GLASNIK MATEMATIČKI Vol. 1(61)(006), 9 30 ON A CERTAIN FAMILY OF QUARTIC THUE EQUATIONS WITH THREE PARAMETERS Volker Ziegler Techniche Univerität Graz, Autria Abtract. We conider the parameterized Thue equation X X 3 Y (ab + (a + b))x Y abxy 3 + a b Y = ±1, with 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 (1.1) F (X, Y ) = m i called a Thue equation in honour of A. Thue [15] who proved that Diophantine equation (1.1) 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 000 Mathematic Subject Claification. 11D59, 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- 8307-MAT. 9
10 V. ZIEGLER approximation to α. Although Thue never tated explicitly anything like thi, Thue [16] actually olved the family of Thue equation (a + 1)X 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 [8] 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 [18] and from Bilu and Hanrot [6]. In 1990 Thoma [1] conidered the family X 3 (n 1)X 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 (1.) X nx 3 Y + (6n + )X Y + nxy 3 + Y = 1 by reducing it to the ytem (n + 1)U 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 (1.3) X nx 3 Y + (6n + )X Y + nxy 3 + Y 6n +. The aim of thi paper i to olve following family of Thue equation: (1.) X X 3 Y (ab + (a + b))x Y abxy 3 + a b Y = µ,
ON A CERTAIN FAMILY OF THUE EQUATIONS 11 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 (1.). In particular we prove the following theorem: Theorem 1.1. Let (X, Y ) be a olution to Thue equation (1.) with Z, a, b 1 Z, a b, a b and 0 ab Z and uppoe > 7.3 1010 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. Table 1. Solution to (1.), 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 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 (1.) i ymmetric in a and b. 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 (1.) 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 1.1 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 1.1, where a, b 1 Z, repectively a, b Z, and give ome example.. Preliminarie We tart with the norm form equation (.1) N K Q (X + αy ) = ±1,
1 V. ZIEGLER 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 1.1]). 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/) <( + b). Therefore neither of ( + a), ( + b) nor ( + a)( + b) i a perfect quare. Becaue of Lemma.1 we aume for the ret of the paper that > ( a + 1/). Moreover we immediately obtain from Galoi theory the conjugate
ON A CERTAIN FAMILY OF THUE EQUATIONS 13 α 1,..., α of α: (.) α: 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 f(x) := X X 3 (ab + (a + b))x abx + a b, i.e. norm form equation (.1) i equivalent to Thue equation (.3) F (X, Y ) := X X 3 Y (ab + (a + b))x 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 known 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 it 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
1 V. ZIEGLER 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. Since C(N) i decreaing with N, thi implie (.5) N n ( 1/ ( + b)( + a) = k n=0 k=0 ( + L n=n+1 )( 1/ n k C(n) a n n 1 ) a k b n k ) n 1 N n ( )( ) 1/ 1/ a k b n k = k n k n 1 n=0 k=0 ) + 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 (3.1) 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, i.e. there i ome real root of F (X, 1), then K i the compoitum of two real quadratic field. Furthermore the equation (3.) 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]), i.e. H(X, Y ) = 1 f f X X Y, G(X, Y ) = 1 f f X Y 1 f 8 H. Y X f Y X H Y
ON A CERTAIN FAMILY OF THUE EQUATIONS 15 We have H(X, Y ) 1 8 Z[X, Y ], G(X, Y ) 1 96 Z[X, Y ] and (cf. [13, Chapter 5, Theorem 1]) (3.3) H 3 g Hf g 3 f 3 = G. Let u put now H = 1 8 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 (3.3) 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 quarefree integer k 1, k and k 3 and quadratic form G i Z[X, Y ], i = 1,, 3 uch that (3.) H 0 r i f = k i G i i = 1,, 3. If (X, Y ) i a olution to (3.1), we obtain from identity (3.) the ytem (3.5) k G k 1G 1 = (r 1 r )m, k 3 G 3 k 1G 1 = (r 1 r 3 )m of Pellian equation. Applying thi procedure to Thue equation (.3) 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 (ab + a + b), 3 σ = 1 ( b + ab + a), 3 σ 3 = 1 ( a + ab + b); 3 G 1 = X aby, G = X +axy +aby, G 3 = X +bxy +aby ; k 1 = 8(a + )(b + ), k = 8(b + ), k 3 = 8(a + ). Thi yield the ytem (3.6) with (a + )U V = µa, (b + )U Z = µb, U = X aby, V = X + axy + aby, Z = X + bxy + aby and µ = ±1.
