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POLYNOMIAL PELL's EQUATION 1. Introduction Let D Be a Nonconstant

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POLYNOMIAL PELL's EQUATION 1. Introduction Let D Be a Nonconstant PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 131, Number 4, Pages 993{1006 S 0002-9939(02)06934-4 Article electronically published on November 6, 2002 POLYNOMIAL PELL'S EQUATION WILLIAM A. WEBB AND HISASHI YOKOTA (Communicated by David E. Rohrlich) Abstract. Consider the polynomial Pell's equation X2 DY 2 =1,where − D = A2 +2C is a monic polynomial in [x]anddegC<deg A.Thenfor Z A; C [x], deg C<2, and B = A=C [x], a necessary and sufficient 2Q 2Q condition for the polynomial Pell's equation to have a nontrivial solution in [x] is obtained. Z 1. Introduction Let D be a nonconstant monic polynomial of even degree with integer coefficients. We consider the polynomial Pell's equation (1) X2 DY 2 =1 − where solutions X; Y are polynomials with integer coefficients. In 1976, Nathanson [5] proved that when D = x2 + d, equation (1) has a nontrivial solution if and only if d = 1; 2. This is a special case of the open problem which asks to determine the polynomials± ± D for which equation (1) has nontrivial solutions, and the special quadratics above is the only class of polynomials for which solutions of (1) have been completely characterized. We will characterize solutions of (1) for a much larger class of polynomials D, which includes all monic D = A2 +2C where deg C 1 and A=C [x]. In particular, this includes all monic quadratic polynomials since≤ they can be2Q written as A2 +2C where deg C =0. As we will see, solving (1) over [x] is relatively easy; determining when solutions in [x] exist is the more difficultQ question. ZWe note that equation (1) has no nontrivial solution if D is a perfect square. For D = A2,wehave1=(X + AY )(X AY ), which implies X = 1;Y =0.So, − ± we assume pD is irrational. We will call W = U + V pD a rational solution of (1) if U 2 DV 2 =1and U; V [x]. We define − 2Q T = U + V pD : U 2 DV 2 =1; sgn U>0; sgn V>0; where U; V [x] f − 2Q g and T0 to be the subset of T such that U; V [x]. If W is any rational solution of (1), so are W and W . Among these four2Z solutions, there is always one for which sgn U>±0andsgn±V>0. Thus to determine all rational solutions of (1), it suffices to find all solutions in T . Received by the editors April 3, 2001. 1991 Mathematics Subject Classification. Primary 11D25, 11A55. Key words and phrases. Polynomial Pell's equation. c 2002 American Mathematical Society 993 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 994 WILLIAM A. WEBB AND HISASHI YOKOTA Among all rational solutions in T ,wesayP +QpD is a minimal (or fundamental) solution if and only if its nonarchimedian absolute value, defined below, satisfies the following condition: P + QpD U + V pD for all U + V pD T: j |≤j j 2 Then we can show (see Lemma 3 below) that a minimal solution is unique, and (see Lemma 4 below) every rational solution W T canbeexpressedasW = W n for 2 0 some n 1, where W0 is the minimal solution. So, to determine the polynomials D for which≥ the polynomial Pell's equation (1) has nontrivial rational solutions, it suffices to find the minimal solution. Let W0 be the minimal solution. We ask the following questions: (1) When is W0 in T0? n (2) Is it possible to have W0 T0 even though W0 T0? 2 n 62 Since T0 T , W T0 implies W = W0 for some n 1, where W0 is the minimal solution.