# \$S\$-integral solutions to a Weierstrass equation

(1)

## W

o

### p. 281-301

<http://www.numdam.org/item?id=JTNB_1997__9_2_281_0>

© Université Bordeaux 1, 1997, tous droits réservés.

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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques

(2)

to a

### equation

par BENJAMIN M.M. DE WEGER

RÉSUMÉ. On détermine les solutions rationnelles de

### diophantienne y2

= x3 - 228x + 848 dont les dénominateurs sont des

de 2. On

une idée de Yuri

### Bilu, qui

évite le recours à des

de Thue et de

et

### qui

permet d’obtenir des

aux

### (S-)

unités à quatre termes dotées de

### propriétés spéciales,

que l’on résout par la théorie des formes linéaires en

### logarithmes

réels et

ABSTRACT. The rational solutions with as denominators powers of 2 to the

### equation y2

= x3 - 228x + 848 are determined. An idea of Yuri Bilu is

### applied,

which avoids Thue and Thue-Mahler

### equations,

and deduces four-term

unit

with

that are solved

### by

linear forms in real and

### logarithms.

1. Introduction

In a recent paper

my

### colleague

R.J. Stroeker and I determined the

### complete

set of solutions in rational

to the

The

### (Q)

can be seen as a model of an

### elliptic

curve. In our solution methods we did not use that

### viewpoint,

but rather worked

N. Tzanakis

### [T]

has shown how the

### logarithms

method of Stroeker and Tzanakis

### designed

for Weierstrass models of

curves

### only,

can be

to the situation of a

model for an

### elliptic

curve, and in fact he chose

as one of the

### examples

to illustrate his ideas.

We stress that the

of

on

### elliptic

curves is not a well defined

### problem,

in contrast to the

of rational

### points.

Bi-rational transformations between different models of the same curve do

the

of rational

but not that of

### point,

in an

This research was supported by the Netherlands Mathematical Research Foundation SWON with financial aid from the Netherlands Organization for Scientific Research NWO.

(3)

essential way. Therefore one should

of

solutions to

rather than of

on

### elliptic

curves.

When Stroeker and I worked on

we also

### (as

kind of a standard

when

with an

the Weierstrass

of the

curve it

### being

We did a limited search for

### solutions,

and found that there is a

but of course

### finite,

number of them. Thus we found it a natural

solutions to

### (W),

and the re-marks in the

show that this

is

### entirely

different from that of

### (Q).

This paper solves this

### problem,

and in-deed a bit more, since it turned out to be not too much additional work to

determine the

### complete

set of rational solutions to

with

### only

powers of 2 in the

i.e. the

### S-integral

solutions for the

2.

For

the

### integral

solutions the method of

also

is

### efficient,

if a basis for the free

### part

of the Mordell-Weil group is known.

a

method for this

lat-ter

has not

been found.

a

of this

### method,

that would be needed for the

is not

### available, mainly

due to the fact that there is not

a

### fully explicit theory

of linear forms in

### logarithms, analogous

to the excellent work of S. David

### [D].

For the rank 1 case however this has

been done

### [RU].

When a lower bound for

can be calculated and the Mordell-Weil basis is

### known,

then it is known how to

see Smart

and

Peth6 and Zimmer

### [GPZ2].

As we have a rank 2 curve

for the

### S-integral

solutions we have

to

### apply

more or less classical ideas. One method would be to translate the

### problem

into Thue- and Thue-Mahler

and

those with

such as in

An alternative

### approach

here is to use a new idea of Yuri Bilu

which uses

linear forms in

but

the Thue

### equations.

This last method is

### practical

in our case, due

to the nice factorization

of the cubic

in

### (W),

and it is this method we use in this paper. This seems to be the first time such a method is used for

### S-integral

solutions.

We did work out the details for the other methods

for the

### integral

solutions and Thue-Mahler

for the

but as this

routine

of the same

### results,

we omit details of these

(4)

As a

### general

conclusion of this paper we

### might

state that Bilu’s method can be

and

also

but

in

cases.

All the

### computations reported

in this paper have been done on a 486

We used our own

for the

### al-gebraic

number field data we used Pari 1.38. We now state our main result.

