ON A CERTAIN FAMILY OF QUARTIC THUE EQUATIONS WITH THREE PARAMETERS

ON A CERTAIN FAMILY OF QUARTIC THUE EQUATIONS WITH THREE PARAMETERS VOLKER ZIEGLER Abtract We conider the parameterized Thue equation X X 3 Y (ab + (a + bx Y abxy 3 + a b Y = ±1, where a, b 1 Z uch that ab Z By the hypergeometric method and a method of Tzanaki we find all olution, if i large with repect to a and b 1 Introduction Let F Z[X, Y ] be a homogeneou, irreducible polynomial of degree d 3 and m a nonzero integer Then the Diophantine equation (1 F(X, Y = m i called a Thue equation in honour of A Thue [15] who proved that Diophantine equation (1 ha only finitely many olution (X, Y Z The proof of thi theorem i baed on Thue approximation theorem Given ε > 0 and an algebraic number α of degree n, then there are only finitely many integer p and q > 0 that atify α p q < q n/ 1 ε Since the proof of thi approximation theorem i not effective we cannot olve Thue equation by exploiting the proof of Thue However, Thue oberved that hi approximation theorem can be made effective, if one can find good approximation to α Although Thue never tated explicitly anything like thi, Thue [16] actually olved the family of Thue equation (a + 1X n ay n = 1, where n 3 i a prime and a i uitable large with repect to n He obtained hi good approximation by conidering uitable differential equation and their related hypergeometric function Mahler [1] wa the firt who tated reult on effective meaurement of algebraic number For Thue equation of degree 3 Chudnovky [] give a detailed tudy on the Thue-Siegel method In the 60 of the previou century, Baker [1, 3] conidered linear form of logarithm In a further paper [], he ued hi reult on linear form in order to how how Thue equation can be olved algorithmically Uing Baker method, Bugeaud and Győry [7] computed upper bound for the olution of a ingle Thue equation Thee bound only depend on the regulator, the degree of the related number field and the degree of the Thue equation Alo efficient algorithm have been developed by everal author The mot famou are from Tzanaki and de Weger [1] and from Bilu and Hanrot [6] 000 Mathematic Subject Claification Primary 11D59; Secondary 11D5, 11D09 Key word and phrae Diophantine equation, parameterized Thue equation, norm form equation, imultaneou Pellian equation The author wa partially upported by the Autrian Science Foundation, project S-307-MAT 1

V ZIEGLER Table 1 Solution to (, provided i large a b X Y a b X Y 17 ± ±1 15 ± ±1 17 ± 1 15 ± 1 5 ± ±1 3 ± ±1 5 3 ± 1 ± 1 1 ±1 ±1 1 ±1 1 In 1990 Thoma [1] conidered the family X 3 (n 1X Y (n + XY Y 3 = 1, where n i ome parameter running through all poitive integer Thi wa the firt time that a family of Thue equation with poitive dicriminant ha been olved Another practical approach to olve Thue equation i the method of Tzanaki [17] who howed how to reduce quartic Thue equation of certain type to a ytem of Pellian equation Uing the method of Tzanaki, Dujella and Jadrijević [9] olved the parametrized Thue equation ( X nx 3 Y + (6n + X Y + nxy 3 + Y = 1 by reducing it to the ytem (n + 1U nv =1, (n U nz = of Pellian equation They olved thi ytem for all rational integer n by the method of Baker and Davenport (cf [] combined with the congruence method (cf [9] and a reult of Bennett [5] about imultaneou approximation of quare root By a refinement of their method, Dujella and Jadrijević (cf [10] olved the Thue inequality (3 X nx 3 Y + (6n + X Y + nxy 3 + Y 6n + The aim of thi paper i to olve following family of Thue equation: ( X X 3 Y (ab + (a + bx Y abxy 3 + a b Y = µ, where µ {1, 1}, a, b 1 Z, with a b and 0 ab Z and Z large with repect to a and b Oberve that for a =, b = 1/ and µ = 1 we obtain Equation ( In particular we prove the following theorem: Theorem 1 Let (X, Y be a olution to Thue equation ( with Z, a, b 1 Z, a b, a b and 0 ab Z and uppoe > 73 10 10 a 9+ 1 Then necearily µ = 1 Furthermore, the only olution are (X, Y = (±1, 0, (X, Y = (0, ±1 if ab = ±1 or thoe lited in Table 1 Oberve that there i no olution in the cae of µ = 1 and ufficiently large Furthermore it i no retriction to aume that a b, ince equation ( i ymmetric in a and b