⊂ Thus2 if the answer to the second question≥ is negative, then every n solution W of the polynomial Pell's equation (1) is expressed as W n or W for ± 0 ± 0 some n 1, where W0 T0. To answer≥ these questions,2 we consider the continued fraction expansion of pD. Note that the continued fraction expansion of pD canbedefinedinmanyways 1 depending on the base field (see [1], [2], [3], [4], [6]). Let = ((x− )) be the field 1 K Q of formal Laurent series in x− over .Thenα implies that Q 2K 1 j α = ajx− ; where aj ;at =0; sgn α = at: 2Q 6 j=t X We define the nonarchimedian absolute value by t α = e− : j j deg A deg B So, A=B = e − for A; B [x]. We use the symbol [α]tomeanthe integerj partj of α: 2Q 0 j t [α]= ajx− = atx− + + a0 [x]: ··· 2Q j=t X Note that for any U + V pD T , U + V pD > 1and U V pD < 1. Hence, 2 j j j − j U = V pD . Also, if W1 and W2 are rational solutions of (1), then so is W1W2. j j j j Write W1 = U1 + V1pD and W2 = U2 + V2pD.Then 2 2 2 2 1=U DV = W1W 1 = W2W 2 = U DV : 1 − 1 2 − 2 Hence (W1W2)(W1W2) = 1 which implies W1W2 is a rational solution of (1). A continued fraction expansion for pD is obtained by putting α0 = pD and, recursively for n 0, putting ≥ An =[αn]andαn+1 =1=(αn An): − The algorithm terminates if, for some n, αn = An. This happens if and only if pD is a rational function. Thus pD canbeexpressedinthefollowingway: 1 pD =[pD]+ α1 1 p =[D]+ 1 : [α1]+ + α2 ··· License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use POLYNOMIAL PELL'S EQUATION 995 For short, we write pD = [pD]; [α1];::: = A0;A1;::: ; where Ai [x]: h i h i 2Q We write convergents to pD as Pn=Qn = A0;A1;:::;An ,where h i Pn Qn An 1 Pn 1 Qn 1 = − − for n 0 Pn 1 Qn 1 10 Pn 2 Qn 2 ≥ − − − − and P 1 Q 1 10 − − = : P 2 Q 2 01 − − Then by looking at the determinant of the above matrix, we have for n 0 ≥ n+1 PnQn 1 Pn 1Qn =( 1) : − − − − We note that since sgn A0 > 0, σ(Pn)=σ(Qn) for all n 0, where σ(A)denote the sign of the leading coefficient of A. ≥ Now write pD as pD = A0;A1;:::;An;An+1;::: = A0;A1;:::;An,αn+1 : h i h i Then αn+1Pn + Pn 1 pD = − : αn+1Qn + Qn 1 − We say αj is reduced if αj > 1and αj < 1. j j j j Suppose P + QpD is the minimal solution. Then we can show (see Lemma 2below)thatP + QpD = λ(Pn + QnpD)forsomeλ . Wenotethatifs 2 2 2Q is the least index satisfying (λPs) D(λQs) =1,thensinceσ(Ps)=σ(Qs), − σ(λQs)(λPs + λQspD) is the minimal solution. Let D = A2 +2C be a polynomial in [x], where A; C [x], deg C<deg A, and B = A=C [x]. Since Z 2Q 2Q 1 1 pD =[pD]+ = A + ; α1 α1 where 1 pD + A pD + A 1 1 α1 = = =[ ]+ = B + pD A 2C 2C α2 α2 − and 1 α2 = pD + A =2A + pD A =2A + ; − α1 then pD = A; B;2A and h i P 2 DQ2 =(AB +1)2 DB2 =(AB +1)2 (A2B2 +2AB)=1: 1 − 1 − − Thus σ(Q1)(P1 + Q1pD) is a nontrivial rational solution in T . Note that this may not be the minimal solution. To see this, notice that 2 2 2 2 2 2 (kP0) D(kQ0) = k (A (A +2C)) = k ( 2C)=1 − − − 2 if and only if 2C = 1=k .ThusW0 = kP0 + kQ0pD with sgn (kP0) > 0isthe − 2 minimal solution if and only if 2C = 1=k ,andW0 = σ(Q1)(P1 + Q1pD)isthe − License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 996 WILLIAM A. WEBB AND HISASHI YOKOTA minimal solution if and only if 2C = 1=k2. Thus we are left to determine when n 6 − W0 T0 for W0 = kP0 + kQ0pD and W0 = σ(Q1)(P1 + Q1pD). We2 will show Theorem 1. Let D = A2+2C be a monic polynomial in [x],wheredeg C<deg A and B = A=C [x]. Suppose either A [x] or 2A Z [x]. Then the following are equivalent: 2Q 2Z 2Z n (1) W T0 for some n 1. 0 2 ≥ (2) W0 T0. 2 A + pD; where A [x]; 2C = 1; 2Z − 2A +2pD; where A [x]; 2A [x]; 8 62 Z 2Z > 2C = 1=4; > 2 − > σ(C)(B C +1+BpD);whereB;C [x]; (3) W0 = > 2 2Z > 2A +1+2ApD; where A [x]; 2C =1; <> 2 2Z σ(C)(B C +1+BpD);whereA [x];B = 2B1; 2Z ± 1 > B1 [x]; sgn C = ; > 2Z ± 2 > 2C [x]; deg C>0: > 2Z > Theorem 2. Let:> D = A2+2C be a monic polynomial in [x],wheredeg C<deg A Z and B = A=C [x].SupposeC = c1x + c0 [x]. Then either A [x] or 2A [x]. 2Q 2Q 2Z 2Z Hence the complete characterization of solutions of (1) for the monic polynomials 2 2 of the form D = A +2C = A +c1x+c0, where deg C<deg A and B = A=C [x]. Before proving these, we need a few notations and lemmas. 2Q i j k k 1 Let νp(m=n)=i j; where (m; n)=1;p m; p n.ForA = x + ak 1x − + − k k − + a1x + a0, define νp(A)=min νp(ai):0 i<k . Denote the coefficient aj of x···j in A by [xj]A and the Gaussianf integer function≤ ofg a by a .
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