Parameterized Norm Form Equations with Arithmetic progressions

Transcription

1 Prmeterized Norm Form Equtions with Arithmetic progressions A. Bérczes,1 A. Pethő b, V. Ziegler c, Attil Bérczes Institute of Mthemtics, University of Debrecen Number Theory Reserch Group, Hungrin Acdemy of Sciences nd University of Debrecen H-4010 Debrecen, P.O. Box 1, Hungry b Attil Pethő Institute of Informtics, University of Debrecen, H-4010 Debrecen, P.O. Box 1, Hungry c Volker Ziegler Institute of Computtionl Number Theory nd Anlysis, Grz University of Technology Steyrergsse 1, A-8010 Grz, Austri Abstrct Let α be zero of the Thoms polynomil X 1X + X 1. We find ll lgebric numbers µ = x 0 + x 1 α + x α Z[α], such tht x 0, x 1, x Z forms n rithmetic progression nd the norm of µ is less thn + 1. In order to find ll progressions we reduce our problem to solve fmily of Thue equtions nd solve this fmily completely. Key words: Arithmetic progressions, norm form equtions, Thue equtions Emil ddresses: berczes@mth.klte.hu A. Bérczes, pethoe@inf.unideb.hu A. Pethő, ziegler@finnz.mth.tugrz.t V. Ziegler. 1 Reserch of the first uthor ws supported in prt by the Hungrin Acdemy of Sciences, by grnts T48791, T4985 nd T85 of the Hungrin Ntionl Foundtion for Scientific Reserch Reserch of the second uthor ws supported in prt by grnts T4985 nd T85 of the Hungrin Ntionl Foundtion for Scientific Reserch The third uthor grtefully cknowledges support from the Austrin Science Fund Preprint submitted to Elsevier Science 7 Februry 006

2 1 Introduction Buchmnn nd Pethő [5] observed tht following lgebric integer α + 8α + 7α + 6α 4 + 5α 5 + 4α 6, with α 7 = is unit. Since the coefficients form n rithmetic progressions they hve found solution to the Diophntine eqution N K/Q x 0 + αx x 6 α 6 = ±1, 1 such tht x 0,..., x 6 Z 7 is n rithmetic progression. A full norm form eqution is defined by N K/Q x 0 + αx x n 1 α n 1 = m, where α is n lgebric integer of degree n, K = Qα, m Z nd x 0, x 1,..., x n 1 Z n. It is well known tht dmits infinitely mny solutions for infinitely mny m [14]. This is lredy true for m = 1. On the other hnd Bérczes nd Pethő [] proved tht hs only finitely mny solutions tht form n rithmetic progression provided β := nαn α is n lgebric α n 1 α 1 number of degree t lest. Moreover they showed tht the solution found by Buchmnn nd Pethő is the only solution to 1. Bérczes nd Pethő lso considered rithmetic progressions rising from the norm form eqution, where α is root of X n, with n nd 100 see []. Let f Z[X], Z be the fmily of simplest cubic polynomils f := X 1X + X 1. Let α = α be root of f nd put K = Qα. It follows from result of Lemmermeyer nd Pethő [9] tht the eqution N K/Q x 0 + x 1 α + x α = m with m + 1, m Z hs infinitely mny solutions x 0, x 1, x Z if nd only if m is cube of n integer or m = ± + 1. By the bove mentioned result of Bérczes nd Pethő [] eqution hs for every Z nd m + 1, m Z only finitely mny solutions x 0, x 1, x Z, which form n rithmetic progression. FWF under project Nr. P18079-N1

3 The im of this pper is to describe completely those solutions, which form n rithmetic progression. A solution x 0, x 1, x Z of is clled primitive, if gcdx 0, x 1, x = 1. With this convention we prove the following theorem: Theorem 1 Let α be root of the polynomil f, with Z. Then the only solutions to the norm form inequlity N K/Q x 0 + x 1 α + x α such tht x 0 < x 1 < x is n rithmetic progression nd x 1, x, x is primitive re either x 1, x, x =, 1, 0, 1, 0, 1 nd 0, 1,, or they re spordic solutions tht re listed in tble 1. Tble 1 Spordic solutions to 4 with 0. m x 0 x 1 x m x 0 x 1 x In tble 1 we only list solutions, where the prmeter is non-negtive. Furthermore m denotes the vlue of the norm, i.e. N K/Q x 0 + x 1 α + x α = m. Lemm 1 will show tht it suffices to study the norm inequlity 4 only for 0 Z. Moreover, Lemm 1 gives correspondence between solutions for nd 1. To prove the min Theorem 1 we trnsform 4 to prmetrized fmily of Thue inequlities 5. From here on we follow essentilly the line of [1]. Although there re lot of prmetrized fmilies of Thue equtions nd inequlities, which were solved completely, our exmple 5 dmits dditionl difficulty, becuse the coefficient of both unknowns depend on α. Therefore we need more precise informtion on the rithmetic of Z[α], especilly we need

