S
AMIR
S
IKSEK
On the diophantine equation x
2= y
p+ 2
kz
pJournal de Théorie des Nombres de Bordeaux, tome 15, n
o3 (2003),
p. 839-846.
<http://www.numdam.org/item?id=JTNB_2003__15_3_839_0>
© Université Bordeaux 1, 2003, tous droits réservés.
L’accès aux archives de la revue « Journal de Théorie des Nombres de Bordeaux » (http://jtnb.cedram.org/) implique l’accord avec les condi-tions générales d’utilisation (http://www.numdam.org/legal.php). Toute uti-lisation commerciale ou impression systématique est constitutive d’une infraction pénale. Toute copie ou impression de ce fichier doit conte-nir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
On the
diophantine
equation x2
= yp + 2k zp
par SAMIR SIKSEK
RÉSUMÉ. Nous étudions
l’équation
du titre en utilisant une courbe deFrey,
le théorème de descente du niveau de Ribet et uneme-thode due a Darmon et Merel. Nous pouvons determiner toutes les solutions entibres x, y, z,
premières
deux àdeux,
si p ~ 7est
premier
et k ~ 2. Decela,
nous déduisons des résultatssur
quelques
cas de cetteéquation qui
ont ete étudiés dans lalitterature. En
particulier,
nous pouvons combiner notre résultatavec les résultats
précédents
de Arif et AbuMuriefah,
et avec ceuxde Cohn pour obtenir toutes les solutions de
l’équation
x2
+2k =
yn pour n ~
3.ABSTRACT. We attack the
equation
of the titleusing
aFrey
curve, Ribet’slevel-lowering
theorem and a method due to Darmon and Merel. We are able to determine all the solutions inpairwise
coprime integers
x, y, z ifp ~ 7
isprime
and 2. From this we deduce some results aboutspecial
cases of thisequation
thathave been studied in the literature. In
particular,
we are able to combine our result withprevious
results of Arif and AbuMuriefah,
and those of Cohn to obtain a
complete
solution for theequation
x2 + 2k = yn for n ~ 3.
1. Introduction
In
[8]
Darmon and Merelstudy
thediophantine equation
x2
=yp
+ z~along
with two other variants of Fermat’sequation.
In this paper we showthat the method of Darmon and Merel can be
adapted
togive
results about the moregeneral equation
(1)
x2 =
yP
+2kZp
where k is a
positive
integer,
and p aprime.
Studying
thisequation
unifiesthe
study
of two well-knownexponential
diophantine equations appearing
in the literature:
x2+2 k
=yn and x2-2 k =yn
(see ~1~,
[3],
[5],
[6], [10]).
We shall call an
integral
solution x, y, z ofequation
(1)
primitive
if x, y, z arepairwise coprime,
and non-trivial if 0. The reader is forewarnedManuscrit recu le 24 juin 2002.
This work is funded by a grant from Sultan Qaboos University, Oman
Theorem 1.
Suppose
k > 2 and that p > 7 isprime.
Then theonl y
non-trivial
primitive
solutionsof
equation
(1)
are k =3,
x =~3,
y = z =1 andp
arbitrary.
In Section 3 we will
give
theimplications
of this theorem to the twoaforementioned
exponential equations.
Inparticular,
we are able tocom-bine our theorem with earlier results of Arif and Abu
Muriefah,
and ofJ.H.E.
Cohn,
togive
acomplete
solution to theequation x2 + 2k
=Our
approach
inproving
Theorem 1 is to associate to eachputative
solu-tion ofequation
(1)
aFrey
curve andapply
Ribet’slevel-lowering
theorem to show that the Galoisrepresentation
on thep-torsion
isisomorphic
to arepresentation
of a much smaller level. This will be sufficient to eliminate all the cases of the theoremexcept
for k = 3 where we have to(slightly)
adapt
the method of Darmon and Merel. Since we are
following
well-troddenpaths
here,
ourexposition
will be rather terse.Apart
from[8]
the readermay wish to compare what follows with
[7]
and[12,
pages397-399].
It hasrecently
come to my attention that W. Ivorra[11]
provesstronger
results for
equations
(1)
and(7)
using
similar methods.2. Proof of Theorem 1
Assume the existence of a non-trivial
primitive
solution 2 andp > 7
prime.
