Integral Points on Certain Elliptic Curves. - Numdam

REND. SEM. MAT. UNIV. PADOVA, Vol. 119 (2008)

Integral Points on Certain Elliptic Curves. HUI LIN ZHU (*) - JIAN HUA CHEN (**)

ABSTRACT - By using algebraic number theory method and p-adic analysis method, we find all integral points on certain elliptic curves y2 (x a)(x2 bx c); a; b; c 2 Z; b2 < 4c: Furthermore, we can find all integer solutions of certain hyperelliptic equations Dy2 Ax4 Bx2 C; B2 < 4AC: As a particular example, we give a complete solution of the equation which was proposed by Zagier y2 x3 9x 28 by this method. In Appendix I and Appendix II, we give the computational method of finding the fundamental unit and factorizing quadratic algebraic number in the subring of a totally complex quartic field, respectively.

1. Introduction. A. Baker [1] developed a method based on linear forms in the logarithms of algebraic numbers, so as to derive an upper bound for the solutions of certain Diophantine equations including elliptic curve. Unfortunately, this upper bound is too large and sometimes beyond the range of computer searching. Recent results on elliptic logarithm methods [2] [3] [4] have allowed the determination of integral points on certain elliptic curves rank as big as 8, but in order to apply it we need the generators of infinite order of the

(*) Indirizzo dell'A.: School of Mathematical Sciences, Xiamen University, Fujian 361005, China. E-mail: [email protected] (**) Indirizzo dell'A.: School of Mathematics and Statistics, Wuhan University, Wuhan 430072, P.R. China. E-mail: [email protected] The project is supported by the National Natural Science Foundation of China (2001AA141010).

2

Hui Lin Zhu - Jian Hua Chen

Mordell-Weil group and the torsion group of the elliptic curve, and the rank should not be too high and the canonical heights of the generators should not be too large either. Mathematicians are asking for simpler method. In this paper, we use an elementary method containing algebraic number theory and p-adic analysis to find all integral points on certain elliptic curves. In section 2, by this method, we find all integral points (x; y) ( 4; 0), ( 1; 6); (9; 26); (764396; 668309460) of the equation proposed by Zagier [5] y2 x3

(1:1)

9x 28:

In section 3, we find all integral points on a class of elliptic curves y2 (x a)(x2 bx c); a; b; c 2 Z;

(1:2)

where the discriminant of x2 bx c; denote as D; satisfies D b2 4c < 0. Furthermore, we can find all integer solutions of certain hyperelliptic Diophantine equations Dy2 Ax4 Bx2 C;

(1:3)

B2 < 4AC:

In section 4, we discuss other cases of equation (1.2) where the discriminant D 0. In Appendix I and Appendix II, we give the computational method of finding the fundamental unit and factorizing quadratic algebraic number in the subring of a totally complex quartic field, respectively.

2. Proof of Main Theorem. THEOREM 1. All integral points of the elliptic curve (1.1) are (x; y) ( 4; 0); ( 1; 6); (9; 26); (764396; 668309460): PROOF.

y2 x3

9x 28 (x 4)(x2

4x 7):

Put z x 4, then we have x2

4x 7 (x 4)2

Let d gcd (z; z2

12(x 4) 39 z2

12z 39:

12z 39), it is easy to obtain dj39. Because

Integral Points on Certain Elliptic Curves

D

12 < 0, we get

(

(2:1)

z du2 z2

12z 39 dv2

3

;

where d 1; 3; 13; 39; gcd (u; v) 1. When d 1; (z 6)2 3 v2 ; [v (u2 6)][v (u2 6)] 3; it is easy to show that there is no solution in rational integer u; v. So equation (2.1) is equivalent to the following three equations ( z 3u2 (2:2) ; z2 12z 39 3v2 ( (2:3)

z 13u2 z2

and

(

(2:4)

12z 39 13v2

z 39u2 z2

12z 39 39v2

:

In the following, we will discuss the three cases, separately. CASE 1: We solve equation (2.2). Reducing mod4 to (2.2), we can get that v is even and u is odd. Put v 2v0 , hence we can write (2.2) as ( z 3u2 : z2 12z 39 3(2v0 )2 Thus we have (3u2

6)2 3 3(2v0 )2 :

We use algebraic number theory method above(see [6][7][8][9]). In the quadratic algebraic h pih 2 pi 3 (3u 6) 3 (3u2 6)

to factor the p equation field Q( 3), we have (

p 2 3) (2AA0 )2 ;

