AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8

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1 To appear in Journal of Symbolic Computation AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER Abstract We describe a procedure implemented in Mathematica r to solve parametrized families of Thue equations F nx, Y = ±, where F n is a binary irreducible form in X and Y of degree d 3 whose coefficients are polynomials in the parameter n This procedure uses Baker s method and asymptotic expansions of the quantities involved with exact remainder terms As an example, we solve a family of degree 8 Introduction Let F Z[X, Y ] be a binary irreducible form of degree d 3 and m 0 an integer The Diophantine equation F x, y = m is called a Thue equation, remembering that A Thue [Thu09] proved that it only has finitely many solutions in integers x, y Nowadays, the solution of a single Thue equation can be found algorithmically using Baker s method [Bak68] and reduction techniques to reduce the usually big upper bound coming from the linear form estimates, cf Bilu and Hanrot [BH96] In 990, E Thomas [Tho90] first considered a parametrized family of Thue equations with positive discriminant Since that time, several such families F n X, Y = m have been solved, where F n Z[n][X, Y ] is a binary irreducible form in X and Y whose coefficients are polynomials in the parameter n, cf for instance [Heu00] or Wakabayashi [Wak02b] or the online survey [Heu] In all these families, there were some polynomial solutions x, y Z[n] Z[n] such that F n x, y = m holds in Z[n] and therefore for every specialization of n to a concrete integer Moreover, there were finitely many extra solutions x, y for finitely many values of the parameter n We will call these solutions sporadic solutions In all papers, a bound n 0 has been found such that for n n 0, there are no sporadic solutions If n 0 was small enough, all sporadic solutions could also be found by solving each of the equations for the remaining values of n separately Many of these families have been solved by using Baker s method combined with direct arguments to exclude solutions of small and medium size In these cases, the explicit calculations used for single equations have been replaced by asymptotic calculations involving the parameter n Whereas these calculations can be done easily for equations of small degree, this becomes more or less impossible to do by hand for larger degrees: it is not sufficient to know the leading terms of the asymptotic expansions, but the knowledge of an explicit error bound is also necessary The aim of the present paper is to provide a procedure to do these calculations automatically It has been implemented in Mathematica r and is available at ~ziegler/publications/automaticsolutionofthueeq We describe the general framework and the principles used in the implementation of the routines We address some particular technical problems which arise and their solutions in our package Of course, we cannot expect to solve every given family 2000 Mathematics Subject Classification Primary D59; Secondary Y50, 68W30 The first author was partially supported by the Austrian Science Foundation, project S-8307-MAT The third author was partially supported by the Austrian Science Foundation, project S-8307-MAT

2 2 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER of Thue equations: it is known cf Lettl [Let] that there are families of Thue equations which have infinitely many sporadic solutions We do not make any attempt to implement the solution via Padé approximations, although they have been used for the solution of some families, too We demonstrate the use of our routines by solving a family of Thue equations of degree 8 To our knowledge, it is the first time that a family of degree > 6 is solved Furthermore, we reconsider some families which have been solved previously by hand and apply our machinery At the present state, we cannot reach the best known constants n 0 as defined above, since our routines do not yet implement all tricks used in the previous papers In particular, we always use the linear form in d logarithms directly, whereas the number of logarithms can sometimes be reduced by a careful study Since the constants in linear form estimates depend on the number of logarithms dramatically, the results can be improved For the family of degree 8, the CPU time for the calculations was considerable around 30 days on a Pentium 4 with 2 GHz running under Linux, which was also due to the fact that the coefficients of the asymptotic expansions were elements of Q 2 The example of degree 5, where the coefficients belong to Q, can be solved in two to three hours In the cubic case, it takes only a few minutes to get the result The remainder of the paper is organized as follows: In Section 2, we recall the general framework avoiding technical details as much as possible Section 3 describes the implementation in more detail The family of degree 8 is solved in Section 4 Section 5 is devoted to the known families of lower degree 2 The procedure We now give an outline of our procedure to solve parametrized families of Thue equations We consider the Thue equation F n X, Y = ±, where F n Z[n][X, Y ] is an irreducible form of degree d 3 and sufficiently large n Let f n X := F n X, and denote the roots of f n by α,, α d We assume that f n is monic and all roots α,, α d are real Let K k = Qα k be the number field generated by α k and let o K k be its ring of algebraic integers k d We call a solution x, y to equation trivial if y Let η,, η r with r = d be a system of independent units in o K, then let η k i denote the k-th conjugate of η i i r, k d Obviously η k,, ηk r is a system of independent units in o K k k d We assume log η k i log n, where g h means that there is some effectively computable constant c such that g < c h Let x, y be a solution to and choose j d such that x α j y = min x α i y We say that x, y is a solution of type j and we define β k := x α k y k d So equation can be rewritten as 2 F n x, y = x α y x α d y = β β d = NQ Kk β k = ±, where NQ Kk denotes the norm Then we have y α i α j x = α i y x α j y x α i y + x α j y 2β i i

