CONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12


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1 CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation. I would like to thank my mentor Avan for introducing and guiding me through this extremely interesting material. I would like to cite Steuding s detailed but slightly flawed book as the main source of learning and Andreescu and Andrica s book as an inspiration for numerous fun experiments I have made this summer. Contents. Continued Fractions 2. Solution to Pell s Equation 9 References 2. Continued Fractions This rather long section gives several crucial tools for solving Pell s equation. Definition.. Let a 0, a, a 2,..., a m be real numbers. Then, a 0 + a + a 2 + a 3 + a a m + a m is called a finite continued fraction and is denoted by [a 0,a,a 2,...,a m ]. If the chain of fractions does not stop, then it is called an infinite continued fraction. Remark.2. From now on, we will use a n as it is defined here. Definitions.3. (a) For n m, [a 0,a,...,a n ] is called nth convergent to [a 0,a,a 2,...,a m ]. (b) Define two sequences of real numbers, (p n ) and ( ), recursively as follows: () p, p 0 a 0, and p n a n p n + p n 2 ; and (2) q 0, q 0, and a n + 2. Remark.4. From now on, we will use p n and as they are defined here. Date: AUGUST 22, 2008.
2 2 SEUNG HYUN YANG Theorem.5. Let [a 0,a,a 2,...,a m ] be a continued fraction. Then, for 0 n m, pn [a 0,a,a 2,...,a n ]. Proof. We proceed by induction. For n, [a 0, a ] a 0 + a a a 0 + a a p 0 + p a q 0 + q p q as desired. Now, suppose the theorem holds for n. [a 0,..., a n+ ] [a 0,..., a n, a n + a n+ ] (a n + a n+ )p n + p n 2 (a n + a n+ ) + 2 (a n+a n + )p n + a n+ p n 2 (a n+ a n + ) + a n+ 2 a n+a n p n + a n+ p n 2 + p n a n+ a n + a n+ 2 + a n+(a n p n + p n 2 ) + p n a n+ (a n + 2 ) + a n+p n + p n a n+ + p n+ + as desired. Remark.6. As a result of this theorem, we will refer to p n as the nth convergent. Theorem.7. Suppose that a 0 is an integer, that a n is a positive integer for each n m, and that a m. Then, p n and are (a) integers and (b) coprime for each n m. Proof. (a) p, q, p 0, and q 0 are integers by definition. a n s are also integers for 0 n m by the given. Since p n and ( n m ) are defined as combinations of multiplication and subtraction of said variables (which are integers), they must be integers as well. (b) We use the following useful lemma: Lemma.8. For n m, following two relations hold: () p n p n ( ) n ; and (2) p n 2 p n 2 ( ) n a n.
3 CONTINUED FRACTIONS AND PELL S EQUATION 3 Proof. () p n p n (a n p n + p n 2 ) p n (a n + 2 ) as desired. (2) p n 2 + a n p n a n p n p n 2 ( )(p n 2 p n 2 )... ( ) n (p 0 q p q 0 ) ( ) n ( ) ( ) n p n 2 p n 2 (a n p n + p n 2 ) 2 (a n + 2 )p n 2 a n p n 2 + p n 2 2 p n 2 2 a n p n 2 a n (p n 2 p n 2 ) () n 2 a n as desired. ( ) n a n To complete the proof of Theorem.7, consider () of Lemma.8. Since there is a linear combination of p n and that is equal to ±, by elementary proposition from number theory, we conclude they are coprime as greatest common divisor between them is at most (equal to). We now move on to use continued fractions to approximate real numbers. Definition.9. Let α be a real number. For n 0,,2,..., define a recursive algorithm as follows: α 0 α, a n α n, and α n a n + α n+. Remark.0. Here, is used as a floor function. That means, by the last equation, α n is positive for all positive n. Observe that α n+ α n  α n. Thus, the left side of the equation must be less than, and therefore α n+ >. Then, by definition, a n+ is a positive integer. Thus, we may conclude that α n,a n for n. Observe also that, given positive m, α [a 0,a,...,a m,α m ]. It is called the mth continued fraction of α. Remark.. Observe that, if α is rational, then the algorithm above is equivalent to Euclidean algorithm, with α n, when reduced to a fraction of coprimes, consisting of nth remainder as numerator and n + th remainder as denominator. That means, the continued fraction of a rational number is finite. On the other hand, if α is irrational, then the continued fraction must be infinite simply because any finite continued fraction is rational (and therefore cannot be equal to an irrational number). Theorem.2. Let α be irrational and p n Then, (.3) α p n (α n+ + ). be a convergent to its continued fraction.