16 V. ZIEGLER. Hypergeometric method In thi ection we want to find an upper bound for U if (U, V, Z) i a olution to ytem (3.6). Let u firt oberve that if (U, V, Z) i a olution to (3.6), then alo (±U, ±V, ±Z) i a olution to (3.6). Therefore we may aume without lo of generality U, V, Z 0. Furthermore U = 0 yield V = ±a and, ince we aume > ( 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 (3.6). Lemma.1. Let (U, V, Z) be a olution to ytem (3.6) 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 a U 1 + a/ = a 1 U ( + a). Diviion with U yield the lemma. Since we aume a b the lemma above how that ( V (.1) max U 1 + a ), Z U 1 + b a 1 U ( + a). Hence we have found a good imultaneou approximation to 1 + a and 1 + 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, i=0,1,
ON A CERTAIN FAMILY OF THUE EQUATIONS 17 then we have where and max i=0,1, ( 1 + a i N p ) i q > (130NΥ) 1 q λ = c 1 q λ, log(33n Υ) λ = 1 + log (1.7N ) 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, i.e. 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.
18 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(1638 a 6 ). We want to have λ <. Therefore we conider the inequality or equivalently 1 < log(116 a 3 ) log(7. ) log(1638 a 6 ) log 116 + log + 3 log a < log 7. + log log 1638 6 log a. The lat inequality hold if log > log 1859376 7. +9 log a, i.e. > 1859376 7. a 9. Therefore we will aume for the ret of thi ection > 1859376 7. a 9. The aumption N > M 9, i.e. > 9 a 9 i now fulfilled and by an application of Theorem., together with Lemma.1, 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)) (.) < 1 λ a U 1 ( + a). ( log 13310 + log + log a 1 ) 3 ( log 7. 1859376 ) < 1 ( log a + 10.71). λ Let u aume > c 0 a 9+r. Then we have 1 λ = 1 1 (.3) log(116 a 3 ) log(7. ) log(1638 a 6 ) log(7. ) log(1638 a 6 ) = log(7. ) log(1638 a 6 ) log(116 a 3 ) = log 6 log a + log ( ) 7. 1638 log 9 log a + log ( ) 7. 1638 116 ( ) 6 9+r log + 6 9+r log c 0 + log ( ) 7. 1638 < ( ) 1 9 9+r log + 9 9+r log c 0 + log ( ) 7. 1638 116 = 1 + r r 9 + r r ) ( 1 log 11533360 ) 1 log c 0 r log ( 199760700 17 17 r log + 9 log c 0 (9 + r) log ( ) 11533360. 17
ON A CERTAIN FAMILY OF THUE EQUATIONS 19 1 The inequality above hold, ince λ i a function increaing with a. Let u conider the econd fraction of the lat line in (.3). A one can eaily compute, the numerator i 0 if 31r 1+ (.) c 0 1 ( ) 1 r 17 1 11 1+ r 6. 5 If c 0 fulfill thi inequality another computation how that the denominator of thi fraction i poitive for 1. Hence we have proved (.5) 1 1 + r λ < r provided (.) hold. By combining (.) and (.5) we obtain (.6) 8 + 8r 1 + r log U < log a + 10.71 r r 8 + 8r 8 + 8r < log (9 + r)r (9 + r)r log c 0 + 10.71 6 + 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 (3.6) yield ( Z = 1 + b ) U ± b ( ) < U 7. 7. 1 + + 1859376 1859376 < (U 1.0001), hence Therefore we have proved log Y < 0.1 + 1 log U. Propoition.3. Let (X, Y ) be a olution to (1.) 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 + r)r (9 + r)r log c 1 + r 0(r) + 10.71 + 0.1. r
