Integral Points on Certain Elliptic Curves

Integral Points on Certain Elliptic Curves.

HUILINZHU(*) - JIANHUACHEN(**)

ABSTRACT- By using algebraic number theory method and p-adic analysis method, we find all integral points on certain elliptic curves

y²ˆ(x‡a)(x²‡bx‡c); a;b;c2Z; b²<4c:

Furthermore, we can find all integer solutions of certain hyperelliptic equations Dy²ˆAx⁴‡Bx²‡C; B²<4AC:

As a particular example, we give a complete solution of the equation which was proposed by Zagier

y²ˆx³ 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 loga- rithms of algebraic numbers, so as to derive an upper bound for the so- lutions of certain Diophantine equations including elliptic curve. Un- fortunately, 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: huilin9200@yahoo.com

(**) Indirizzo dell'A.: School of Mathematics and Statistics, Wuhan Univer- sity, Wuhan 430072, P.R. China.

E-mail: chenjh_ecc@163.com

The project is supported by the National Natural Science Foundation of China (2001AA141010).

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)ˆ( 4;0), ( 1;6); (9;26);(764396;668309460) of the equation proposed by Zagier [5]

y²ˆx³ 9x‡28:

(1:1)

In section 3, we find all integral points on a class of elliptic curves y²ˆ(x‡a)(x²‡bx‡c); a;b;c2Z;

(1:2)

where the discriminant of x²‡bx‡c; denote as D; satisfies Dˆb² 4c<0. Furthermore, we can find all integer solutions of certain hyperelliptic Diophantine equations

Dy²ˆAx⁴‡Bx²‡C; B²<4AC:

(1:3)

In section 4, we discuss other cases of equation (1.2) where the dis- criminantD0.

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)ˆ( 4;0);( 1;6);(9;26);(764396;668309460):

PROOF.

y²ˆx³ 9x‡28ˆ(x‡4)(x² 4x‡7):

Putzˆx‡4, then we have

x² 4x‡7ˆ(x‡4)² 12(x‡4)‡39ˆz² 12z‡39:

Let dˆgcd (z;z² 12z‡39), it is easy to obtain dj39. Because

Dˆ 12<0, we get

zˆdu²

z² 12z‡39ˆdv²; (

(2:1)

wheredˆ1;3;13;39; gcd (u;v)ˆ1.

When dˆ1; (z 6)²‡3ˆv²; [v (u² 6)][v‡(u² 6)]ˆ3; it is easy to show that there is no solution in rational integeru;v. So equation (2.1) is equivalent to the following three equations

zˆ3u²

z² 12z‡39ˆ3v²; (

(2:2)

zˆ13u²

z² 12z‡39ˆ13v² (

(2:3) and

zˆ39u²

z² 12z‡39ˆ39v²: (

(2:4)

In the following, we will discuss the three cases, separately.

CASE1: We solve equation (2.2). Reducing mod4 to (2.2), we can get that vis even anduis odd. Putvˆ2v0, hence we can write (2.2) as

zˆ3u²

z² 12z‡39ˆ3(2v₀)²: (

Thus we have

(3u² 6)²‡3ˆ3(2v₀)²:

We use algebraic number theory method to factor the equation above(see [6][7][8][9]). In the quadratic algebraic fieldQ( 

p 3

), we have h(3u² 6)‡ 

p 3ih

(3u² 6) 

p 3i

ˆ ( 

p 3

)²(2AA⁰)²;

where A;A⁰ are two algebraic integers in the quadratic algebraic field Q( 

p 3

) andv0ˆAA⁰. The class number ofQ( 

p 3

) is one and we get (3u² 6) 

p 3

ˆ 

p 3

wⁿ(2A²);

(2:5)

wherewˆ 1‡ 

p 3

2 is a root of unity inQ( 

p 3

) andw³ˆ1,nˆ0;1 or 2:

From (2.5), we obtain

(3u)² 6 

p 3

wⁿA²ˆ183 

p 3

; (3u)²2 

p 3

wⁿ( 

p 3

A)²ˆ183 

p 3 : PutA₁ˆ 

p 3

A, then we have (3u)² 2 

p 3

wⁿA²₁ˆ18 3 

p 3

; (2:6)

or

(3u)²‡2 

p 3

wⁿA²₁ˆ18 3 

p 3 : (2:7)

CASE1.1: First, we solve equation (2.6). We write equation (2.6) as (3u)² 2 

p 3

A²₂ˆ183 

p 3

ˆ15 6wor 21‡6w;

(2:8)

whereA²₂ˆA²₁wⁿ, 

p 3

ˆ2w‡1.