4 bsis of its unit group. Fortuntely this is known by the result of Thoms [17]. The pln of the proof is s follows. First Section we show how our problem is connected with fmily of Thue inequlities. In order to solve this fmily we hve to do lot of symbolic computtions nd we therefore need good pproximtions to the roots of the relevnt polynomil 7 see Section. The proof of the min Theorem 1 is split into four steps. The first step is to find n upper bound 0 for the prmeter such tht there re no further solutions if 0. This bound is found by n ppliction of vrint of Bker s method combined with technicl computtions see Sections 4 nd 5. In prticulr we use liner forms in two logrithms nd pply powerful theorem due to Lurent, Mignotte nd Nesterenko [8]. The bound which is found in the previous step is too big to solve ll remining Thue inequlities. We hve to consider essentilly two different cses occurring from the liner forms of logrithms used in Section 5. The first cse is treted in Section 6 by method due to Mignotte [11]. For n ppliction of this method we hve to reconsider the liner forms treted in Section 5. The method of Bker nd Dvenport see [1] is used to tke cre of the other cse see Section 7. In order to pply this method we hve to use once gin Bkers method. This time we re fced with liner forms in three logrithms. This liner forms will be estimted from below by theorem due to Mtveev [10]. After the ppliction of the methods of Bker, Dvenport nd Mignotte we re left to solve 1000 Thue inequlities. This is done by PARI. For detils see Section 8. Nottions nd Thue Equtions Let us prove first tht we my ssume 0. Lemm 1 Let α denote zero of f x nd put K = Qα. Then N K/Q x 0 + x 1 α + x α = m holds if nd only if N K 1/Q x x 1 α 1 x 0 α 1 = m. In prticulr ech solution to 4 for yields solution for 1. 4

5 1 Proof: It is esy to see tht α is root of f x if nd only if is root α of f 1 x. As N K/Q α = 1 the ssertion follows immeditely. Next, we wnt to trnsform the norm form inequlity 4 into Thue inequlity. Since x 0, x 1, x form n rithmetic progression we my write x 0 = X Y, x 1 = X nd x = X + Y. Using this nottion in 4 we obtin N K/Q X1 + α + α Y 1 α + 1. Expnding the norm on the left side to polynomil in X nd Y we obtin the Thue inequlity + + 7X + + 7XY + 1Y Since we hve the restrictions x 0 < x 1 < x nd x 0, x 1, x is primitive, we re only interested in solutions with Y 1 nd X, Y is primitive. For the rest of this pper we will use the following nottions: We denote by f Z[X] the Thoms polynomil, which is defined s follows: f X := X 1X + X 1. Let α := α 1 > α > α be the three distinct rel roots of f. Furthermore we define γ := 1 + α + α, δ := 1 α nd ɛ := δ/γ nd denote by γ 1 := γ, γ, γ, δ 1 := δ, δ, δ nd ɛ 1 := ɛ, ɛ, ɛ their conjugtes respectively. Moreover we define G Z[X, Y ] nd g Z[X] by G X, Y := + + 7X + + 7XY + 1Y, 6 g X :=G X, 1 = + + 7X + + 7X Let us remrk tht ɛ 1, ɛ nd ɛ re exctly the roots of g. If X, Y is solution to 5 then we define β := Xγ Y δ nd we denote by β 1 := β, β, β the conjugtes of β. As one cn esily see β i is n element of the order Z[α i ] for ll i = 1,...,. In fct the orders Z[α i ] re ll the sme see [15, 17, 18] or Section 4. There re lot of well known fcts bout the number fields K := Qα, which we will stte in Section 4. We will use the following vrint of the usul O-nottion: For two functions gt nd ht nd positive number t 0 we will write gt = L t0 ht if gt ht for ll t with bsolute vlue t lest t 0. We will use this nottion in the middle of n expression in the sme wy s it is usully done with the O- nottion. Sometimes we omit the index t 0. This will hppen only in theoreticl results, nd it mens tht there exists computble t 0 with the desired property. 5