ON A CERTAIN FAMILY OF THUE EQUATIONS 3 The ret of the paper i organized a follow In Section we preent ome preliminary reult and invetigate aymptotic expanion of the relevant root How to reduce Thue equation ( to a ytem of Pellian equation i demontrated in Section 3 For olution to thi ytem we will find an upper bound by the hypergeometric method (cf Section In order to obtain a lower bound we will ue Padé approximation in Section 5 The proof of Theorem 1 will be finihed in Section 6, where we conider the remaining cae Y = 1 In the lat ection we will tate pecial cae of Theorem 1, where a, b 1 Z repectively a, b Z and give ome example We tart with the norm form equation Preliminarie (5 N K Q (X + αy = ±1, where α = ( + a + ( + b + ( + a( + b +, with Z and a, b ( 1 Z, uch that ab Z, a b and a b Obviouly, α i an element of the ( ( ( compoitum K := Q + a Q + b In any cae, K i Galoi, ince K i the compoitum of two field that are Galoi over Q If > max( a, b, then K i a real field and moreover K i quartic if and only if none of the quantitie ( + a, ( + b and ( + a( + b i a perfect quare Lemma 1 Aume > ( a + 1/ Then K i Galoi, real and quartic with Galoi group G Z/Z Z/Z Proof From the dicuion above ( we know that K i Galoi and real Let u aume ( + a( + b i ( ( ( not a perfect quare Then Q + a Q + b = Q and therefore K i quartic and we know from Galoi theory that the Galoi group i of the wanted form (cf [11, Chapter VI, Theorem 11] So we are left to prove that neither ( + a, ( + b nor ( + a( + b i a perfect quare Aume (+a(+b i a perfect quare From the aumption on a and b we find that (+a(+b i the quare of an integer On the other hand, we have ( + (a + b =( + a( + b + (a b >( + a( + b, ( ( + (a + b 1/ =( + a( + b + a + b (a b 1 16 ( + a( + b ( ( a + 1/ <( + a( + b, a contradiction, hence ( + a( + b i not a perfect quare Similarly we find and ( + a >( + a, ( + a 1/ <( + a, ( + b >( + b, ( + b 1/ <( + a Therefore neither of ( + a, ( + b nor ( + a( + b i a perfect quare

V ZIEGLER Becaue of Lemma 1 we aume for the ret of the paper that > ( a + 1/ Moreover we immediately obtain from Galoi theory the conjugate α 1,,α of α: (6 α = α 1 = + ( + a + ( + b + ( + a( + b; α = ( + a + ( + b ( + a( + b; α 3 = + ( + a ( + b ( + a( + b; α = ( + a ( + b + ( + a( + b Therefore we are able to compute the minimal polynomial f Q[X] of α: f(x := X X 3 (ab + (a + bx abx + a b, ie that norm form Equation (5 i equivalent to Thue equation (7 F(X, Y := X X 3 Y (ab + (a + bx Y abxy 3 + a b Y = ±1 Furthermore we have proved that α i an algebraic integer Next we want to invetigate the aymptotic of the α a Becaue of the tructure of the α, we only have to conider the aymptotic of ( + a, ( + b and ( + b( + a The following expanion i well know 1 + a = n=0 ( 1/ a n n n and it i valid for > a Thi implie ( 1/ a n ( + a = n n 1, n=0 ( 1/ b n ( + b = n n 1, n=0 n ( ( 1/ 1/ a k b n k ( + b( + a = k n k n 1, n=0 k=0 where all three expanion are valid if > a The following variant of the uual O-notation i ued For two function g( and h( we write g( = L(h( if g( h( Thi notation i ued in the middle of an expreion in the ame way a i uually done with O-notation With thi L-notation we obtain N ( ( 1/ a n ( + a = n n 1 + L ( 1/ a n n n 1 ( = n=0 N n=0 n=n+1 ( ( ( 1/ a n 1/ n n 1 + L N + 1 a N+1 N a For an exact aymptotic of ( + b( + a we conider the N-th coefficient of it expanion By elementary calculation we oberve that ( N ( ( C(N := 1/ 1/ 1/ = if N ; N k N k k=0 1 if N = 0, 1