Putuˆ 

2 

p 3

p ;u is an algebraic integer in a totally complex quartic field and satisfies u²ˆ2 

p 3

ˆ2‡4w. Define the subring Rˆ

ˆZ[1;u;w;uw]:

For simplicity, we denoteaˆa‡bu‡cw‡duwasaˆ(a;b;c;d) and denote the conjugates ofaas the following:

a ˆa bu‡cw duwˆ(a; b;c; d);

a⁰ˆa‡bu⁰‡cw²‡du⁰w²; a⁰ ˆa bu⁰‡cw² du⁰w²; where u⁰ˆ 

2 

p 3

p is the complex conjugate of u. And we denote the coefficients of a as aˆ(a)₀;bˆ(a)₁;cˆ(a)₂;dˆ(a)₃: Denote by jzj the complex absolute value of the complex number z and denote kak ˆmax (jaj;ja j;ja⁰j;ja⁰ j) as the greatest absolute value ofa;a ;a⁰and a⁰.

Because u is an algebraic number in a totally complex quartic field, according to Dirichlet's unit theorem, there is only one independent fun- damental unit inQ(u). By direct computation, the fundamental unit inR (Appendix I) is

eˆ1‡2u‡4w‡2uw:

From (2.8), we get

(3u‡uA₂)(3u uA₂)ˆ15 6wor 21‡6w:

Thus we have

3u‡uA₂ˆ ae^kw^t; u;k;t2Z;tˆ0;1;2:

(2:9)

If algebraic number j in the fieldQ(u) hasjˆaa⁰ˆbb⁰ and bˆae, whereeis the unit in the field Q(u), we callaandb as relevant factors.

Otherwise, we call a;a⁰;b and b⁰ as irrelevant factors. By computation (Appendix II), the irrelevant factors of 15 6wand 21‡6ware:

a₁ ˆ(3;2;0;1); a₁ ˆ(3; 2;0; 1);

a2ˆ(9;4;6; 1); a2 ˆ(9; 4;6; 1):

Now we usep-adic analysis method[10] to solve (2.9). From (2.9), we get

3u uA₂ˆ a e^kw^t; u;k;t2Z;tˆ0;1;2:

(2:10)

(2:9)(2:10), sinceee ˆ1;(3u)² u²A²₂ˆ(a2a2 )(ee )^kv^2t;we gettˆ0.

BecauseA₂2Q( 

p 3

);we putA₂ˆb‡dw; b;d2Z, then we have (3u‡uA₂)₂ˆ( ae^k)₂ˆ0:

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 primepfrom the prime factors of gcd ((e^k)₁;(e^k)₂;(e^k)₃). Here we choosepˆ109 to take p-adic analysis.

Ifaˆa₂ ora₂ ,

ee ˆ1; (3u)² u²A²₂ˆ(a2a2 )(ee )^k ˆa2a2 ˆ (21‡6w);

it leads to contradiction, soaˆa₁ ora₁ : By direct computation,

e⁶ˆ( 40223; 45604; 89232; 35828) 1(mod 13);

(a1e)₂ 1(mod 13); (a1e²)₂ 1(mod 13); (a1e³)₂ 6(mod 13);

(a₁e⁴)₂4(mod 13); (a₁e⁵)₂1(mod 13):

So we assumekˆ6m;writekˆ108n‡s;0s107;where actually sˆ0;6;12;18;24;30;36;42;48;54;60;66;72;78;84;90;96;102:

By direct computation, we know that onlysˆ0 meets (a₁e^s)₂0(mod 109):

In the following, we will prove equation (1.1) has the solutions (x;y)ˆ

ˆ( 1;6) whensˆ0.

Put 109^rkn:Because

e¹⁰⁸(1;9374;0;8720)(mod 11881ˆ109²);

denote

e¹⁰⁸ˆ1‡109(0;86;0;80)‡109²h:

a1e¹⁰⁸ⁿˆa1(1‡109((0;86;0;80)‡109h))ⁿˆa1(1‡109n((0;86;0;80)‡109h)‡ );

0(a1e¹⁰⁸ⁿ)₂ (a1)₂‡109n(a1(0;86;0;80))₂(mod 109^r‡2);

a1(0;86;0;80)ˆ( 480;258;36;240); 109y36;

we getnˆ0;uˆ 1;(x;y)ˆ( 1;6): Similarly, we deal withaˆa₁ and get (x;y)ˆ( 1;6):