6 This L-nottion will help us to stte symptotic results in comfortble wy. Asymptotic expnsions Due to Thoms [17] we know tht α 1, α 1, α 1/. We pply Newton s method to the polynomil f with strting points, 1 nd 0. After 4 steps of Newton s method nd n symptotic expnsion of the resulting expressions we get α 1 := α 1, α := α, 8 α := α. We consider the quntities f α i + e i / 5 f α i e i / 5 with e 1 = 10, e = 8 nd e = 18. These quntities re ll positive provided tht 8, 7 nd 10 respectively, hence α 1 = α = , 5 α = ,. 9 Since α 1 + α + α = 1 is n integer we lso obtin α = In order to keep the error terms low from now on we ssume tht Using these symptotic expnsions we obtin for the γ s 5. γ 1 = , 4 γ = , 5 γ = ,

7 nd similrly for the δ s nd for the ɛ s δ 1 = + t δ = , 5 δ = ,, ɛ 1 = , 6 ɛ = , 5 ɛ = We will lso use the symptotic expnsions of the logrithms of the α s. Therefore we recll simple fct from nlysis: If t > r then N r/t i r N+1 1 log t + r = log t i=1 i t N + 1 t. t r We hve omitted the index t 0 since this index depends on the L-Term of the quntity r. Let us write =:r {}}{ =:t {}}{ α = We cn write similr expressions for α nd α, too. Using the bove formul we get log α 1 = log log α = , log α = log , 1 4 Auxiliry results Let us recll first some well known fcts bout the number field K = Qα, where α is root of the Thoms polynomil f these results cn be found in [9, 15, 17, 18]. 7

8 Lemm Let α be root of the polynomil f. Then we hve the following fcts: 1 The polynomils f re irreducible for ll Z. Moreover ll roots of f re rel. The number fields K = Qα re cyclic Glois extensions of degree three of Q for ll Z. The roots of f re permuted by the mp α 1 1 α. 4 Any two of α 1, α, α form fundmentl system of units of the order Z[α], where α 1, α, α denote the conjugtes of α. 5 Let 0. If N K/Q γ + 1 then γ is either ssocited to rtionl integer or ssocited to conjugte of α 1. Proof: Proofs of these sttements cn be found in [15, 17, 18, 9] except sttement 5 in the cse of = 0 nd = 1. The cse = 0 is trivil. So let us consider the cse = 1. If γ fulfills N K/Q γ nd if γ is not unit of Z[α] then γ or γ. According to [7, Chpter I, Proposition 5] we hve = p 1 with p 1 = α = α 1 1 nd = p, where p 1 nd p re prime idels. Therefore γ is multiple of α 1 1 or. Computing the norms yields tht γ is ssocited to α 1 1 or is 0. Therefore we hve proved the sttement for = 1. Prt 5 of Lemm shows tht we only hve to consider lgebric integers, tht re ssocited to rtionl integer or ssocited to conjugte of α 1. Let us exclude the cse tht γ = nɛ with n ±1 Z nd ɛ Z[α] nd γ yields solution to 4. Since γ = x 0 + x 1 α + x α with unique x 0, x 1, x Z, lso ɛ = x 0 n + x 1 n α + x n α yields solution to 4. Therefore n x 0, x 1, x. However, x 0, x 1, x is primitive, thus γ cnnot be ssocited to rtionl integer ±1. We hve to solve the Diophntine inequlity 5, therefore we strt to exclude ll smll vlues of Y. Lemm Let X,Y be solution to 5 such tht Y = 1, then X, Y only yields solutions stted in Theorem 1. Proof: We insert Y = 1 into 5 nd obtin + + 7X 1X If we ssume X, respectively X, then X 1X

9 yields contrdiction. Therefore X 1 nd we only obtin solutions stted in Theorem 1. Now we investigte pproximtion properties of solutions X, Y to 5. We distinguish three types of solutions. We sy tht X, Y is of type j, if X Y ɛ j = min X i=1,, Y ɛ i. A specific cse j will be clled by its romn number. Let us ssume tht X, Y is solution of type j. Then we hve remember β i = Xγ i Y δ i β i γ i β i γ i + β j γ j = X Y ɛ i + X Y ɛ j Y ɛ i ɛ j. Since β 1 β β + 1 by the bove inequlity we obtin β j + 1 i j β i Y i j γ i ɛ j ɛ i or equivlently β j γ j Y N K/Q γ i j ɛ j ɛ i =: c 1 14 Y nd we lso get signyɛ j c 1 Y X Y signyɛ j + c 1 Y, hence β i γ i = Y ɛ c1 c1 j ɛ i = Y ɛ Y0 j ɛ i, 15 Y0 where Y 0 is some lower bound for Y. Becuse of Lemm we my ssume Y 0. Using the symptotic expnsions 9, 10, 11 nd 1 we find c 1 = if j = 1; c 1 = if j = ; c 1 = if j = ; Now we cn prove new lower bound Y 0 for Y. Lemm 4 If 1000 nd X, Y is primitive solution to 5 such tht Y > 1 then Y..01 9