3 AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8 3 for i j This implies together with equation 2 3 β j 2 d y d i j αi α j = 2 d y d f n αj We further assume f n αj α i α j n for i j This implies log β i x = log α j y α i α j y = log y + log α i α j β + log j yα i α j = log y + log α i α j β j y 4 α i α + O j n 2 for i j Since β k is a unit by 2, there are integers b,, b r and I with [ ] I o :, η k K k,, ηk r such that 5 log β k b η = I log k + + b r η I log k r, k j Solving this system of linear equations by Cramer s rule we obtain 6 R b k I = u k log y + v k β j y w k + r k for k r, where u k = det log v k = det log w k = det log r k = det log R = det log η i η i η i η i η i,, log,, log,, log,, log,, log η i k η i k η i k η i k η i k,, log η i k+,, log η r i, i j, log α i α j η i, log k+,, log η, α i α, log i j k+,, log η r i, O, log η i,, log η i, log n 2 η i k, log k+ η i k+,, log r η r i η r i i j i j i j, i j, log r n = O, In the next section we will compute the value of O/n 2 in r k more explicitly We take some constant integers λ 0, λ,, λ r and consider r r r b := λ 0 I + λ k b k, u := λ k u k, v := λ 0 R + λ k v k, From 6 we deduce that k= w := k= r λ k w k, r := k= r λ k r k k= k= n 2 7 R b I = u log y + v βj y w + r

4 4 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER We try to choose λ 0,, λ r in such a way that 8 n u logr n n v logr n n signu = sign u log y 0 + v w 2f nα j + r =, where y 0 is a lower bound for nontrivial y y > From the definition one can always choose y 0 = 2 In Section 35, we will discuss how to obtain a better lower bound y 0 for y For y y 0 this implies R b I > 0, hence b By a theorem of Friedman [Fri89] we further obtain R I R [o :, η k K k,, ηk r ] = R R /RegK k = RegKk > 02 and so I 5 R From this inequality we also obtain Rb I > 02 Using 7, we solve R b/i > 02 for log y and we obtain 9 log y n log r n, if y is nontrivial y > Let Hn be an upper bound for the coefficients of F n Since the coefficients of F n X, Y are polynomials in n, we have log Hn log n Since we assume log log n we obtain RegK k log n Using a theorem of Bugeaud and Győry [BG96] we obtain η k i 0 log y log 2r n log log n, a contradiction to 9 So we have n Since all bounds are effectively computable one can give an explicit bound n 0, such that has only solutions x, y of type j with y for n n 0 We will compute n 0 in the next section The upper bound obtained from the theorem of Bugeaud and Győry can be improved using Baker s method directly From 3 we get x α j y c y d, with c = 2 d f nα j The c, are all effectively computable constants depending on n, α k and η k i for k d, i r From this inequality we obtain signyα j c y d < x y < signyαj + c y d, hence 2 y α j α i c y d < βi < y α j α i + c y d