4 4 SEUNG HYUN YANG Proof. Let n be a positive number. Then, α [a 0,a,...,a n,α n+ ]. Then, α p n α n+p n + p n α n+ + p n p n + α n+ p n α n+ p n p n (α n+ + ) p n p n (α n+ + ) ( )(p n p n ) (α n+ + ) ( ) n (α n+ + ) by the Lemma.8. Remark.4. Observe, for each positive n, a n α n by a property of floor function. Also, since q, q 0, and a n (n > 0) are positive integers, the same must be true for (n > 0) by definition. Then, α p n < (α n+ + ) (a n+ + ). + By definition, a n + 2. Since a n and 2 > 0, we conclude that is strictly increasing as n increases. Then, we may conclude that the continued fraction of a number converges to that number by the inequality just given here. In other words, α lim n p n [a 0,a,a 2,... ]. Corollary.5. Let α be a real number with convergent p n. Then, α p n < α p n. Proof. By Theorem.2, α p n α p n (α n+ + ) α n+ + Similarly, α p n α n + 2
5 CONTINUED FRACTIONS AND PELL S EQUATION 5 It now suffices to prove the following inequality: < α n+ + α n + 2 (a n + α n+ ) + 2 a n + α n+ + 2 α n+ a n α n α n+ a n α n+ + + α n+ 2 < α 2 n+ + α n+ α n+ ( a n + 2 ) + < αn+ 2 + α n+ α n+ + < α n+ ( α n+ + ) which is true by Remark.0. < α n+ We are now ready to move onto two extremely beautiful and important approximation theorems involving convergents. Theorem.6. (The Law of Best Approximations) Let α be a real number with convergent pn and n 2. If p,q are integers such that 0 < q and p q p n, then α p n < qα p. Moreover, a reduced fraction p q with q q 2 that satisfies the latter inequality is a convergent. Proof. By Corollary.5, we have already proven the case in which p q is a convergent. Suppose q. Then, p p n. p q p n p p n. α p n < + < 2 because + 3 if n 2 ( is strictly increasing for n and q q 0 ). By the Triangle Inequality, we have α p q p q p n α p n > 2 2 > α p n
6 6 SEUNG HYUN YANG Multiplying both sides by q, we obtain the desired inequality. Now suppose 0 < q <. We may set up following system of two equations with two variables X and Y: (.7) (.8) p n X + p n Y p X + Y q A series of basic manipulations from high school mathematics yields the following unique solution (x,y): x y p qp n p n p n p qp n p n p n By Lemma.8, the denominators reduce to ±: x ±(p qp n ) y ±(p qp n ) Therefore, x and y are nonzero (otherwise p q is a convergent) integers. By the equation.8, since > q, we conclude that x and y have opposite signs. By Lemma.8, α p n alternates signs. That is, α p n and α p n have opposite signs. This implies that α p n and α p n have opposite signs as well. Therefore, x( α p n ) and y( α p n ) have the same sign. Thus, qα p ( x + y)α (p n x + p n y) x( α p n ) + y( α p n ) qα p x( α p n + y( α p n ) qα p > α p n > α p n as desired. In particular, this inequality holds for < q <, which, by Induction Principle, implies only convergents satisfy the said inequality. Theorem.9. (a) Given any two consecutive convergents to a real number α, there is at least one satisfying α p q < 2q 2 (b) Any reduced fraction that satisfies the above inequality is a convergent. Proof. (a) By Lemma.8, given any two consecutive convergents, exactly one is greater than or equal to α, and the other is less than or equal to α. Thus, (.20) p n+ p n + α p n+ + p n α +
7 CONTINUED FRACTIONS AND PELL S EQUATION 7 Now, suppose the said inequality is not true for some consecutive convergents p n and pn+ +. By Lemma.8, p n+ p n p n+ p n + α p n+ + p n α + 2q 2 n+ 2q 2 n q 2 n ( + ) q2 n q 2 n+ 2 Because is strictly increasing for positive n, this may be true only if n 0 and q q 0 a. Thus, by contradiction, (a) is true for all positive n, and we only need to check for the case n 0 (the two consecutive convergents being p0 q 0 and p q ): 0 < p q α a a 0 + α a 0 + α a 0 + [a 0,, a 2, a 3,... ] < + a 2 a 2 a which satisfies the statement of the theorem. (b) Suppose p q satisfies the said inequality. By the law of best approximations, it suffices to show p q is a best approximation to α. Let P Q be such that P Q p α q and Qα P qα p q p < 2q. Then, q Thus, p q qq is a best approximation. pq P q qq p q P Q α p q + P Q α < 2q 2 + 2qQ q + Q 2q 2 Q < q + Q 2q 2q < q + Q q < Q
