SERIJA III

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1 SERIJA III I. Gaál, B. Jadrijević and L. Remete Totally real Thue inequalities over imaginary quadrati fields Aepted manusript This is a preliminary PDF of the author-produed manusript that has been peer-reviewed and aepted for publiation. It has not been opyedited, proofread, or finalized by Glasnik Prodution staff.

2 TOTALLY REAL THUE INEQUALITIES OVER IMAGINARY QUADRATIC FIELDS István Gaál, Borka Jadrijevi and László Remete Abstrat. Let F (x, y) be an irreduible binary form of degree 3 with integer oeients and with real roots. Let M be an imaginary quadrati eld, with ring of integers Z M. Let K > 0. We desribe an eient method how to redue the resolution of the relative Thue inequalities F (x, y) K (x, y Z M ) to the resolution of absolute Thue inequalities of type F (x, y) k (x, y Z). We illustrate our method with an expliit example. 1. Introdution Let F (x, y) Z[x, y] be an irreduible binary form of degree 3 and let a Z \ {0}. There is an extensive literature of Thue equations of type F (x, y) = a in x, y Z. In 1909 A. Thue [10] proved that these equations admit only nitely many solutions. In 1967 A. Baker [1] gave eetive upper bounds for the solutions. Later on authors onstruted numerial methods to redue the bounds and to expliitly alulate the solutions, see [6] for a summary. Let M be an algebrai number eld with ring of integers Z M. Let F (x, y) Z M [x, y] be an irreduible binary form of degree n 3 and let µ Z M \ {0}. As a generalization of Thue equations onsider relative Thue equations of type F (x, y) = µ in x, y Z M. 000 Mathematis Subjet Classiation. 11D59, 11D57. Key words and phrases. relative Thue equations, Thue inequalities. 1

3 I. GAÁL, B. JADRIJEVI AND L. REMETE Using Baker's method S. V. Kotov and V. G. Sprindzuk [8] were rst to give eetive upper bounds for the solutions of relative Thue equations. Their theorem has been extended by several authors. Applying Baker's method, redution and enumeration algorithms I. Gaál and M. Pohst [7] gave an eient algorithm for solving relative Thue equations (see also [6]). Let M be an imaginary quadrati number eld. Assuming in addition that the roots of F (x, 1) are all real, in the present paper we give an eient algorithm to redue the resolution of relative Thue inequalities of the type F (x, y) K in x, y Z M to the resolution of (absolute) Thue inequalities of the type F (x, y) k in x, y Z. To nd the solutions of the above absolute Thue inequality one an use Kash [4] or Magma [] whih admit eient algorithms for solving (absolute) Thue equations F (x, y) = k for k Z with k k. For an eient method for alulating "small" solutions of Thue inequalities we refer to [9]. Our method is illustrated with an expliit example.. The main result Let F (x, y) be a binary form of degree n 3 with rational integer oeients. Assume that f(x) = F (x, 1) has leading oeient 1 and distint real roots α 1,..., α n. Let 0 < ε < 1, 0 < η < 1 and let K 1. Set A = min α i α j, B = min α j α i, i j i j i { } K C = max (1 ε) n 1 B, 1, { } { } K 1/n K 1/n C 1 = max εa, (C)1/(n ), C = max εa, C1/(n ), ( ) 1/n K D = η(1 ε) n 1, E = (1 + η)n 1 K AB (1 ε) n 1. Let m 1 be a squarefree positive integer, and set M = Q(i m). Consider the relative Thue inequality (.1) F (x, y) K in x, y Z M.

4 TOTALLY REAL THUE INEQUALITIES 3 If m 3 (mod 4), then x, y Z M an be written as x = x 1 + x 1 + i m = (x 1 + x ) + x i m, 1 + i m y = y 1 + y = (y 1 + y ) + y i m with x 1, x, y 1, y Z. If m 1, (mod 4), then x, y Z M an be written as with x 1, x, y 1, y Z. x = x 1 + x i m, y = y 1 + y i m Theorem.1. Let (x, y) Z M be a solution of (.1). Assume that (.) y > C 1 if m 3 (mod 4), (.3) Then y > C if m 1, (mod 4). (.4) x y 1 = x 1 y. I. Further, if m 3 (mod 4), then the following holds: IA1. If y 1 + y = 0, then x 1 + x = 0 and (.5) F (x, y ) n K ( m) n. IA. If y 1 + y D, then (.6) F (x 1 + x, y 1 + y ) n E. IB1. If y = 0 then x = 0 and (.7) F (x 1, y 1 ) K. IB. If y m D, then (.8) F (x, y ) n ( m) n E. II. If m 1, (mod 4), then the following holds: IIA1. If y 1 = 0 then x 1 = 0 and (.9) F (x, y ) K ( m) n.