10 Proof: We hve to distinguish between three cses j = 1, j = nd j =. We find from 14 nd 1: 1.00 X Y Y,.00 X Y L Y, 1.00 X Y Y. Some strightforwrd clcultions yield X + Y Y + Y 1.00 X Y + Y.00 X Y Y + Y 1.00 < 1.51Y <.01Y < 1.51Y We conclude tht X + Y = 0, X Y = 0 or X = 0 if Y <. But if.01 X + Y = 0, X Y = 0 or X = 0 we get contrdiction, hence Y..01,,. Let σ be the utomorphism of K = Qα tht is induced by α 1 1. α Then we hve α i = σ i 1 α. From prt 5 of Lemm we know tht β is either unit, ssocited to rtionl integer or ssocited to conjugte of α 1 1. By the discussion fter Lemm we know tht β is not ssocited to rtionl integer 1. Furthermore α 1 nd α form fundmentl system of units of the relevnt order Z[α], hence the liner system log β i = b 1 log σ i 1 α 1 + b log σ i 1 α + log σ i 1 µ i j 16 with µ ssocited to one of 1, α 1 1, α 1 or α 1, hs unique integrl solution b 1, b. Solving 16 by Crmer s rule we find B := mx{ b 1, b } mx i j log β i log σ i 1 µ mx i=1,, log α i Regα 1, α := mx log β i log σ i 1 µ c i j γ log mx i i j σ log Y c 1 + i 1 ɛj ɛ µ i + c 1 Y0 log Y 0 := log Y c 17 We will compute the quntity c in Section 5, when we hve better lower bound Y 0 Y. 10

11 Now we will investigte Siegel s identity. Therefore choose i, k {1,, } such tht i, j, k re ll pirwise distinct. We consider the quntity β i γ i ɛ j ɛ k + β j γ j ɛ k ɛ i + β k γ k ɛ i ɛ j = 0. Tking into ccount 14 nd 15 we find fter some mnipultions tht β j γ j β i γ i ɛk ɛ i ɛ k ɛ = 1 j c 1 Y ɛj ɛ i ɛ j ɛ k ɛ k ɛ i ɛ k ɛ j 1 Y ɛ j ɛ i c 1 β k γ k β i γ i Y0 := c 4 Y. 18 By the symptotic expnsions 9, 10, 11 nd 1 together with the bounds for c 1 nd Lemm 4, we see tht for ny choice of i, j, k except i, j, k =,, 1 we hve c provided tht In the exceptionl cse we get c Note tht this exceptionl cse, will not occur in this pper. 5 A first bound for the prmeter In this section we will derive first upper bound for such tht 5 hs no primitive solution X, Y with Y > 1. First we consider β k Λ i,j,k := log γ k β i ɛj ɛ i ɛ γ j ɛ k i γ i = log ɛj ɛ i γ k ɛ j ɛ k + b σ k 1 α 1 log σ i 1 α + b σ k α log σ i α + log σ k 1 µ σ i 1 µ. From Siegel s identity 18 nd the fct tht log x < 1 x provided tht 1 x < 1/ we obtin Λ i,j,k < 1 β k γ k β i γ i ɛj ɛ i ɛ j ɛ c 4 k Y. 19 Let θ i,j,k := γ i γ k ɛj ɛ i ɛ j ɛ k. We wnt to write Λ i,j,k s liner combintion of the σ logrithms of θ k 1 µ i,j,k, α σ i 1 µ 1 nd α. Therefore we hve to distinguish between 11