CASE 1.2: Second, we solve equation (2.7). Let uˆ 

2 

p 3

p , wˆ

ˆ 1‡ 

p 3

2 ,RˆZ[1;u;w;uw], we haveu²ˆ 2 4w:We use the similar method above and get

3u‡uA2ˆ ae^kw^t; u;k;t2Z:

By computation, the fundamental unit inRis eˆ(3;0;4;2);

the irrelevant factors of 15 6wand 21‡6winRare:

a1ˆ(3;1;0; 1); a1 ˆ(3; 1;0;1);

a₂ˆ(3;1;6; 1); a₂ ˆ(3; 1;6;1);

a3ˆ(3;5; 6;1); a3 ˆ(3; 5; 6; 1);

a₄ˆ(3;7; 12; 1); a₄ ˆ(3; 7; 12;1):

We can get tˆ0 by the same method with Case1.1. Because ee ˆ1;a2a2 ˆ (15 6w);a3a3 ˆ (15 6w); it results in contra- diction, thenaˆa₁;a₁ ;a₄ ora₄ :

Ifaˆa₁,

e⁶ˆ(49009; 9776;89232;35828) 1(mod 13);

(a₁e)₂ 1(mod 13); (a₁e²)₂ 1(mod 13); (a₁e³)₂ 6(mod 13);

(a₁e⁴)₂ 4(mod 13); (a₁e⁵)₂1(mod 13):

So we assumekˆ6m;writekˆ108n‡s;0s107;where actually sˆ0;6;12;18;24;30;36;42;48;54;60;66;72;78;84;90;96;102: By direct computation, we know that onlysˆ0 meets

(a₁e^s)₂0(mod 109):

Put 109^rkn. Because

e¹⁰⁸(1;654;0;3161)(mod 11881ˆ109²);

denote

e¹⁰⁸ˆ1‡109(0;6;0;29)‡109²h:

a₁e¹⁰⁸ⁿˆa₁(1‡109((0;6;0;29)‡109h))ⁿˆa₁(1‡109n((0;6;0;29)‡109h)‡ );

0(a₁e¹⁰⁸ⁿ)₂(a₁)₂‡109n(a₁(0;6;0;29))₂(mod 109^r‡2);

a1(0;6;0;29)ˆ(138;18; 36;87); 109y36;

we get nˆ0;uˆ 1;(x;y)ˆ( 1;6):Similarly, we deal withaˆa1

and get (x;y)ˆ( 1;6):Similarly, we deal withaˆa₁ ,a₄anda₄ , so we get (x;y)ˆ( 1;6):

CASE2: We solve equation (2.3). By taking modula 8, we can get thatvis even anduis odd. It can be written as

[(13u² 6)‡ 

p 3

][(13u² 6) 

p 3

]ˆ( 1‡2 

p 3

)( 1 2 

p 3

)(2AA⁰)²: We have

(13u² 6) 

p 3

ˆ 2( 1‡2 

p 3

)wⁿA²; where wˆ 1‡ 

p 3

2 is a root of unity in Q( 

p 3

) and w³ˆ1;

nˆ0;1 or 2;u2Z;A;A⁰are two algebraic integer inZ[w] andvˆ2AA⁰. LetA₁ˆ( 1‡2 

p 3

)A, we can get (2:11) (13u)² 2( 1 2 

p 3

)wⁿA²₁ˆ7813 

p 3

ˆ65 26wor 91‡26w;

or

(2:12) (13u)²‡2( 1 2 

p 3

)wⁿA²₁ˆ7813 

p 3

ˆ65 26wor 91‡26w:

CASE 2.1: First, we solve equation (2.11). Let A²₂ˆwⁿA²₁;uˆ

ˆ 

2 4 

p 3

p ;thenu²ˆ 6 8w:uis an algebraic integer in a totally complex quartic field. We define a subringRˆZ[1;u;w;uw]:

From (2.11), we get

13u‡uA₂ˆ ae^kw^t; u;k;t2Z; tˆ0;1;2:

(2:13)

By computation, the fundamental unit inRis eˆ(11;3;4;4);

the irrelevant factors of 65 26wand 91‡26winRare:

a₁ˆ(13;4;0;3); a₁ ˆ(13; 4;0; 3);

a₂ˆ(1;4;10;1); a₂ ˆ(1; 4;10; 1);

a3 ˆ(3;1;4; 3); a3 ˆ(3; 1;4; 3);

a₄ˆ(1;3;10;1); a₄ ˆ(1; 3;10; 1);

a5ˆ(17;3;14;7); a4 ˆ(17; 3;14; 7):

We can getuˆ 1;(x;y)ˆ(9;26) by p-adic analysis method similar to Case 1.