ON A CERTAIN FAMILY OF THUE EQUATIONS 5 Since C(N i decreaing with N, thi implie N n ( ( 1/ 1/ a k b n k ( + b( + a = k n k n 1 n=0 k=0 (9 N n ( ( 1/ 1/ a k b n k = k n k n 1 n=0 k=0 + L ( n=n+1 C(n a n n 1 + L (C(N + 1 a N+1 N a 3 From Thue Equation to Pellian Equation In 1993, Tzanaki [17] conidered Thue equation of the form (10 F(X, Y := a 0 X + a 1 X 3 Y + 6a X Y + a 3 XY 3 + a Y = m uch that F(X, Y Z[X, Y ], m Z and a 0 > 0 Furthermore the correponding number field K ha to be Galoi and non-cyclic If we aume K i not totally complex, ie there i ome real root of F(X, 1, then K i the compoitum of two real quadratic field Furthermore the equation (11 σ 3 g σ g 3 = 0 ha three ditinct rational root σ 1, σ and σ 3 ; here g and g 3 are invariant of the following form: g =a 0 a a 1 a 3 + 3a, g 3 = det a 0 a 1 a a 1 a a 3 a a 3 a Let H(X, Y and G(X, Y be the quartic and extic covariant of F(X, Y repectively (cf [13], ie H(X, Y = 1 f f X X Y, G(X, Y = 1 f f X Y 1 f H We have Y X f Y H(X, Y 1 1 Z[X, Y ], G(X, Y 96 Z[X, Y ] and (cf [13, Theorem 1 Chapter 5] (1 H 3 g Hf g 3 f 3 = G Let u put now H = 1 H 0, G = 1 96 G 0, σ i = 1 1 r i (i = 1,, 3, then H 0, G 0 Z[X, Y ] and r i Z In view of Equation (1 we have (H 0 r 1 f(h 0 r f(h 0 r 3 f = 3G 0 Since H and f are relatively prime (cf [17, Propoition 1] there exit quare-free integer k 1, k and k 3 and quadratic form G i Z[X, Y ], i = 1,, 3 uch that (13 H 0 r i f = k i G i i = 1,, 3 If (X, Y i a olution to (10, we obtain from identity (13 the ytem (1 of Pellian equation k G k 1 G 1 = (r 1 r m, k 3 G 3 k 1 G 1 = (r 1 r 3 m X H Y

6 V ZIEGLER Applying thi procedure to Thue Equation (7 we obtain a 0 =1, a 1 =, a = g =a b ab + 1 3 (ab + (a + b, ab + (a + b, a 3 = ab, a =a b ; 3 g 3 = (ab + (a + b(ab + (b a(ab + (a b; 7 σ 1 = 1 3 (ab + a + b, σ = 1 3 ( b + ab + a, σ 3 = 1 ( a + ab + b; 3 G 1 = X aby, G = X + axy + aby, G 3 = X + bxy + aby ; Thi yield the ytem (15 with and µ = ±1 k 1 = (a + (b +, k = (b +, k 3 = (a + (a + U V = µa, (b + U Z = µb, U = X aby, V = X + axy + aby, Z = X + bxy + aby Hypergeometric Method In thi ection we want to find an upper bound for U if (U, V, Z i a olution to ytem (15 Let u firt oberve that if (U, V, Z i a olution to (15, then alo (±U, ±V, ±Z i a olution to (15 Therefore we may aume without lo of generality U, V, Z 0 Furthermore U = 0 yield V = ±a and, ince we aumed > ( a + 1/, we have V < 1, hence V = 0 and a = 0, a contradiction Similar argument apply to V and Z, therefore we may aume U, V, Z > 0 In order to prove an upper bound for U we will dicu firt ome approximation propertie of olution (U, V, Z to (15 Lemma Let (U, V, Z be a olution to ytem (15 with U, V, Z > 0 Then V U 1 + a a 1 U ( + a ; Z U 1 + b b 1 U ( + b Proof We only prove the firt inequality The proof of the econd inequality i analogou One jut ha to replace Z by V and b by a Since U, V > 0 we have 1 + a U V + U 1 + a and therefore 1 V U + a V U 1 + a = V + U 1 + a V + U 1 + a Diviion through U yield the lemma a U 1 + a/ = a 1 U ( + a