5 AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8 5 Putting B := max b i and solving 5 with Cramer s rule we get the estimate 3 where are the cofactors of R The j i B I r max i j j i max i j log β i R j i = log η l k k i,l j c 2 max i j log β i, can be estimated by Hadamard s inequality: Lemma Let A = a ij i,j, n a n n-matrix with real entries then 4 with Using 2 and 3 one obtains c 3 = c 2 max i j det A 2 n j= i= B I c 3 log y, n a ij 2 α log j α i + c y0 + d log y 0 For k l {, 2,, d} \ {j} one obtains by Siegel s identity and 2 5 αj α k β l α j α l β k = α k α l β j α j α l β k c 4 y d, with c 4 = α k α l c α j α l α j α k c y0 d Next we will use a theorem of Matveev [Mat00, Corollary 23] Lemma 2 Denote by α,, α n algebraic numbers, not 0 or, by log α,, log α n determinations of their logarithms, by D the degree over Q of the number field K = Qα,, α n, and by b,, b n rational integers Define B = max{ b,, b n }, and A i = max{dhα i, log α i, 06} i n, where hα denotes the absolute logarithmic Weil height of α Assume that the number Λ = b log α + + b n log α n does not vanish; then Λ exp{ Cn, κd 2 A A n loged logeb}, where κ = if K R and κ = 2 otherwise and { κ } Cn, κ = min κ 2 en 30 n+3 n 35, 2 6n+20 Applying this theorem to 6 Λ = I log α j α k α j α l + u log η l η k + + u r log η l r η k r

6 6 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER and using the estimate log x < 2 x for x 3 together with 4 and 5 it results exp log I c 5 logeb log α j α k α j α l + u I log η l η k + + u r I log η l r η r k 2c 4 y d = exp B c 6 c 7, I where c 5 comes from the theorem of Matveev Lemma 2, c 6 = log 2 + log c 4 and c 7 = d c 3, for B > I From the inequality B 7 log I + c 5 + c 5 log B > c 7 I c 6 one obtains an upper bound c 8 for B and by 3 and 4 an upper bound c 9 for log y with r c 9 = c 8 log η k α i log j α k c 2 d i= The computation of the quantities c, will be described in Section 34 3 Implementation This section describes the basic ideas of implementing the procedure described above 3 Exact O-Notation One of the main problems is that the roots α,, α d are not known explicitly But it suffices to know an asymptotic approximation of the roots This can be done by some symbolic steps of Newton s method In the following we use the L-notation Let c be a real number, assume fx, gx and hx are real functions and hx > 0 for x > c We will write fx = gx + L c hx for gx hx fx gx + hx The use of the L-notation is like the use of the O-notation but with the advantage to have an explicit bound for the error term The following lemma is obvious from the definition and some power series expansions of elementary functions Lemma 3 Let hx and gx be real functions and let fx, f x and f 2 x be non-negative real functions for x > c, x > c and x > c 2 respectively Then 2 hx + L c f x + gx + L c2 f 2 x = hx + gx + L maxc,c 2f x + f 2 x hx + L c f x gx + L c2 f 2 x 3 Assume 0 fx < hx if x > c, then loghx + L c fx = loghx + L c 4 Assume 0 fx < hx for x > c, then = hxgx + L maxc,c 2 gx f 2 x + hx f x + f xf 2 x hx + L c fx = hx + L c fx hx fx fx hx hx fx