12 severl cses. In prticulr, we consider the three liner forms: Λ 1 := log α 1 + B log α + log θ σµ,1, σ i, j, k =, 1,, 0 µ Λ := log α 1 + B log α + log θ σ µ 1,, i, j, k = 1,,, 1 µ Λ := log α 1 + B log α + log θ σµ 1,, i, j, k = 1,,, µ where :=b 1 b B :=b 1 b in cse of Λ 1, := b 1 + b B := b 1 b in cse of Λ, :=b 1 + b B := b 1 + b in cse of Λ. Let us find reltions between B nd B. These will be used in view of 17. Below we will distinguish between the cse of = 0 nd 0. Let us consider cse I: Since B = mx{ b 1, b } we hve trivilly B B. If we ssume = 0 then we hve b 1 = b nd therefore B = b 1. Inserting this reltion in the eqution for B we get B = b 1, hence B = B. The two other cses re similr nd the reltions re given in tble. Tble Reltions between B nd B. Cse I Cse II Cse III 0 B B B B B B = 0 B = B B = B B = B We hve to distinguish between 1 cses three liner forms nd for ech liner form four possible choices for µ. Since ll 1 cses cn be treted similrly, we only consider the cse of Λ 1 nd µ being ssocited to α 1. We choose this cse becuse it is representtive for most of the other cses. The computed quntities for the other cses re presented in tbles. To sy tht µ is ssocited to some quntity α we use the nottion µ α. By 19 nd 0 we find Λ 1 = log B c 4 Y log θ σµ,1, σ µ By this inequlity we see tht B hs to be lrge with respect to, except the min terms of log α 1 nd log θ,1, σµ/σ µ cncel. We wnt to choose 1

13 µ such tht cnceltion my only occur if = 0. Since θ,1, = log + we hve to choose µ such tht µ α 1 nd σµ/σ µ = O1. With this constrints we choose µ = α 1α 1. The other choices for µ re given in tble. Tble Choices for µ. µ 1 µ α 1 1 µ α 1 µ α 1 Cse I 1 α 1 1 α 1α 1 α 1 α 1 Cse II α 1 α 1 1 α α 1α 1 α α 1α 1 Cse III 1 α α 1 1α 1 α 1 α α 1α α Now we distinguish between two further cses: = 0 nd 0. In the cse of = 0 we hve Λ 1 = B log = L. Solving this eqution for B, we obtin B = log + log In the cse of 0 we similrly determine the quntity B = log + log + log log + 5 The results obtined in the other cses re listed in tble log. 4 Looking t tble 4 we see tht in the cse of = 0 two different phenomen occur. In the cses I µ α 1, II µ 1, II µ α 1 1, II µ α 1 nd III µ α 1 the quntity B is of the form constnt plus some error term, while in the other cses B is constnt times log plus lower terms. We re interested in the former cses. In cse I µ α 1, II µ α 1 nd III µ α 1 B cnnot be n integer if 500. However, by definition B is n integer, so we hve contrdiction. In the cses of II µ 1 respectively II µ α 1 1 we hve B = 1 respectively B = 5 provided 500. Therefore we hve the following two liner systems: b 1 + b = 0, b 1 b = 1, nd b 1 + b = 0, b 1 b = 5. 1

14 Tble 4 The quntities B nd B /. Cse I µ 1 B = log + log B = log + log + log + log 1 µ α 1 1 B = log 4 + log B = log + log + log 4+log 1 µ α 1 B = log + log B = log + log µ α 1 B = B = log + log 1 Cse II µ 1 B = B + log + log 5 = log + log + 5 µ α 1 1 B = B = log + log + 1 µ α 1 B = B = log + log log log log log log log log µ α 1 B = log 4 + log B = log + log + log 4+log + 9 Cse III µ 1 B = log + log B = log + log + log + log +4 µ α 1 1 B = log 4 + log B = log + log µ α 1 B = B + log 4+log + 1 = log + log log µ α 1 B = log + log B = log + log + log + log log log log log Solving these systems we find b 1 = 1/, b = / nd b 1 = 5/, b = 10/. By definition b 1 nd b hve to be integers, hence we hve gin contrdiction. Therefore we my exclude the cses I µ α 1, II µ 1, II µ α 1 1, II µ α 1 nd III µ α 1, if we ssume = 0. Next, we wnt to estimte the quntity c nd find lower bound for log Y. 14

15 From nd 4 we find B = log + log B = log + log + log + log log 5+L 6 6., respectively. Let us estimte the quntity c. From 17 nd 1 we find c Now we re redy to estimte the quntity c log. Put γ log mx i i j c := 1 + σ i ɛj ɛ µ i + c 1 Y 0 log Y 0. Using Lemm together with the symptotic expnsions from Section we obtin c log log log nd from the bound for c we find c.006 log log log log log Since we hve lower bounds for B, hence lso for B, nd upper bounds for c, using tble nd inequlity 17 we find tht: log Y if = 0, log Y if 0. Computing gin c using this time insted of Lemm the new bounds found for log Y we get better results. Iterting this procedure four times yields: c log c.0008 log nd log Y if = 0, respectively nd log Y if 0. The bounds for c nd log Y tht re obtined in the other cses re listed in tble 5 nd tble 6. In the next step we use powerful theorem on lower bounds for liner forms in two logrithms due to Lurent, Mignotte, nd Nesterenko [8]. Lemm 5 Let α 1 nd α be two multiplictively independent elements in number field of degree D over Q. For i = 1 nd i =, let log α i be ny determintion of the logrithm of α i, nd let A i > 1 be rel number stisfying log A i mx{hα i, log α i /D, 1/D}, 15