THEOREM 1. The

### only

rational solutions to the

### of

which the denominators

### of

x and y are powers

are the

43:

2. Preliminaries

We rewrite the

### problem

of rational solutions to

### (W)

with powers of 2 as denominators to the

### where x,

y E Z and k E We may assume that

since if is a solution with

then also

## (ix,ky,k -1)

is a solution.

The

hand side of

a square, thus

Let d be the

of x -

such that the

of d

the

of x -

### 22k+2.

Then d is also the

Of X2

(5)

for some u, v E

+

53 .

45 .

hence :J:d E

the

first

### equation

into the second we find

is of the

=

If d E

then this

has no

as is

### easily

shown.

If d 0 and k = 0 then

+

x

which

leaves the

solutions with x

### = -16, -14, -11, -2, 2, 4.

It follows that if k = 0 then we

may assume that x &#x3E;

+ = 13.

If d E

### {-6, -2,10, 30}

and k &#x3E; 1 then x -

= du2 and

### imply

that u is odd. It follows that

+

=

hence the left

hand side of

to 2. But

is

### odd,

which is

If d = -5 and k &#x3E; 1 then we find from

that

### 16).

Since -5 is not a

residue

this

and

### (1).

If d = 3 and k &#x3E; 1 then

hence

= 3w. Then

to

+

=

### (1)

we infer that u is

and then

we see that 1 -

which is

### Finally,

if d = 1 then we derive from

### (2)

that

where a, b E Z with 0 a

and 0 b 3. This

is

This

### immediately

shows that I a 4k + 2.

If

### 2 a

41~ + 1 then u is even, hence

If a = 1 then

which

and

thus k = 0.

the

### only

solution in this case is

=

### (3, 3),

to the solution with x = 13. Hence we may assume that a = 41~ +

### 2,

and

thus

If b = 0 then U2 =- 2

which is

If b = 2 then

so

=

Then we obtain

=

which is

If b = 1 then

so we

=

to

(6)

### integer a

&#x3E; 0 such that t = :1:1 +

This

which is

seen to have

### only

one solution: a = =

to t =

so

that

=

to x =

### 97/22.

If b = 3 then we have

22~ -

### 1),

and it follows that +

### By u

&#x3E; 0 there is an

### integer

a &#x3E; 0 such that u = :1:1 +

and this

## a~,

which is

seen to

have

### only

three solutions: a =

= 0 and a =

=

both

### (7, 53),

and

to the solution with x =

and a =

=

### 863),

to the solution with x =

This

### completes

the treatment of the case d = 1.

3. Bilu’s idea

We now have the

situation:

for some u, v E

### Z&#x3E;o,

and with either k =

d E

or

k &#x3E;

d

+

x

We factor

over

a root of

=

### 6,

and one of its factors can be written as

where a is

d is the

### part

of and a is determined

up to squares of the fundamental unit 5 +

Thus there are

a few

### possibilities

for a, and we have to find them all.

In

we denote

a

We have the

with

and

non-associated

Let 7r be a

### dividing

a. If it divides also a, then it divides

- 2

k

so 7r = 2

or 7r = 3

### + ç.

If 7r does not divide

then

= we see that

has the same

as

which is

since a is

### squarefree.

Thus divides d. Hence we

also that

for p, q,

E

with

Since d is the

of Na =

we find that

(7)

### (where

we sometimes take the freedom of

a

of 2

We

### get

rid of the cases a = 8 ~

a == 18 d=

= ~3 +

that the

obtained from

the coefficients

+

= A

in

is

For

### example,

in the case a = ~3 +

this

is

6AB +

which is

to

+

+

=

which is

modulo 5.

For a =

9 -

we take in

the

in

Then we have to

### (which

is allowed because it is a

and

the

a

their

and

### -~.

Thus we are left with

### only

four values for a,

+

### 4~,

and for each of them we have the case

as it

and the case

### of ~ replaced by -~.

Now we eliminate x from

and

and we find

the ±

### corresponding

to the cases as described in the

Bilu’s idea

now is that this

factors over the

field K =

satisfies

### 7P2

= da. Then from

the four different

### conjugates

of these factors we can eliminate the variables u E Z and

E

### ~(~),

and obtain a four-term

unit

with

4. The

### quartic

field for d = 3

Let 0 be a root of

= 0.