8 8 SEUNG HYUN YANG We now move on to the discussion of quadratic irrationals. We first present two wellknown theorems, Lagrange s and Galois, without proof. Both may be proved with what we have demonstrated so far (plus some basic knowledge of polynomials), and have quite long proofs that would only make this alreadytoolong paper even longer. Definitions.2. Let α be an irrational number. α is called a quadratic irrational if it is a root of integer polynomial of degree two. The other root β is called a conjugate of α. Definitions.22. Let [a 0,a,a 2,... ] be a continued fraction such that a n a n+l for all sufficiently large n and a fixed positive integer l. Then, it is periodic and l is called a period. Theorem.23. (Lagrange s Theorem) An irrational number is quadratic irrational if and only if its continued fraction is periodic. Definition.24. Let α be a quadratic irrational and β be its conjugate. Then, α is reduced if α > and  < β < 0. Theorem.25. (Galois Theorem) Let α be irrational. Then, α is purely periodic if and only if α is reduced. If α [a 0, a, a 2,..., a l ] and β is its conjugate, then  β [a l,..., a 2, a ]. We finally conclude this section with the following simple theorem: Theorem.26. Let d be a nonsquare natural number. Then, there exist integers a, a 2,..., a n such that [ d,a, a 2,..., a 2, a, 2 d ] is the continued fraction of d. Also, a, a 2,..., a 2, a is a palindrome. Proof. Observe that  < α d  d < 0 and < β d + d. (x α)(x β) x 2 (α + β)x + αβ x d x + d 2 d which is an integer polynomial of degree two. Thus, α and β are quadratic irrationals and conjugates of each other. More, β is reduced. By Galois Theorem, β is purely periodic. β d + d [2 d, a, a 2,..., a l ] On the other hand, observe: d d [2 d, a, a 2,..., a l, 2 d ] d [ d, a, a 2,..., a l, 2 d ] d + d [a l,..., a 2, a, 2 d ] by Galois Theorem. That means, d [ d, d d ] [ d, a l,..., a 2, a, 2 d ] Thus, we have obtained the desired palindrome.
9 CONTINUED FRACTIONS AND PELL S EQUATION 9 2. Solution to Pell s Equation Definition 2.. Let d be a natural number. Then, Pell s equation is X 2 dy 2. Remarks 2.2. The case in which d is a perfect square is trivial. Let d m 2. X 2 dy 2 (X my )(X + my ) X my ± X + my ± Solving this linear system of equations yields the only solutions: (±,0). Note that these are solutions regardless of d. Observe that if (x,y) is a solution, then ( x, y) and (±x, y) are also solutions. Thus, it suffices to consider only positive solutions. For convenience, if (x,y) is a solution, then we call x + y d a solution as well. Theorem 2.3. (a) Let d not be a perfect square and α d. Note that α n is as defined in Definition.9. For nonnegative n, there exist integers Q n and P n such that α n P n+ d Q n and that d Pn 2 0 (mod Q n ). (b) For each n 2, we have: (2.4) p 2 n dq 2 n ( ) n Q n Proof. (a) We proceed by induction. For n 0, α 0 d, Q 0, and P 0 0. For n, α d d d d d d 2 Now assume the statement for n. α n+ Q d d 2 and P d α n a n P n + d Q n a n Q n P n + d a n Q n Q n(p n a n Q n d) (P n a n Q n ) 2 d : P n+ + d Q n+ P n+ a n Q n P n, which is integral, and Q n+ d (P n a n Q n ) 2 Q n d P 2 n Q n + 2a n P n a 2 nq n,