ON A CERTAIN FAMILY OF THUE EQUATIONS 7 Since we aume a b the lemma above how that ( V (16 max U 1 + a, Z U 1 + b a U 1 ( + a Hence we have found a good imultaneou approximation to 1 + a 1 and + b The following dicuion will how that thi approximation i in ome ene too good We tart with a theorem of Bennett [5, Theorem 3] Theorem If a i, p i, q and N are integer for 0 i with a 0 < a 1 < a, a j = 0 for ome 0 j, q nonzero and N > M 9, where M = max a i, then we have i=0,1, ( max 1 + a i i=0,1, N p i q > (130NΥ 1 q λ = c 1 q λ, where and log(33n Υ λ = 1 + log (17N 0 i<j (a i a j (a a 0 (a a 1 if a a 1 a 1 a 0, a Υ = a 0 a 1 (a a 0 (a 1 a 0 if a a 1 < a 1 a 0 a 1 + a a 0 We want to apply Theorem to a i = a and a j = b with ome i, j {0, 1, }, i j, a = a /, b = b / and N = Firt let u etimate Υ Therefore we have to ditinguih between 6 cae (remind that we alway aume a b, hence a b Suppoe a > b > 0 and a b b 0, then we have a b and Υ = (a 0 (a b a b 0 = a (a b a b 3 a 3 Let a > b > 0 and a b < b 0 The lat inequality i a < b and therefore Υ = (a 0 (b 0 b + a 0 = a b a + b 3 a 3 Aume a > 0 > b Since a b we have a a 1 a 1 a 0, ie Υ = (a b (a 0 a b 0 = (a b a a b 3 a 3 Provided b > 0 > a we have a a 1 a 1 a 0, hence in both cae < and = we find Υ = (b a (0 a 0 + b a = (b a a a + b 3 a 3 In the cae of 0 > b > a and 0 b b a we have b a and obtain following etimation Υ = (0 b (0 a 0 b a = b a b a 3 a 3 At lat we conider the cae 0 > b > a and 0 b < b a Therefore b < a and Υ = (0 a (b a b + 0 a = a (b a a + b 3 a 3

V ZIEGLER All cae together yield the etimation Υ 3 a 3 = 56 3 a 3 Hence we have c c := 13310 a 3, λ λ log(116 a 3 = 1 + 3 log(7 log(163 a 6 We want to have λ < Therefore we conider the inequality or equivalently 1 < log(116 a 3 log(7 log(163 a 6 log 116 + log + 3 log a < log 7 + log log 163 6 log a The lat inequality hold if log > log 159376 7 + 9 log a, ie > 159376 7 a 9 Therefore we will aume for the ret of thi ection > 159376 7 a 9 The aumption N > M 9, ie > 9 a 9 i now fullfilled and by an application of Theorem, together with Lemma, we obtain c 1 V U λ < max( U 1 + a, Z U 1 + b Taking logarithm and olving for log U yield log U < 1 (log λ c + log a 1 log(( + a (17 < 1 λ < 1 ( log a + 1071 λ Let u aume > c 0 a 9+r, then we have 1 λ = 1 log(116 a 1 3 (1 a U 1 ( + a ( log 13310 + log + log a 1 3 ( log 7 159376 log(7 log(163 a 6 log(7 log(163 a 6 = log(7 log(163 a 6 log(116 a 3 = log 6 log a + log ( 7 163 log 9 log a + log ( 7 163 116 ( 6 9+r log + 6 9+r log c 0 + log ( 7 163 < ( 1 9 9+r log + 9 9+r log c 0 + log ( 7 163 116 1 + r = 9 + r r r 1 logc 0 r log ( 199760700 17 ( 1 log 11533360 17 r log + 9 logc 0 (9 + rlog ( 11533360 17 i a function increaing with a Let u conider the econd 1 The inequality above hold, ince λ fraction of the lat line in (1 A one eaily can compute, the numerator i 0 if 31r 1+ (19 c 0 1 ( 17 5 1 r 1 11 1+ r 6 If c 0 fullfill thi inequality another computation how that the denominator of thi fraction i poitive for 1 Hence we have proved 1 1 + r (0 < λ r

ON A CERTAIN FAMILY OF THUE EQUATIONS 9 provided (19 hold By combining (17 and (0 we obtain (1 + r log U < r + r < (9 + rr 1 + r log a + 1071 r + r log (9 + rr log c 0 + 1071 1 + r r Next we want to find an upper bound for Y Firt let u aume ab < 0 Then we have U = X aby Y and therefore we find log Y 1 log U Let now ab > 0 then Z = X + bxy + aby = (X + by + (ab b Y 3 Y On the other hand, the econd Pellian equation of (15 yield ( Z = 1 + b U ± b ( < 7 7 U 1 + + 159376 159376 < (U 10001, hence Therefore we have proved log Y < 01 + 1 log U Propoition 1 Let (X, Y be a olution to ( and aume > c 0 (r a 9+r with r > 0 and ( 1 r 31r 1+ 17 1 c 0 (r := 1 11 1+ r 6 5 Then log Y < + r + r log (9 + rr (9 + rr log c 1 + r 0(r + 1071 + 01 r 5 Approximation Propertie of α In the previou ection we have found an upper bound for log Y if i large with repect to a and b In thi ection we find a lower bound for Y provided Y > 1 Thi bound will be found by uing approximation propertie of the root α i, 1 i We further aume > 159376 7 a 9 Firt we prove the following lemma: Lemma 3 Let (X, Y be a olution to (, then at leat one of the following cae occur X α 1 Y < 6399 Y 3 3 ; X α Y < X α 3 Y < 99 Y 3 ; X α Y < Proof From (6, ( and (9 we find ( α 1 = + a + b + L 3 a a ( ( α 3 = b + L 3 a a ; α = L Let j be choen uch that X α j Y = ( ; α = a + L 3 a ( 3 a a min X α iy, then we have i=1,,3, 399 Y 3 ; 399 Y 3 a Y α i α j X α i Y + X α j Y X α i Y ;