16 Tble 5 Upper bounds for c. c µ 1 µ α 1 1 µ α 1 µ α 1 Cse I = log log log log log log log Cse II = log log Cse III = log log Tble 6 Lower bounds for log Y log log log log log log log log log Y µ 1 µ α 1 1 µ α 1 µ α 1 Cse I = Cse II = Cse III = where hα i denotes the bsolute logrithmic Weil height of α i. Further, let b 1 nd b be two positive integers. Define Then b = { b 1 b + nd log b = mx log b, 1/D, 1 }. D log A D log A 1 b log α b 1 log α 1 exp 0.9D 4 log b log A 1 log A. Before we pply this result we hve to compute some heights: Lemm 6 Let h denote the bsolute logrithmic Weil height, then hα 1 = hα = hα log 7 nd h σµ θ,1, σ µ 4 log + log , 8 16

17 where µ = α 1α 1. The estimtions for H := h σ θ k 1 µ i,j,k σ i 1 µ in the other cses re given in tble 7. Tble 7 Estimtions for the bsolute logrithmic Weil height H := h θ i,j,k. σk 1 µ σ i 1 µ Cse I µ 1 H 4 log + log µ α 1 1 H 4 log + log µ α 1 H 4 log + log µ α 1 H 4 log + log Cse II µ 1 H log + log µ α 1 1 H 5 log + log µ α 1 H 4 log µ α 1 H 5 log + log t Cse III µ 1 H log µ α 1 1 H 4 log + log µ α 1 H 5 log µ α 1 H 5 log + log Proof: We strt with the proof of 7. Since α 1, α, α re conjugte, we only hve to check the lst inequlity. hα 1 = 1 i=1,, mx0, log α i = therefore we obtin the first prt of the lemm. 1 log log 4, Since θ,1, nd σµ re not integers in generl we lso hve to compute their σ µ denomintors, which cn be estimted by θ :=N K/Q γ 1 ɛ ɛ 1 = + 1 µ :=N K/Q α 1 = + 1. respectively, With this preliminry result we obtin h σµ θ,1, σ µ 1 log θ µ + σµ mx 0, log σj θ,1, = j=1,, σ µ 4 log + log

18 Now we pply Lemm 5 to the liner form 0. We distinguish between the cse of = 0 nd 0. In the cse of = 0 we cn pply Lemm 5 t once. In the nottion of Lemm 5 we hve b = 1 log + B 4 log + log log + log + log log + log log Inserting the vrious bounds we obtin log Λ 1 > 84.log log log log 4 log + log On the other hnd we hve from 19 log Λ 1 < log c 4 Y < log log + log log. Compring the upper nd lower bound for log Λ 1 yields contrdiction for lrge. In prticulr, if we hve contrdiction. Since hs to be n integer we know tht we my hve solutions with Y only if 0 := 590. Now we investigte the cse 0. In this cse we do not hve liner form in two logrithms. But we cn study the liner form Λ 1 = log α B σµ 1 1 θ,1, σ µ + B log α. Since hxy hx + hy we hve h α σµ 1 θ,1, σ µ B1 hα 1 + h σµ θ,1, σ µ nd becuse of Lemm 6 we choose b = 1 log + B log + 4 log + log log + B log 1 log + log + log + log + log log log

19 By Lemm 5 we find log Λ 1 > 84.log log B1 log + 4 log + log log log B B 5 log + log > 84. log log B 5 log log log + log log log log On the other hnd log Λ 1 < log c 4 log Y log log B c log log B 1 B c log log log B log + log log log log If we compre these bounds for log Λ 1 we see tht B cncels, nd we obtin n inequlity which cnnot hold for Tht is, if there is solution not found yet for this cse, then 0 := In tble 8 one finds the other upper bounds 0 of the prmeter for the remining cses. Tble 8 Upper bounds 0 for the prmeter. µ 1 µ α 1 1 µ α 1 µ α 1 Cse I = 0 0 = = = = = = = Cse II = 0 0 = = = = = Cse III = 0 0 = = = By tble 8 we hve: 0 0 = = = = Proposition 1 There re no other solutions to 4 thn those listed in Theorem 1 if >

