equation (.) has only finitely many solutions; in the former case because both summands on the left-hand side become strictly positive, and in the latter case because the left-hand side factorises over the integers. We limit ourselves to considering solutions when d > 0, not a perfect square. An exercise in [] poses the problem of demonstrating that (.) has no solutions when N = and d 3 (mod 4). This can be seen by reducing the equation modulo 4 to give x 2 + y 2 3 (mod 4). Thus without loss of generality a solution would require x 2 0 (resp.) (mod 4) and y 2 3 (resp.) 2 (mod 4). However, there are no elements of the ring Z/4Z that square to either 2 or 3, and so the equation has no solution. A similar argument shows that there is no solution when d 0 (mod 4). Indeed, for any n, if there is no element of the ring Z/nZ which squares to n, then there is no solution when d 0 (mod n). A similar problem is posed in [2, question 5]: Problem 3.. Show that if d is an odd squarefree integer then a necessary condition for the equation x 2 dy 2 = to have an integer solution is that all the primes dividing d must be of the form 4n +. Solution. Let p = 4n + 3 be a prime, where n is a non-negative integer, such that p d. Reducing x 2 dy 2 = modulo p gives x 2 (mod p), whence o Z p (x) = 4, where o Z p (x) denotes the order of x in the multiplicative group Z p. Then Lagrange s theorem implies that 4 (p ), which contradicts the definition of p. Hence all the primes dividing d must be of the form 4n Lagrange s Method. Lagrange developed a method for solving (.) in the general case, which he describes in [3]. The process is recursive and is rather involved and so is not detailed here for the sake of brevity; however, the terminating case is considered to allow comparison with the method of solution described in section 4., infra. The form given here is based on that included in [4]. Theorem 3.2 (Lagrange). Consider Pell s equation ±E = r 2 Bs 2, (3.) where B > 0 and 0 < E < B. Construct sequences {E i }, {ɛ i } and {λ i } thus: E 0 = E E i E i+ = B ɛ 2 i, i 0 B E < ɛ0 < B, E (B ɛ 2 0) ɛ i = λ i E i ɛ i, i B + ɛi B + ɛi < λ i <, i. E i E i If no such ɛ 0 exists, the equation is insoluble. Otherwise, the sequence {E i } has a periodic tail beginning after E 0. Let the length of the periodic part be µ. Then there exists at least one m, m µ such that E m =. If all terms in the sequence {E i } are equal to unity, take m = 0 as a special case. Then for each even (resp. odd) such value of m when the upper (resp. lower) sign in (3.) is taken, the equation has a family of solutions given by r + s B = (R + S B)(X + Y B) n for all non-negative integers n, where R, S, X and Y are given by R = βl m + l m 2 S = l m X = βl µ + l µ 2 Y = l µ, where β is the greatest integer less than B and the sequence {l i } is given by l = 0, l 0 = and l i = λ i l i + l i 2 for i. In the case m = 0, the solution is given by (r, s) = (, 0) if the upper sign is taken in (3.); if the lower sign is taken, no solutions exist. If no such values of m exist, the equation is insoluble.

4 4 I. GREGOR DICK In the course of this project, the algorithm described in Theorem 3.2 was implemented, finding the smallest solution to (.) for arbitrary values of d and N. 4. An Alternative to Lagrange s Method 4.. Method of Continued Fractions. A method for solving equation (.) using continued fractions is given in [5], as follows. Theorem 4.. Consider the equation x 2 dy 2 =. (4.) Let ξ i be the i-th convergent to d, and write ξ i = pi with gcd(p i, ) = for all integers i. Then there is a least integer k such that (x, y) = (p k, q k ) is a solution to (4.), and the general solution is given by for all positive integers n. x + y d = (p + q d) n, Lemma 4.2. The equation x 2 dy 2 = has infinitely many solutions if the length of the periodic part of the continued fraction expansion of d is odd, and none if it is even. In [5] is also given the following more general theorem, accommodating larger values on the right-hand side of (4.). Theorem 4.3. Let d in (.) have continued fraction expansion [a 0 ; a, a 2,..., a r, 2a 0 ] for some positive integer r, which form is attained by the result of Theorem 2.5. Let ξ i = pi be defined as in Theorem 4.. Then, if N < d, there exists an integer k with 0 k r such that (x, y) = (p k, q k ) is a solution to (.). Furthermore, if (x, y) = (r, s) is any solution to x 2 dy 2 =, then } x = p k r ± dq k s (4.2) y = p k s ± q k r is also a solution of (.). Remark 4.4. The k in Theorem 4.3 need not be unique. In this case, there is a family of solutions of the form given in (4.2) for each such k. The following problem from [2, question 6] illustrates an application of Lemma 4.2. Problem 4.5. Let d = 4k 2 +k, where k > is an integer. Show that x 2 dy 2 = has no integer solutions. Solution. The truth of the proposition can be demonstrated by applying the iterative scheme for the partial quotients a i of the continued fraction expansion of d given in Theorem 2.5. Observing that 4k 2 + k = 2k for all positive integers k, we tabluate a i, m i and as defined in Theorem 2.5: i a i m i 0 2k 0 4 2k k 2 4k 2k 3 4 2k k Observing that a i, m i and are determined entirely by a i, m i and for any i, it follows that the period of the continued expansion fraction of d is 2, and hence x 2 dy 2 = has no solutions by Lemma 4.2.

5 PELL S EQUATION AND CONTINUED FRACTIONS Equivalence To Lagrange s Method. Although not immediately apparent, this method for solving Pell s equation can be shown to be equivalent to that of Lagrange by comparing the sequences {E i }, {λ i } and {ɛ i } in the latter to { }, {a i } and {m i } respectively in Theorem 2.5, and by noting that Lagrange s expressions for the solutions to (3.) in terms of these sequences corresponds to the formulæ given for the numerators and denominators of successive convergents in, for example, []. References [] Niven, I. et al. An Introduction to the Theory of Numbers, (3rd Edition). John Wiley & Sons, New York, 99. [2] Boyd, D. Mathematics 437/537 Final Examination. The University of British Columbia, Vancouver, [3] Lagrange, J. L. Œuvres, II, Chez Courcier, Paris, 868. [4] Dickson, L. E. History of the Theory of Numbers, Vol. II. Chelsea Publishing Company, New York, 97. [5] Chrystal, G. Algebra: an Elementary Text-Book. Adam and Charles Black, London, 900.

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