10 V ZIEGLER Hence 1 (3 X α j Y = i j X α iy Y 3 i j α j α i Some elementary computation yield lower bound for i j α j α i and from thee we obtain the lemma Propoition Let (X, Y be a olution to ( with Y > 1 and > 35 10 9 a 9 According to the four cae in Lemma 3 we have Y > 136 a 6; Y > 096 a 7; Y > 3 096 a 7; Y > 7107 a Hence, in all cae we have Y > 096 a 7 Proof From ( together with Lemma 3 we obtain X ( + a + by < Y 0751 a X + ay < Y 0751 a X + by < Y 0751 a X < Y 0751 a + 6399 Y 3 < Y 075 a, 3 + + + 399 Y 3 99 Y 3 399 Y 3 < Y 077 a, < Y 0919 a, < Y 077 a, according to the four poible cae Since the left hand ide are 1 Z, we conclude that they vanih if ( Y 300 a, Y 350 a, Y 3676 a, Y 350 a Auming ( we obtain X = ( + a + by, X = ay, X = by and X = 0 repectively Inerting thee relation in F(X, Y = ±1 we deduce F(Y ( + a + b, Y = Y 16 (a ab + b + (a 3 + a b + ab + b 3 +(a + a 3 b + 3a b + ab 3 + b > Y 1199 a > 1, F( Y a, Y = Y a (a b Y > 1, F( Y b, Y = Y b (a b Y 9 16 9 > 1, F(0, Y = Y > 1 In any cae we get a contradiction and therefore we may aume (5 Y > 300 a, Y > 350 a, Y > 3676 a, Y > 350 a Now we plit the proof into the four cae according to Lemma (3 Cae 1: We ue the approximation (6 α 1 = + a + b a ab + b + a3 a b ab + b 3 16 ( 037 a + L 3

ON A CERTAIN FAMILY OF THUE EQUATIONS 11 Let u denote by ᾱ 1 the approximation (6 with omitted L-term Uing Padé approximation we find polynomial P and Q uch that ᾱ 1 P Q = (a3 a b ab + b 3 16 = a6 a 5 b 3a b + 10a 3 b 3 3a b ab 5 + b 6 16, where Q :=16 (a ab + b + (3a 3 a b ab + 3b 3 + (a + 3a 3 b 5a b + 3ab 3 + b, P :=(a ab + b + (a 3 a b ab + b 3 By elementary calculation we obtain P < 1 a + 6 a 3 < 1001 a Uing Lemma 3 we obtain (PX QY Y (Pᾱ 1 Q Y P(α 1 ᾱ 1 < Some elementary calculation together with (5 yield (7 P 6399 Y 3 3 PX QY < Y Pᾱ 1 Q + Y P(α 1 ᾱ 1 + Y P 10 a 7 < Y ( a 6 + a 6 + 19 a 10 6 < Y 3 a 6 Since PX QY 1 56 Z, we have PX QY = 0 provided Y 136 a 6 Inerting thi relation in F(X, Y = ±1 yield ( Y (56 A + R 1 = ±P, where A = ( a ab + b 3 (a 6 3a 5 b + 9a b 13a 3 b 3 + 9a b 3ab 5 + b 6 For a 0, a 1 > 0 we have ( (9 X a 0 XY + a 1 Y a0 = X ( ( a 1 Y + X 1 a 0 X 1 a 0 a 1 a 1 a 1 Therefore we find the etimation A = ( a ab + b 3 (a 6 3a 5 b + 9a b 13a 3 b 3 + 9a b 3ab 5 + b 6 > 7 6 a 6 (a ab + 5b (a 013ab + 01b (a 11ab + b > a 1 00667 Furthermore, R 1 i ome expreion of lower term that one can etimate by R 1 < 19955 3 a 13 + 11916 a 1 + 3690 a 15 + 6597 a 16 < 10 5 3 a 13 Therefore we receive from ( and (5 On the other hand Y 56 A + R 1 > Y 1709 a 1 > 00 a P < 073 a which yield a contradiction to ( if > 1 a Therefore we have proved Y > 136 a 6, ie the firt cae The proof of the other cae i imilar and we will dicu them le detailed Cae : For thi cae we ue the approximation α = :=ᾱ {}}{ a + a ab + a3 + a b + ab 16 + 5a a 3 b a b + ab 3 6 3 +L ( 0161 a 5