5 4 I. GAÁL, B. JADRIJEVI AND L. REMETE IIA. If y 1 D, then (.10) F (x 1, y 1 ) E. IIB1. If y = 0 then x = 0 and (.11) F (x 1, y 1 ) K. IIB. If y D m, then (.1) F (x, y ) E ( m) n. Our result is a far reahing generalization of an idea of [5]. 3. Proof of the main result In the proof of Theorem.1 we shall use the following Lemma. Lemma 3.1. Let x, y Z, y 0. Assume that α i 0 x y d y n for some i 0 (1 i 0 n) and d > 0. If ( ) 1/n d y, ηa then F (x, y) d(1 + η) n 1 α j α i0. Proof. By our assumption, we have α j x y α j α i0 + α i 0 x y (1 + η) α j α i0 for j i 0. Therefore n α j x y = α i0 x y j=1 whih implies our assertion. n α j x y d y n (1 + η)n 1 α j α i0, Now we turn to the proof of our main result Theorem.1.

6 TOTALLY REAL THUE INEQUALITIES 5 Proof. Let (x, y) Z M be an arbitrary solution of (.1) with y 0. Let β j = x α j y, j = 1,..., n, then the inequality (.1) an be written as (3.13) β 1 β n K. Let i 0 be the index with Then β i0 K 1 n β i0 = min j β j. and together with (.) and (.3) we get β j β j β i0 β i0 α j α i0 y K 1 n (1 ε) αj α i0 y for j i 0. From the previous inequality and (3.13), we have (3.14) β i0 with By (3.14) we obtain hene whih implies = K β j y n 1 K (1 ε) n 1 α j α i0. α i 0 xy y = α i0 x y y n, α i0 y xy y n, Im(xy) y n. Note that y < 1 n and y < 1 for m 3 (mod 4) and m 1, (mod n 4), respetively, aording to (.) and (.3). Therefore Im(xy) = 1 x y 1 x 1 y m < 1 and Im(xy) = x y 1 x 1 y m < 1 for m 3 (mod 4) and m 1, (mod 4), respetively. Hene in both ases we have (.4). I. Let m 3 (mod 4). IA. The inequality (3.14) implies Re(β i0 ) y, i.e. n 1 (3.15) (x 1 + x ) α i0 (y 1 + y ) y n 1. IA1. If y 1 + y = 0, then (3.15) yields x 1 + x = 0, and the inequality (.1) has the form ( F x i m, y i ) m K whene we get (.5).

7 6 I. GAÁL, B. JADRIJEVI AND L. REMETE IA. If y 1 + y 0, then (x 1 +x ) α i0 (y 1 +y ) Sine we have assumed Lemma 3.1 implies y n 1 = y 1 + y (y 1 + y ) + y i m ( n ) 1/n, ηa F (x 1 + x, y 1 + y ) n (1 + η) n 1 n 1 α j α i0 whene we get (.6). IB. By the inequality (3.14), we have Im(β i0 ) y, i.e. n 1 (3.16) m x α i0 y y n 1. n y 1 + y n 1. IB1. If y = 0, then (3.16) implies x = 0 and the inequality (.1) has the form ( F x1, y ) 1 K whene we get (.7). IB. If y 0, then x α i0 y Sine Lemma 3.1 implies whih implies (.8). m y n 1 = m (y 1 + y ) + y i m ( n ) 1/n y (, m) n ηa n 1 F (x, y ) n ( (1 + η)n 1 α m) n j α i0 II. Let m 1, (mod 4). IIA. The inequality (3.14) implies Re(β i0 ) y, i.e. n 1 (3.17) x 1 α i0 y 1 y n 1. n ( m) n y n 1.

