Table Of ContentA REFINED MODULAR APPROACH TO THE DIOPHANTINE
EQUATION x2+y2n =z3
0
1 SANDERR.DAHMEN
0
2
n
a
Abstract. Let n be a positive integer and consider the Diophantine equa-
J
tion of generalized Fermat type x2+y2n = z3 in nonzero coprime integer
9
unknownsx,y,z. UsingmethodsofmodularformsandGaloisrepresentations
2
forapproachingDiophantineequations,weshowthatforn∈{5,31}thereare
no solutions to this equation. Combiningthis with previouslyknown results,
]
thisallowsacompletedescriptionofallsolutionstotheDiophantineequation
T
aboveforn≤107. Finally,weshowthattherearealsonosolutionsforn≡−1
N
(mod6).
.
h
t 1. Introduction
a
m Since the proof of Fermat’s Last Theorem [21], [20] and the establishment of
[ the full Shimura-Taniyama-Weil conjecture [3], many Diophantine equations were
solved using a strategy similar to that of the proof of FLT. Amongst them are
1
various so-called generalized Fermat equations. These are Diophantine equations
v
0 of the form
2 (1) axp+byq =czr, x,y,z Z, xyz =0, gcd(x,y,z)=1
0 ∈ 6
0 where a,b,c are nonzero integers and p,q,r are integers 2. The nature of the
. ≥
2 solutions depends very much on the quantity
0 1 1 1
0 χ(p,q,r):= + + 1.
p q r −
1
: If χ(p,q,r) > 0, then there are either no solutions to (1) or infinitely many. In
v
i the latter case,there existfinitely many triples (X,Y,Z)of binary forms X,Y,Z
X Q[u,v] satisfying aXp+bYq =cZr such that every solution (x,y,z) to (1) can b∈e
r obtainedbyspecializingthevariablesu,vtointegersforoneofthesetriples;see[2].
a
If χ(p,q,r)=0, then the determinationof the solutions to (1) basicallyboils down
to finding rationalpoints oncurvesofgenus one. If χ(p,q,r)<0, then there exists
a curve C of genus 2 and a covering φ : C P1, both defined over a number
≥ →
field K, such that for every solution (x,y,z) to (1) we have [axp :czr] φ(C(K)).
∈
Since by Faltings’ theorem C(K) is finite, there are only finitely many solutions to
(1) in this case; see [10].
Of special interest is the generalized Fermat equation with a = b = c = 1. In
thecaseχ(p,q,r)>0,allthe (parameterized)solutionsareknown;see[2]and[11].
If χ(p,q,r) = 0, the only solution, up to sign and permutation, is given by the
Date:January29,2010.
2000MathematicsSubjectClassification. 11D41(primary),11F11,11F80,11G05(secondary).
Key words and phrases. generalized Fermat equation, modular form, Galois representation,
ellipticcurve.
1
2 SANDERR.DAHMEN
Catalan solution 23 +16 = 32. If χ(p,q,r) < 0, the only solutions known, up to
sign and permutation, are given by 23+1q =32 (q 7) and
≥
173+27 =712, 22134592+14143 =657,
153122832+92623=1137, 762713+177 =210639282;
15490342+338 =156133, 962223+438 =300429072;
132+73 =29;
72+25 =34, 114+35 =1222.
Foralistcontainingmany(familiesof)triples(p,q,r)forwhich(1)hasbeensolved
in this case, we refer to [17, Table 1]. In this paper we will focus on the special
case (p,q,r) = (2,2n,3) for n Z , i.e. we are concerned with the Diophantine
>0
∈
equation
(2) x2+y2n =z3, x,y,z Z, xyz =0, gcd(x,y,z)=1.
∈ 6
Previously known results. For n = 1,2 we have χ(2,2n,3) > 0, and in both cases
there are infinitely many solutions to (2). By factoring x2 + y2 = z3 over the
Gaussian integers as (x+iy)(x iy) = z3 one readily gets, that a solution to (2)
−
for n=1 satisfies
(x,y,z)=(u(u2 3v2),v(3u2 v2),u2+v2)
− −
for some u,v Z with gcd(u,v) = 1 and uv = 0. By demanding that v(3u2 v2)
∈ 6 −
is a square, one can obtain parameterized solutions to (2) for n = 2. This was
carriedout by Zagierand reported in [2]; up to sign there are 4 parameterizations,
3 of which have coefficients in Z. In [11] parameterized solutions to (2) for n = 2
were obtained by a different method, the same parameterizations as in [2] were
found, except that the one with non-integer coefficients was replaced by one with
coefficients in Z.