0 where quadratic algebraic field pA; A are two 0 algebraic integers in thep 3) is one and we get Q( 3) and v0 AA . The class number of Q(

(2:5)

(3u2

6)

p p n 3 3w (2A2 );

4


p p 1 3 is a root of unity in Q( 3) and w3 1, n 0; 1 or 2: where w 2 From (2.5), we obtain p n 2 p 3w A 18 3 3; (3u)2 6 p n p 2 p (3u)2 2 3w ( 3A) 18 3 3: p Put A1 3A, then we have p n 2 p (2:6) 3w A1 18 3 3; (3u)2 2 or (2:7)

(3u)2 2

p n 2 p 3w A1 18 3 3:

CASE 1.1: First, we solve equation (2.6). We write equation (2.6) as p 2 p (2:8) 3A2 18 3 3 15 6w or 21 6w; (3u)2 2 p where A22 A21 wn , 3 2w 1. p p Put u 2 3; u is an algebraic integer in a totally complex quartic p 3 2 4w. Define the subring R field and satisfies u2 2 Z[1; u; w; uw]: For simplicity, we denote a a bu cw duw as a (a; b; c; d) and denote the conjugates of a as the following: a a

bu cw

duw (a; b; c; d);

a0 a bu0 cw2 du0 w2 ; a0 a bu0 cw2 du0 w2 ; p p 2 3 is the complex conjugate of u. And we denote the where u0 coefficients of a as a (a)0 ; b (a)1 ; c (a)2 ; d (a)3 : Denote by jzj the complex absolute value of the complex number z and denote kak max (jaj; ja j; ja0 j; ja0 j) as the greatest absolute value of a; a ; a0 and a0 . Because u is an algebraic number in a totally complex quartic field, according to Dirichlet's unit theorem, there is only one independent fundamental unit in Q(u). By direct computation, the fundamental unit in R (Appendix I) is e 1 2u 4w 2uw: From (2.8), we get (3u uA2 )(3u

uA2 ) 15

6w or 21 6w:


5

Thus we have 3u uA2 aek wt ; u; k; t 2 Z; t 0; 1; 2:

(2:9)

If algebraic number j in the field Q(u) has j aa0 bb0 and b ae, where e is the unit in the field Q(u), we call a and b as relevant factors. Otherwise, we call a; a0 ; b and b0 as irrelevant factors. By computation (Appendix II), the irrelevant factors of 15 6w and 21 6w are: a1 (3; 2; 0; 1); a2 (9; 4; 6; 1);

a1 (3; 2; 0; 1); a2 (9; 4; 6; 1):

Now we use p-adic analysis method [10] to solve (2.9). From (2.9), we get (2:10)

3u

uA2 a ek wt ;

u; k; t 2 Z; t 0; 1; 2:

1; (3u)2 u2 A22 (a2 a2 )(ee )k v2t ; we get t 0. (2:9) (2:10), since eep 3); we put A2 b dw; b; d 2 Z, then we have Because A2 2 Q( (3u uA2 )2 ( aek )2 0: It is the main reason that we take p-adic analysis to solve equations (2.6). To take p-adic analysis, we need to find the prime p from the prime factors of gcd ((ek )1 ; (ek )2 ; (ek )3 ). Here we choose p 109 to take p-adic analysis. If a a2 or a2 , ee 1; (3u)2

u2 A22 (a2 a2 )(ee )k a2 a2

(21 6w);

it leads to contradiction, so a a1 or a1 : By direct computation, e6 ( (a1 e)2

40223; 45604; 89232; 35828)

1(mod 13); (a1 e2 )2 (a1 e4 )2 4(mod 13);

1(mod 13);

1(mod 13); (a1 e3 )2

6(mod 13);

(a1 e5 )2 1(mod 13):

So we assume k 6m; write k 108n s; 0 s 107; where actually s 0; 6; 12; 18; 24; 30; 36; 42; 48; 54; 60; 66; 72; 78; 84; 90; 96; 102: By direct computation, we know that only s 0 meets (a1 es )2 0(mod 109): In the following, we will prove equation (1.1) has the solutions (x; y) ( 1; 6) when s 0.

