Pell Equation x 2 Dy 2 = 2, II

Irish Math Soc Bulletin 54 2004 73 89 73 Pell Equation x 2 Dy 2 2 II AHMET TEKCAN Abstract In this paper solutions of the Pell equation x 2 Dy 2 2 are formulated for a positive non-square integer D using the solutions of the Pell equation x 2 Dy 2 1 Moreover a recurrence relation on the solutions of the Pell equation x 2 Dy 2 2 is obtained Furthermore the solutions of the equation x 2 n Dyn 2 2 n are obtained using the solutions of the equation x 2 Dy 2 2 1 Introduction A real binary quadratic form f or just a form is a polynomial in two variables of the type fx y ax 2 + bxy + cy 2 with real coefficients a b c The discriminant of f is defined by the formula b 2 4ac and denoted by D Let D be a non-square discriminant Then the Pell form is defined by the formula { x f D x y 2 D 4 y2 if D 0 mod 4 x 2 + xy D 1 11 4 y2 if D 1 mod 4 Let P elld { x y Z 2 : f D x y 1 } and P ell ± D { x y Z 2 : f D x y ±1 } Then P elld is infinite The binary operation x 1 y 1 x 2 y 2 x 1 x 2 + Dy 1 y 2 x 1 y 2 + y 1 x 2 2000 Mathematics Subject Classification 11E15 11E18 11E25 Key words and phrases Pell forms Pell equation solution of the Pell equation

74 Ahmet Tekcan is a group law on P elld for which P elld {±1} Z Let d be a positive non-square discriminant and K Q d be the quadratic number field Then every element α of K can be represented as α a + b d c for a b c Z The conjugate of α is denoted by α a b d c The trace and norm of α are given by and T rα α + α 2a c Nα αα a2 db 2 c 2 respectively It is easy to show that for α β K and T rα + β T rα + T rβ Nαβ NαNβ If d 1 mod 4 then the elements of K are of the form α a + b d 12 2 where a b Z and if d 2 3 mod 4 then the elements of K are of the form α a + b d 13 where a b Z As in the case of rationals the set of integers of K forms a ring which we will be denoted as O K the maximal order of K If every integer α O K can be uniquely expressed as α a 1 w 1 + a 2 w 2 where a i Z and w i O K then we call w 1 w 2 an integral basis for K and we denote O K by the Z-module [w 1 w 2 ] w 1 Z + w 2 Z Every algebraic number field has an integral basis and in the quadratic fields it is especially easy to give one

Pell Equation x 2 Dy 2 2 II 75 If d 1 mod 4 then from Eq 12 it is seen that w 1 1 and w 2 1 + d 2 is an integral basis If d 2 3 mod 4 then from Eq 13 it is seen that w 1 2 and w 2 d is an integral basis The discriminant of K is defined as DK w 2 1 w 1 w 2 w 2 hence DK D for d 1 mod 4 and DK 4d for d 2 3 mod 4 Therefore DK 4d for { 2 d 1 mod 4 r 1 d 2 3 mod 4 and {1 w} is an integral basis of K for see [2] The D-order O D where ρ D r 2 w r 1 + d r is defined to be the ring It is clear from definition that O D 14 O D {x + yρ D : x y Z} 15 { D 4 if D 0 mod 4 1+ D 2 if D 1 mod 4 is a subring of Q D Lemma 11 [1] Let α O D Then α is a unit in O D if and only if Nα ±1 The unit group OD is defined to be the group of units of the ring O D Let OD1 {α OD : Nα +1} and for D > 0 OD+ {α OD : α > 0}