20 6 The method of Mignotte In this section we wnt to eliminte the cse of = 0. We hve lredy discussed the cses I µ α 1, II µ 1, II µ α 1 1, II µ α 1 nd III µ α 1. We know tht B hs to be n integer therefore let us compute B to higher symptotic order in the remining cses: B = log log cse I µ 1 B = log 4 1 log 4 cse I µ α log B = log cse I µ α log B = log + cse II µ α 1 B = log log cse III µ 1 B = log log 4 cse III µ α log B = log + cse III µ α 1 54 log log 4 54 log log 4 54 log log 4 54 log e e e e e e e.4 Since B hs to be n integer, for ech cse we hve criteri wether there exists solution such tht = 0 for one specific. For exmple, the cse I µ α 1 yields following criteri: Lemm 7 Let denote the distnce to the nerest integer. If 4 hs solution, which is not found yet, tht coresponds to the cse I µ α 1 such tht = 0, then 10 log log 54 log e The other cses yield similr criteri. Therefore, in the cse of = 0 nd 0

21 I µ 1, I µ α 1 1, I µ α 1, II µ α 1, III µ 1, III µ α 1 1 or III µ α 1 we check for ech wether the corresponding criteri is fulfilled or not. A computtion in MAGMA see Section 8 yields: Proposition If X, Y is solution to 5 with Y 1 which yields solution to 4 tht is not listed in Theorem 1, then Moreover the solution X, Y yields 0 or < Remrk 1 This method is clled Mignotte s method, becuse Mignotte [11] used similr trick to solve the fmily of Thue equtions completely. X n 1X Y n + XY Y = 1 7 The method of Bker nd Dvenport We cnnot use the method described bove to solve the cse of 0, becuse we hve found n upper bound for the quntity B but not for B itself, which would be essentil. So we re forced to use nother method. We choose the method of Bker nd Dvenport [1]. In prticulr we dpt lemm of Mignotte, Pethő nd Roth [1] to our needs. In order to use the method of Bker nd Dvenport, we hve to find n bsolute lower bound for B. Therefore we hve to revise the liner forms Λ 1, Λ nd Λ. This time we do not consider them s liner combintions of two logrithms but s three logrithms. So we cnnot use the theorem of Lurent, Mignotte nd Nesterenko [8] nd hve to pply result of Mtveev [10]: Lemm 8 Denote by α 1,..., α n lgebric numbers, not 0 or 1, by log α 1,..., log α n determintions of their logrithms, by D the degree over Q of the number field K = Qα 1,..., α n, nd by b 1,..., b n rtionl integers. Furthermore let κ = 1 if K is rel nd κ = otherwise. Define log A i = mx{dhα i, log α i } 1 i n, where hα denotes the bsolute logrithmic Weil height of α nd B = mx{1, mx{ b j A j /A n : 1 j n}}. Assume tht b n 0 nd log α 1,..., log α n re linerly independent over Z; then log Λ CnC 0 W 0 D Ω, 1

22 with Ω = loga 1 loga n, Cn = Cn, κ = 16 1 κ n!κ en n κn + 4n + 1 n+1 en, C 0 = log e 4.4n+7 n 5.5 D loged, W 0 = log1.5eb D loged. We lredy hve computed ll relevnt heights in Lemm 6 respectively tble 7. We combine Siegel s identity 18 with Mtveev s lower bound Lemm 8 nd obtin for our stndrd cse I µ α 1: B log log 8.07 log < log log + log log.6688 B. The only not strightforwrd step is to compute B. Therefore let us rerrnge σ the terms of Λ j such tht the term θ k 1 µ i,j,k is the lst one. Since in ny cse σ i 1 µ B > nd B > 1000 we hve B log = B B. The 4 log +log + 4 inequlity 9 yields contrdiction if B is lrge, i.e. B c 5, where c 5 is some quntity depending on. In view of n bsolute lower bound for B the worst cse occurs, if is s lrge s possible. Therefore we insert 0 insted of into the inequlity bove nd by solving this inequlity we obtin B > The lower bounds for B in the other cses cn be found in tble 9. Tble 9 Absolute lower bounds for B B > µ 1 µ α 1 1 µ α 1 µ α 1 Cse I Cse II Cse III Now we find by the method of Bker nd Dvenport [1] criteri for which there re no solutions. Lemm 9 Suppose nd put δ 1 := log θi,j,k σk 1 µ σ i 1 µ log α nd δ := log α 1 log α, where i nd k re chosen ccording to 0, 1 nd. Further let δ 1 nd 9

