J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
B
ENJAMIN
M. M.
DE
W
EGER
Sintegral solutions to a Weierstrass equation
Journal de Théorie des Nombres de Bordeaux, tome
9, n
o_{2 (1997),}
p. 281301
<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
Sintegral
solutions
to aWeierstrass
equation
par BENJAMIN M.M. DE WEGER
RÉSUMÉ. On détermine les solutions rationnelles de
_{l’équation}
diophantienne y2
= x3  228x + 848 dont les dénominateurs sont des_{puissances }
de 2. On_{applique }
une idée de YuriBilu, qui
évite le recours à deséquations
de Thue et deThueMahler,
et_{qui}
permet d’obtenir des
_{équations }
aux(S)
unités à _{quatre }termes dotées de_{propriétés spéciales, }
_{que }l’on résout _{par }la théorie des formes linéaires en_{logarithmes }
réels et_{padiques.}
ABSTRACT. The rational solutions with as denominators _{powers} of 2 to the
_{elliptic diophantine }
_{equation y2 }
= x3  228x + 848 are determined. An idea of Yuri Bilu isapplied,
which avoids Thue and ThueMahler_{equations, }
and deduces fourterm_{(S)}
unit_{equations }
with_{special properties, }
that are solved_{by }
linear forms in real and_{padic }
_{logarithms.}
1. Introduction
In a recent _{paper }
[SWI] ,
_{my }colleague
R.J. Stroeker and I determined thecomplete
set of solutions in rational_{integers }
to the_{diophantine }
_{equation}
The
_{quartic diophantine equation }
_{(Q) }
can be seen as a model of anelliptic
curve. In our solution methods we did not use that_{viewpoint, }
but rather worked_{algebraically. Recently }
N. Tzanakis_{[T] }
has shown how the_{elliptic}
logarithms
method of Stroeker and Tzanakis_{[ST], }
_{originally }
_{designed }
for Weierstrass models of_{elliptic }
curvesonly,
can beadapted
to the situation of aquartic
model for anelliptic
_{curve, }and in fact he choseequation
(Q)
as one of theexamples
to illustrate his ideas.We stress that the
_{problem }
of_{integral }
_{points }
onelliptic
curves is not a well definedproblem,
in contrast to the_{problem }
of rational_{points. }
Birational transformations between different models of the same curve do
respect
the_{concept }
of rational_{point, }
but not that of_{integral }
_{point, }
in anThis research was _{supported }_{by }the Netherlands Mathematical Research Foundation SWON with financial aid from the Netherlands _{Organization }for Scientific Research NWO.
essential _{way. }Therefore one should
speak
of_{integral }
solutions to_{elliptic}
equations,
rather than of_{integral }
_{points }
onelliptic
curves.When Stroeker and I worked on
_{equation }
(Q),
we alsocomputed
(as
kind of a standardprocedure
whendealing
with anelliptic
curve)
the Weierstrassequation
of the_{elliptic }
curve it_{represents, }
_{being}
We did a limited search for
_{integral }
_{solutions, }
and found that there is alarge,
but of coursefinite,
number of them. Thus we found it a naturalquestion
to ask for all the_{integral }
solutions to_{equation }
_{(W), }
and the remarks in the_{previous }
_{paragraph }
show that this_{problem }
is_{entirely }
different from that of_{solving equation }
_{(Q). }
This paper solves thisproblem,
and indeed a bit _{more, }since it turned out to be not too much additional work todetermine the
_{complete }
set of rational solutions to_{(W) }
with_{only }
powers of 2 in the_{denominators, }
i.e. the_{Sintegral }
solutions for the_{prime }
2.For
_{determining }
the_{integral }
solutions the method of_{elliptic }
_{logarithms}
[ST] (see
also_{[GPZ1], [S]) }
is_{efficient, }
if a basis for the free_{part }
of the MordellWeil group is known.However,
ageneral
method for thislatter
_{problem }
has not_{yet }
been found._{Moreover, }
a_{padic }
_{analogue }
of thismethod,
that would be needed for the_{Sinteger }
_{solutions, }
is not_{yet }
_{fully}
available, mainly
due to the fact that there is not_{yet }
a_{fully explicit theory}
of linear forms in_{padic elliptic }
_{logarithms, analogous }
to the excellent work of S. David_{[D]. }
For the rank 1 case however this has_{recently }
been done[RU].