76 Ahmet Tekcan be the subgroup of positive units By Eq 11 and Lemma 11 there is a bijection ψ : P ell ± D OD given by ψx y x + yρ D Since it is in bijection with a commutative group P ell ± D itself is a group for every non-square discriminant D The mapping ψ is used to transport the group law from OD so that by definition ab ψ 1 ψaψb for every a b P ell ± D ie the product u vu V of two elements u v U V P ell ± D is defined by the rule u vu V x y 16 where x + yρ D u + vρ D U + V ρ D It follows by a calculation that { uu + D u vu V 4 vv uv + vu if D 0 mod 4 uu + D 1 4 vv uv + vu + vv if D 1 mod 4 17 The group structure on P ell ± D has been defined as ψ : P ell ± D O D and is an isomorphism of groups Moreover Eq 17 is a group law on P ell ± D with identity element 1 0 for all non-zero discriminants D Lemma 12 [1] OD+ Z for every positive non-square discriminant D From Lemma 12 we obtain Lemma 13 [1] Let D be a non-square discriminant and let ε D be the smallest unit of O D that is greater than 1 and let { εd if Nε τ D D +1 if Nε D 1 then and ε 2 D P ell ± D O D {±ε n D; n Z} {±1} Z P elld O D1 {±τ n D; n Z} {±1} Z

Pell Equation x 2 Dy 2 2 II 77 The fundamental unit ε D is defined to be the smallest unit of O D that is greater than 1 The Pell equation is the equation x 2 dy 2 1 and the negative Pell equation is the equation x 2 dy 2 1 where d is a positive non-square integer The set of all solutions of the Pell equation is infinite The first solution x 1 y 1 of the Pell equation is called the fundamental solution One may rewrite the Pell equation as x 2 dy 2 x + y d x y d 1 so that finding a solution comes down to finding a nontrivial unit in the ring Z [ d ] of norm 1; here the norm Z [ d ] Z {±1} between unit groups multiplies each unit by its conjugate and the units ±1 of Z [ d ] are considered trivial This reformulation implies that once one knows a solution fundamental solution of the Pell equation one can find infinitely many others More precisely if the solutions are ordered by magnitude then the n th solution x n y n can be expressed in terms of the fundamental solution by x n + y n d x 1 + y 1 d n 18 for n 2 In [3] Jacobson and Williams considered the solutions of the consecutive Pell equations X0 2 D 1Y0 2 1 and X1 2 DY1 2 1 where D and D 1 are not perfect square They also proved that ρd log X 1 log X 0 could be arbitrary large for integers X 1 and X 2 In [4] McLaughlin considered the solutions of the multi-variable polynomial Pell equation Ci 2 F ihi 2 1n 1 where n is a positive integer {F i } a finite collection of multi-variable polynomials C i and H i are multi-variable polynomials with integral coefficients In [5] Lenstra gave the solution of the Pell equation using the ring Z [ d ] He considered the solvability of the Pell equation as a special case of Dirichlet s unit theorem from algebraic number theory which describes the structure of the group of units of a general ring of algebraic integers; for the ring Z [ d ] it is product of {±1} and an infinite cyclic group In [6] Li proved that the Pell equation x 2 dy 2 1 has infinitely positive solutions If x 1 y 1 is the fundamental solution then for n 2 3 4 x n + y n d x1 + y 1 d n The pairs x n y n are all the positive solutions of the Pell equation The x n s and y n s are