0 V. ZIEGLER 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 > 1859376 7. a 9. Firt we prove the following lemma: Lemma 5.1. Let (X, Y ) be a olution to (1.). Then at leat one of the following cae occur 8 X α 1 Y < 63.99 Y 3 3 ; X α 8 Y < 3.99 Y 3 ; 8 X α 3 Y <.99 Y 3 ; X α 8 Y < 3.99 Y 3. Proof. From (.), (.) and (.5) we find (5.1) ( ) 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 = Hence ( ; α = a + L 3 a ( 3 a a ). a ) ; min X α iy. Then we have i=1,,3, Y α i α j X α i Y + X α j Y X α i Y. 1 (5.) X α j Y = i j X α iy 8 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 5.. Let (X, Y ) be a olution to (1.) with Y > 1 and > 3.5 10 9 a 9. According to the four cae in Lemma 5.1 we have Y > 136.8 a 6 ; Y > 8089.6 a 7 ; Y > 3 ; Y > 8089.6 a 7 7107 a 8. Hence, in all cae we have Y > 8089.6 a 7.
ON A CERTAIN FAMILY OF THUE EQUATIONS 1 Proof. From (5.1) together with Lemma 5.1 we obtain X ( + a + b)y < Y 0.751 a X + ay < Y 0.751 a X + by < Y 0.751 a X < Y 0.751 a 8 + 63.99 Y 3 < Y 0.75 a, 3 + + + 8 3.99 Y 3 8.99 Y 3 8 3.99 Y 3 < Y 0.877 a, < Y 0.919 a, < Y 0.877 a, according to the four poible cae. Since the left hand ide are 1 Z, we conclude that they vanih if (5.3) Y 3.008 a, Y 3.508 a, Y 3.676 a, Y 3.508 a. Auming (5.3) we obtain X = (+a+b)y, 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 11.99 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 > 3.008 a, Y > 3.508 a, Y > 3.676 a, Y > 3.508 a. Now we plit the proof into the four cae according to Lemma 5.1. Cae 1: We ue the approximation (5.5) α 1 = + a + b a ab + b + a3 a b ab + b 3 16 ( 0.37 a + L Let u denote by ᾱ 1 the approximation (5.5) with omitted L-term. 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, 3 ). Uing
V. ZIEGLER 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 < 1.001 a. Uing Lemma 5.1 we obtain (P X QY ) Y (P ᾱ 1 Q) Y P (α 1 ᾱ 1 ) < Some elementary calculation together with (5.) yield (5.6) 8 P 63.99 Y 3 3. P X QY < Y P ᾱ 1 Q + Y P (α 1 ᾱ 1 ) + Y P 10. a 8 ) 7 < Y ( a 6 +.8 a 6 + 1.9 a 10 6 < Y.83 a 6. Since P X QY 1 56 Z, we have P X QY = 0 provided Y 136.8 a 6. Inerting thi relation in F (X, Y ) = ±1 yield (5.7) 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 (5.8) X a 0 XY + a 1 Y = Therefore we find the etimation ( a0 a 1 X a 1 Y ( ) X 1 a 0. a 1 ) ( ) + X 1 a 0 a 1 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 + 5.8b )(a 0.183ab + 0.18b )(a 1.818ab + b ) > a 1 0.0667. 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 (5.7) and (5.) On the other hand Y 56 A + R 1 > Y 17.09 a 1 > 0.08 8 a. P < 073 a 8
ON A CERTAIN FAMILY OF THUE EQUATIONS 3 which yield a contradiction to (5.7) if > 18 a. Therefore we have proved Y > 136.8 a, i.e. the firt cae. The proof of the other cae i imilar and 6 we will dicu them le detailed. Cae : For thi cae we ue the approximation :=ᾱ {}}{ α = a + a ab + a3 + a b + ab 16 + 5a a 3 b a b + ab 3 ( ) 6 3 0.161 a 5 + L. For ᾱ we obtain a Padé approximation Q P ᾱ with Q :=16 a(a ab + b ) a(11a 3 + 9a b + 9ab + b 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 ). Similarly a in the firt cae we compute (5.9) P X QY < Y P ᾱ Q + Y P (α ᾱ ) + Y P 303.6 a 8 ( ) 5 85 a 8 < Y 3 3 + 7.88 a 7 + 1576 a 10 3 < Y 7.9 a 7. Therefore P X QY = 0 provided Y 8089.6 a 7. From F (X, Y ) = ±1 we obtain now (5.10) Y ( 5 A + R 1 ) = ±P, with A =10a (a b) (a ab + b ) 1a 3 + 9a b + 7ab + 5b 3 81 56 a 10, R 1 550016 a 16 + 331360 3 a 17 + 1588768 a 18 + 18008 a 19 + 19663 a 0 < a 16 5.51 10 6, P 5.31 10 6 8 a 8. Comparing the bound from the right hand ide and left hand ide of (5.10), we find 0.00 9 a < 5.31 10 6 8 a 8, which i a contradiction for >.66 10 9 a 6.