6


Put 109r kn: Because e108 (1; 9374; 0; 8720)(mod 11881 1092 ); denote e108 1 109(0; 86; 0; 80) 1092 h: a1 e108n a1 (1 109((0; 86; 0; 80) 109h))n a1 (1 109n((0; 86; 0; 80) 109h) ); 0 (a1 e108n )2 (a1 )2 109n(a1 (0; 86; 0; 80))2 (mod 109r2 ); a1 (0; 86; 0; 80) (

480; 258; 36; 240);

we get n 0; u 1; (x; y) ( and get (x; y) ( 1; 6):

109 y 36;

1; 6): Similarly, we deal with a a1

p p Second, we solve equation (2.7). Let u 2 3, w CASEp1.2: 1 3 , R Z[1; u; w; uw], we have u2 2 4w: We use the similar 2 method above and get 3u uA2 aek wt ; u; k; t 2 Z: By computation, the fundamental unit in R is e (3; 0; 4; 2); the irrelevant factors of 15

6w and 21 6w in R are:

a1 (3; 1; 0; 1);

a1 (3; 1; 0; 1);

a2 (3; 1; 6; 1);

a2 (3; 1; 6; 1);

a3 (3; 5; 6; 1);

a3 (3; 5; 6; 1);

a4 (3; 7; 12; 1);

a4 (3; 7; 12; 1):

We can get t 0 by the same method with Case1.1. Because ee 1; a2 a2 (15 6w); a3 a3 (15 6w); it results in contradiction, then a a1 ; a1 ; a4 or a4 : If a a1 , e6 (49009; 9776; 89232; 35828) (a1 e)2

1(mod 13); (a1 e2 )2 (a1 e4 )2

4(mod 13);

1(mod 13);

1(mod 13); (a1 e3 )2 (a1 e5 )2 1(mod 13):

6(mod 13);


7

So we assume k 6m; write k 108n s; 0 s 107; where actually s 0; 6; 12; 18; 24; 30; 36; 42; 48; 54; 60; 66; 72; 78; 84; 90; 96; 102: By direct computation, we know that only s 0 meets (a1 es )2 0(mod 109): Put 109r kn. Because e108 (1; 654; 0; 3161)(mod 11881 1092 ); denote e108 1 109(0; 6; 0; 29) 1092 h: a1 e108n a1 (1 109((0; 6; 0; 29) 109h))n a1 (1 109n((0; 6; 0; 29) 109h) ); 0 (a1 e108n )2 (a1 )2 109n(a1 (0; 6; 0; 29))2 (mod 109r2 ); a1 (0; 6; 0; 29) (138; 18; 36; 87);

109 y 36;

we get n 0; u 1; (x; y) ( 1; 6): Similarly, we deal with a a1 and get (x; y) ( 1; 6): Similarly, we deal with a a1 ,a4 and a4 , so we get (x; y) ( 1; 6): CASE 2: We solve equation (2.3). By taking modula 8, we can get that v is even and u is odd. It can be written as p p p p 3][(13u2 6) 3] ( 1 2 3)( 1 2 3)(2AA0 )2 : [(13u2 6) We have

p p n 2 (13u2 6) 3 2( 1 2 3)w A ; p p 1 3 is a root of unity in Q( 3) and w3 1; where w 2 0 0 n 0; 1 or 2; u 2 Z; pA; A are two algebraic integer in Z[w] and v 2AA . 3)A, we can get Let A1 ( 1 2 p n 2 p 3)w A1 78 13 3 65 26w or 91 26w; (2:11) (13u)2 2( 1 2 or (2:12) (13u)2 2(

1

2

p n 2 p 3)w A1 78 13 3 65

26w or 91 26w:

C ASE 2.1: First, we solve equation (2.11). Let A22 wn A21 ; u p p 2 4 3; then u2 6 8w: u is an algebraic integer in a totally complex quartic field. We define a subring R Z[1; u; w; uw]:

8


From (2.11), we get (2:13)

13u uA2 aek wt ;

u; k; t 2 Z; t 0; 1; 2:

By computation, the fundamental unit in R is e (11; 3; 4; 4); the irrelevant factors of 65


a1 (13; 4; 0; 3);

a1 (13; 4; 0; 3);

a2 (1; 4; 10; 1);

a2 (1; 4; 10; 1);

a3 (3; 1; 4; 3);

a3 (3; 1; 4; 3);

a4 (1; 3; 10; 1);

a4 (1; 3; 10; 1);

a5 (17; 3; 14; 7);

a4 (17; 3; 14; 7):

We can get u 1; (x; y) (9; 26) by p-adic analysis method similar to Case 1. C ASE 2.2: Second, we solve equation (2.12). Let A22 wt A21 ; u p p 24 3; then u2 6 8w: u is an algebraic integer in a totally complex quartic field. We define a ring R Z[1; u; w; uw]: From (2.12), we get (2:14)

13u uA2 aek wt ;

u; k; t 2 Z; t 0; 1; 2:

By computation, the fundamental unit in R is e (3; 1; 2; 0); but there are no irrelevant factors of 65 is no solution.