CASE 2.2: Second, we solve equation (2.12). Let A²₂ˆw^tA²₁;uˆ

ˆ 

2‡4 

p 3

p ; then u²ˆ6‡8w: u is an algebraic integer in a totally complex quartic field. We define a ringRˆZ[1;u;w;uw]:

From (2.12), we get

13u‡uA2ˆ ae^kw^t; u;k;t2Z; tˆ0;1;2:

(2:14)

By computation, the fundamental unit inRis eˆ(3;1;2;0);

but there are no irrelevant factors of 65 26wand 91‡26winR, so there is no solution.

CASE3: We solve equation (2.4). It can be written as [(39u² 6)‡ 

p 3

][(39u² 6) 

p 3

]ˆ( 6‡ 

p 3

)( 6 

p 3

)(AA⁰)²: We have

(39u² 6) 

p 3

ˆ ( 6‡ 

p 3

)wⁿA²; where wˆ 1‡ 

p 3

2 is a root of unity in Q( 

p 3

) and w³ˆ1;

nˆ0;1 or 2;u2Z;A;A⁰are two algebraic integers inZ[w] andvˆAA⁰:

LetA1ˆ( 6‡ 

p 3

)A, we get (2:15) (39u)² ( 6 

p 3

)wⁿA²₁ˆ23439 

p 3

ˆ195 78wor 273‡78w;

or

(2:16) (39u)²‡( 6 

p 3

)wⁿA²₁ˆ23439 

p 3

ˆ195 78wor 273‡78w:

CASE 3.1: First, we solve (2.15). Let A²₂ˆwⁿA²₁; uˆ 

6 

p 3

p ;

then u²ˆ 7 2w:u is an algebraic integer in a totally complex quartic field. We define a subringRˆZ[1;u;w;uw]:

From (2.15), we get

39u‡uA₂ˆ ae^kw^t; u;k;t2Z:

(2:17)

By computation, the fundamental unit inRis eˆ(2;1; 1;1);

the irrelevant factors of 195 78wand 273‡78winRare:

a₁ˆ(0;5;0; 2); a₁ ˆ(0; 5;0;2);

a2ˆ(6;7;24;2); a2 ˆ(6; 7;24; 2);

a₃ ˆ(9;4; 3; 1); a₃ ˆ(9; 4; 3;1):

We can getuˆ0;140;so (x;y)ˆ( 4;0);(764396;668309460) by p-adic analysis method similar to Case 1.

CASE 3.2: Second, we solve equation (2.16). Let A²₂ˆw^tA²₁;uˆ

ˆ 

6‡ 

p 3

p ; then u²ˆ7‡2w: u is an algebraic integer in a totally complex quatic field. We can define a ringRˆZ[1;u;w;uw]:

We get

39u‡uA₂ˆ ae^kw^t; u;k;t2Z;tˆ0;1;2:

By computation, we know that the fundamental unit inRis eˆ(5;2;3;1);

the irrelevant factors of 195 78wand 273‡78winRare:

a₁ ˆ(0;5;0; 2); a₁ ˆ(0; 5;0;2);

a2ˆ(6;1;24;8); a2 ˆ(6; 1;24; 8);

a₃ˆ(12;5;9;7); a₃ ˆ(12; 5;9; 7):

We can get there is no solution by p-adic analysis.

Theorem 1 has been proved.

3. Method Discussion.

By using algebraic number theory and p-adic analysis method, we can find all integral points on certain elliptic curves

y²ˆ(x‡a)(x²‡bx‡c); a;b;c2Z;

where the discriminant of x²‡bx‡c; denote as D; satisfies Dˆb² 4c<0. Furthermore, we can solve all integer solutions of certain hyperelliptic equations

Dy²ˆAx⁴‡Bx²‡C; B²<4AC:

Now we give the complete solution of equation (1.2). Putzˆx‡a;from (1.2), we get

x²‡bx‡cˆz²‡(b 2a)z‡(a² ab‡c);

zˆdu²

z²‡(b 2a)z‡(a² ab‡c)ˆdv²; (

(3:1)

where dˆgcd (z;z²‡(b 2a)x‡(a² ab‡c)), so dj(a² ab‡c). Actu- ally, whendhas a square factor, we can extract it touandv. For some cases of d, we can solve the equations by elementary method. Next we use al- gebraic number theory method and p-adic analysis method to solve all in- tegral points on certain elliptic curves (1.2).