When a lower bound forpadic elliptic logarithms
can be calculated and the MordellWeil basis is_{known, }
then it is known how to_{proceed, }
see Smart_{[S], }
and_{Gebel, }
Peth6 and Zimmer_{[GPZ2].}
As we have a rank 2 curve
here,
for theSintegral
solutions we haveto
_{apply }
more or less classical ideas. One method would be to translate the_{problem }
into Thue and ThueMahler_{equations, }
and_{treating }
those with_{diophantine approximation methods, }
such as in[TWI],
[TW2].
An alternative_{approach }
here is to use a new idea of Yuri Bilu[B],
[BH],
which uses_{’ordinary’ }
(nonelliptic)
linear forms inlogarithms,
butbypasses
the Thue_{(Mahler) }
_{equations. }
This last method is_{practical }
in our _{case, }dueto the nice factorization
_{over Q }
of the cubic_{polynomial }
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_{Sintegral }
solutions.We did work out the details for the other methods
_{(elliptic }
_{logarithms}
for the_{integral }
solutions and ThueMahler_{equations }
for the_{Sintegral}
solutions)
too,
but as thisyields only
routineproofs
of the sameresults,
we omit details of these_{proofs.}
As a
general
conclusion of this _{paper }we_{might }
state that Bilu’s method can be_{practical }
and_{efficient, }
also_{padically, }
but_{only }
in_{special }
cases.All the
_{computations reported }
in this _{paper }have been done on a 486personal
computer.
We used our ownmultiprecision
_{routines; }
for thealgebraic
number field data we used Pari 1.38. We now state our main result.THEOREM 1. The
_{only }
rational solutions to the_{Weierstrassequation}
of
which the denominators_{of }
x and _{y are }_{powers }of 2,
are thefollowing
43:2. Preliminaries
We rewrite the
_{problem }
of rational solutions to_{(W) }
with _{powers }of 2 as denominators to the_{equation}
where x,
y E Z and k E We may assume thatsince if is a solution with
4 ~ x
_{1, }
then also(ix,ky,k 1)
is a solution.The
_{right }
hand side of_{(~V’) }
_{being }
a _{square, }thusbeing >
_{0, }
implies
Let d be the
_{squarefree }
_{part }
of x 22k+2
such that the_{sign }
of d_{equals }
thesign
of x 22k+2.
Then d is also the_{squarefree }
_{part }
Of X2_{f 22k+2x  53 · 24k+2~}
for some _{u, v }_{E }
_{Z>o. }
_{Further, }
22~+3)(x  22k+2) _ (x2
_{+ }2~+~ 
53 .24k+2) =
45 ._{22k+2, }
hence :J:d E_{f 1, }
_{2, 3, 5, 6,10,15, }
_{30~. }
_{Substituting }
thefirst
_{equation }
into the second we findEquation
(2)
is of the_{type X2  6Y’ }
=dZ2.
If d E
_{f 30, 10, 3, 1,2,5,6,15}, }
then this_{equation }
has nosolutions,
as is
_{easily }
shown.If d 0 and k = 0 then
(2
_{+ }6B/6)
x4,
whichonly
leaves thesolutions with x
_{= 16, 14, 11, 2, 2, 4. }
It follows that if k = 0 then wemay assume that x >
f  2
+ = 13.If d E
_{{6, 2,10, 30} }
and k > 1 then x 22k+2
= du2 and(1)
imply
that u is odd. It follows that
_{ord2(3  }
22k+1
+_{dU2) }
=1,
hence the lefthand side of
_{(2) }
has 2adic order_{equal }
to 2. But_{ord2(dV2) }
is_{odd, }
which is_{contradictory.}
If d = 5 and k > 1 then we find from
(2)
that25u’ = 5v2
(mod
16).
Since 5 is not aquadratic
residue(mod 16),
thisimplies
2/u
and4~v,
leading
to a contradiction with(1).