7 AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8 7 For computing expressions we want to compute determinants with entries in L-notation it is useful to keep the L-term as simple as possible We define: Definition The quantity z is said to be given in simple L-form, if there are some c R, a, b Z and Rn, log n such that z = Rn, log n + Lc n a log b n However, Lemma 3 does not give simple L-forms, so we have to simplify the results of Lemma 3 to that form Subroutine Simplify L-form : Given g RX, Y, find a, b Z and c, m R such that gn, log n = L m c n a log b n and the L-term has still the same order of magnitude as g To find a, b, c is rather easy Let gx, Y = fx, Y hx, Y, where fx, Y and hx, Y are polynomials Let f = c X a Y b and h = c 2 X a2 Y b2 be the monomials of highest degree lexicographically of fx, Y and hx, Y respectively, further let f 2 = c 3 X a3 Y b3 be the monomial of highest degree f of f such that sign c 3 = sign c if no such monomial exists set c 3 = 0 Then set a = a a 2, b = b b 2, c = c + c3 c and c = c c /c 2 In practice one will get numerical problems to calculate m if c + c 3 /c is too close to So we will set c = max, c + c 3 / c To get m we have to find an upper bound for the real solutions of cn a log b n gn, log n = 0 if such solutions exist, otherwise set m = 0 So we have reduced our problem to finding an upper bound for the largest root of fn, log n = 0 for some given polynomial f We will use two routines to get that upper bound We will substitute log n = q and treat q as an independent variable Let p = + deg q f, let f i q be the coefficient of n i and let d i be the leading coefficient of f i We will construct a new function fn such that fn fn, log n for n m as follows: a If d i > 0 let m i be the largest real solution of f iq = d i if it doesn t exist set m i = 0, set f i := d i and m i = expm i b If d i < 0 let m i be the largest real solution of log n = n/p if such exist which is the case for p 3 and set m i = 0 otherwise Set all coefficients in f iq which are > 0 to 0 and substitute q = n /p Set m i = m ip Let fn be the function obtained by these substitutions and let m = max i m i, then for all n > m we have fn fn, log n for n m Let m 2 be an upper bound for the largest real root of f then m = maxm, m 2 is an upper bound for n such that fn, log n = 0 We can compute m 2, since the substitution n n p transforms f into a polynomial Since there are algorithms as implemented in Mathematica or Pari to find all roots, in particular the largest real root of a polynomial, we are done 2 Similar to the first routine we will construct a function fn such that fn fn, log n for n m 0 with q = log n To obtain fn we set all coefficients of f that are positive to 0 except the leading term in lexicographical order Then we substitute q = n /p where p = deg q f Let m 0 be the largest real root of log n = n/p if no real root exists let m 0 = 0 and let m 0 = m 0p We obtain fn fn, log n for n m 0

8 8 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER Let m be the largest real root of fn = 0 then m = maxm 0, m is an upper bound for the root n of fn, log n = 0 Now we take the minimum of the bounds obtained by these two routines Next we want to obtain a simple L-form of the logarithm of a simple L-form This can be done by using Lemma 3 and the following result Lemma 4 d logrn + Lc/n k = a q + log b + Qn + L n l, where bn a is the main part of the Laurent expansion at of Rn, Qn R[n, /n] with Qn = o, l = a + k and d is some effectively computable constant depending on R Proof: Using the power series for log and Lemma 3 one obtains log Rn + Lc/n k c = logrn + L n k = logbn a + T n + Lc /n a+k Rn c with T n = Rn bn a = orn The power series expansion of log Rn at cn a gives the lemma 32 Calculation of the necessary quantities in simple L-form We will use the index of the L-notation only for concrete computations and will omit it for brevity in most cases We assume that the roots α j are given in simple L-form We further assume a system of independent units η i,, ηi r is given by rational functions R i x such that R i α = η i Using Lemma 3 and the described procedure to get simple L-terms, we can easily compute all units η i and their conjugates in simple L-form Using Lemma 4 one gets the matrix log log η η d log log η r η r d, where the entries are given in simple L-form Given the type j of the solution a similar computation gives the matrices considered in a quantitative form and hence the determinants R, u k, v k and w k for k d Next we want to compute the determinants r k k r Since y 2 and β j y = L 2 d yd f n αj = L 2 f n αj, by 3 and log + x = x x 2 2 x2 + = x + L 2 x2 + x + = x + L 2 x