= 3 - 82

### ç2 =

6. The has discriminant 27648 = 210

basis

### 83~,

trivial class group, Galois

group

(8)

and

=

### 5+2~

is the fundamental unit of the

subfield

We have the

relevant

into

### primes:

There is a nontrivial

Q of

defined

### by

u0 = -0. I1

acts as follows on the numbers defined above:

From

with

in the

hand side we

and k =

### 0,

that

and hence we find for its two factors

for some

E e Z . On

### multiplying

it follows at once that 2c =

which is

From

with 1

in the

hand side we

and I~ =

that

and hence

for some

### a, b, c, d E

Z. It follows at once that a =

b =

or

### (0, 1) , n

= -m - 3. Because of

we may

without loss of

### generality

one of the cases for

So we take c =

d =

### 0,

and hence we obtain

=

=

and

(9)

5. The

field for d = 10

Let 0 be

0.

-89 +

-4 +

= l0a =

### ~), ~2

= 6. The field K =

### Q(0)

has

discriminant 92160 =

basis

### {1, B, B2, 83~,

trivial class

group, Galois group

### D4,

a set of fundamental units is

and q = 5 +

### 2~

is the fundamental unit of the

subfield

We

have the

relevant

into

### primes:

There is a nontrivial

Q of

defined

### by

u0 = -0. It

acts as follows on the numbers defined above:

From

with

in the

hand side we

and k =

that

and hence

for some a,

c,

e E

### 7~ &#x3E;_ o, l ,

m, n E Z . It follows at once that 2d =

which

is

with 1

in the

hand side we

and k =

### 0,

that

and hence we find

for some

E

### 7~ ~ o, l , m, n

E Z. Now it follows at once that a =

or

or

d =

e =

(10)

loss of

one of the cases

for

### ( b, c) .

So we take either b =

= 0 or b =

=

### 1,

and hence we obtain

11 =

= -710 - 820 +

or

12 =

= 970 - 3420 -

### + 23483.

Notice that we intend to show

that there are

the

solutions to

### (10):

Solutions for y = ~y2 occur in

because al2 =

so if

is a

then so is

6. The

### quartic

field for d = 30

Let 0 be a root of

+ 30 = 0.

-120

=

= 30a =

+

### 6), ç2

= 6. The field K =

=

### Q (0)

has discriminant 276480 =

basis

### 83~,

trivial class

group, Galois group

### D4,

a set of fundamental units is

and

= 5 +

### 2~

is the fundamental unit of the

subfield

We have the

relevant

into

### primes:

There is a nontrivial

Q of

defined

### by

QB = -B. It

acts as follows on the numbers defined above:

From

with

in the

hand side we

and k =

(11)

and hence

for some

### a, b, c, d

E E Z. It follows at once that 2c =

which

is

From

with 1

in the

hand side we

and k =

### 0,

that

and hence we find

for some a,

### E z~o, 1,

m, n 6 Z. It follows

at once that a =

b =

c =

### = 1,

m = 0. Hence we obtain

=

= 690 +

### 1629 -12092 - 6(J3.

Notice that we intend to

show that there are

the

solutions to

### (12):

Solutions occur in

because ay =

so if

is a

then so

7. The

### quartic

field for d = -15

Let 0 be a root of

+ 4x - 2 = 0.

= 2

=

+

= 6. The

field K =

### Q(1b) = Q(0)

has discriminant -8640 =

basis

### 82, 83~,

trivial class group, Galois group

### D4,

a set of fundamental units is

and

= 5 +

### 2~

is the fundamental unit of the

subfield

### Q(~).

(12)

There is a nontrivial

of

defined

### by

u0 =1- 0. It

acts as follows on the numbers defined above:

From

with

in the

hand side we

### (6),

that

and hence we find

for some

### a, b, c, d, e

E E Z. It follows at once that 2d =

which

is

From

with 1

in the

hand side we

### (6),

that

and hence we find

for some a,

c,

### d,

e m, n E 7~ . It follows at once that

e =

### 1,

m = -k + 1. Because of

### symmetry

we may assume a b. Note that

and that u is odd

of

and

### (3)),

and a + b &#x3E; 6

k &#x3E;

that

=

### 2,

it follows that a =

### 2,

b = 4k. Hence we obtain

with 7 =

-81 + 1200 -

and 7r =

-2 +

### satisfying

xux = 4. Notice that we intend to show that there are

### only

the

(13)

8. Linear forms in real

In the cases d =

we need

’real’