If d = 3 and k > 1 then
clearly
3~v,
hence_{put v }
= 3w. Then(2)
leadsto
_{(2 2k+l }
_{+ }_{u2)2  6(22k+1)2 }
=3w2.
_{By }
(1)
_{we }infer that u isodd,
and thenwe see that 1 
3w2
_{(mod }
_{8), }
which is_{impossible.}
Finally,
if d = 1 then we derive from(2)
thatwhere _{a, }b E Z with 0 a
4k + 3
and 0 b 3. This_{system }
is_{equivalent}
This
_{immediately }
shows that I a 4k + 2.If
_{2 a }
41~ + 1 then u is even, hence4 ~ x,
which contradicts(1).
If a = 1 then0 v = 3b  24k+133b,
whichimplies
24k+1
32b3 C 27,
andthus k = 0.
_{Clearly }
theonly
solution in this _{case }is(u , v)
=(3, 3),
leading
to the solution with x = 13. Hence _{we }_{may }assume that a = 41~ +
2,
andthus
If b = 0 then U2 = 2
(mod 3),
which isimpossible.
If b = 2 then
_{3~u, }
so_{put u }
=_{3t, }
Then _{we }obtain2 2k+l + 3t2
=3.2 4k+l + 1,
which is
_{impossible }
_{(mod }
_{3).}
If b = 1 then
again
3~u,
so we_{put u }
=3t,
leading
to3(t2 
1) =
integer a
> 0 such that t = :1:1 +a22k.
Thisimplies
22k1=
3a~ «~,
which is_{easily }
seen to haveonly
one solution: a = =1,
leading
to t =_{3, }
_{so}that
_{(u, v) }
=(9, 87),
leading
to x =97/22.
If b = 3 then we have
U2  1 = 3 
2 2k+l (9
22~ _{1), }
and it follows that +_{1)(u  }
_{1). }
By u
> 0 there is an_{integer }
a > 0 such that u = :1:1 +a2~~,
and this_{implies }
22k1 =
a~,
which is_{easily }
seen tohave
_{only }
three solutions: a =6, k
= 0 and a =8, k
=_{0, }
bothimplying
(u, v) =
(7, 53),
and_{leading }
to the solution with x =53,
and a =7, k
=_{1,}
implying
(u,
v) = (29,
863),
leading
to the solution with x =857/22.
This
_{completes }
the treatment of the case d = 1.3. Bilu’s idea
We now have the
following
situation:for some _{u, v E }
_{Z>o, }
and with either k =_{0, }
d _{E }~3,10, 30~, x >
13,
_{or}k >
_{1, }
d_{= 15, (2 }
+6~)22~
x2 2k+2.
We factor_{equation }
_{(4) }
overQ(~)
for ~
a root ofx2
=6,
and one of its factors can be written aswhere a is
squarefree,
d is the_{squarefree }
_{part }
of and a is determinedup to squares of the fundamental unit 5 +
2~.
Thus there areonly
a fewpossibilities
for _{a, }and we have to find them all.In
_{Q(~) }
we denote_{conjugation by }
a_{bar, }
so Z =
_{~. }
We have the_{following}
decompositions:
with
_{2 + ~, 3 + ~, 1 + ~ }
and_{1  ç being }
nonassociated_{primes.}
Let 7r be a_{prime }
_{dividing }
a. If it divides also _{a, }then it divides 2
k
2+,
so 7r = 2+ g
_{or 7r }= 3+ ç.
If _{7r }does _{not }divide_{a, }
thenby
dV2
= _{we see }_{that }ord1r(d)
has the same_{parity }
asord1r(a),
which is_{1, }
since a is_{squarefree. }
Thus divides d. Hence wefind, using
also thata > 0, that
for _{p, q, }
_{r, s, t }
E_{{O, 1}, }
with_{(s, t) ~ ( 1,1 ~. }
Since d is the_{squarefree }
_{part }
of Na =2q3r( 5)s+t,
_{we }_{find that }possibilities
(where
we sometimes take the freedom of_{adding }
a_{multiple }
of 2_{to p):}
We
_{get }
rid of the cases a = 8 ~3~,
_{a }== 18 d=7 ç, a
= ~3 _{+ }2~,
by noting
that the
_{quadratic equation }
obtained from_{comparing }
the coefficients_{of ~}
in x+2 2k+l
+ 22k+13ç =
+_{B~)2 }
_{(from }
inserting
~3
= A_{+ B~ }
in_{(5)) }
isimpossible
(mod
5).