8 TOTALLY REAL THUE INEQUALITIES 7 IIA1. If y 1 = 0, then (3.17) yields x 1 = 0 and the inequality (.1) has the form whene we get (.9). IIA. If y 1 0, then Sine we have assumed Lemma 3.1 implies F (i mx, i my ) K, x 1 α i0 y 1 y n 1 = y 1 + i my n 1 y 1 n 1. y 1 ( ) 1/n, ηa F (x 1, y 1 ) (1 + η) n 1 α j α i0 whene we get (.10). IIB. By the inequality (3.14) we have Im(β i0 ) y, i.e. n 1 (3.18) m x α i0 y y n 1. IIB1. If y = 0, then (3.18) implies x = 0 and the inequality (.1) has the form whih is just our assertion (.11). IIB. If y 0, then Sine x α i0 y Lemma 3.1 implies F (x 1, y 1 ) K = m y n 1 y 1 + i my n 1 ( m) n y. n 1 F (x, y ) whene we get (.1). ( ) 1/n y (, m) n ηa ( (1 + η)n 1 α m) n j α i0

9 8 I. GAÁL, B. JADRIJEVI AND L. REMETE 4. How to apply Theorem.1 In this setion we give useful hints for a pratial appliation of Theorem.1. Using the same notation let us onsider again the relative Thue inequality (.1). We desribe our algorithm in the ase I (for m 3 (mod 4)) sine the ase II is ompletely similar. 1. If y C 1 then we have only nitely many possible values for y and hene for y 1, y, as well. For eah possible y and for all integers µ Z M with µ K we alulate the roots of the equation F (x, y) µ = 0 in x. For suh a root x we alulate the orresponding x 1, x. If x 1, x are integers, then x Z M and (x, y) is a solutions of (.1). Alternatively, by β i0 K 1 n we obtain x K 1 n + max α j C 1. We an simply enumerate and test the nitely many possible values of x 1, x and y 1, y.. Assume that y > C 1. (a) If y 1 + y < D, then (i) If y < D/ m, then we have only nitely many values for y 1, y, we proeed as in 1. (ii) If y D/ m, then we use IB. We solve F (x, y ) = k for all k Z with k n E/( m) n. We determine the possible values of y 1 whih satisfy y 1 + y < D. We substitute x, y 1, y into x y 1 = x 1 y to see if there exist orresponding integer x 1. (b) If y 1 + y D, then we use IA. We alulate the solutions X = x 1 + x, Y = y 1 + y of F (X, Y ) = k for all k Z with k n E. (i) If y < D/ m then there are only nitely many possible values for y. We determine y 1 from Y. Using X = x 1 +x we set x = X x 1, substitute x = X x 1, y 1, y into x y 1 = x 1 y and test if there is a orresponding x 1 in Z. (ii) If y D/ m we use IB. We solve F (x, y ) = k for k n E/( m) n. We determine x 1, y 1 from x, y and X, Y. For solving absolute Thue equations F (x, y) = k for ertain values k Z one an eiently apply Kash [4] and Magma []. We remark that an appropriate hoie of the parameters ε, η of Thereom.1 makes the resolution muh easier. It is worthy to keep C 1, C and also D small, to avoid extensive tests of small possible solutions. On the other hand, if E is small, then there are fewer Thue equations (over Z) to be solved. Of ourse we an not make all these onstants simultaneously small, therefore we need to make a ompromise, taking into onsideration also the value of K