From (3.1), we have

d²u⁴‡(b 2a)du²‡(a² ab‡c)ˆdv²: It can be written as the form of equation (1.3)

Dy²ˆAx⁴‡Bx²‡C; B²<4AC;

whereuˆx;vˆy;Aˆd²;Bˆ(b 2a)d;Cˆa² ab‡c;Dˆd:

Equation (1.3) can be written as

D1y²ˆx⁴₁‡B1x²₁‡C1; (3:2)

whereD1ˆA³D;B1ˆAB;C1ˆA³C;x1ˆAxandB²₁<4C1. So we have 4D1y²ˆ(2x²₁‡B1)²‡(4C1 B²₁):

Letf ˆ4C1 B²₁ˆlg²>0 (lis square-free positive integer), we have 4D₁y²ˆ(2x²₁‡B₁)²‡lg²:

(3:3)

We factorize equation (3.3) in complex quadratic fieldQ(pg) to get (2x²₁‡B₁‡lpg)ˆ(KLM²);

(3:4)

where K is the common divisor of ideals (2x²₁‡B1‡lpg) and (2x²₁‡B₁ lpg),Lis the ideal factorization of 4D₁andN(L)ˆ4D₁.

We suppose the ideal classes inQ(pg) areI₁;I₂; ;I_hand their re- presentatives areJ₁;J₂; ;J_h:There is no loss of generality by assuming thatKLI₁¹;MI₂¹. From (3.4), we have

J1J²₂(2x²₁‡B1‡lpg)ˆJ1KL(J2M)²: (3:5)

Since J1KL and J2M are principle ideals, J1J₂² is a principle ideal. Let J1KLˆS1;J2MˆS2;J1J²₂ˆS3, we have

S₃(2x²₁‡B₁‡lpg)ˆS₁S²₂: (3:6)

Suppose the conjugation ofS₃isS⁰₃, andS₃S⁰₃ˆa. From (3.6), we get 2ax²₁‡B2ˆS⁰S²₂;

(3:7)

whereB2ˆa(B1‡lpg);S⁰ˆS1S⁰₃2Q(pg) is an algebraic number in Q(pg). From (3.7), we obtain

2ax²₁ S⁰S²₂ˆ B₂; (2ax₁)² 2aS⁰S²₂ˆ 2aB₂: (3:8)

REMARK1. As long as 2aS⁰in (3.8) has a square factorb2Q(pg), we putu²ˆ2aS⁰

b² ;S⁰₂ˆbS₂. Define the subringRˆZ[1;u;w;uw]:u is an al- gebraic integer in a totally complex quartic field. We choose the basis 1;u;w;uwin order thatkukandkekin the subringRis by far least andp which we choose to take p-adic analysis is the smallest possible.

From (3.8), we have

2ax₁‡uS⁰₂ˆ w^te^ka;

(3:9)

wherewis a root of unity,eis a fundamental unit in the subringR,ais an irrelevant factor of 2aB₂ in Q(u)(the irrelevant factors a of 2aB₂ are finite).

SinceS22Q(pg), we putS⁰₂ˆb‡du²; b;d2Z. From (3.9), we get 2ax1‡bu‡du³ˆ w^te^ka:

(3:10)

From (3.10), we have

2ax1 bu du³ˆ w^te^ka : (3:11)

(3:10)(3:11), we gettˆ0:The coefficient ofu²is 0 in the left of (3.10), so (2ax1‡bu‡du³)₂ˆ(e^ka)₂ˆ0:

(3:12)

From equation (3.12), we know there must be another equation corre- sponding to the unknownk. 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)ˆAx⁴‡Bx³y‡Cx²y²‡Dxy³‡Ey⁴ˆ m;

wherep(x;1)ˆ0 has two real roots and a couple of complex roots. Suppose u is a root of p(x;1)ˆ0, 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.

REMARK2. It is necessary to point out that the method works theo- retically. 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 unite, even though we find a properuin order thatRˆZ[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

y²ˆx³ 30x‡133 (3:13)

are: (x;y)ˆ( 7;0);( 3;4);(2;9);(6;13);(5143326;116644498677).