For_{example, }
in the case a = ~3 +_{2~ }
thisequation
is2A2 +
6AB +12B2 =
_{22k+13, }
which is_{equivalent }
to_{(2A }
+_{3B)2 }
+15B2
=22k+23,
which is_{impossible }
modulo 5.For a =
3  ~, 4 + ~, 6  ~,
9 4~
we take in(5)
theconjugate
inQ(~).
Then we have to_{replace ~3 }
by #
_{(which }
is allowed because it is avariable),
and_{replace }
the_{parameters }
aand ~ by
theirconjugates Q
and~.
Thus we are left with
only
four values for _{a, }namely a = 3 + ç, 4  ç, 6 +
~, 9
+4~,
and for each of them we have the caseof ~ remaining
as it_{is, }
and the caseof ~ replaced by ~.
Now we eliminate x from
(3)
and(5),
and we findthe ±
_{corresponding }
to the cases as described in thepreceding paragraph.
Bilu’s idea_{(cf. }
_{[B], [BH]) }
now is that thisequation,
multiplied by d,
factors over thequartic
field K =~(~),
where 1b
satisfies7P2
= da. Then fromthe four different
_{conjugates }
of these factors we can eliminate the variables u E Z and_{(3 }
E_{~(~), }
and obtain a fourterm(S)
unitequation
withspecial
properties.
4. The
_{quartic }
field for d = 3Let 0 be a root of
x4  6x2 + 3
= 0.Then 0 =
6(} + (}3, ç
= 3  82satisfy
~2 = 3a = 3(3 
~),
ç2 =
6. The has discriminant 27648 = 21033,
integral
basisf 1,
0,
(}2,
83~,
trivial class _{group, }Galoisgroup
and
_{r¡lX1 }
=5+2~
is the fundamental unit of thequadratic
subfieldQ(~).
We have the
_{following }
relevant_{decompositions }
into_{primes:}
There is a nontrivial
Q(~)automorphism
Q ofK,
definedby
u0 = 0. I1acts as follows on the numbers defined above:
From
_{(7) }
with_{1  ~ }
in the_{right }
hand side wefind, by
(6)
and k =0,
that
and hence we find for its two factors
for some
_{a, b, c }
E e Z . On_{multiplying }
it follows at once that 2c =1,
which isimpossible.
From
_{(7) }
with 1_{+ ~ }
in the_{right }
hand side we_{find, by }
(6)
and I~ =0,
that
and hence
for some
_{a, b, c, d E }
Z. It follows at once that a =2,
b =4,
(c, d) = ( 1, 0)
or(0, 1) , n
= m  3. Because of_{symmetry }
we _{may }disregard
without loss of_{generality }
one of the cases for(c, d).
So we take c =l,
d =0,
and hence we obtain
wherey
=p2q4r,X3
=48I9B968239B3,
and77/X
= 129~4B2~283.
to(8):
5. The
_{quartic }
field for d = 10Let 0 be
_{a root of x4  8x2 f 10 = }
0._{Then 0= }
89 +83, ~ _
4 +BZ
satisfy
1/;2
= l0a =10(4 
~), ~2
= 6. The field K =Q(1b) =
Q(0)
hasdiscriminant 92160 =
211325,
integral
basis{1, B, B2, 83~,
trivial classgroup, Galois _{group }
_{D4, }
a set of fundamental units isand _{q }= 5 _{+ }
2~
is the fundamental unit of thequadratic
subfieldQ(~).