10 0 SEUNG HYUN YANG which is integral because d P 2 n 0 (mod Q n ) by assumption. Finally, as desired. (b) Q n d (P n a n Q n ) 2 Q n+ d P n+ Q n+ d P 2 n+ 0 (mod Q n ) d α n p n + p n 2 α n + 2 (P n + d)p n + p n 2 Q n (P n + d) + 2 Q n d((p n + d) + 2 Q n ) (P n + d)p n + p n 2 Q n d(p n + Q n 2 + d ) p n d + Pn p n + p n 2 Q n (P n + Q n 2 ) d + d p n d + Pn p n + p n 2 Q n p n P n + 2 Q n and d P n p n + p n 2 Q n Multiply the first by p n and the second by, and obtain: p 2 n dq 2 n Q n (p n 2 p n 2 ) p 2 n dq 2 n ( ) n Q n by Lemma.8. Definition 2.5. Given Pell s equation with some d, let (x,y) be the positive solution with the minimum x. Such solution is called the minimal solution. We finally give the complete solution to Pell s equation: Theorem 2.6. Let variables be as defined in the previous theorem. Let l be the minimal period of the continued fraction of d. (a) The minimal solution to Pell s equation is: { (p l, q l ) if l is even (x,y ) (p 2l, q 2l ) if l is odd (b) All the solutions (x,y) to Pell s equation are up to sign given by the powers of the minimal solution (x,y ): x + y d ±(x + y d) ±n (n 0,,2,... )
11 CONTINUED FRACTIONS AND PELL S EQUATION Proof. (a) We first establish that all the solutions to Pell s equation X 2 dy 2 must consist of convergents to d. Let (x,y) be a solution. Observe first that: x d y d x y y dy 2 x 2 y(y d + x) y 2 ( d + x y ) < (since d > and x > y) 2y2 By Theorem.9, we conclude (x,y) is a convergent to d. We now use (2.4) to show that: () Q n for every positive integer n; and (2) Q n if and only if n is a multiple of l. Thus, with regard to (2.4), the minimal solution is obtained by taking the smallest positive odd multiple of l minus, as (a) states. To show (), observe that α α n+ if and only if n is a multiple of l because l is minimal. Let Q n. Then, by Theorem 2.3, α n P n + d. Being quadratic irrational, by Lagrange s Theorem, α n has purely periodic continued fraction expansion. By Galois Theorem, α n is reduced: < β n P n d < 0 d < P n < d P n d Thus, α d + d. This implies that α n+ α, and n is a multiple of l. On the other hand, if n is a multiple of l, n kl, then: P kl + d Q kl α kl [0, a, a 2,..., a l ] d d Q kl Q kl, as desired. For (2), suppose Q n. By Theorem 2.3, α n P n d. α n is quadratic irrational, and thus is reduced by Galois Theorem: < P n + d < 0 and < P n d d < P n < d + d < 2 This is contradiction. Thus, Q n for all n, as desired. (b) We observed before that, given a solution (x,y), signs of x and/or y may be changed without contradicting the statement x 2 dy 2. We then concluded that it sufficed to find positive solutions. With regard to this, observe that ± x + y d ±(x y d) x 2 dy 2 ±x y d (x 2 dy 2 ) Thus, we may change the signs of x and y only by taking powers and multiplying by . Thus, it now suffices to show this theorem for only positive solutions.
12 2 SEUNG HYUN YANG Let ɛ : x + y d and (x,y) be a positive solution to Pell s equation. Then, there exists n such that ɛ n x + y d < ɛ n+. Now, let X + Y d ɛ n (x + y d). Observe that, due to the definition of ɛ, X + Y d. It suffices to show that it is equal to. We do this by contradiction; suppose ɛ > X + Y d > (the first inequality is due to the definition of epsilon also). By rules of elementary algebra, we get the following conjugation to this quadratic irrational: X Y d ɛ n (x y d). Then, (X + Y d)(x Y d) X 2 dy 2 ɛ n ɛ n (x y d)(x + y d) x 2 dy 2. We then have: 0 < ɛ n < (X + Y d) X Y d < { 2X (X + Y d) + (X Y d) > + ɛ > 0 2Y d (X + Y d) (X Y d) > 0 Thus, (X, Y ) is a positive solution with X + Y d < x + y d. That is, X < x. This is a contradiction to the minimality of (x, y ), and thus our assumption (X + Y d > ) is false, as desired. Remarks 2.7. In fact, (a) alone suffices to give the complete set of solutions to Pell s equation. The set of (positive) solutions deduced from (a) would be: { (p kl, q kl ) if l is even (x k, y k ) (p 2kl, q 2kl ) if l is odd (b) shows that the complete set of solutions to Pell s equation is infinite cyclic group generated by the minimal solution. References [] Jörn Steuding. Diophantine Analysis. Chapman and Hall/CRC [2] Titu Andreescu and Dorin Andrica. An Introduction to Diophantine Equations. Gil Publishing House
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