26w and 91 26w in R, so there

CASE 3: We solve equation (2.4). It can be written as p p p 3][(39u2 6) 3] ( 6 3)( 6 [(39u2 6) We have

p 3)(AA0 )2 :

p p n 2 (39u2 6) 3 ( 6 3)w A ; p p 1 3 is a root of unity in Q( 3) and w3 1; where w 2 n 0; 1 or 2; u 2 Z; A; A0 are two algebraic integers in Z[w] and v AA0 :


Let A1 ( (2:15) (39u)2

9

p 3)A, we get p n 2 p ( 6 3)w A1 234 39 3 195 78w or 27378w;

6

or

p n 2 p 3)w A1 234 39 3 195 78w or 27378w: p p 6 3; CASE 3.1: First, we solve (2.15). Let A22 wn A21 ; u 2 then u 7 2w: u is an algebraic integer in a totally complex quartic field. We define a subring R Z[1; u; w; uw]: From (2.15), we get (2:16) (39u)2 ( 6

(2:17)

39u uA2 aek wt ; u; k; t 2 Z:

By computation, the fundamental unit in R is e (2; 1; 1; 1); the irrelevant factors of 195


a1 (0; 5; 0; 2); a2 (6; 7; 24; 2); a3 (9; 4; 3; 1); We can get u 0; 140; so (x; y) ( analysis method similar to Case 1.

a1 (0; 5; 0; 2); a2 (6; 7; 24; 2); a3 (9; 4; 3; 1): 4; 0); (764396; 668309460) by p-adic

C ASE 3.2: Second, we solve equation (2.16). Let A22 wt A21 ; u p p 6 3; then u2 7 2w: u is an algebraic integer in a totally complex quatic field. We can define a ring R Z[1; u; w; uw]: We get 39u uA2 aek wt ; u; k; t 2 Z; t 0; 1; 2: By computation, we know that the fundamental unit in R is e (5; 2; 3; 1); the irrelevant factors of 195


a1 (0; 5; 0; 2);

a1 (0; 5; 0; 2);

a2 (6; 1; 24; 8);

a2 (6; 1; 24; 8);

a3 (12; 5; 9; 7);

a3 (12; 5; 9; 7):

We can get there is no solution by p-adic analysis. Theorem 1 has been proved.

10


3. Method Discussion. By using algebraic number theory and p-adic analysis method, we can find all integral points on certain elliptic curves y2 (x a)(x2 bx c); a; b; c 2 Z; where the discriminant of x2 bx c; denote as D; satisfies D b2 4c < 0. Furthermore, we can solve all integer solutions of certain hyperelliptic equations Dy2 Ax4 Bx2 C; B2 < 4AC: Now we give the complete solution of equation (1.2). Put z x a; from (1.2), we get

(3:1)

x2 bx c z2 (b ( z du2 z2 (b

2a)z (a2

2a)z (a2

ab c);

ab c) dv2

;

where d gcd (z; z2 (b 2a)x (a2 ab c)), so dj(a2 ab c). Actually, when d has a square factor, we can extract it to u and v. For some cases of d, we can solve the equations by elementary method. Next we use algebraic number theory method and p-adic analysis method to solve all integral points on certain elliptic curves (1.2). From (3.1), we have d2 u4 (b

2a)du2 (a2

ab c) dv2 :

It can be written as the form of equation (1.3) Dy2 Ax4 Bx2 C; B2 < 4AC; where u x; v y; A d2 ; B (b 2a)d; C a2 Equation (1.3) can be written as (3:2)

ab c; D d:

D1 y2 x41 B1 x21 C1 ;

where D1 A3 D; B1 AB; C1 A3 C; x1 Ax and B21 < 4C1 . So we have 4D1 y2 (2x21 B1 )2 (4C1 Let f 4C1 (3:3)

B21 ):

B21 lg2 > 0 (l is square-free positive integer), we have 4D1 y2 (2x21 B1 )2 lg2 :


We factorize equation (3.3) in complex quadratic field Q( p (2x21 B1 l (3:4) g ) (KLM2 );