9 AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8 9 for x < we obtain log β i α = log y + log i α j β j 8 y L β j α i α j + 4 2f n 2 α j α i α j 2 f n αj α i α j for y > α i α With 8 we can compute the determinants r j k K r in simple L-form 33 Computation of the lower bound The computation of the λ s can be done by solving the equations obtained by comparing coefficients Once the λ s are computed it is easy to obtain a lower bound for log y by solving the inequality 02 < R I u log y + v w βj y + r w obtained from 7 Since βj y appears as coefficient of w, we use the estimate β j y compute a lower bound for log y < w 2f n αj 34 Calculation of the upper bound Computing c 8 amounts essentially to solve an equation of the form hx := cx a log x = 0 This can be done by using one step of Newton s method starting at x 0 > /c, since h x > 0 and h x > 0 for x > /c Hence applying one step of Newton s method will give an upper bound for the root x of hx = 0 Lemma 5 Let a, c R +, 0 < ε < and x > 0 Then for all x > a ε log cε εc + cε we have hx = cx a log x > 0 to Proof: Apply one step of Newton s method starting at the point x 0 = εc In the implementation we used the value ε = 0 35 Finding trivial lower bounds for y Assume x, y is a nontrivial y > solution of type j of the Thue equation F n X, Y = ± and α j = P n + Qn c n k + L n k+ with P Z[X], Q R[X], deg Q < k, c R and 0 < k Z From we obtain y y {}}{ Qn n k c n k+ c 2 d < x P ny < y Since x P ny is an integer we have x P ny = 0, if 9 y < y 0 := min, y y 2 {}}{ Qn n k + c n k+ + c 2 d Assume x = P ny Substitute P ny for x in the Thue equation to obtain y d T n = ±, y 2

10 0 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER where T n Z[X] Hence the only possibility for y to satisfy this equation is y = and since x, y is nontrivial we obtain y y 0 Wakabayashi [Wak02a] showed how to obtain further trivial bounds, using continued fraction expansions of α j and a generalization of Legendre s Theorem For details see [Wak02a], Section 6 4 An equation of degree 8 As an example for the use of the procedure described above, we consider the parametrized familiy of Thue equations of degree 8 20 F n X, Y := X 8 8nX 7 Y 28X 6 Y 2 +56nX 5 Y 3 +70X 4 Y 4 56nX 3 Y 5 28nX 2 Y 6 +8nXY 7 +Y 8 = ± We first want to construct these polynomials in order to understand the structure Let ε := + 2 and let ε A = ε Then A is of order 8 in the group PGL 2 Q 2, since ε 0 A 2, A 3, A ε 4 0 A 6, A 7 ε ε We consider the usual action of PGL 2 Q 2 on C given by a b z = az + b c d cz + d, A 8 Let ϑ i = A i ϑ, i =,, 8 Writing them out, we have ϑ = ϑ, ϑ 2 = εϑ ϑ + ε, ϑ 3 = ϑ ϑ +, ϑ 4 = ϑ ε εϑ +, ϑ 5 = ϑ, ϑ 6 = ϑ ε εϑ ε, A 5 ε 0 0 ϑ 7 = ϑ ϑ, ϑ 8 = εϑ ϑ ε Since ϑ i = ϑ i+4, i =,, 4, we have 8 i= ϑ i = Therefore ϑ is a root of the octic polynomial P X = X 8 a X 7 + a 2 X 6 a 3 X 5 + a 4 X 4 a 5 X 3 + a 6 X 2 a 7 X +, where a = 8 i= ϑ i, a 2 = i<j ϑ iϑ j etc Shen [She9] showed that 2 P X = X 8 a X 7 28X 6 + 7a X X 4 7a X 3 28X 2 + a X + = X 8 28X X 4 28X 2 + a XX 2 X 2 ε 2 X 2 ε 2 Since P > 0 and P ε < 0 there is a real root of P X = 0 The construction of the polynomial shows that P X = 0 has eight distinct real roots satisfiying ϑ ε,, ϑ 2, ε, ϑ 3 ε,, ϑ 4 0, ε, ϑ 5 ε, 0, ϑ 6, ε, ϑ 7 ε,, ϑ 8, ε They are all units in the ring of algebraic integers of the field Qϑ, 2, if a is an algebraic integer of the field Q 2 Shen could prove the following proposition Proposition in [She9],