Forn=3therearenosolutionsto(2). Thisfollowsreadilyfromthewell-known
fact that the only rational points on the elliptic curve given by Y2 = X3 1 are
−
(X,Y)=(1,0) and the point at infinity.
Bydemandingthaty isasquarefortheparameterizedsolutionsto(2)forn=2,
one obtains genus2 curvessuchthat everysolutionto (2) for n=4 correspondsto
a rational point on one of these curves. This, together with the determination of
all rational points on these curves using effective Chabauty methods, was carried
out in [4]. The result is that, up to sign, the only solution to (2) for n=4 is given
by 15490342+338 =156133.
We see that it suffices to deal with n a prime > 3. In [7], the equation is
studiedusingmethodsofmodularformsandGaloisrepresentationsforapproaching
Diophantineequations. Inparticular,anexplicitcriterionisgivenforshowingthat
(2) has no solution for a given prime n>7. This criterion is verified for all primes
7 < n < 107 except n = 31, thereby showing that (2) has no solutions for these
values of n. The nonexistence of solutions for n =7 follows from the work of [17],
so the only small values of n left to deal with are n=5,31.
At this point we should mention that, for m Z , the Diophantine equation
>1
∈
x2+y3 =zm, x,y,z Z, xyz =0, gcd(x,y,z)=1
∈ 6
ismuchhardertodealwiththan(2)is(apartfromsomesmallvaluesofm),evenif
we just consider even m. This is because of the Catalan solution 32+( 2)3 =1m.
−
A REFINED MODULAR APPROACH TO x2+y2n=z3 3
For m 5 there are infinitely many solutions. The parameterization for m = 2
≤
is again very easy to obtain, for m = 3,4,5 we refer to [11] (m = 3 was earlier
done by Mordell and m=4 by Zagier). The case m=6 is classical, it amounts to
determining the rational points on the elliptic curve given by Y2 = X3+1. The
cases m=7,8,9,10 are solved in [17], [5], [6], [19] respectively.
New results. In this paper, we will extend the criterion mentioned above to all
primes n > 3 and use it to show that there are no solutions for n = 5. This is
basicallydone byshowingthataFrey curveassociatedtoa (hypothetical)solution
hasirreduciblemod-n representationforn=5,7. Byusing extralocalinformation
obtainedfromclassicalalgebraicnumbertheory,weobtainarefinedcriterion,which
is used to show that (2) has no solutions for n=31. So from a Diophantine point
of view our main result is the following.
Theorem 1. If n 5,31 , then (2) has no solutions.
∈{ }
Although we focus on very specific equations, we feel that the methods we use
to overcome the earlier difficulties, could definitely be used in other cases as well.
Combiningourmainresultwiththe previouslyknownresultsmentionedabove,we
arrive at a description of all solutions for n 107.
≤
Corollary 2. Let n Z with n 107. If n 1,2,4 , then (2) has solutions,
>0
∈ ≤ ∈ { }
all of which are described above. If n 1,2,4 , then (2) has no solutions.
6∈{ }
Finally, by applying a key idea from[8], we solve (2) for infinitely many (prime)
values of n.
Theorem 3. If n Z with n 1 (mod 6), then (2) has no solutions.
>0
∈ ≡−
2. A modular approach to x2+y2l =z3
Throughoutthissection,letldenoteaprime>3. Weshallexplainhowmodular
forms and Galois representations can be used to study (2) for n = l. We want to
stress that the methods and results described in this section are not new and can
be found essentiallyin[7]. For a generalintroductionto usingmethods ofmodular
formsandGaloisrepresentationsforapproachingDiophantineequations,onecould
consult e.g. Chapter 2 of the author’s Ph.D. thesis [9].