11

p g ) to get

p where K is the common divisor of ideals (2x21 B1 l g ) and p g ), L is the ideal factorization of 4D1 and N(L) 4D1 . (2x21 B1 l p We suppose the ideal classes in Q( g ) are I1 ; I2 ; ; Ih and their representatives are J1 ; J2 ; ; Jh : There is no loss of generality by assuming that KL I1 1 ; M I2 1 . From (3.4), we have p J1 J22 (2x21 B1 l (3:5) g ) J1 KL(J2 M)2 : Since J1 KL and J2 M are principle ideals, J1 J22 is a principle ideal. Let J1 KL S1 ; J2 M S2 ; J1 J22 S3 , we have p S3 (2x21 B1 l (3:6) g ) S1 S22 : Suppose the conjugation of S3 is S03 , and S3 S03 a. From (3.6), we get 2ax21 B2 S0 S22 ; p 0 p g ); S S1 S03 2 Q( g ) is an algebraic number in where B2 a(B1 l p Q( g ). From (3.7), we obtain (3:7)

2ax21 (3:8)

(2ax1 )2

S0 S22 2aS0 S22

B2 ; 2aB2 :

p g ), we REMARK 1. As long as 2aS0 in (3.8) has a square factor b 2 Q( 0 2aS put u2 2 ; S02 bS2 . Define the subring R Z[1; u; w; uw]: u is an alb gebraic integer in a totally complex quartic field. We choose the basis 1; u; w; uw in order that kuk and kek in the subring R is by far least and p which we choose to take p-adic analysis is the smallest possible. From (3.8), we have (3:9)

2ax1 uS02 wt ek a;

where w is a root of unity, e is a fundamental unit in the subring R, a is an irrelevant factor of 2aB2 in Q(u)(the irrelevant factors a of 2aB2 are finite). p g ), we put S02 b du2 ; b; d 2 Z. From (3.9), we get Since S2 2 Q( (3:10)

2ax1 bu du3 wt ek a:

From (3.10), we have (3:11)

2ax1

bu

du3 wt ek a :

12


(3:10) (3:11), we get t 0: The coefficient of u2 is 0 in the left of (3.10), so (3:12)

(2ax1 bu du3 )2 ( ek a)2 0:

From equation (3.12), we know there must be another equation corresponding to the unknown k. Finally we solve it directly by p-adic analysis method. In [11][12], Ljunggren and Tzanakis proved that solving equation (1.3) is equivalent to solve the integer solutions of a class of quartic Thue equation p(x; y) Ax4 Bx3 y Cx2 y2 Dxy3 Ey4 m; where p(x; 1) 0 has two real roots and a couple of complex roots. Suppose u is a root of p(x; 1) 0, then by Dirichlet's unit theorem there are two independent fundamental units e1 ; e2 in Q(u). It is complicate to compute them. In [6], Ljunggren's method need compute a relative unit in a quartic field. Obviously, their methods are not easy. REMARK 2. It is necessary to point out that the method works theoretically. Our computations involved can be carried out successfully but in very exceptional cases. The first point is that there is no control on the size and the computation time of the fundamental unit e, even though we find a proper u in order that R Z[1; u; w; uw] is a domain. The second point, is that computing a set of representative of ideal class group of an imaginary field can take a lot of time. That is to say, we can not multiply the examples at will. But in fact, many equations can be solved by this method. We compute all integral points on another elliptic curve which Don.Zagier proposed (3:13)

y2 x3

30x 133

are: (x; y)( 7; 0); ( 3; 4); (2; 9); (6; 13); (5143326; 116644498677). Many famous problems are the examples of equation (3.2), for example[13][2][14]: x(x 1) 2 y(y 1) (3:14) ; 2 2 (3:15) (3:16) (3:17)

6y2 x3 5x 6 (x 1)(x2 31y2 x3

1;

y2 (x 337)(x2 3372 ):

x 6);


13

4. Other Cases' Discussion. When the discriminant D b2 4c 0 in equation (1.2), we have b2 c , then b is even and c is a square number. Let b 2l; l 2 Z, then 4 c l2 , so y2 (x a)(x2 2lx l2 ) (x a)(x l)2 :

(4:1)

So x a is a square number. Put x a m2 ; m 2 Z, then we get y2 m2 (x l)2 :

(4:2) Thus we get the solution (4:3)

(

x m2

a2

y m(m2

a l)

:

When the discriminant D b2 4c > 0 in equation (1.2), we have b2 c < , then x2 bx c 0 has two different real roots x1 and x2 . 4 b n and Especially when D b2 4c n2 ; n 2 N, we have x1 2 bn x2 . Hence equation (1.2) can be written as 2 b n b n y2 (x a) x (4:4) x : 2 2 We can refer to Pell Equation's method in Zagier [5]. When D b2 4c is not a square number, we refer to linear forms in the logarithms of algebraic number[1] and elliptic logarithm method [2].