11 AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8 Proposition The octic polynomial PX in equation 2 is irreducible over the field Q for a Z \ {0, ±6, ±5} Let y := 2 ϑ + ϑ 5, z := 4 ϑ + ϑ 3 + ϑ 5 + ϑ 7 and let 8n = a, with this substitution we get f n X then Shen proved: Proposition 2 The minimal polynomial of y over Q is X 4 4nX 3 6X 2 + 4nX +, and hence Qy is a simplest quartic field 2 The minimal polynomial of z over Q is X 2 2nX, and hence Qz = Q n We have Qz = Q 2 if n S := {a Z : a + b 2 = ε 2n+, n N} 4 For n S the field K n := Qρ is a totally real cyclic octic field, whose Galois group GK n Q =< σ > Z/8Z 5 The units ρ, ρ 2, ρ 3, ρ 4, y, y 2 and ε in the ring of algebraic integers o Kn are independent 6 The regulator R of K n has the lower bound 2 6 log ε log 6 n Since the algebraic data required for solving the family is known for n S only, we will restrict our attention to this case Let ρ be the largest root of f n X = 0 and let ρ i = σ i ρ for i =,, 8, where σ GQρ Q is determined by ρ ερ ρ + ε Since n S we have 2 Qρ and hence ε Qρ and so σ is indeed an automorphism Note that ρ i = ϑ i for i =, 2, 5, 6 but ρ 3 = ϑ 7, ρ 4 = ϑ 8, ρ 7 = ϑ 3, ρ 8 = ϑ 4 We have a different ordering since < σ Qz >= GQ 2 Q and hence σε = ε As above let y := 2 ρ + ρ 5 = 2 ρ /ρ and z := 4 ρ + ρ 3 + ρ 5 + ρ 7 = 2 y /y Hence we obtain two equations ρ 2 2yρ = 0 and y 2 2yz = 0 It is easy to compute ρ i, i =,, 8 by solving these quadratic equations recursively As n one obtains: ρ 8n, ρ 2 ε, ρ 3, ρ 4 ε, ρ 5 0, ρ 6 ε, ρ 3, ρ 4 ε We apply Newton s method three times starting at x i = lim n ρ i n for 2 i 8 and x = 8n We obtain:

12 2 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER ρ = 8n + 2 8n + L ρ 2 = ε 05 n 3 ; ε 2 2n + ε 2 6 2n L 2 ρ 3 = 4n 32n 2 + L n 3 ; ; ; ρ 4 = ε ε 2 2n ε 2 6 2n L 2 ρ 5 = n + L n 3 ; ρ 6 = ε n 3 ε 2 2n ε 2 6 2n L 2 ρ 7 = 4n + 32n 2 + L ρ 8 = ε n 3 ε 2 2n + ε 2 6 2n 2 L n ; n n 3 ; So all input is collected to use the procedure to solve the family F n X, Y = ± We remark that the coefficients of the asymptotic expansions of the ρ i contain ε and therefore 2 This makes the computation lengthier Before applying the procedure we collect some other useful facts First we prove a lemma about the type of a solution Lemma 6 If x, y is a solution of type j =, 2, 3, 4, then y, x is a solution of type j + 4 for n > 002 Proof: We have ρ i = /ρ i+4 Using the L-form representation of the ρ i we obtain min ρ i ρ j > 03 for i j and n > 0 We further obtain by computing c from Section 2 x/y ρ j < 6n < for n > 002 These two inequalities prove the lemma Since y i = y i+4 for i =, 2, 3, 4 we can choose l and k from the linear form 6 such that l = k + 4 and we obtain a linear form in only five logarithms So we get upper bounds type : log y < log 8 n, type 2: log y < log 8 n, type 3: log y < log 8 n, type 4: log y < log 8 n, Calculating the determinants from Section 2 we get lower bounds