1 V ZIEGLER For ᾱ we obtain a Padé approximation Q P ᾱ with Similarly a in the firt cae we compute (30 Q :=16 a(a ab + b a(11a 3 + 9a b + 9ab + 1b 3 a(13a + 3a 3 b + 31a b + 17ab 3 + 7b, P := 16 (a ab + b + a(10a 3 + 11a b + 7ab + b 3 + (5a + 30a 3 b + 9a b + 1ab 3 + b PX QY < Y Pᾱ Q + Y P(α ᾱ + Y P 3036 a ( 5 5 a < Y 3 3 + 7 a 7 + 1576 a 10 3 < Y 79 a 7 Therefore PX QY = 0 provided Y 096 a 7 From F(X, Y = ±1 we obtain now (31 Y ( 5 A + R 1 = ±P, with A =10a (a b (a ab + b 1a 3 + 9a b + 7ab + 5b 3 1 56 a 10, R 1 550016 a 16 + 331360 3 a 17 + 1576 a 1 + 100 a 19 + 19663 a 0 < a 16 551 10 6, P 531 10 6 a Comparing the bound from the right hand ide and left hand ide of (31, we find which i a contradiction for > 66 10 9 a 6 Cae 3: Now we ue the following approximation α 3 = b + 000 9 a < 531 10 6 a, :=ᾱ 3 { }} { ab + b + a b + ab b 3 16 + a3 b a b b 3 a + 5b 6 3 +L Applying Padé theorem to ᾱ 3 we obtain an approximation Q P ᾱ 3 with Similarly a in the firt cae, we obtain (3 Q :=16 b(a ab + b b(a 3 + 9a b + 9ab + 11b 3 b(7a + 17a 3 b + 31a b + 3ab 3 + 13b, P := 16 (a ab + b + a(a 3 + 7a b + 11ab + 10b 3 + (a + 1a 3 b + 9a b + 30ab 3 + 5b PX QY < Y Pᾱ 3 Q + Y P(α 3 ᾱ 3 + Y P 57 a ( 5 5 a < Y 3 3 + 7 a 7 + 35 a 10 3 < Y 79 a 7 ( 0161 a 5 Therefore PX QY = 0 provided Y 096 a 7 If we inert thi relation in F(X, Y = ±1, we get (33 Y ( 5 A + R 1 = ±P,

ON A CERTAIN FAMILY OF THUE EQUATIONS 13 with A =10b (a b (a ab + b 5a 3 + 7a b + 9ab + 1b 3 15 65536 a 10, R 1 550016 a 16 + 331360 3 a 17 + 1576 a 1 + 100 a 19 + 19663 a 0 < a 16 551 10 6, P 531 10 6 a Thi time we deduce from (33 a contradiction, provided > 35 10 9 a 6 000153 9 a < 531 10 6 a, Cae : In the lat cae we ue the method of Padé approximation twice Firt we ue the following approximation :=ᾱ {}}{ ab α = a b + ab ( 035 a 16 +L 3 and by Padé theorem we obtain an approximation Q P ᾱ with Similarly a in the firt cae we obtain (3 Q :=ab P := + a + b PX QY < Y Pᾱ Q + Y P(α ᾱ + Y P 3036 a ( 5 a < Y + 095 a + 119 a < Y 11 a Therefore PX QY = 0 provided Y 1936 a Thi relation together with F(X, Y = ±1 yield (35 Y ( A + R 1 = ±P, where From (35 we deduce A =16a b (a ab + b 1 a, R 1 a 7 + 9 a > 001 a 7, P 57 00 6 < 57 a 6, a contradiction provided > 57 a 3 Therefore we may aume Y > 1936 a For the econd application of Padé approximation we ue α = :=ᾱ {}}{ ab a b + ab 16 + a3 b + a b + ab 3 6 3 5a b + a 3 b + a b 3 + 5ab 56 +L Therefore we find an approximation Q P ᾱ with Q :=ab(a ab + b + ab(a 3 a b ab + b 3 ( 0131 a 6 P :=16 (a ab + b + (3a 3 a b ab + 3b 3 + (a + 3a 3 b 5a b + 3ab 3 + b 5