Appendix I p Find pa fundamental unit in R Z[1; u; w; uw], where u 4 12; w 1 3 . 2 If e a bu cw duw is a fundamental unit in R, where a; b; c; d 2 Z, then e a bu cw duw is a conjugate of e. So we have ee (a bu cw duw)(a (a2

2b2 8bd

c2

2d2 ) (2ac

bu cw

4b2 4bd

duw)

c2 2d2 )w F Gw:

N(e) N(e ) N(F Gw) (F Gw)(F Gw2 ) F 2

FG G2 1:

14


Thus we get FG 1 F2 G2 2jFGj: From FG > 0; FG 0; FG < 0; we obtain (F; G) ( 1; 1), ( 1; 0), (0; 1); where 8 < F a2

()

: G 2ac

2b2 8bd 4b2 4bd

c2

2d2

c2 2d2

:

There is no loss of generation of assuming c 6 0, otherwise c 0 is the easier condition. From G 2ac we have a

4b2 4bd

c2 W ; where W 4b2 2c c2 W 2 2c

c2 2d2 ; 2d2 G: Then we obtain

4bd

2b2 8bd

c2

2d2 F:

Thus we have 3c4 c2 (8b2

32bd 8d2

2W 4F)

W 2 0:

We regard this equation as a quadratic equation with one unknown c1 c2 , then 2

c1 c

B

p B2 12W 2 ; where B 8b2 6

32bd 8d2

2W 4F:

We limit the ranges of b and d to find the integers a; b; c; d which meet ( ). We suppose jbj 5; jdj 5, then in the ranges we search c1 c2 . If there exist two integers b0 and d0 to get c1 c20 ; c0 2 Z, we can compute a0 c2 W . If a0 is an integer, we have already found a unit in R. If from a 2c we can not find a; b; c; d in the ranges, we should extend the ranges of b and d till we find a unit e0 a0 b0 u c0 w d0 uw: Now we will compute the fundamental unit e a bu cw duw from e0 a0 b0 u c0 w d0 uw, where a; b; c; d 2 Z: If ke0 k is not the least, we have e0 et ; jtj 2:


Denote

0

1 0 e 1 Be C B1 B 0 CB @e A @1 e0 1

then

u

0

w w w2 w2

u u

0

10 1 0 1 uw a a CB b C 4 B b C uw CB C M B C; 0 @c A u w0 2 A@ c A d d u w2

0 1 a Bb C B CM @c A d 0

From M

u

1

1w

B B1 w B 1 B B u 4w 2 B B 1 B @ 1 u

1w

0

1 e B C 1B e C @ e0 A : e0 w

w

1w w u u0 1

1 1 u

15

1 u0

1 u0

1 C wC C u0 C C, we obtain C 1 C C A

8 1 > > a [(1 w)e (1 w)e we0 we0 ] > > > 4w 2 > > > 1 w > w i > 1 h1w w > > e e 0 e0 e0 1 > 0 0 > (e e c e e ) > > > 4w 2 > > > > 1 e e e0 e0 > > :d 4w 2 u u u0 u0 8 8 t > e e1=t > e e 0 0 > > > > > > > > < e e1=t < e0 e t 0 From , thus we obtain (where jtj 2), we get 0 0t 01=t > > 0 e e > > e e 0 > > 0 > > > : > : 0 01=t e00 e0t e e0 1 jaj (j1 wkej j1 wke j jwke0 j jwke0 j) 4w 2 1 1=t 1=t 1=t 1=t (j1 wke0 j j1 wke0 j jwke00 j jwke00 j ) 4w 2 1 1=2 4 ke0 k ; 4w 2

16


jbj

w w 1 w 1 1 w jej je j 0 je0 j 0 je0 j 4w 2 u0 u0 u0 u0 w 1 1 w 1=t 1 w 1=t w 1=t 1=t je0 j je0 j 0 je00 j 0 je00 j 4w 2 u0 u0 u0 u0