Wehave the
_{following }
relevant_{decompositions }
into_{primes:}
There is a nontrivial
Q(~)automorphism
Q ofK,
definedby
u0 = 0. Itacts as follows on the numbers defined above:
From
_{(7) }
with_{1  ç }
in the_{right }
hand side wefind, by
(6)
and k =0,
that
and hence
for some _{a, }
_{b, }
_{c, }_{d, }
e E_{7~ >_ o, l , }
m, n E Z . It follows at once that 2d =1,
whichis
_{impossible.}
From (7)
with 1_{+ ~ }
in the_{right }
hand side we_{find, by }
_{(6) }
and k =0,
that
and hence we find
again
(9)
for some_{a, b, c, d, e }
_{E }_{7~ ~ o, l , m, n }
_{E }Z. Now it follows at once that a =_{4, }
(b, c) = (2, 0)
or(1, 1)
or(o, 2),
d =_{1, }
_{e = }_{1,}
loss of
_{generality }
one of the cases(2, 0), (0, 2)
for( b, c) .
So we take either b =_{2, c }
= 0 _{or }b =1, c
=1,
and hence we obtainwhere 1
equals
11 =P4 q2rl r2
= 710  820 +31082 
6683
_{or }12 =
p4qlq2rlr2
= 970  3420 61082
+ 23483.
Notice that _{we }intend to showthat there are
only
thefollowing
solutions to_{(10):}
Solutions for y = _{~y2 }occur in
_{pairs, }
because _{al2 }=12f3X3,
so if(1, n)
is asolution,
then so is(3  1, 3  n).
6. The
_{quartic }
field for d = 30Let 0 be a root of
X4  12X2
+ 30 = 0.Then 0=
120_{+ (}3, ç }
=6  BZ
satisfy
1/;2
= 30a =30(6
+6), ç2
= 6. The field K =Q(1b)
=Q (0)
has discriminant 276480 =
211335,
integral
basis{I,
0,
02,
83~,
trivial classgroup, Galois group
D4,
a set of fundamental units isand
_{1]1 }
= 5 _{+ }2~
is the fundamental unit of thequadratic
subfieldQ(~).
We have the
_{following }
relevant_{decompositions }
into_{primes:}
There is a nontrivial
Q(~)automorphism
Q ofK,
definedby
QB = B. Itacts as follows on the numbers defined above:
From
_{(7) }
with_{1  ~ }
in the_{right }
hand side wefind, by
(6)
and k =0,
and hence
for some
a, b, c, d
E E Z. It follows at once that 2c =_{1, }
whichis
_{impossible.}
From
_{(7) }
with 1_{+ ~ }
in the_{right }
hand side wefind, by
(6)
and k =0,
that
and hence we find
_{again }
(11)
for some _{a, }b,
c, d
E z~o, 1,
_{m, n }6 Z. It followsat once that a =
4,
b =4,
_{c }=l, d
= 1,
_{m }= 0. Hence _{we }obtainwhere /
=p4q4TIT2
= 690 +1629 12092  6(J3.
Notice that we intend toshow that there are
only
thefollowing
solutions to(12):
Solutions occur in
_{pairs, }
because _{ay }=’If.3,
so if(1,
n)
is asolution,
then sois (3  l, n).
7. The
_{quartic }
field for d = 15Let 0 be a root of
X4 
2X3  3x2
+ 4x  2 = 0.Then 0= 17+300+
1292 
_{803, ~ }
= 2+ {}  ()2 satisfy
1/;2
=15(9
_{+ }4~), ~2
= 6. Thefield K =
_{Q(1b) = Q(0) }
has discriminant 8640 =2~3~5,
integral
basis82, 83~,
trivial _{class group, Galois group }_{D4, }
a set of fundamental units isand
c’
= 5 _{+ }_{2~ }
is the fundamental unit of thequadratic
subfieldQ(~).