We will show that this forces k = 3and y
= z = 1. It ishelpful
togive
acomplete
list ofassumptions
we make about x, y, z,k,
p(1)
Equation
(1)
is satisfied.(2)
k > 2.(3)
p > 7 isprime.
(4)
x, y, z arepairwise coprime, odd,
and non-zero.(5)
x - 3(mod 4).
There is no loss of
generality
inmaking
theseassumptions;
for the fourthstatement above we need to absorb any power of 2
dividing z
into the2k
inequation
(1).
For the fifth one we need toreplace x by -x
if necessary. 2.1. TheFrey
Curve. Denoteby
E the‘Frey
curve’In standard notation we
have,
/A-2 ~~.~~3
Let
Amin
be the minimal discriminant ofE,
and N be its conductor.Fi-nally,
if a is a non-zerointeger
then denoteby
rad(a)
theproduct
of distinctprimes
dividing
a.Lemma 1. The values
of
Amin
and N aregiven by
thefollowing
table:& , .
Proof.
It is easy to show that c4 and A are notsimultaneously
divisibleby
any oddprime.
We deduce thus that the curve is minimal and hasmulti-plicative
reduction at allprimes
dividing
yz and at no other oddprimes.
This shows that the table entries are correct modulo powers of 2. It
re-mains to
study
the reduction at 2. Recall thatx2
Thus wemay rewrite the
equation
for E asSuppose
first that k > 6.Replace
Xby
4X - x and Yby
8Y + 4X. Aftersimplifying
we reach the modelI . 1 B
This model is
integral
since we assumed that x = 3(mod 4).
It is minimal at 2 since the ’newc4’
is4x2 -
3y~
which is odd. Hence this model isglobal
minimal. The formulas forAmin
for the cases k = 6 and k > 6 arenow immediate on
noting
that thechange
of variable above forcesAmin
=2-12 Ll.
.If k = 6 then
clearly
2 does not divideAmin
and so it does not divide N. Thus suppose k > 7. The reduction modulo 2 of the model(4)
isIn either case the
singularity
at(0,0)
is a node. This shows that 2 dividesthe conductor ~V
precisely
once.For k =
2,3,4,5,
it is clear that the curve isalready
minimal at 2 since212
does not divide A. The rest now follows from Tate’salgorithm.
Forthe benefit of the reader who would like to
verify
the details wepoint
outthat it is best to make the
following
changes
of variable in the model(3)
2.2. The Galois
Representation.
Now letbe the Galois
representation
on thep-torsion
of E. To be able toapply
Ribet’s
level-lowering theorem,
andcomplete
ourproof,
we will need toshow that p is irreducible. For this we need two
lemmas,
the first of which iselementary.
Lemma 2. The
equation a2
= :1:1 :I: 2~ has no solutionsfor
m > 2except
for32=1+23.
_
Lemma 3. The
j-invariant of
E isnon-integral except i f
k = 3 and y = z=1.Proof. Suppose
that thej-invaxiant
is
integral.
It is easy to use this and theassumptions
listed at the outsetof the
proof
aboutx, y, x, k, p
to showthat y
== ±1 and z = ±1. Fromequation
(1)
we see thatx2
= +1 ±2k,
and this forces k = 3and y
= z = 1by
theprevious
Lemma. 0Lemma 4. p is irreducible.
Proof.
If we are in the case k = 3and y
= z = 1 then the curve E iscurve 32A4 in Cremona’s tables
[4]
and so we know that it does not haveisogenies
of odddegree.
Thus p is irreducible.Suppose
we are not in thiscase. We note the
following,
~ E has a
point
of order 2. This is clear from the model(2)
given
forE. ’
~ The
j-invariant
of E isnon-integral
(by
theprevious
lemma).
~ E has some oddprime
ofmultiplicative
reduction.Otherwise,
fromthe formulas for the conductor
given
in Lemma 1 we seethat y
=and z
= ::1:1,
which isimpossible
as above.These three
properties
areenough
to force theirreducibility
of p forp > 7;
for this see the
proof
of[8,
Theorem7].
02.3. Conclusion of the Proof. It has been shown that all
elliptic
curvesover Q
are modular[2].