Lemma 4. Suppose that (x,y,z) is a solution to (2) for n=l. Then
(3) (x,yl,z)=(u(u2 3v2),v(3u2 v2),u2+v2)
− −
for some u,v Z with gcd(u,v)=1, uv =0, 3v and 2uv. Furthermore,
∈ 6 | |
(4) 3v=rl and 3u2 v2 =3sl
−
for some r,s Z with gcd(r,s)=1 and rs=0.
∈ 6
Proof. That (3) holds for some nonzero coprime integers u,v follows immediately
from the parameterization of solutions to (2) for n = 1, given in Section 1. From
v(3u2 v2) = yl we see that up to primes dividing gcd(v,3u2 v2), both v and
− −
3u2 v2 must be l-thpowers. One easily checksthat gcd(v,3u2 v2) equals either
− −
1 or 3.
First suppose that gcd(v,3u2 v2) = 1. Then v = rl and 3u2 v2 = sl
− −
for some nonzero r,s Z. Furthermore, r,s and u are pairwise coprime and
∈
(r2)l+sl =3u2. However,[1, Theorem 1.1] tells us that for an integer n>3 there
4 SANDERR.DAHMEN
are no nonzero (pairwise) coprime integers a,b,c satisfying an +bn = 3c2, which
contradicts gcd(v,3u2 v2)=1. So suppose now that gcd(v,3u2 v2)=3. Then
− −
3v and of course 3∤u, so 3 3u2 v2. We get that 3v =rl and 3u2 v2 =3sl for
| || − −
some nonzero r,s Z. Finally, if both u,v are odd, then v(3u2 v2) 2 (mod 4),
∈ − ≡
but this contradicts that v(3u2 v2) is an l-th power. (cid:3)
−
We have reduced (2) for n = l to the equation v(3u2 v2) = yl with u,v,y as
−
in the lemma above. This equation can be approached using methods of modular
forms and Galois representations. In fact, one can use almost exactly the same
methods as used in [12] for studying the equation a3+b3 =cl in nonzero coprime
integers a,b,c.
Suppose (x,y,z)is a solutionto (2) for n=l, letu,v,r,s be asin Lemma 4 and
consider the following Frey curve associated to this solution
X3+2uX2+ v2X if u is even;
(5) E :Y2 = 3
( X3 uX2+ v2X, u 1 (mod 4) if u is odd.
± 12 ± ≡
Because 3v and 2v if u is odd, we see that the model for E above actually has
| |
coefficients in Z. Using Tate’s algorithm,or the in practice very handy to use [16],
onefinds thatthemodelforE isminimalateveryprimep,exceptatp=2whenu
isodd. Furthermore,theconductorN andtheminimaldiscriminant∆ofE satisfy
(see also [7, Propositio 8] and [1, Lemma 2.1])
∆=2α 3 3v4(3u2 v2)=2α 3 6 r4s l, α:=6 if 2u, α:= 12 if 2∤u
− −
· − · | −
N =2β 3rad 2,3 (∆)=2β 3rad 2,3(cid:0) (rs(cid:1)), β :=5 if 2u, β :=1 if 2∤u,
· { } · { } |
whereforafinitesetofprimesS andanonzeron Z,rad (n)denotestheproduct
S
∈
of prime p with pn and p S.
| 6∈
Remark 5. From x2 z3 = y2l, an obvious choice, up to quadratic twist, for a
− −
Frey curve associated to the solution (x,y,z) would be
E :Y2 =X3 3zX+2x.
′
−
Thismodelhas(notnecessarilyminimal)discriminant 26 33(x2 z3)=26 33y2l,
− · − ·
which has no primes >3 in commonwith c =24 32z. Using (3), we can write E
4 ′
·
as
E :Y2 =X3 3(u2+v2)X +2u(u2 3v2),
′
− −
fromwhichweobtainthatE hasarational2-torsionpointbecausethe right-hand
′
side factors as (X +2u)(X2 2uX +u2 3v2). The 2-isogenous elliptic curve
− −
associatesto the 2-torsionpoint(X,Y)=( 2u,0)is simply,uptoquadratictwist,
−
the Frey curve E above. For both theory and practical computation, it makes
no difference whether E or E is used, except for establishing irreducibility of the
′
mod-5 representation, where the computations actually become cleaner if one uses
E instead of E. The only reason for introducing E above, is to follow [7] more
′
closely.