There is a nontrivial
Q(g)automorphism u
of_{K, }
defined_{by }
u0 =1 0. Itacts as follows on the numbers defined above:
From
_{(7) }
with_{1  ~ }
in the_{right }
hand side wefind, by
(6),
thatand hence we find
for some
_{a, b, c, d, e }
E E Z. It follows at once that 2d =1,
whichis
_{impossible.}
From
_{(7) }
with 1_{+ ~ }
in the_{right }
hand side wefind, by
(6),
thatand hence we find
_{again }
(13)
for some _{a, }b,
_{c, }d,
e _{m, n E }7~ . It follows at once that_{a + b = 4k + 2, }
_{c = 4, d = 1, }
e =_{1, }
m = k + 1. Because ofsymmetry
we _{may }assume a b. Note thatand that u is odd
_{(because }
of_{(1) }
and_{(3)), }
and a + b > 6_{(because }
k >_{1).}
Comparing pladic values,
noting
that_{ordpl (30) }
=2,
it follows that a =2,
b = 4k. Hence _{we }obtainwith _{7 }=
p2 q4rlr2’6 =
81 + 1200 3982  8483,
and _{7r }=p2E 1=
2 _{+ }82,
satisfying
xux = 4. Notice that _{we }intend to show that there are_{only }
the8. Linear forms in real
_{logarithms}
In the cases d =
3,10, 30
we need_{only }
’real’_{arguments. }
We treat _{}equations
_{(8), (10) }
and_{(12) }
as follows. Write them aswhere (
=q/x, h
= _{m }if d =_{3, }
and
=x, h
= n if d_{=10, 30. }
We_{apply}
the_{Q(~)automorphism }
a to_{(15), }
_{using d, u }
E_{Z, # }
E_{Q(~), }
so that and =d, uu
=~~cr/? = ,Q.
On furthernoting
that1j;, a£.
=sE1,
o~~
=where s = 1 if d =
3,30
and _{s }= 1 if d =_{10, }
andwriting 6
=u7, we obtain
Adding
(15)
to_{(16), }
_{using }
that I+ h
is_{always }
odd in the case d = 10 ands =1 in the cases d =
3, 30,
so thatsj+h
_{= s, }we findand thus we have eliminated the unknown
¡3.
Now we notice that the
quartic
field K istotally
real. We choose twoembeddings
of K into_{R, }
_{denoting }
elements in one of theseembeddings
without_{primes, }
and in the other one with_{primes, }
such that 0 8 01. Note that_{by u : 0 }
 8 this determines all fourembeddings,
and also notethat now an
embedding
ofQ(~)
into R has been fixed.We thus have two different manifestations in R of
_{equation }
_{(17), }
one written_{again exactly }
as(17),
and the other one_{being}
Hence now we can eliminate the unknown u from
(17)
and(18),
and thusget
For convenience we write this four term unit
_{equation }
asI +II
= III+ IV.
Now an
important
observation is that I x Il =sy6
and III _{x }IV =sy’6’
are constant. Bilu’s
_{original }
idea is to solve I from the_{quadratic }
_{equation}
I2 
_{(III }
+_{7V)7 }
+ lð
= 0. We _{use a }somewhatsimpler idea,
and workdirectly
with_{(19). }
_{Namely, }
of the_{pair I, II only }
one can be_{large }
in absolute_{value, }
and the same holds for the_{pair }
_{III, }
IV. So we either have two of the four terms_{being large }
and two_{small, }
in which case we can_{apply}
the_{theory }
of linear forms in_{logarithms, }
or all four terms arebounded,
in which case we caneasily
determine the solutions_{by }
hand.In each of the cases we look
carefully
at the_{signs }
of_{I, II, III, IV, }
and thus determine which one is thelargest
in absolute value. We find:In each case we
distinguish
four subcases:Note that half of the subcases lead to
_{contradictions, }
_{e.g. if d }=3, 1 odd,
then
_{~7j~ > III, }
so thenonly
subcases 3 and 4 occur. Put cl =1761
_{+ }1,’b’l.
In subcase 1 we now haveThen we find
Analogously,
in subcase 2 we findAnd in subcase 3 we find
Finally,
in subcase 4 we findThe left hand side of each of these
_{inequalities }
_{(20)  (27) }
is of the formleA 
11 [
~ ~ ~ ,
where A is a linear form in_{logarithms }
of_{algebraic }
numbers,
namely
where
_{Al }
=llog
+hlog
1(/(’1,
A2
=llog
_{+ }hlog
1((’1.