78 Ahmet Tekcan strictly increasing to infinity and satisfy the recurrence relations x n+2 2x 1 x n+1 x n y n+2 2x 1 y n+1 y n Li also proved that the fundamental solution of the Pell equation is obtained by writing d as a simple continued fraction It turns out that 1 d a0 + a 1 + 1 a 2 + where a 0 [ d ] and a 1 a 2 is a periodic positive integer sequence The continued fraction will be denoted by a 0 a 1 a 2 The kth convergent of a 0 a 1 a 2 is the number p k q k a 0 a 1 a 2 a k with p k and q k relatively prime numbers Let a 0 a 1 a 2 a m be the period for d Then the fundamental solution of the Pell equation x 2 dy 2 1 is { pm 1 q x 1 y 1 m 1 if m is even p 2m 1 q 2m 1 if m is odd and the other solutions are x n + y n d x 1 + y 1 d n for n 2 In [7] Matthews considered the solutions of the equation x 2 Dy 2 N for D > 0 He shoved that a necessary condition for the solvability of x 2 Dy 2 N with gcdx y 1 is that the congruence u 2 D mod Q 0 shall be soluble where Q 0 N In [8] Mollin gave a formula for the solutions of the equations both X 2 DY 2 c and x 2 Dy 2 c using the ideals I [Q P + D] for positive non-square integer D In [9] Mollin Cheng and Goddard considered the solutions of the Diophantine equation ax 2 by 2 c in terms of the simple continued fraction expansion of a 2 b and they explored a criteria for the solvability of AX 2 BY 2 C for given integers A B C N In [10] Mollin Poorten and Williams considered the equation x 2 Dy 2 3 They obtained a formula for the solutions of this equation using the continued fraction expansion of D and using the ambiguous ideals I [Q P + D] ie I I where I denotes the conjugate of I

Pell Equation x 2 Dy 2 2 II 79 In [11] Stevenhagen considered the solutions of the negative Pell equation x 2 Dy 2 1 for a positive non-square integer D He stated a conjecture for the solutions of the equation x 2 Dy 2 1 In [12] we considered the solutions of the Pell equation x 2 Dy 2 2 using the fundamental element of the field Q In the present paper the solutions of the Pell equation x 2 Dy 2 2 for positive non-square integer and a recurrence relation for the solutions of the Pell equation x 2 Dy 2 2 are obtained using the solutions of the Pell equation x 2 Dy 2 1 2 The Pell Equation x 2 Dy 2 2 First consider the fundamental solution of the Pell equation x 2 Dy 2 2 Theorem 21 If x 1 y 1 a 1 is the fundamental solution of the Pell equation x 2 Dy 2 1 then the fundamental solution of the Pell equation x 2 D 1y 2 2 is X 1 Y 1 a 1 Proof Since x 1 y 1 a 1 is the fundamental solution of the Pell equation x 2 Dy 2 1 we have a 2 D 1 Hence by basic calculation it is easily seen that X 2 1 D 1Y 2 1 a 2 D 1 a 2 D + 1 2 Therefore X 1 Y 1 a 1 is the fundamental solution of the Pell equation x 2 Dy 2 2 Theorem 22 If D k 2 2 k 2 then the fundamental solution of the Pell equation x 2 Dy 2 1 is x 1 y 1 a b k 2 1 k and the fundamental solution of the Pell equation x 2 Dy 2 2 is X 1 Y 1 b 1 Proof Note that x 1 y 1 a b k 2 1 k is the fundamental solution of the Pell equation x 2 Dy 2 1 since x 2 1 Dy 2 1 a 2 Db 2 k 2 1 2 k 2 2k 2 1 For X 1 Y 1 b 1 we get X 2 1 DY 2 1 b 2 D k 2 k 2 2 2 Therefore X 1 Y 1 b 1 is the fundamental solution of the Pell equation x 2 Dy 2 2