V. ZIEGLER Cae 3: Now we ue the following approximation :=ᾱ 3 {}}{ ab + b α 3 = b + + a b + ab b 3 16 + a3 b a b b 3 a + 5b ( ) 6 3 0.161 a 5 + L. Applying Padé theorem to ᾱ 3, we obtain an approximation Q P ᾱ 3 with 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 ). Similarly a in the firt cae, we obtain (5.11) P X QY < Y P ᾱ 3 Q + Y P (α 3 ᾱ 3 ) + Y P 88.57 a 8 ( ) 5 85 a 8 < Y 3 3 + 7.88 a 7 + 35 a 10 3 < Y 7.9 a 7. Therefore P X QY = 0 provided Y 8089.6 a 7. If we inert thi relation in F (X, Y ) = ±1, we get (5.1) Y ( 5 A + R 1 ) = ±P, with A =10b (a b) (a ab + b ) 5a 3 + 7a b + 9ab + 1b 3 185 65536 a 10, R 1 550016 a 16 + 331360 3 a 17 + 1588768 a 18 + 18008 a 19 + 19663 a 0 < a 16 5.51 10 6, P 5.31 10 6 8 a 8. Thi time we deduce from (5.1) 0.00153 9 a < 5.31 10 6 8 a 8, a contradiction, provided > 3.5 10 9 a 6. Cae : In the lat cae we ue the method of Padé approximation twice. Firt we ue the following approximation α = :=ᾱ {}}{ ab a b + ab ( ) 0.35 a 16 +L 3
ON A CERTAIN FAMILY OF THUE EQUATIONS 5 and by Padé theorem we obtain an approximation Q P ᾱ with Q :=ab, P := + a + b. Similarly a in the firt cae we obtain (5.13) P X QY < Y P ᾱ Q + Y P (α ᾱ ) + Y P 303.6 a 8 ( ) 5 a < Y + 0.95 a + 11.9 a 8 < Y 1.1 a. Therefore P X QY = 0 provided Y 19.36 a. Thi relation together with F (X, Y ) = ±1 yield (5.1) Y ( A + R 1 ) = ±P, where From (5.1) we deduce A =16a b (a ab + b ) 1 a, R 1 a 7 + 9 a 8 >.001 a 7, P 57. 0.08 6 < 57 a 6, a contradiction provided > 57 a 3. Therefore we may aume Y > 19.36 a. For the econd application of Padé approximation we ue :=ᾱ {}}{ ab α = a b + ab 16 + a3 b + a b + ab 3 6 3 5a b + a 3 b + a b 3 + 5ab ( ) 56 0.131 a 6 + L. 5 Therefore we find an approximation Q P ᾱ with Q :=ab(a ab + b ) + ab(a 3 a b ab + b 3 ), P :=16 (a ab + b ) + (3a 3 a b ab + 3b 3 ) + (a + 3a 3 b 5a b + 3ab 3 + b ). Similarly a in the firt cae we obtain P X QY < Y P ᾱ Q + Y P (α ᾱ ) + Y P 81670 a 16 9 < Y (1.0 a 8 3 + 5.91 a 8 3 + 1.36 107 a 18 ) (5.15) 7 < Y 6.9 a 8 3.