Many famous problems are the examples of equation (3.2), for ex- ample[13][2][14]:

x(x 1)

2 ₂

ˆy(y 1)

2 ;

(3:14)

6y²ˆx³‡5x‡6ˆ(x‡1)(x² x‡6);

(3:15)

31y²ˆx³ 1;

(3:16)

y²ˆ(x‡337)(x²‡337²):

(3:17)

4. Other Cases' Discussion.

When the discriminant Dˆb² 4cˆ0 in equation (1.2), we have cˆb²

4, thenb is even andc is a square number. Let bˆ2l; l2Z, then cˆl², so

y²ˆ(x‡a)(x²‡2lx‡l²)ˆ(x‡a)(x‡l)²: (4:1)

Sox‡ais a square number. Putx‡aˆm²;m2Z, then we get y²ˆm²(x‡l)²:

(4:2)

Thus we get the solution

xˆm² a²

yˆ m(m² a‡l): (

(4:3)

When the discriminant Dˆb² 4c>0 in equation (1.2), we have c<b²

4, then x²‡bx‡cˆ0 has two different real roots x₁ and x₂. Especially when Dˆb² 4cˆn²; n2N, we have x1ˆ b n

2 and

x2ˆ b‡n

2 . Hence equation (1.2) can be written as y²ˆ(x‡a)

x b n 2

x b‡n

2 : (4:4)

We can refer to Pell Equation's method in Zagier [5]. WhenDˆb² 4cis not a square number, we refer to linear forms in the logarithms of alge- braic number[1] and elliptic logarithm method [2].

Appendix I

Find a fundamental unit in RˆZ[1;u;w;uw], where uˆ 

12 p4

;wˆ

ˆ 1‡ 

p 3

2 .

If eˆa‡bu‡cw‡duw is a fundamental unit in R, where a;b;c;d2Z, thene ˆa bu‡cw duwis a conjugate ofe. So we have

ee ˆ(a‡bu‡cw‡duw)(a bu‡cw duw)

ˆ(a² 2b²‡8bd c² 2d²)‡(2ac 4b²‡4bd c²‡2d²)wˆF‡Gw:

N(e)ˆN(e )ˆN(F‡Gw)ˆ(F‡Gw)(F‡Gw²)ˆF² FG‡G²ˆ 1:

Thus we get

FG1ˆF²‡G²2jFGj:

From FG>0;FGˆ0;FG<0; we obtain (F;G)ˆ( 1;1), (1;0), (0;1);where

Fˆa² 2b²‡8bd c² 2d² Gˆ2ac 4b²‡4bd c²‡2d²: 8<

() :

There is no loss of generation of assumingc6ˆ0, otherwisecˆ0 is the easier condition. From

Gˆ2ac 4b²‡4bd c²‡2d²;

we haveaˆc²‡W

2c ;whereWˆ4b² 4bd 2d²‡G:Then we obtain c²‡W

2c ₂

2b²‡8bd c² 2d²ˆF:

Thus we have

3c⁴‡c²(8b² 32bd‡8d² 2W‡4F) W²ˆ0:

We regard this equation as a quadratic equation with one unknownc1ˆc², then

c1ˆc²ˆ B 

B²‡12W² p

6 ; whereBˆ8b² 32bd‡8d² 2W‡4F:

We limit the ranges ofbanddto find the integersa;b;c;dwhich meet (). We supposejbj 5;jdj 5, then in the ranges we search c1ˆc². If there exist two integersb0andd0to getc1ˆc²₀;c02Z, we can computea0

fromaˆc²‡W

2c . Ifa₀is an integer, we have already found a unit inR. If we can not finda;b;c;din the ranges, we should extend the ranges ofband dtill we find a unite0ˆa0‡b0u‡c0w‡d0uw:

Now we will compute the fundamental uniteˆa‡bu‡cw‡duwfrom e₀ ˆa₀‡b₀u‡c₀w‡d₀uw, wherea;b;c;d2Z:Ifke₀kis not the least, we havee₀ˆ e^t;jtj 2:

Denote ee e⁰ e⁰ 0 BB

@ 1 CC

Aˆ

1 u w uw

1 u w uw

1 u⁰ w² u⁰w² 1 u⁰ w² u⁰w² 0

BB

@

1 CC A

ab cd 0 BB

@ 1 CC Aˆ⁴ M

ab cd 0 BB

@ 1 CC A;

then a

bc d 0 BB

@ 1 CC

AˆM ¹

ee e⁰ e⁰ 0 BB

@ 1 CC A:

FromM ¹ˆ 1 4w‡2

1‡w 1‡w w w

1‡w u

1‡w u

w u⁰

w u⁰

1 1 1 1

1 u

1 u

1 u⁰

1 u⁰ 0

BB BB BB BB

@

1 CC CC CC CC A

, we obtain

aˆ 1

4w‡2[(1‡w)e‡(1‡w)e ‡we⁰‡we⁰] bˆ 1

4w‡2 h1‡w

u e‡ 1‡w u

e ‡w

u⁰e⁰‡ w u⁰

e⁰ i cˆ 1

4w‡2(e‡e e⁰ e⁰ ) dˆ 1

4w‡2 e

u e

u e⁰ u⁰‡e⁰

u⁰

: 8>

>>

>>

>>

>>

>>

><

>>

>>

>>

>>

>>

>>

:

From

e₀ˆ e^t e₀ ˆ e^t e⁰₀ˆ e^0t e⁰₀ ˆ e^0t 8>

>>

><

>>

>>

:

(wherejtj 2), we get

eˆ e^1=t₀ e ˆ e^1=t₀ e⁰ˆ e^01=t₀ e⁰ ˆ e^01=t₀ 8>

>>

>>

><

>>

>>

>>

:

, thus we obtain

jaj 1

4w‡2(j1‡wkej ‡ j1‡wke j ‡ jwke⁰j ‡ jwke⁰ j)

ˆ 1

4w‡2(j1‡wke0j^1=t‡ j1‡wke0 j^1=t‡ jwke⁰₀j^1=t‡ jwke⁰₀ j^1=t) 4 1

4w‡2ke₀k¹⁼²;

jbj 1

4w‡21‡w

u₀ jej ‡1‡w

u₀ je j ‡w

u⁰₀je⁰j ‡w u⁰₀je⁰ j

ˆ 1

4w‡21‡w u0

je0j^1=t‡1‡w u0

je0 j^1=t‡w

u⁰₀je⁰₀j^1=t‡w

u⁰₀je⁰₀ j^1=t 2 1

4w‡21 u₀‡1

u⁰₀ ke₀k¹⁼²;

jcj 4 1

4w‡2je₀k¹⁼²; jdj 2 1

4w‡21 u₀‡1

u⁰₀ ke₀k¹⁼²:

If we can not find integers a;b;c;d in the ranges, then e₀ˆa₀‡

‡b₀u‡c₀w‡d₀uwis just the fundamental unit.

If we can find integers a⁰₀;b⁰₀;c⁰₀;d⁰₀, then we get a unit e1ˆ

ˆa⁰₀‡b⁰₀u‡c⁰₀w‡d⁰₀uw whose absolute value ke1k is smaller than ke0k.

We use the same method for e₁ till we find the fundamental unit. By computation, the fundamental unit of the subring RˆZ[1;u;w;uw] is eˆ1‡2u‡4w‡2uw:

The method of computation here is more effective than the ones in [15] [16].

Appendix II

Factorize the quadratic algebraic numberjˆ15 6wandj⁰ˆ21‡6w inRˆZ[1;u;w;uw], wherewˆ 1‡ 

p 3

2 ;uˆ 

4 12 p .

Let jˆ15 6wˆ21‡6w²ˆ aa , j⁰ˆ21‡6wˆ15 6w²ˆ a⁰a⁰ . The factorization is only defined ``up to multiplication by units''. If there are two factors a₁ˆ(a₁;b₁;c₁;d₁) and a₂ˆ(a₂;b₂;c₂;d₂) which meet above condition, anda1

a2is an algebraic integer and is a uniteˆa1

a2(N(e)ˆ 1), then they are relevant algebraic numbers. We have

N(j)ˆN(j⁰)ˆ(15 6w)(21‡6w)ˆ351ˆaa a⁰a⁰ ; and we have an important inequality[10]

kak jN(j)j¹⁼⁴ 

pkek

;

whereeis the fundamental unit in a totally complex quartic fieldQ(u). In the following, we will give a simple proof for above inequality.