However,
we note inpassing
that ourFrey
curveE has semistable reduction away from 3 and 5
(by
Lemma1),
and soits
modularity
follows from an earlier theorem of Diamond[9]
which is aLet
N(p)
be the Serre conductor of p(see
[15,
page180]
for thedef-inition).
It follows from theproperties
of the Serre conductor and ourLemma 1 that
For this see
[15,
page207]
or[7,
page264].
Applying
Ribet’slevel-lowering
theorem
[14]
we see that pcorresponds
to amod p
eigenform
ofweight
2and level
2,4,8
or 16 ifk ~
3 and of level 32 if k = 3. Since there are noweight
2 cusp forms of levels2, 4, 8,16
weimmediately
deduce that k = 3.We will henceforth assume that k =
3,
and tocomplete
ourproof
wemust show that y = z = 1. The rest of our
proof
will mimic theapproach
made
by
Darmon and Merel[8,
pages88-99]
for theequation
xP+ yP
=z2
(which
really corresponds
to the case k = 0 of ourequation
1).
We sawabove that p
corresponds
to a mod peigenform
ofweight
2 and level32,
and we know that ourFrey
curve E possesses apoint
of order 2. This isexactly
the same situation as in Darmon-Merel for their
equation
mentioned above.Their
arguments
show that if p - 1(mod 4)
then E haspotentially
good
reduction at all oddprimes
(see
theproof
of[8,
Proposition
4.2]).
As before this isjust saying that y
= tl and z = ~1. Weimmediately
seethat y
= 1and z = 1
(since
k =3).
Wemay thus assume that p - 3
(mod 4).
Proposition
4.3 of[8]
shows that theimage
of p isisomorphic
to the normalizer of thenon-split
Cartansubgroup.
Then Theorem 8.1 of[8]
shows that thej-invariant
of Ebelongs
toZ[~].
Moreover, p
does notdivide yz
by
the sameargument
as in theproof
of[8,
Corollary
4.4].
Hencewe know
that j
is inZ,
and from Lemma 3 above we knowthat y
= z = 1as
required.
2.4.
Exceptions
to Theorem1;
the cases p 7 and k = 1. Thereader will
probably
bewondering
where ourproof
fails for the cases p 7and k = 1.
For p
7 we are unable toprove
theirreducibility
of the Galoisrepresentation
p. For k = 1 the Serre conductorof p
is128,
and we knowthat the dimension of
SZ(ro(128))
is 9. So far as weknow,
there are asof
yet
no methodsdeveloped
to deal with cases where the dimension afterlevel-lowering
isgreater
than 1.3.
Special
Cases3.1. The
equation x2
+2 k
=yn.
In[5]
J.H.E. Cohnshows,
for kodd,
that the
equation
for a > 0.
The case with k even has
proved
to be more troublesome. In[1]
Arifand Abu Muriefah made a
plausible conjecture
as to the solutions with k = 2m. Cohn[6]
proves the
conjecture
for most values of m less than1000, including
all those less than 100.Equation
(5)
is of course aspecial
case of(1)
corresponding
to the case z = -1provided
that n =p is an
odd
prime.
It turns out that our Theorem 1 isenough
tocomplete
the resolution of(5).
Theorem 2.
(Conjecture
of
Arif
and AbuMuriefah)
If n >
3, k
=2m,
the
diophdntine equations
(5)
hasprecisely
twofamilies of
solutionsgiven
for
and11-23M
3M + 1.Proof.
We shall use the results of~1~, [6]
to reduce the theorem to aspecial
case of Theorem 1. It is shown in[1,
page300]
that n must beodd,
and that there are no further solutions with x, y even. It isenough
therefore toshow that there are no solutions to the
equation
for m >
0,
p > 3prime,
and x, y oddintegers.
However Cohn showed thatany such solution to
equation
(6)
mustsatisfy
m > 100([6,
page462])
andp =
1, 4, 7
(mod 9) ([6,
Lemma3]).
Inparticular,
any such solution will bea solution to
equation
(1)
with p >7, z
= -1 and k = 2m > 200. Thiscontradicts Theorem 1. 0
Remark. M. Le has
just published
a solution toequation
(6)
using
linear forms inlogarithms
(see [13]).