We treat

equa-tions

and

### (12)

as follows. Write them as

=

= m if d =

=

= n if d

We

the

a to

E

E

so that and =

=

On further

that

=

### o~~

=

where s = 1 if d =

### 3,30

and s = -1 if d =

and

=

u7, we obtain

to

that I

is

### always

odd in the case d = 10 and

s =1 in the cases d =

so that

### sj+h

= s, we find

and thus we have eliminated the unknown

### ¡3.

Now we notice that the

field K is

### totally

real. We choose two

of K into

### denoting

elements in one of these

without

### primes,

and in the other one with

### primes,

such that 0 8 01. Note that

### by u : 0

- -8 this determines all four

and also note

that now an

of

### Q(~)

into R has been fixed.

We thus have two different manifestations in R of

one written

as

### (17),

and the other one

### being

Hence now we can eliminate the unknown u from

and

and thus

### get

For convenience we write this four term unit

as

= III

Now an

### important

observation is that I x Il =

and III x IV =

### sy’6’

are constant. Bilu’s

### original

idea is to solve I from the

+

### + lð

= 0. We use a somewhat

and work

with

of the

one can be

in absolute

### value,

and the same holds for the

### III,

IV. So we either have two of the four terms

and two

### small,

in which case we can

the

### theory

of linear forms in

### logarithms,

or all four terms are

### bounded,

in which case we can

### easily

determine the solutions

### by

hand.

(14)

In each of the cases we look

at the

of

### I, II, III, IV,

and thus determine which one is the

### largest

in absolute value. We find:

In each case we

### distinguish

four subcases:

Note that half of the subcases lead to

e.g. if d =

then

so then

### only

subcases 3 and 4 occur. Put cl =

+

### 1,’b’l.

In subcase 1 we now have

Then we find

### Analogously,

in subcase 2 we find

And in subcase 3 we find

### Finally,

in subcase 4 we find

(15)

The left hand side of each of these

is of the form

### ~ ~ ~ ,

where A is a linear form in

of

where

=

+

=

+

Put N =

The theorem of

### implies

the existence of an absolute

constant C such that

We will show that

this with the

from

-

### (27)

leads to an absolute upper bound for N. For

if d =

subcase

### 1,

then I is even, and we have

&#x3E;

### IIIII

&#x3E; From these

### inequalities

it follows at once that there are

### only finitely

many solutions with 1 &#x3E;

and if 1 G

then N =

and

0.5. This

last

so

and

&#x3E;

and N = -l we

constants

such that

In

for each of the

we find

and

for

that

with c3 =

Hence we obtain

which at once

### implies

an absolute upper bound for N.

A similar

### procedure

works in all cases. We

### give

some details for each case below. We

### give

in each case a condition to ensure that the

hand side of the

from

is

which fails for

many

determined

### 1, h.

Sometimes we have to make a further distinction between N =

and N =

### lhl.

In the column marked

### ’inequ.’

we indicate which

### inequality

we used to derive

&#x3E;

for X =

Y =

### III,

IV. The constant C is

from

(16)

field of

8. The constant

### No

is the upper bound for N

### following

from

(17)

9. Real reduction

To reduce the upper bound

for N we use

as in

### previous

sections. Let us write the linear form A as A =

+ +

We take C =

which is a bit

than

### N02.

We introduce the lattice F =

and the

y

Here

stands for

to an

### integer.

Then we define A E Z

We

to sufficient

To 10 decimal

### places they

are:

Note that for the 16 different linear forms there are

### only

6 different lattices.

the Euclidean

we can

### compute

a lower bound for the distance

### y)

from y to the nearest lattice

### point

in F. Then we have

notice that

~ , /

If

is

this

an

### explicit

lower bound for which

(18)

The details for each case are as follows.