By a logarithmic mappingl, we can establish the relation between j and the fundamental unit e. That is to say, from nˆr₁‡2r₂ ˆ

ˆ10‡22ˆ4;we suppose

h₁ˆl(e)ˆ(2 logjej;2 logje⁰j); h₂ ˆ(2;2);

l(a)ˆ(2 logjaj;2 logja⁰j)ˆx₁h₁‡x₂h₂ˆx₁l(e)‡x₂h₂; so we have

2 logjaj ˆ2x₁logjej ‡2x₂ 2 logja⁰j ˆ2x₁logje⁰j ‡2x₂; (

whereeande⁰aree's conjugates whose absolute values are different,aand a⁰area's conjugates whose absolute values are different. Hence we obtain 2 logjaj ‡2 logja⁰j ˆ[2(x₁logjej)‡2x₂]‡[2(x₁logje⁰j)‡2x₂]ˆ

ˆx₁logjee⁰j²‡4x₂ˆ4x₂: So we have

x₂ˆ1

4 logjaj²ja⁰j²ˆ1

4 logjN(a)j ˆ1

4logjN(j)j:

Therefore we get

jaj ˆ jx₂jje^x1j ˆ jN(j)j¹⁼⁴je^x1j:

Put x1ˆm1‡y1;jy1j 1

2;m12Z: From aˆ~ae^m1; we have ~aˆae ^m1; then

j~aj jN(j)j¹⁼⁴max 1

pjej; 

p jej

ˆ jN(j)j¹⁼⁴ke0k;

wheree0is the fundamental unit inRˆZ[1;u;w;uw]:Because we factorize the quadratic algebraic numberjin the subringR, thea's that we will find are stable under multiplication by powers ofeand actually we compute a set of representatives~amodulo this action.

REMARK3. If there is a factorization ofj, then up to changingaby a relevant integer, the important inequality is fulfilled. Of course it is clear the inequality does not work for any algebraic integer ``relevant'' toa.

If

a a a⁰ a⁰ 0 BB BB B@

1 CC CC CAˆ

1 u w uw

1 u w uw

1 u⁰ w² u⁰w² 1 u⁰ w² u⁰w² 0

BB BB B@

1 CC CC CA

a b c d 0 BB BB B@

1 CC CC CAˆM

a b c d 0 BB BB B@

1 CC CC CA;

then

a b c d 0 BB BB B@

1 CC CC CAˆM ¹

a a a⁰ a⁰ 0 BB BB B@

1 CC CC CA:

We get aˆ 1

4w‡2[(1‡w)a‡(1‡w)a ‡wa⁰‡wa⁰] bˆ 1

4w‡2

h1‡w u

a‡ 1‡w

u

a ‡w

u⁰a⁰‡ w u⁰

a⁰ i cˆ 1

4w‡2(a‡a a⁰ a⁰ ) dˆ 1

4w‡2 a

u a

u a⁰ u⁰‡a⁰

u⁰

; 8>

>>

>>

>>

>>

>>

<

>>

>>

>>

>>

>>

>:

jaj 1 2w‡1

j1‡wj ‡ jwj

kak 2

p3 

4351 p 

pkek

ˆ13:5294 jbj 1

2w‡11‡w u ‡w

u⁰

kak 2

p3



351 p4



12 p4 

pkek

ˆ7:26913 jcj 2

2w‡1kak 2

p3 

4351 p 

pkek

ˆ13:5294 jdj 1

2w‡11 u‡1

u⁰

kak 1

p3 2

12 p4 

4351 p 

pkek

ˆ7:26913 : 8>

>>

>>

>>

>>

>>

><

>>

>>

>>

>>

>>

>>

:

In the range of

jaj 13;jbj 7;jcj 13;jdj 7;

there are somea;b;c;d2Zsatisfying

jˆ (15 6w)ˆaa ˆ(a‡bu‡cw‡duw)(a bu‡cw duw)

ˆ(a‡cw)² u²(b‡dw)²ˆ(a²‡2acw‡c²w²) (4w‡2)(b²‡2bdw‡d²w²)

ˆ(a² 2b²‡8bd c² 2d²)‡(2ac 4b²‡4bd c²‡2d²)wˆF‡Gw:

If there are two factors a₁ˆ(a₁;b₁;c₁;d₁) and a₂ˆ(a₂;b₂;c₂;d₂) which meet above condition, anda1

a₂is an algebraic integer and is a unit. We should filter one of them(for examplea₂) and remain another(for examplea₁). We do the procedure till all remained factorsaofjare irrelevant.

By computation, we get the irrelevant factors of jˆ15 6w and j⁰ˆ21‡6wfactoring in the subringRare:

a1ˆ(3;2;0;1); a⁰₁ˆ(3; 2;0; 1);

a₂ˆ(9;4;6; 1); a⁰₂ˆ(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. Satge for his valuable suggestions and Giovanni Gerotto for his patient work.

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[12] N. TZANAKIS, On the Diophantine equation x² dy⁴ˆk, 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.

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Manoscritto pervenuto in redazione l'8 gennaio 2006