80 Ahmet Tekcan Now we consider the solutions of the Pell equation x 2 Dy 2 2 To get this we have the following theorem Theorem 23 Let X 1 Y 1 k l be the fundamental solution of the Pell equation x 2 Dy 2 2 Then the other solutions of the Pell equation x 2 Dy 2 2 are X n Y n where Xn k ld xn 1 21 Y n l k y n 1 for n 2 Proof From above equalities we get Xn k ld xn 1 Y n l k y n 1 Hence it is easily seen that kxn 1 + ldy n 1 lx n 1 + ky n 1 X 2 n DY 2 n kx n 1 + ldy n 1 2 D lx n 1 + ky n 1 2 k 2 x 2 n 1 + 2klDx n 1 y n 1 + l 2 D 2 y 2 n 1 Dl 2 x 2 n 1 + 2klx n 1 y n 1 + k 2 y 2 n 1 k 2 x 2 n 1 Dy 2 n 1 Dl 2 x 2 n 1 Dy 2 n 1 x 2 n 1 Dy 2 n 1k 2 Dl 2 2 22 since x 2 n 1 Dyn 1 2 1 and X 1 Y 1 k l be the fundamental solution of the Pell equation x 2 Dy 2 2 ie k 2 Dl 2 2 From Theorem 23 the following corollary can be obtained Corollary 24 The solutions X n Y n of the Pell equation x 2 Dy 2 2 satisfy the following relations Xn+1 axn + bdy n a bd Xn Y n+1 bx n + ay n b a Y n for n 1 Proof We know from Eq 22 that Xn kxn 1 + ldy n 1 Y n lx n 1 + ky n 1 Hence x n 1 klx n + 2 k 2 Y n 2l and y n 1 lx n + ky n 23 2

Pell Equation x 2 Dy 2 2 II 81 On the other hand from Eq 21 n xn a bd 1 y n b a 0 n 1 a bd a bd 1 b a b a 0 a bd xn 1 b a y n 1 axn 1 + bdy n 1 bx n 1 + ay n 1 Using Eq 23 and Eq 24 we find Xn+1 k ld xn Y n+1 l k y n k ld axn 1 + bdy n 1 l k bx n 1 + ay n 1 xn 1 ak + bld + y n 1 bkd + ald x n 1 al + bk + y n 1 bld + ak Applying Eq 23 and Eq 25 it follows that and X n+1 x n 1 ak + bld + y n 1 bkd + ald klxn + 2 k 2 Y n ak + bld 2l lxn + ky n + bkd + ald 2 2alX n + 2bDlY n 2l ax n + bdy n Y n+1 x n 1 al + bk + y n 1 bld + ak klxn + 2 k 2 Y n al + bk 2l lxn + ky n + bld + ak 2 2blX n + 2alY n 2l bx n + ay n 24 25

82 Ahmet Tekcan Hence Xn+1 axn + bdy n Y n+1 bx n + ay n a bd Xn b a Y n Example 21 Let D 2 Then the fundamental solution of the Pell equation x 2 2y 2 1 is x 1 y 1 3 2 and the other solutions are 2 x2 3 4 1 17 y 2 2 3 0 12 3 x3 3 4 1 99 y 3 2 3 0 70 4 x4 3 4 1 577 y 4 2 3 0 408 5 x5 3 4 1 3363 y 5 2 3 0 2378 6 x6 3 4 1 19601 y 6 2 3 0 13860 7 x7 3 4 1 114243 y 7 2 3 0 80782 8 x8 3 4 1 665857 y 8 2 3 0 470832 x9 y 9 3 4 2 3 9 1 0 3880899 2744210 The fundamental solution of the Pell equation x 2 2y 2 2 is X 1 Y 1 k l 2 1 and the other solutions are X2 2 2 3 10 Y 2 1 2 2 7 X3 2 2 17 58 Y 3 1 2 12 41 X4 2 2 99 338 Y 4 1 2 70 239 X5 2 2 577 1970 Y 5 1 2 408 1393

Pell Equation x 2 Dy 2 2 II 83 X6 2 2 3363 11482 Y 6 1 2 2378 8119 X7 2 2 19601 66922 Y 7 1 2 13860 47321 X8 2 2 114243 390050 Y 8 1 2 80782 275807 X9 2 2 665857 2273378 Y 9 1 2 470832 1607521 X10 2 2 3880899 13250218 Y 10 1 2 2744210 9369319 a bd k ld Now we give a relation between and To b a l k get this set { } r s SL 2 2 R : r s t u R ru st 2 t u Then we have Theorem 25 There exists an element A SL 2 2 R such that a bd k ld A b a l k r s Proof Let A for r s t u R Then we have t u a bd r s ar + bdt as + bdu k ld b a t u br + at bs + au l k Hence from the two equations and ar + bdt k br + at l as + bdu ld bs + au k