13 AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE 8 3 type : log y > n log 2 n, type 2: log y > n log 2 n, type 3: log y > n log 2 n, type 4: log y > n log 2 n, Comparing these bounds we obtain a bound for n 0 Theorem The Thue equation 20 has only trivial solutions for n n 0 and n S with n 0 = Looking at the structure of S one obtains { S = an := + 2 2n + 2 2n } : n N \ {0} 2 A quick computation shows that there are only 45 elements in S that are smaller than A straight forward calculation shows: Lemma 7 Let x, y, c Z Suppose F n x, y = c, then F n x + y, x + y = F n x y, x + y = 6c 2 Suppose F n x, y = 6c, then F n x+y 2, x+y 2 = F n x y 2, x+y 2 = c One observes that F n x, y = 6c implies that x y mod 2 and so x+y 2 and x y 2 are integers All together gives the following corollary Corollary Let an := n + 2 2n and let F n X, Y := X 8 8 anx 7 Y 28X 6 Y anx 5 Y X 4 Y 4 56 anx 3 Y 5 28X 2 Y anxy 7 + Y 8 Let n 45 then the Thue equation F n X, Y = c with c {±, ±6} has only the integer solutions {±, 0, 0, ±} if c =, {±,, ±, } if c = 6 and there are no integer solutions for c {, 6} 5 Further examples In this section we will reconsider some other examples that have been solved before In the particular cases, our results are worse than those obtained by other authors, since they used algebraic relations to reduce the linear form in logarithms 6 to a linear form in fewer logarithms They also exploited the Galois group of the polynomial fx = F X, to get better estimates 5 The equation of Thomas and Mignotte We will now consider the Thue equation X 3 n X 2 Y n + 2XY 2 Y 3 = ± It has been solved for n > by Thomas [Tho90] and for all n by Mignotte [Mig93] Let α be the largest root of f n x := F n X, = x 3 n x 2 n + 2x = 0, then Qα/Q is a cyclic Galois extension and α, /α + are fundamental units of Qα This was proved by Thomas [Tho79] If x, y is a solution of type j, then y, x + y is a solution of

14 4 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER type j + mod 3 + Hence it suffices to consider only one type We treated the type By using Newton s method we see that the roots of f n x = 0 are α = n + 2 n 3 n 2 + L n 3, α 2 = 3 n + L n 3, α 3 = n + 3 n 2 + L n 3 Applying the procedure from Section 2 we obtain that there are only trivial solutions for n > n 0, where n 0 = If we take into account the fact that α, /α + is a system of fundamental units, hence I =, and that Qα/Q is cyclic we get the better result n 0 = The fact that Qα/Q is cyclic leads to a better result since we know the structure of the Galois group, hence we can compute the quantity c 5 more effectively 52 An equation of degree 4 The next example is the Thue equation F n X, Y = X 4 nx 3 Y X 2 Y 2 + nxy 3 + Y 4 = ± This equation was first treated by Pethő [Pet9] He proved that for n > there are only trivial solutions The equation was solved in 996 for n 3 by Mignotte, Pethő and Roth A system of fundamental units is given by α, α, α +, where α is the largest root of f n X := F n X, = 0 By Newton s method we obtain α = n n 3 + L n 4, α 2 = + 2n 8n 2 + 2n 3 + L n 4, α 3 = + 2n + 8n 2 + 2n 3 + L α 4 = n + L n 4 n 4, If we use all information we have and use the procedure from Section 2 we obtain that there are only trivial solutions x, y for n > n 0, where n 0 = if x, y is of type j = The other three cases j = 2, 3, 4 do not satisfy the Assumption 8 Since α 2 = R 2 n + L/n 4, α 3 = R 3 n + L/n 4 and α 4 = R 4 n + L/n 4 with R 2, R 3, R 4 ZX we can compute a trivial lower bound for y and obtain j is the type of solution: log y log n 4 if j = 2, 2 log y log n 4 if j = 3, 3 log y log n 2 if j = 4 Using these trivial bounds we obtain n 0 = if j = 2, 2 n 0 = if j = 3, 3 n 0 = if j = 4 Hence the Thue equation has only trivial solutions for n n 0 with n 0 =