6 V. ZIEGLER From (5.15) we deduce P X QY = 0, if Y 3 7107 a 8. Let u inert thi relation into the original Thue equation F (X, Y ) = ±1. Then (5.16) 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 ) > 17.1 a 1, R 1 19955 3 a 17 + 11916 a 18 + 3690 a 19 + 6597 a 0 < 3 a 17 10 5, P 5.31 10 6 8 a 8. Thi, combined with (5.16), implie 1.1 10 1 a < 5.31 10 6 8 a 8, a contradiction, ince we aume > 3.5 10 9 a 9 > 58 a 3. If a and b are given, one may obtain better reult than thoe proved in Propoition 5.. 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 = 3.5 10 9 in Propoition 5. 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. (5.17) and Theorem 7.1). Corollary 5.3. Let (X, Y ) be a olution to (1.). Then Y 1 provided > 7.3 10 10 a 9+ 1. Proof. Suppoe (X, Y ) i a olution to (1.) with Y > 1. Let u compare the bound from Propoition.3 and 5. and aume > c 1 a 9+r with c 1 > max(3.5 10 9, c 0 (r)) (cf. Propoition.3). Then we obtain ( 7 ) log + 7 9 + r 9 + r log c 1 log 8089.6 < log Y + r < (9 + r)r (log log c 1) + 10.71 1 + r + 0.1 r and therefore (5.17) r + 7r 11r + log + r(r + 9) r + 9 log c 1 < 10.71 1 + r + 0.1 + log 8089.6. r The coefficient of log i poitive if r > 7+ 1 and (5.17) fail provided i large enough. Suppoe now r = 7+ 1. Then the coefficient of log
ON A CERTAIN FAMILY OF THUE EQUATIONS 7 i zero and (5.17) alo fail if c 1 i large enough, i.e. c 1 > 7.3 10 10 > max(3.5 10 9, c 0 (r)). 6. The cae of Y = 1 In view of Corollary 5.3 it remain to conider the cae of Y 1 in order to prove Theorem 1.1. 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 (1.) 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 (1.) reduce to P (X) = X X 3 (ab + (a + b))x 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 cloe together. Therefore we want to prove that the root of P (X) lie in the dijoint interval I 1 :=( + a + b 1/8, + a + b + 1/8), I :=( a 1/8, a + 1/8), I 3 :=( b 1/8, b + 1/8) I :=( 1/8, 1/8). Let u conider the quantitie P ( + a + b 1 8 )P ( + a + b + 1 8 ) = 66 + 18(a + b) 5 + P ( a 1 8 )P ( a + 1 8 ) = ( 1+6a )( 1+6(a b) ) 1638 + > 63 3 1638 a + and P ( b 1 8 )P ( b + 1 8 ) = ( 1+6b )( 1+6(a b) ) 1638 + > 168 1638 a + P ( 1 8 )P ( 1 8 ) = ( 1+6a )( 1+6b ) 1638 + > 63 3 1638 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 (1.) 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, i.e. >.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:
8 V. ZIEGLER Propoition 6.1. Let (X, Y ) be a olution to (1.) with Y = 1 and let > 130. a 3. Then the olution (X, Y ) i lited in Theorem 1.1 or Table 1. Combining Corollary 5.3 and Propoition 6.1 we immediately obtain Theorem 1.1. 7. Some example Firt let u tate a theorem that one may obtain by recomputing the proof of Theorem 1.1. Theorem 7.1. If a, b 1 Z, repectively a, b Z. Then Theorem 1.1 alo hold for > 6.33 10 7 a 9+ 1, repectively > 1.07 10 5 a 9+ 1. Let u now conider four example to illutrate Theorem 1.1 and 7.1: Let a = and b = 1/. Then we have the Thue equation (7.1) X X 3 Y + (6 + )X Y + XY 3 + Y = ±1. Let (X, Y ) be a olution to (7.1). Then (X, Y ) = (0, ±1) or (X, Y ) = (±1, 0) if > 1. 