2

1 1 1 1=2 0 ke0 k ; 4w 2 u0 u0

jcj 4

1 1=2 je0 k ; 4w 2

jdj 2

1 1 1 1=2 0 ke0 k : 4w 2 u0 u0

If we can not find integers a; b; c; d in the ranges, then e0 a0 b0 u c0 w d0 uw is just the fundamental unit. If we can find integers a00 ; b00 ; c00 ; d00 , then we get a unit e1 a00 b00 u c00 w d00 uw whose absolute value ke1 k is smaller than ke0 k. We use the same method for e1 till we find the fundamental unit. By computation, the fundamental unit of the subring R Z[1; u; w; uw] is e 1 2u 4w 2uw: The method of computation here is more effective than the ones in [15] [16].

Appendix II 0 Factorize the quadratic algebraic number p j 15 6w and j 21 6w p 1 3 4 ;u 12. in R Z[1; u; w; uw], where w 2 0 Let j 15 6w 21 6w2 aa , j 21 6w 15 6w2 a0 a0 . The factorization is only defined ``up to multiplication by units''. If there are two factors a1 (a1 ; b1 ; c1 ; d1 ) and a2 (a2 ; b2 ; c2 ; d2 ) which meet above a1 a1 condition, and is an algebraic integer and is a unit e (N(e) 1), then a2 a2 they are relevant algebraic numbers. We have

N(j) N(j 0 ) (15

6w)(21 6w) 351 aa a0 a0 ;

and we have an important inequality[10] kak jN(j)j1=4

p kek;


17

where e is the fundamental unit in a totally complex quartic field Q(u). In the following, we will give a simple proof for above inequality. By a logarithmic mapping l, we can establish the relation between j and the fundamental unit e. That is to say, from n r1 2r2 1 0 2 2 4; we suppose h1 l(e) (2 log jej; 2 log je0 j); h2 (2; 2); l(a) (2 log jaj; 2 log ja0 j) x1 h1 x2 h2 x1 l(e) x2 h2 ; so we have

(

2 log jaj 2x1 log jej 2x2 2 log ja0 j 2x1 log je0 j 2x2

;

where e and e0 are e's conjugates whose absolute values are different, a and a0 are a's conjugates whose absolute values are different. Hence we obtain 2 log jaj 2 log ja0 j [2(x1 log jej) 2x2 ] [2(x1 log je0 j) 2x2 ] x1 log jee0 j2 4x2 4x2 : So we have x2

1 1 1 log jaj2 ja0 j2 log jN(a)j log jN(j)j: 4 4 4

Therefore we get jaj jx2 jjex1 j jN(j)j1=4 jex1 j: 1 Put x1 m1 y1 ; jy1 j ; m1 2 Z: From a ~aem1 ; we have ~a ae 2 then 1 p j~ aj jN(j)j1=4 max p ; jej jN(j)j1=4 ke0 k; jej

m1

;

where e0 is the fundamental unit in R Z[1; u; w; uw]: Because we factorize the quadratic algebraic number j in the subring R, the a's that we will find are stable under multiplication by powers of e and actually we compute a set of representatives ~ a modulo this action. REMARK 3. If there is a factorization of j, then up to changing a by a relevant integer, the important inequality is fulfilled. Of course it is clear the inequality does not work for any algebraic integer ``relevant'' to a.

18


If

0

a

1

0

1

C B C B1 C B CB C B1 A @ 0 1 a

B Ba B B 0 Ba @ then

u

w u

u

w

0

w2 u

0

w2

0 1 10 1 a a B C CB C Bb C uw CB b C B C CB C M B C; C B C 0 Bc C Bc C u w2 C @ A A@ A 0 2 uw d d

uw

0 1 a B C Bb C B C B CM Bc C @ A d

0

a

B Ba B 0 Ba @

1B

1 C C C C: C A

a0

We get 8 1 > > a [(1 w)a (1 w)a wa0 wa0 ] > > 4w 2 > > > 1 w > > 1 h 1 w w > > a a 0 a0 1 >c > (a a a0 a0 ) > > 4w 2 > > > > a a > a0 a0 > :d 1 4w 2 u u u0 u 0

w 0 i a u0

;