84 Ahmet Tekcan we obtain r l a2 l + abk b s k a2 k + abld b t al bk u ak bld Hence A l a 2 l+abk b al bk k a 2 k+abld b ak bld It is easily seen that l a 2 l + abk deta ak bld b k a 2 k + abld al bk b bk2 Dl 2 b k 2 Dl 2 2 Therefore A SL 2 2 R Now we would like to obtain a recurrence relation on the solutions of the Pell equation x 2 Dy 2 2 To get this using Eq 21 we obtain x1 a 26 y 1 b x2 2a 2 1 y 2 2ab x3 4a 3 3a y 3 x4 y 4 b4a 2 1 8a 4 8a 2 + 1 b8a 3 4a

Pell Equation x 2 Dy 2 2 II 85 x5 x6 x7 y 5 y 6 y 7 x8 y 8 16a 5 20a 3 + 5a b16a 4 12a 2 + 1 32a 6 48a 4 + 18a 2 1 b32a 5 32a 3 + 6a 64a 7 112a 5 + 56a 3 7a b64a 6 80a 4 + 24a 2 1 128a 8 256a 6 + 160a 4 32a 2 + 1 b128a 7 192a 5 + 80a 3 8a and hence X2 X3 X4 X5 X6 X7 X8 X9 Y 2 Y 3 Y 4 Y 5 Y 6 Y 7 Y 8 Y 9 ka + lbd 27 la + kb k2a 2 1 + 2ablD l2a 2 1 + 2abk k4a 3 3a + bld4a 2 1 l4a 3 3a + kb4a 2 1 k8a 4 8a 2 + 1 + bld8a 3 4a l8a 4 8a 2 + 1 + kb8a 3 4a k16a 5 20a 3 + 5a + bld16a 4 12a 2 + 1 l16a 5 20a 3 + 5a + kb16a 4 12a 2 + 1 k32a 6 48a 4 + 18a 2 1 l32a 6 48a 4 + 18a 2 1 + bld32a 5 32a 3 + 6a + kb32a 5 32a 3 + 6a k64a 7 112a 5 + 56a 3 7a l64a 7 112a 5 + 56a 3 7a + bld64a 6 80a 4 + 24a 2 1 + kb64a 6 80a 4 + 24a 2 1 k128a 8 256a 6 + 160a 4 32a 2 + 1 l128a 8 256a 6 + 160a 4 32a 2 + 1 + bld128a 7 192a 5 + 80a 3 8a + kb128a 7 192a 5 + 80a 3 8a

86 Ahmet Tekcan Using Eq 27 it is easily seen that X 4 2k 2 1X 3 X 2 + X 1 28 X 5 2k 2 1 X 4 X 3 + X 2 X 6 2k 2 1 X 5 X 4 + X 3 X 7 2k 2 1 X 6 X 5 + X 4 X 8 2k 2 1 X 7 X 6 + X 5 X 9 2k 2 1 X 8 X 7 + X 6 Eq 28 is also satisfied for Y n Hence we have Conjecture 26 The solutions of the Pell equation x 2 Dy 2 2 satisfy the following recurrence relations for n 4 X n 2k 2 1 X n 1 X n 2 + X n 3 Y n 2k 2 1 Y n 1 Y n 2 + Y n 3 Now we obtain a formula for the solutions of the equation x 2 n Dy 2 n 2 n using the solutions of the Pell equation x 2 Dy 2 2 Theorem 27 Let X 1 Y 1 k l be the fundamental solution of the Pell equation x 2 Dy 2 2 Let n Un k ld 1 29 l k 0 for n 1 Then for n 1 V n U 2 n DV 2 n 2 n Proof We prove the Theorem by induction on n For n 1 we have U 2 1 DV 2 1 k 2 Dl 2 2 since X 1 Y 1 k l is the fundamental solution of the Pell equation x 2 Dy 2 2 Let us assume that the equation Un 2 DVn 2 2 n is satisfied for U n 1 V n 1 ie U 2 n 1 DV 2 n 1 2 n 1