Put N =The theorem of
_{[BW] }
_{implies }
the existence of an absoluteconstant C such that
We will show that
_{combining }
this with the_{appropriate equation }
from_{(lU)}

(27)
leads to an absolute _{upper }bound for N. Forexample,
if d =_{3,}
subcase
_{1, }
then I is _{even, }and we haveIII
>_{IIIII }
> From theseinequalities
it follows at once that there areonly finitely
_{many }solutions with 1 >_{4, }
and if 1 G_{5, }
then N =1,
andcIIII1IIIII1
0.5. Thislast
_{inequality implies }
_{1.39~  1) ~ }
so(20)
and(28)
imply
Using
J7Y7J
>_{JIVI }
and N = l we_{compute }
constants_{C2, C’3 }
such thatIn
_{general, }
for each of the_{inequalities }
_{(20)  (27) }
we findby
(29)
and(30),
for_{large enough }
_{N, }
thatwith c3 =
+ log(1.39ci).
Hence we obtainby
(28)
which at once
_{implies }
an absolute _{upper }bound for N.A similar
_{procedure }
works in all cases. Wegive
some details for each case below. We_{give }
in each case a condition to ensure that theright
hand side of the_{appropriate }
_{inequality }
from_{(20)  (27) }
is_{0.5, }
which fails for_{only}
finitely
manyeasily
determined1, h.
Sometimes we have to make a further distinction between N =III
and N =lhl.
In the column marked’inequ.’
we indicate which
inequality
we used to derive_{log I X Y }
_{I }
>c2N 
for X =I, II,
Y =III,
IV. The constant C iscomputed directly
from[BW],
field of
_{degree }
8. The constantNo
is the upper bound for Nfollowing
from9. Real reduction
To reduce the _{upper }bound
_{No }
for N we usecomputational diophantine
approximation techniques,
as in_{previous }
sections. Let us write the linear form A as A =_{00 }
_{+ } _{+ }_{h02  }
We take C =1044,
which is a bitlarger
thanN02.
We introduce the lattice F =EZ2 }
and thepoint
_{y }given by
Here
_{[.] }
stands for_{rounding }
to aninteger.
Then we define A E Z_{by}
We
_{computed 00, ~1~2 }
to sufficient_{precision. }
To 10 decimal_{places they}
are:Note that for the 16 different linear forms there are
only
6 different lattices.Using
the Euclidean_{algorithm }
we can_{easily }
_{compute }
a lower bound for the distance_{d(r, }
_{y) }
from _{y }to the nearest lattice_{point }
in F. Then we haveVd(f,y)2  NJ.
Further,
notice that~ , /
If
_{d(1’, y) }
is_{large enough, }
this_{gives }
anexplicit
lower bound for whichThe details for each case are as follows.
I , , I ,, I  Y ,
Finally
we checked_{equation }
(19)
directly
for all N 30. This revealedonly
the known_{solutions, }
listed_{already }
in_{previous }
sections. Thiscompletes
the treatment of the cases d =3, 10, 30.
10. Linear forms in
_{padic }
_{logarithms}
In the case d = 15 we need almost
_{only ’padic’ }
_{arguments. }
We startwith
_{equation }
_{(14), }
_{analogous }
to the_{beginning }
of our treatment in theprevious sections,
as follows. Weapply
theQ(~)automorphism
Q to_{(14),}
using u
E_{Z, 0 }
E_{Q(~), }
so that Qu =/3.
On further_{noting }
thatcro = 1b,
a7r =47r1,
=~1,
andwriting b
= _{Q7, }_{we }obtainAdding
(14)
to_{(33), }
we findand thus we have eliminated the unknown
_{/3.}
At this
_{point }
the main difference with the_{previous }
sections becomesapparent,
since _{now, }due to the fact that 7r is not aunit,
theproduct
of thetwo factors in the
_{right }
hand side of_{(34) }
is not_{constant, }
but is a constanttimes
4k .