8 1 p 2 p 4 > > p jaj j1 wj jwj kak 351 kek 13:5294 > > 2w 1 > 3 > > p > 1 1 w w > > 2 4 351 p > > kek 7:26913 0 kak p p < jbj 2w 1 u u 3 4 12 : 2 p > 2 p 4 > > jcj kak p 351 kek 13:5294 > > > 2w 1 3 > > > 1 1 p > 1 1 2 p > > 4 351 kek 7:26913 : jdj 0 kak p p 4 2w 1 u u 3 12 In the range of jaj 13; jbj 7; jcj 13; jdj 7; there are some a; b; c; d 2 Z satisfying j (15

6w) aa (a bu cw duw)(a

bu cw

duw)


(a cw)2 (a2

u2 (b dw)2 (a2 2acw c2 w2 )

2b2 8bd

c2

2d2 ) (2ac

19

(4w 2)(b2 2bdw d2 w2 )

4b2 4bd

c2 2d2 )w F Gw:

If there are two factors a1 (a1 ; b1 ; c1 ; d1 ) and a2 (a2 ; b2 ; c2 ; d2 ) which a1 meet above condition, and is an algebraic integer and is a unit. We should a2 filter one of them(for example a2 ) and remain another(for example a1 ). We do the procedure till all remained factors a of j are irrelevant. By computation, we get the irrelevant factors of j 15 6w and j 0 21 6w factoring in the subring R are: a1 (3; 2; 0; 1); a2 (9; 4; 6; 1);

a01 (3; 2; 0; 1); a02 (9; 4; 6; 1):

Acknowledgments. We express our gratitude to everyone who takes part in discussion of this paper in School of Mathematics and Statistics in Wuhan University. Especially, we thank Dr. Jing-bo Xia for his kind help. The first author express his gratitude to Guo-tai Deng, Yong-wei Yu and Wen-bo Wang for their help in programming. Finally, we thank the referee Professor Ph. SatgeÂ for his valuable suggestions and Giovanni Gerotto for his patient work. REFERENCES [1] A. BAKER, Linear forms in the logarithms of algebraic number I, Mathematika, 13 (1966), pp. 204-216; II: ibid, 14 (1967), pp. 102-107; III: ibid, 14 (1967), pp. 220-228; IV: ibid, 15 (1968), pp. 204-216. [2] R. J. STROEKER - N. TZANAKIS, Solving Elliptic Diophantine Equations by Estimating Linear Forms in Elliptic Logarithms, Acta Arith., 29(2) (1994), pp. 177-196. [3] R. J. STROEKER - N. TZANAKIS, On the Elliptic Logarithm Method for Elliptic Diophantine Equations: Reflections and an Improvement, Experimental Mathemetics, 8(2) (1999), pp. 135-149. [4] R. J. STROEKER - N. TZANAKIS, Computing All Integer Solutions of a Genus 1 Equation, Math. Comp., 72 (2003), pp. 1917-1933. [5] D. ZAGIER, Large Integral Point on Elliptic Curves, Math. Comp., 48(177) (1987), pp. 425-536. [6] W. LJUNGGREN, A Diophantine Problem, Jour London Math. Soc., 3(2) (1971), pp. 385-391. [7] J. H. CHEN, A Note on the Diophantine Equation x2 1 dy4 , Abh. Math. Sem. Univ. Hamburg, 64 (1994), pp. 1-10. [8] K. Q. FENG, Algebraic Number Theory, Science Press, 2000 (Chinese). [9] L. G. HUA, Introduction to Number Theory, Science Press, 1979 (Chinese).

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[10] R. J. STROEKER - N.TZANAKIS, On the Application of skolem's p-adic Method to the solution of Thue Equations, Journal of Number Theory, 29 (1988), pp. 166-195. [11] W. LJUNGGREN, Solution complete de quelquls equations du sixieme, degre a deux indeterminees, Arch. Math., 48(7) (1946), pp. 26-29. [12] N. TZANAKIS, On the Diophantine equation x2 dy4 k, Acta Arith., 46(3) (1986), pp. 257-269. [13] L. J. MORDELL, Diophantine Equations, London: Academic Press, 1969. [14] Z. F. CAO - S. Z. MU - X. L. DONG, A New Proof of a Conjecture of Antoniadis, Jour Number Theory, 83 (2000), pp. 185-193. [15] J. BUCHMANN, A generalization of Voronoi's unit algorithm (I II), Jour. Number Theory, 20 (1985), pp. 177-209. [16] J. BUCHMANN, The Computation of the Fundamental Unit of Totally Comples Quartic Orders, Math. Comp., 48(177) (1987), pp. 39-54. Manoscritto pervenuto in redazione l'8 gennaio 2006