Pell Equation x 2 Dy 2 2 II 87 We want to show that the equation Un 2 DVn 2 2 n is also satisfied for U n V n From Eq 29 it is easily seen that Hence Un V n n k ld 1 l k 0 n 1 k ld k ld 1 l k l k 0 k ld Un 1 l k V n 1 kun 1 + ldv n 1 lu n 1 + kv n 1 U 2 n DV 2 n ku n 1 + ldv n 1 2 D lu n 1 + kv n 1 2 U 2 n 1k 2 Dl 2 + U n 1 V n 1 2klD 2klD +V 2 n 1l 2 D 2 k 2 D Un 1k 2 2 Dl 2 Dk 2 Dl 2 Vn 1 2 k 2 Dl 2 Un 1 2 DVn 1 2 2 2 n 1 2 n Thus U n V n is also a solution of the equation U 2 n DV 2 n 2 n Example 22 Consider the Pell equation x 2 2y 2 2 The fundamental solution is X 1 Y 1 k l 2 1 Using Eq 29 we obtain U1 U2 U3 V 1 V 2 V 3 U4 V 4 1 2 2 1 2 1 2 0 1 2 2 2 1 6 1 2 0 4 3 2 2 1 20 1 2 0 14 4 2 2 1 68 1 2 0 48

88 Ahmet Tekcan U5 U6 V 5 V 6 Hence it is easily seen that 5 2 2 1 232 1 2 0 164 6 2 2 1 792 1 2 0 560 U1 2 2V1 2 2 U2 2 2V2 2 4 U3 2 2V3 2 8 U4 2 2V4 2 16 U5 2 2V5 2 32 U6 2 2V6 2 64 References [1] D E Flath Introduction to Number Theory Wiley 1989 [2] M J Jacobson Computational Techniques in Quadratic Fields MSc Thesis University of Manitoba 1995 [3] M J Jacobson H C Williams The Size of the Fundamental Solutions of the Consecutive Pell Equations Experimental Math Vol 9 2000 631 640 [4] J McLaughlin Multi-Variable Polynomial Solutions to Pell s Equation and Fundamental Units in Real Quadratic Fields Pacific J Math 210 2003 335 349 [5] H W Lenstra Jr Solving the Pell Equation Notices Amer Math Soc 49 2002 182 192 [6] K Y Li Pell s Equation Mathematical Excalibur 6 2001 1 4 [7] K Matthews The Diophantine Equation x 2 Dy 2 N D > 0 Expositiones Math 18 2000 323 331 [8] R A Mollin A Simple Criterion for Solvability of both X 2 DY 2 c and x 2 Dy 2 c New York J Math 7 2001 87 97 [9] R A Mollin K Cheng B Goddard The Diophantine Equation AX 2 BY 2 C Solved via Continued Fractions Acta Math Univ Comenianae 71 2002 121 138 [10] R A Mollin A J Poorten H C Williams Halfway to a Solution of x 2 Dy 2 3 Journal de Theorie des Nombres Bordeaux 6 1994 421 457 [11] P Stevenhagen A Density Conjecture for the Negative Pell Equation Computational Algebra and Number Theory Math Appl 325 1992 187 200

Pell Equation x 2 Dy 2 2 II 89 [12] A Tekcan O Bizim M Bayraktar Solving the Pell Equation Using the Fundamental Element of the Field Q To be printed in Southeast Asian Bulletin of Mathematics Ahmet Tekcan Department of Mathematics Faculty of Science University of Uludag Görükle 16059 Bursa Turkey fahmet@uludagedutr Received on 19 March 2004