I , , I ,,- I - Y ,

we checked

### directly

for all N 30. This revealed

the known

listed

in

sections. This

### com-pletes

the treatment of the cases d =

### 3, 10, 30.

10. Linear forms in

### logarithms

In the case d = -15 we need almost

We start

with

to the

### beginning

of our treatment in the

as follows. We

the

Q to

E

E

so that Qu =

On further

that

a7r =

=

and

= Q7, we obtain

to

### (33),

we find

and thus we have eliminated the unknown

At this

### point

the main difference with the

sections becomes

### apparent,

since now, due to the fact that 7r is not a

the

### product

of the

two factors in the

hand side of

is not

### constant,

but is a constant

times

The idea now is

this is

### large

in the archimedean

sense, it is small in the 2-adic sense, so that we can

### try

to transform the

(19)

It is not difficult to show that the Galois closure of the

### quartic

field K can be embedded into

### Q2(~),

so that we can embed K into the field

### Q2(~)

in two different ways

is defined

=

### 6).

We

denote elements in one of these

without

### primes,

and in the other one with

### primes.

Then we have two different manifestations in

of

one written

as

### (34),

and the other one

### Again

we can eliminate the unknown u from

and

and thus

### get

a four term S-unit

I + II = III +

use the notation

-We choose

as follows:

Then the

is true:

And if x E

### Q2(~)

is written as x = a +

for

then x’ = a -

so that x - x’ =

and

=

With x =

= IV

it follows that

=

2k

hence

where

## - 2 -

### - 2 .

Notice that here we have a linear form in 2-adic

in

later on.

Now we

the

### theory

of linear forms in

of

numbers. In

.. I. 1 1

(20)

omit

### details,

but note that we used the result Yu mentions in his Section

on page 242 of

### [Y]).

To obtain an upper bound for the variables k

from

and

we also need to

the

of K. So now we

as an

in

where we choose

as follows:

So we have

E

and

E C are

=

= 1 we find

since

=

This

38.618’

### 2k,

and we obtain the

From

= 0.93 ... ,

= 2.17...

= 7.39 ... we

### immedi-ately

obtain that if k &#x3E; 6 then -k n

hence

= k.

and

thus

### immediately imply

11.

reduction

In this final section we reduce the bound

use of

to

Put A’ =

### 1092 1/11"

If k &#x3E; 1 then

### implies

Notice that A’ = + k

+

where we

So if we

A =

= +

+ n then it

that

are in

and are

in fact

(21)

For a

M, let

be the

rational

0 2~ and

### 0(p)) ~!

/I- Consider the lattice r = E

and the

y

### given by

Then it follows that if and

### only

if

for a z E Z. So the distance

### y)

from y to the lattice r can be at most

which is bounded

On the other

for

### each p

this distance can be

the Euclidean

In

### fact,

we can reach a contradiction when we choose p so that 2~‘ is of the

### magnitude

of the square of the upper bound.

We took p =

and

&#x3E; 2.0409 x

### (39),

and it follows that 1 =

which with

a reduced upper

k 102.

Next we

=

and

&#x3E;

### 153.05,

which

contradicts k 102. Hence the upper bound reduces further to k 9.

### Finally

it’s trivial to determine the solutions with k

and our

is

REFERENCES

Yu.

the method of

Univ. Bordeaux 2

### [BH]

Yu. BILU AND G. HANROT,

equa-tions

Baker’s

to appear.

A. BAKER AND G.

forms and group

### vari-eties",

J. reine angew. Math. 442

19-62.

### [D]

S. DAVID, Minorations de

linéaires de

### logarithmes elliptiques,

Mém. Soc. Math. de

Num. 62

J.

### GEBEL,

A. PETHÖ AND H.G. ZIMMER,

on

Acta Arith. 68

171-192.

J.

### GEBEL,

A. PETHÖ AND H.G.

on

in: H. COHEN

Number

Lecture Notes in

Science VOl.

1122,

Berlin

pp. 157-171.

G. REMOND AND F.

de loga-rithmes

J. Number Th. 57

133-169.

(22)

N.P.

on

### elliptic

curves" , Math. Proc.

Phil. Soc. 116

391-399.

### [ST]

R.J. STROEKER AND N. TZANAKIS,

equa-tions

linear forms in

Acta Arith. 67

177-196.

### [SW1]

R.J. STROEKER AND B.M.M. DE

"On a

Proc.

Math. Soc. 39

97-115.

N.

lin-ear forms in

The case of

Acta Arith. 75

165-190.

### [TW1]

N. TZANAKIS AND B.M.M. DE WEGER, "On the

### practical

solution of the Thue

J. Number Th. 31

99-132.

### [TW2]

N. TZANAKIS AND B.M.M. DE

"How to

### explicitly

solve a Thue-Mahler

Math. 84

223-288.

K.R.

"Linear forms in

Math. 91

241-276.

M.M. DE WEGER

30

2924 XN

### Krimpen

aan den IJssel The Netherlands

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