10 11. Note that Thue equation (7.1) 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 (7.) X X 3 Y + X Y + XY 3 + Y = ±1 provided > 1.07 10 5. Let a = 5/, b = and > 1.71 10 1. Then the Diophantine equation (7.3) X X 3 Y (10 + 18)X 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 (7.) X X 3 Y + (6 3)X Y + 5XY 3 + 169Y = ±1 if we aume > 3.6 10 17. All four example have been olved uing Theorem 1.1, repectively Theorem 7.1. In mot cae one could obtain harper bound 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 (7.1). In order to apply thi powerful method, we tart with ytem (3.6) and multiply the firt equation by b and
ON A CERTAIN FAMILY OF THUE EQUATIONS 9 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 ( (7.5) X k + ab ) ( Y k ab ) = µab. Note that the coefficient of (7.5) need not to be integer. The congruence method eentially depend on finding a fundamental olution to (7.5), i.e. in the cae above we have to find a fundamental olution correponding to X (k 1)Y = 1, X (k 1)Y = 1, X (k 5)Y = 1, X (16k 676)Y = 1, repectively. Note that the Pell equation above have integral coefficient. The firt two cae yield the fundamental olution k + k 1. For the other cae no parameterized fundamental olution are known and we cannot apply the powerful congruence method in thoe cae. A cloe look on (7.5) 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. In all other cae no parameterized fundamental olution are known. 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 (1967), 0-8. [] 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 (1967/1968), 173-191. [3] A. Baker, Linear form in the logarithm of algebraic number IV, Mathematika 15 (1968), 0-16. [] A. Baker and H. Davenport, The equation 3x = y and 8x 7 = z, Quart. J. Math. Oxford Ser. () 0 (1969), 19-137. [5] M. A. Bennett, On the number of olution of imultaneou Pell equation, J. Reine Angew. Math. 98 (1998), 173-199. [6] Yu. Bilu and G. Hanrot, Solving Thue equation of high degree, J. Number Theory 60 (1996), 373-39. [7] Y. Bugeaud and K. Győry, Bound for the olution of Thue-Mahler equation and norm form equation, Acta Arith. 7 (1996), 73-9. [8] G. V. Chudnovky, On the method of Thue-Siegel, Ann. of Math. () 117 (1983), 35-38. [9] A. Dujella and B. Jadrijević, A parametric family of quartic Thue equation, Acta Arith. 101 (00), 159-170. [10] A. Dujella and B. Jadrijević, A family of quartic Thue inequalitie, Acta Arith. 111 (00), 61-76. [11] S. Lang, Algebra, Springer-Verlag, New York, 00. [1] K. Mahler, Zur Approximation algebraicher Zahlen (I), Math. Ann. 107 (1933), 691-730.
30 V. ZIEGLER [13] L. J. Mordell, Diophantine equation, Academic Pre, London, 1969. [1] E. Thoma, Complete olution to a family of cubic Diophantine equation, J. Number Theory 3 (1990), 35-50. [15] A. Thue, Über Annäherungwerte algebraicher Zahlen, J. Reine und Angew. Math. 135 (1909), 8-305. [16] A. Thue, Berechnung aller Löungen gewier Gleichungen von der Form ax r by r = f, Vid. Skrivter I Mat.- Naturv. Klae (1918), 1-9. [17] N. Tzanaki, Explicit olution of a cla of quartic Thue equation, Acta Arith. 6 (1993), 71-83. [18] N. Tzanaki and B. M. M. de Weger, On the practical olution of the Thue equation, J. Number Theory 31 (1989), 99-13. Intitut für Mathematik A Techniche Univerität Graz Steyrergae 30 A-8010 Graz Autria E-mail: ziegler@finanz.math.tugraz.at Received: 15..005. Revied: 3.8.005.