The idea now is_{that, though }
this is_{large }
in the archimedeansense, it is small in the 2adic sense, so that we can
_{try }
to transform theIt is not difficult to show that the Galois closure of the
_{quartic }
field K can be embedded intoQ2(~),
so that we can embed K into the fieldQ2(~)
in two different _{ways }_{(remember }
_{that 6 }
is defined_{by 62 }
=6).
Wedenote elements in one of these
embeddings
without_{primes, }
and in the other one withprimes.
Then we have two different manifestations in_{Q2(~)}
of_{equation }
_{(34), }
one writtenagain exactly
as(34),
and the other one_{being}
Again
we can eliminate the unknown u from(34)
and(35),
and thus_{get }
a four term Sunit_{equation }
I + II = III +IV,
namely
For the 2adic number
_{~o+~i2+~22~+... we }
use the notation_{O.aoala2 }
We choose_{conjugates }
as follows:Then the
_{following }
is true:And if x E
_{Q2(~) }
is written as x = _{a }_{+ }b~
fora, b E Q2,
then x’ = a _{b~,}
so that x  x’ =
2b~,
andord2(x 
x’)
=ord2(b)
~ 2.
With x =II, x’
= IVit follows that
_{ord2(I  III) }
=ord2(IV  II) >
2k+ 2,
hencewhere
_{l~o }
_{ 2  }
_{ord(y’) }
_{ 2 . }
Notice that here we have a linear form in 2adic_{logarithms }
in_{disguise. }
It will be made_{explicit }
later on.Now we
apply
thetheory
of linear forms inpadic
logarithms
ofalgebraic
numbers. In_{fact, using}
.. I. 1 1
(we
omit_{details, }
but note that we used the result Yu mentions in his Section0.1,
on _{page }242 of[Y]).
To obtain an _{upper }bound for the variables k
and inj
from(37)
and(38),
we also need to_{study }
the_{complex embeddings }
of K. So now weinterpret
equation
(36)
as anequation
incomplex
numbers,
where we chooseconjugates
as follows:So we have
I, II
E_{R, }
and_{III, IV }
E C are_{complex conjugates. }
_{Using}
11["’1
=2,
= 1 we findsince
_{/6 }
=90(V6  1).
Thisinequality implies
3.37812~
38.618’2k,
and we obtain theinequality
From
_{log Jr /2 }
= 0.93 ... ,logq
= 2.17...log(y)
= 7.39 ... weimmediately
obtain that if k > 6 then k n_{0, }
hence_{max~k, }
= k.Inequalities
(37)
and_{(38) }
thus_{immediately imply}
11.
_{padic }
reductionIn this final section we reduce the bound
(39)
by making
use ofpadic
computational diophantine approximation techniques, applied
to_{inequality}
(37).
Put A’ =1092 1/11"
If k > 1 then(37)
implies
Notice that A’ = _{+ k }
log2
7r/7r’
+nlog2 T//T/’,
where wecomputed
So if we
_{put }
A =A’/ log2 7//7/
= _{+ }kol
_{+ n then }ithappens
that/>1
are in
Q2
and are_{integral, }
in factFor a
positive integer
_{M, }let0(14)
be the_{unique }
rational_{integer }
satisfying
0 2~ and
_{0(p)) ~! }
/I Consider the lattice r = _{E}Z2 }
and the_{point }
_{y }_{given by}
Then it follows that if and
_{only }
iffor a z E Z. So the distance
_{d(r, }
_{y) }
from _{y to }the lattice r can be at mostýk2 + n2,
which is bounded_{by }
_{(39). }
On the other_{hand, }
for_{each p }
this distance can be_{computed easily by }
the Euclidean_{algorithm. }
Infact,
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 }=
204,
andcomputed
k2 + n2 >
d(r,
y)
_{> } 2.0409 _{x}103°.
This contradicts_{(39), }
and it follows that 1 =_{203,}
which with
_{(41) }
_{yields }
a reduced _{upper }bound, namely
k 102.Next we
took u
=18,
andcomputed
d(r, y)
_{> }153.05,
whichcontradicts k 102. Hence the upper bound reduces further to k 9.
Finally
it’s trivial to determine the solutions with k_{9, }
and ourproof
is_{complete.}
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