3.2. The
equation x2 -
2k
=yn.
Theequation
has been considered
by
Bugeaud
[3]
andby
Guo and Le[10].
Inparticular,
Bugeaud
showed thatif x,
y,k, n
is a solutionsatisfying
the above conditionsthen,
o n and k are
odd;
this was shownusing elementary
factorizationar-guments.
o n 7.3 x
105;
this was establishedusing
lower bounds for linear formsin
logarithms.
While we are unable to resolve
equation
(7)
completely,
our Theorem 1Theorem 3.
If
theequations
(7)
has a solutionsatisfying
the abovecondi-tions then either k = 1 or n =
3a5b for
some a, b.We note in
passing
that itmaybe
possible
to eliminate the case n =3a5b
from the theorem as follows.
Clearly
it is sufficient to solve theequations
Now k must be odd
by
the resultquoted
above,
and this leads us to fac-torize the left-hand side of eachequation
over We can reduce eachequation
to a set of Thue-Mahlerequations,
and these can be solvedusing
standard methods as in
[16,
Chapter
VIII],
although
we have notattempted
to do this. We do not see how theequation x2 -
2 =yn
(corresponding
tok =
1)
can be solvedusing
existing
methods.Acknowledgment.
I amdeeply
indebted to the referee and ProfessorBjorn
Poonen for many useful comments andcorrections,
andparticularly
for thesuggestion
that my method can deal withequation
(1)
and notjust
theequation
(5).
References
[1] S. A. ARIF, F. S. ABU MURIEFAH, On the diophantine equation x2 + 2k = yn. Internat.
J. Math. & Math. Sci. 20 no. 2 (1997), 299-304.
[2] C. BREUIL, B. CONRAD, F. DIAMOND, R. TAYLOR, On the modularity of elliptic curves
over Q: wild 3-adic exercises. J. Amer. Math. Soc. 14 (2001), 843-939.
[3] Y. BUGEAUD, On the diophantine
equation x2 -
2m = ±yn. Proc. Amer. Math. Soc.125 (1997), 3203-3208.
[4] J.E. CREMONA, Algorithms for modular elliptic curves (second edition). Cambridge
Uni-versity Press, 1996.
[5] J. H. E. COHN, The diophantine equation = yn. Arch. Math. 59 (1992), 341-344.
[6] J. H. E. COHN, The diophantine equation x2+2k = yn, II. Internat. J. Math. & Math.
Sci. 22 no. 3 (1999), 459-462.
[7] H. DARMON, The equations xn +yn = x2 and xn + yn = z3. International Mathematics
Research Notices 10 (1993), 263-274.
[8] H. DARMON, L. MEREL, Winding quotients and some variants of Format’s Last Theorem.
J. Reine Angew. Math. 490 (1997), 81-100.
[9] F. DIAMOND, On deformation rings and Hecke rings. Ann. Math. 144 no. 1 (1996),
137-166.
[10] Y. Guo, M. LE, A note on the exponential diophantine equation x2 - 2m = yn. Proc.
Amer. Math. Soc. 123 (1995), 3627-3629.
[11] W. IVORRA, Sur les équations xP +
203B2yp
= z2 et xP +203B2 yp
= 2z2. To appear in ActaArith.
[12] A. W. KNAPP, Elliptic curves. Mathematical Notes 40, Princeton University Press, 1992.
[13] M. LE, On Cohn’s conjecture concerning the Diophantine equation x2 + 2m = yn, Arch.
Math. 78 no. 1 (2002), 26-35.
[14] K. RIBET, On modular representations of arising from modular forms. Invent.
Math. 100 (1990), 431-476.
[15] J.-P. SERRE, Sur les répresentations modulaires de degré 2 de Duke Math. J. 54 (1987), 179-230.
[16] N.P. SMART, The algorithmic resolution of diophantine equations. LMS Student Texts
[18] A. WILES, Modular elliptic curves and Fermat’s Last Theorem. Ann. Math. 141 (1995),
443-551. Samir SIKSEK
Department of Mathematics and Statistics
College of Science
Sultan Qaboos University
P.O. Box 36
Al-Khod 123, Oman