1 V ZIEGLER Similarly a in the firt cae we obtain (36 PX QY < Y Pᾱ Q + Y P(α ᾱ + Y P 1670 a 16 9 < Y (10 a 3 + 591 a 3 + 136 107 a 1 7 < Y 69 a 3 From (36 we deduce PX QY = 0, if Y 3 7107 a Let u inert thi relation into the original Thue equation F(X, Y = ±1, then (37 Y ( A + R 1 = ±P, with A =56a b ( a ab + b 3 (a 6 3a 5 b + 9a b 13a 3 b 3 + 9a b 3ab 5 + b 6 > 171 a 1, R 1 19955 3 a 17 + 11916 a 1 + 3690 a 19 + 6597 a 0 < 3 a 17 10 5, P 531 10 6 a Thi, combined with (37, implie 11 10 1 a < 531 10 6 a, a contradiction, ince we aumed > 35 10 9 a 9 > 5 a 3 If a and b are given, one may obtain better reult than thoe proved in Propoition Indeed, one only ha to apply the method of Padé approximation ucceively to obtain reult of the form Y > µ c(a, b, µ for ome µ > In the general cae the difficulty to find optimal or even ueful etimation rapidly grow Oberve that the technical bound c 0 = 35 10 9 in Propoition i quite large In the cae of a, b Z the technical bound actually exceed the bound that one obtain by comparing lower and upper bound for log Y (cf (3 and Theorem 3 Corollary 1 Let (X, Y be a olution to (, then Y 1 provided > 73 10 10 a 9+ 1 Proof Suppoe (X, Y i a olution to ( with Y > 1 Let u compare the bound from Propoiton 1 and and aume > c 1 a 9+r with c 1 > max(35 10 9, c 0 (r (cf Propoition 1 Then we obtain ( 7 log + 7 9 + r 9 + r log c 1 log 096 < log Y + r < (9 + rr (log log c 1 + 1071 1 + r + 01 r and therefore (3 r + 7r r(r + 9 log + 11r + r + 9 log c 1 < 1071 1 + r + 01 + log 096 r The coefficient of log i poitive if r > 7+ 1 and (3 fail provided i large enough Suppoe now r = 7+ 1, then the coefficient of log i zero and (3 alo fail if c 1 i large enough, ie c 1 > 73 10 10 > max(35 10 9, c 0 (r

ON A CERTAIN FAMILY OF THUE EQUATIONS 15 6 The cae of Y = 1 In view of Corollary 1 it remain to conider the cae of Y 1 in order to prove Theorem 1 The cae Y = 0 yield the trivial olution (X, Y = (±1, 0 Therefore we are left to the four cae Y = ±1 and µ = ±1 with mixed ign Since with (X, Y alo ( X, Y i a olution to ( we only have to check the cae Y = 1 and µ = ±1 Let u conider firt the cae Y = 1 and µ = 1 Thue equation ( reduce to P(X = X X 3 (ab + (a + bx abx + a b 1 = 0 Since F(X, 1 = P(X + 1 we deduce, that the root of P(X and F(X, 1 are very near Therefore we want to prove that the root of P(X lie in the dijoint interval Let u conider the quantitie I 1 :=( + a + b 1/, + a + b + 1/, I :=( a 1/, a + 1/, I 3 :=( b 1/, b + 1/ and I :=( 1/, 1/ P( + a + b 1 P( + a + b + 1 = 66 + 1(a + b 5 + P( a 1 P( a + 1 = ( 1+6a ( 1+6(a b 163 + > 63 3 163 a + P( b 1 P( b + 1 = ( 1+6b ( 1+6(a b 163 + > 16 163 a + P( 1 P(1 = ( 1+6a ( 1+6b 163 + > 63 3 163 a + which yield that if i large enough then each root of P lie in one of the interval I 1, I, I 3 or I A more detailed analyi yield that > 130 a 3 i adequate (alo for the cae µ = 1 Therefore the only integral olution may be X = + a + b, a, b or X = 0 Inerting in ( yield 16(a ab + b + (a 3 + a b + ab + b 3 + (a + a 3 b + 3a b + ab 3 + b = 1, a (a b = 1, b (a b = 1, a b = 1, repectively The firt equation fail if i too large, ie > 36 a The other equation only yield olution lited in Table 1 Similar argument apply for µ = 1 Oberve that in thi cae there are no olution Therefore we have proved the following propoition: Propoition 3 Let (X, Y be a olution to ( with Y = 1 and let > 130 a 3, then the olution (X, Y i lited in Theorem 1 or Table 1 Combining Corollary 1 and Propotion 3 we immediately obtain Theorem 1 7 Some Example Firt let u tate a theorem, that one may obtain by recomputing the proof of Theorem 1 Theorem 3 If a, b 1 Z, repectively a, b Z then Theorem 1 alo hold for > 633 107 a 9+ 1 repectively > 107 10 5 a 9+ 1 Let u now conider four example to illutrate Theorem 1 and 3: Let a = and b = 1/ Then we have the Thue equation (39 X X 3 Y + (6 + X Y + XY 3 + Y = ±1,