23 δ be rtionls such tht δ 1 δ 1 < nd δ δ < nd ssume there exists convergent p/q in the continued frction expnsion of δ, with q 10 0 nd q q δ 1 > c 6 log, then there is no solution for the cse corresponding to j, µ nd 0. The quntities c 6 re listed in tble 10. Tble 10 Absolute lower bounds for B c 6 = µ 1 µ α 1 1 µ α 1 µ α 1 Cse I Cse II Cse III Proof: We give the detils for our stndrd cse I µ α 1. The other cses re similr. Assume tht there is solution corresponding to cse I µ α 1 such tht 1. From 19 we hve δ 1 + δ + B Multipliction by q yields c 4 Y 0 log α exp1.487 < q δ 1 + qδ 1 δ 1 + δ q p + qδ δ + p + B q < nd therefore q δ 1 < q δ q p + q By nother multipliction with q we get q q δ 1 < q q δ q p + q < Tble 4 nd tble 9 together with some estimtions yield q q δ 1 < B log < log

24 Using Lemm 9 we find: Proposition There re no primitive solutions X, Y to 5 with Y > 1, provided Proof: In ech cse nd ech µ from tble we check by computer for ech vlue of in question whether the criteri given in Lemm 9 is fulfilled or not. Combining the result of this computer serch with Proposition we obtin the sttement of the proposition. For more detils on the implementtion see Section 8. By prt 5 of Lemm nd Proposition it is left to solve the Thue equtions X XY Y + 1 = ±1, X XY Y + 1 = ± + 1, for Solving these 996 Thue equtions with PARI yields no further solution. Therefore we hve proved our min Theorem 1. 8 Computer Serch The computtions needed to prove Proposition vi Lemm 7 nd to prove Proposition vi Lemm 9 were implemented in MAGMA. The running times on n Intel Xeon PIII 700MHz processor re collected in tble 11. Finlly, we hve solved the corresponding equtions in the cse both in MAGMA nd in PARI. For references concerning the computer lgebr pckges used in this work see [4], [16] nd [1]. Tble 11 Running times in seconds. µ 1 µ α 1 1 µ α 1 µ α 1 Cse I = Cse II = Cse III =

25 References [1] A. Bker nd H. Dvenport. The equtions x = y nd 8x 7 = z. Qurt. J. Mth. Oxford Ser., 0:19 17, [] A. Bérczes nd A. Pethő. Computtionl experiences on norm form equtions with solutions forming rithmetic progressions. submitted. [] A. Bérczes nd A. Pethő. On norm form equtions with solutions forming rithmetic progressions. Publictiones Mthemtice. [4] W. Bosm, J. Cnnon, nd C. Plyoust. The Mgm lgebr system. I. The user lnguge. J. Symbolic Comput., 4-4:5 65, [5] J. Buchmnn nd A. Pethő. Computtion of independent units in number fields by Dirichlet s method. Mthemtics of Computtion. [6] C. Heuberger, A. Pethő, nd R. F. Tichy. Thoms fmily of Thue equtions over imginry qudrtic fields. J. Symbolic Comput., 45:47 449, 00. [7] S. Lng. Algebric number theory, volume 110 of Grdute Texts in Mthemtics. Springer-Verlg, New York, second edition, [8] M. Lurent, M. Mignotte, nd Y. Nesterenko. Formes linéires en deux logrithmes et déterminnts d interpoltion. J. Number Theory, 55:85 1, [9] F. Lemmermeyer nd A. Pethő. Simplest cubic fields. Mnuscript Mth., 881:5 58, [10] E. M. Mtveev. An explicit lower bound for homogeneous rtionl liner form in logrithms of lgebric numbers. II. Izv. Ross. Akd. Nuk Ser. Mt., 646:15 180, 000. [11] M. Mignotte. Verifiction of conjecture of E. Thoms. J. Number Theory, 44:17 177, 199. [1] M. Mignotte, A. Pethő, nd R. Roth. Complete solutions of fmily of qurtic Thue nd index form equtions. Mth. Comp., 651:41 54, [1] M. B. Mongn, K. O. Geddes, K. M. Hel, G. Lbhn, S. M. Vorkoetter, nd J. McCrron. Mple 6 Progrmming Guide. Wterloo Mple Inc., Wterloo, Cnd, 000. [14] W. M. Schmidt. Diophntine pproximtion, volume 785 of Lecture Notes in Mthemtics. Springer, Berlin, [15] D. Shnks. The simplest cubic fields. Mth. Comp., 8: , [16] The PARI Group, Bordeux. PARI/GP, version.1.5, 004. vilble from [17] E. Thoms. Fundmentl units for orders in certin cubic number fields. J. Reine Angew. Mth., 10: 55, [18] E. Thoms. Complete solutions to fmily of cubic Diophntine equtions. J. Number Theory, 4:5 50,