15 AUTOMATIC SOLUTION OF FAMILIES OF THUE EQUATIONS AND AN EXAMPLE OF DEGREE An equation of degree 5 We now consider a Thue equation of degree 5: XX 2 Y 2 X 2 n 2 Y 2 + Y 5 = ± This equation was first solved by Heuberger [Heu98], who proved that there exist only trivial solutions for n n 0 with n 0 = Using Newton s method we obtain: α = n + 2n n 6 + L n 7, α 2 = n + 2n n 6 + L n 7, α 3 = 2n 2 7 8n n 6 + L n 7, α 4 = 2n 2 8n n 6 + L α 5 = n 2 + n 6 + L 2 n 7 2 n 7, Heuberger could also prove that α, α +, α, α n is a system of fundamental units Using the procedure described in Section 2 we get that there are no non trivial solutions for n n 0 with n 0 = if the solution is of type, 2 n 0 = if the solution is of type 2, 3 n 0 = if the solution is of type 3, 4 n 0 = if the solution is of type 4 For the type 5 the Assumption 8 is not true But estimating a trivial lower bound for y we get log y > 2 log n 29 Using this bound we can use the procedure described in Section 2 also in this case and obtain for solutions of type 5 that n 0 = Hence there are no trivial solutions of this equation for n n 0 with n 0 = References [Bak68] A Baker Contribution to the theory of Diophantine equations I On the representation of integers by binary forms Philos Trans Roy Soc London Ser A, 263:73 9, 968 [BG96] Y Bugeaud and K Győry Bounds for the solutions of Thue-Mahler equations and norm form equations Acta Arith, 743: , 996 [BH96] Yu Bilu and G Hanrot Solving Thue equations of high degree J Number Theory, 60: , 996 [Fri89] E Friedman Analytic formulas for the regulator of a number field Invent Math, 983: , 989 [Heu] C Heuberger Parametrized Thue equations A survey available at ~cheub/publications/thue-surveypdf [Heu98] C Heuberger On a family of quintic Thue equations J Symbolic Comput, 262:73 85, 998 [Heu00] C Heuberger On general families of parametrized Thue equations In F Halter-Koch and R F Tichy, editors, Algebraic Number Theory and Diophantine Analysis Proceedings of the International Conference held in Graz, Austria, August 30 to September 5, 998, pages Walter de Gruyter, 2000 [Let] G Lettl Parametrized solutions of Diophantine equations Preprint [Mat00] E M Matveev An explicit lower bound for a homogeneous rational linear form in logarithms of algebraic numbers II Izv Ross Akad Nauk Ser Mat, 646:25 80, 2000 [Mig93] M Mignotte Verification of a conjecture of E Thomas J Number Theory, 442:72 77, 993 [Pet9] A Pethő Complete solutions to families of quartic Thue equations Math Comp, 5796: , 99 [She9] Y Y Shen Unit groups and class numbers of real cyclic octic fields Trans Amer Math Soc, 326:79 209, 99

16 6 CLEMENS HEUBERGER, ALAIN TOGBÉ, AND VOLKER ZIEGLER [Tho79] E Thomas Fundamental units for orders in certain cubic number fields J Reine Angew Math, 30:33 55, 979 [Tho90] E Thomas Complete solutions to a family of cubic Diophantine equations J Number Theory, 342: , 990 [Thu09] A Thue Über Annäherungswerte algebraischer Zahlen J Reine Angew Math, 35: , 909 [Wak02a] I Wakabayashi Cubic Thue inequalities with negative discriminant J Number Theory, 972:222 25, 2002 [Wak02b] I Wakabayashi On families of cubic Thue equations In K Matsumoto and C Jia, editors, Analytic Number Theory, volume 6 of Developments in Mathematics, pages Kluwer Academic Publishers, 2002 Institut für Mathematik B, Technische Universität Graz, Steyrergasse 30, A-800 Graz, Austria address: clemensheuberger@tugrazat Purdue University North Central, 40 S, US 42, Westville IN 4639, USA address: atogbe@pncedu Institut für Mathematik A, Technische Universität Graz, Steyrergasse 30, A-800 Graz, Austria address: ziegler@finanzmathtugrazat