10 TOTALLY REAL THUE INEQUALITIES 9 (whih also determines the number of Thue equations to be solved). Usually it is worthy to try several values of ε, η before we start solving (.1). Let M = Q(i 5), and let and onsider the solutions of 5. An example F (x, y) = x 4 9x 3 y 1x y + 88xy y 4 (5.19) F (x, y) 0 in x, y Z M. The polynomial F (x, y) is irreduible and the roots of F (x, 1) are approximately 3.471, ,.7581, We may set A =.978, B = Further, let ε = 0.1 and η = 0.1. We are in ase II. Calulating the onstants, Theorem.1 gives: Assume y > 7.9. Then: IIA1. If y 1 = 0, then x 1 = 0 and F (x, y ) IIA. If y , then F (x 1, y 1 ) IIB1. If y = 0, then x = 0 and F (x 1, y 1 ) 0. IIB. If y , then F (x, y ) First we onsider the values with y C = 7.9. We have x max α j C = Enumerating and testing all possible x = x 1 + i 5x and y = y 1 + i 5y satisfying these bounds we obtain the solutions (x 1, x, y 1, y ) = (0, 0, 0, 0), (1, 0, 0, 0), (, 0, 0, 0), (1, 0,, 0), (, 0, 4, 0), up to sign. If y 1 = 0 then by IIA1 we have x 1 = 0 and F (x, y ) 0.8, whene F (x, y ) = 0, x = 0, y = 0. If y = 0 then by IIB1 we have x = 0 and F (x 1, y 1 ) 0. Using Magma we solve F (x 1, y 1 ) = k for 0 k 0. We obtain the solutions (x 1, y 1 ) = (0, 0), (1, 0), (1, ), (, 0), (, 4), up to sign. These bring the above solutions (x 1, x, y 1, y ) again. From now on we assume that y 1 0 and y 0. If y and y then by IIA we have F (x 1, y 1 ) and by IIB we have F (x, y ) In addition to the above alulation we solve F (x 1, y 1 ) = k for 1 k 36 but we do not get any further solutions. Hene the solutions of F (x 1, y 1 ) are (x 1, y 1 ) = (0, 0), (1, 0), (1, ), (, 0), (, 4), up to sign. Also the solutions of F (x, y )

11 10 I. GAÁL, B. JADRIJEVI AND L. REMETE are (x 1, y 1 ) = (0, 0), (1, 0), (1, ), up to sign. Testing these possible (x 1, x, y 1, y ) we do not get any new solutions. If either y 1 < or y < then y 1 = 0 or y = 0 whih ases we have already onsidered. Hene all solutions of (5.19) are (x, y) = (0, 0), (1, 0), (, 0), (1, ), (, 4), up to sign. The alulation takes just a few seonds. Aknowledgements. Researh of the rst author is supported in part by K from the Hungarian National Foundation for Sienti Researh and by the EFOP projet. The projet is o-naned by the European Union and the European Soial Fund. Researh of the seond author is supported in part by the Croatian Siene Foundation under the projet no. 64. Researh of the third author is supported by the ÚNKP-17-3 new national exellene program of the Ministry of human apaities. Referenes [1] A. Baker, Transendental Number Theory, Cambridge, [] W. Bosma, J. Cannon and C. Playoust, The Magma algebra system.i. The user language, J. Symboli Comput. 4(1997), [3] B. W. Char, K. O. Geddes, G. H. Gonnet, M. B. Monagan, S. M. Watt (eds.), MAPLE, Referene Manual, Watom Publiations, Waterloo, Canada, [4] M. Daberkow, C. Fieker, J. Klüners, M. Pohst, K.Roegner and K.Wildanger, KANT V4, J. Symboli Comput. 4(1997), kant/ [5] I. Gaál, Computing elements of given index in totally omplex yli sexti elds, J. Symboli Comput. 0(1995), [6] I. Gaál, Diophantine equations and power integral bases, Birkhäuser, Boston, 00. [7] I. Gaál and M. Pohst, On the resolution of relative Thue equations, Math. Comput. 71(00), [8] S. V. Kotov and V. G. Sprindzuk, An eetive analysis of the Thue-Mahler equation in relative elds (Russian), Dokl. Akad. Nauk. BSSR 17(1973), , 477. [9] A. Peth, On the resolution of Thue inequalities, J. Symboli Comput. 4(1987), [10] A. Thue, Über Annäherungswerte algebraisher Zahlen, J. Reine Angew. Math. 135(1909), István Gaál University of Debreen Mathematial Institute H400 Debreen Pf.400. Hungary gaal.istvan@unideb.hu

12 TOTALLY REAL THUE INEQUALITIES 11 Borka Jadrijevi University of Split Faulty of Siene Ružera Bo²kovi a Split Croatia borka@pmfst.hr László Remete University of Debreen Mathematial Institute H400 Debreen Pf.400. Hungary remete.laszlo@siene.unideb.hu