16 V ZIEGLER Let (X, Y be a olution to (39, then (X, Y = (0, ±1 or (X, Y = (±1, 0 if > 1 10 11 Note that Thue equation (39 ha been olved for all 0 in the cae of the +-ign by Dujella and Jadrijević [9] Let a = 1 and b = 1 then (±1, 0 and (0, ±1 are the only olution to the Thue equation (0 X X 3 Y + X Y + XY 3 + Y = ±1 provided > 107 10 5 Let a = 5/, b = and > 171 10 1 then the Diophantine equation (1 X X 3 Y (10 + 1X Y 0XY 3 + 5Y = ±1 ha only the olution (±1, 0, (, 1 and (, 1 Let a = and b = 13 then (X, Y = (±1, 0 i the only olution to Thue equation ( X X 3 Y + (6 3X Y + 5XY 3 + 169Y = ±1 if we aume > 36 10 17 All four example have been olved uing Theorem 1 repectively Theorem 3 In mot cae one could obtain harper bounde for, if one would apply the method of Padé approximation everal time more However Dujella and Jadrijević [9] ued the congruence method in order to obtain a harp etimate for in the cae of equation (39 In order to apply thi powerful method we tart with ytem (15 and multiply the firt equation by b and put k = b + ab and imilarly we multiply the econd equation by a and put k = a+ ab In both cae we obtain a Pellian equation of the form ( (3 X k + ab ( Y k ab = µab Note that the coefficient of (3 need not be integer The congruence method eentially depand on finding a fundamental olution to (3, ie in the cae above we have to find a fundamental olution correponding to X (k 1Y = 1, X (k 1Y = 1, X (k 5Y = 1, X (16k 676Y = 1, repectively Note that the Pell equation above have integral coefficient The firt two cae yield the fundamental olution k + k 1 The other cae do not have parameterized fundamental olution and we cannot apply the powerful congruence method in thoe cae A cloe look on (3 how, that only in the cae of ab = 1 and a, b 1 Z or ab = and a, b Z we can find parameterized fundamental olution Reference [1] A Baker Linear form in the logarithm of algebraic number I, II, III Mathematika 13 (1966, 0-16; ibid 1 (1967, 10-107; ibid, 1:0, 1967 [] A Baker Contribution to the theory of Diophantine equation I On the repreentation of integer by binary form Philo Tran Roy Soc London Ser A, 63:173 191, 1967/196 [3] A Baker Linear form in the logarithm of algebraic number IV Mathematika, 15:0 16, 196 [] A Baker and H Davenport The equation 3x = y and x 7 = z Quart J Math Oxford Ser (, 0:19 137, 1969 [5] M A Bennett On the number of olution of imultaneou Pell equation J Reine Angew Math, 9:173 199, 199 [6] Y Bilu and G Hanrot Solving Thue equation of high degree J Number Theory, 60(:373 39, 1996

ON A CERTAIN FAMILY OF THUE EQUATIONS 17 [7] Y Bugeaud and K Győry Bound for the olution of Thue-Mahler equation and norm form equation Acta Arith, 7(3:73 9, 1996 [] G V Chudnovky On the method of Thue-Siegel Ann of Math (, 117(:35 3, 193 [9] A Dujella and B Jadrijević A parametric family of quartic Thue equation Acta Arith, 101(:159 170, 00 [10] A Dujella and B Jadrijević A family of quartic Thue inequalitie Acta Arith, 111(1:61 76, 00 [11] S Lang Algebra, volume 11 of Graduate Text in Mathematic Springer-Verlag, New York, third edition, 00 [1] K Mahler Zur Approximation algebraicher Zahlen (I Math Ann, 107:691 730, 1933 [13] L J Mordell Diophantine equation Pure and Applied Mathematic, Vol 30 Academic Pre, London, 1969 [1] E Thoma Complete olution to a family of cubic Diophantine equation J Number Theory, 3(:35 50, 1990 [15] A Thue Über Annäherungwerte algebraicher Zahlen J Reine und Angew Math, 135: 305, 1909 [16] A Thue Berechnung aller Löungen gewier Gleichungen von der Form ax r by r = f VidSkrivter I Mat- Naturv Klae, :1 9, 191 [17] N Tzanaki Explicit olution of a cla of quartic Thue equation Acta Arith, 6(3:71 3, 1993 [1] N Tzanaki and B M M de Weger On the practical olution of the Thue equation J Number Theory, 31(:99 13, 199 Intitut für Mathematik A, Techniche Univerität Graz, Steyrergae 30, A-010 Graz, Autria E-mail addre: ziegler@finanzmathtugrazat