Foranelliptic curveEoverQ,letρE :Gal(Q/Q) Gl (F )denotethestandard
l → 2 l
2-dimensionalmod-l Galois representationassociatedto E (induced by the natural
actionofGal(Q/Q)onthel-torsionpointsE[l]ofE). SupposethatρE isirreducible.
l
Thenbymodularity[3]andlevellowering[18](notethatρE isfiniteatl),weobtain
l
thatρE is modularoflevel N :=2β 3,weight2 andtrivialcharacter. This means
l 0 ·
A REFINED MODULAR APPROACH TO x2+y2n=z3 5
that ρE ρf for some (normalized) newform f S2(Γ (N )), where of course ρf
l ≃ l ∈ 0 0 l
denotes the standard 2-dimensional mod-l Galois representation associated to f.
If u is odd, then N = 6 and since there are no newforms in S2(Γ (6)) we have
0 0
reached a contradiction in this case. If u is even, then N = 96 and S2(Γ (96))
0 0
contains 2 newforms, both rational and quadratic twists over Q(i) of each other.
So ρE ρf ρE0 for some elliptic curve E over Q with conductor N = 96. As
l ≃ l ≃ l 0 0
is well-known, we get from ρE ρE0 by comparing traces of Frobenius, that for
l ≃ l
primes p
a (E ) a (E) (mod l) if p∤N
p 0 p
≡
a (E ) a (E)(1+p) (1+p) (mod l) if pN and p∤N =25 3.
p 0 p 0
≡ ≡± | ·
Note that we explicitly know E up to isogeny and quadratic twist. So if E is
0 0′
any elliptic curve over Q with conductor 96, e.g. given by Y2 = X3 X2 2X,
± −
then a (E )2 = a (E )2 for all primes p. Therefore, for every prime p, we can
p 0 p 0′
effectively compute the uniquely determined value a (E )2. We summarize some
p 0
of the Diophantine information obtained so far.
Lemma 6. Suppose that (x,y,z) is a solution to (2) for n = l, let u,v be as in
Lemma4, lettheelliptic curveE begiven by(5), let E beanyelliptic curveover Q
0
withconductor96andletp>3beprime. IfρE isirreducibleanda (E )2 (p+1)2
l p 0 6≡
(mod l), then p∤v(3u2 v2) and a (E )2 a (E)2 (mod l).
p 0 p
− ≡
We see that in order to obtain a contradiction and conclude that (2) has no
solutions for n = l, it suffices (still assuming the irreducibility of ρE) to find a
l
prime p>3 such that a (E )2 (p+1)2 (mod l) and a (E )2 a (E)2 (mod l).
p 0 p 0 p
6≡ 6≡
A priori possible values of a (E) can be obtained by plugging all values of u,v
p
(mod p) with p ∤ v(3u2 v2) into the equation for E. However, this will never
−
lead to a contradiction, since plugging the values u,v with (u, v ) = (2,3) into
| | | |
the equationfor E,will giveus elliptic curveswith conductor96. Now by reducing
(4) modulo p, one obtains extra information on u,v (mod p) whenever one knows
that the nonzero l-th powers modulo p are strictly contained in F . This happens
∗p
exactly when p 1 (mod l), say p=kl+1, in which case the nonzero l-th powers
≡
modulo p are given by
µ (F ):= ζ F ζk =1 .
k p p
{ ∈ | }
Heuristically speaking, the bigger k is, the more possibilities for u,v (mod p) we
expect, hence the more a priori possibilities for a (E) we expect, hence the less
p
likely it is to conclude that a (E)2 a (E )2 (mod l). From a computational
p p 0
6≡
point of view, it is desirable to only consider the possibilities for u/v (mod p), this
indeedsuffices,sinceu/v (mod p)determinesthereductionofE modulop,denoted
E˜(F ), up to quadratic twist and hence determines a (E)2 uniquely.
p p
Continuing our more formal discussion, let indeed k Z be such that p :=
>0
∈
kl+1 is prime (so (p+1)2 4 (mod l)) and suppose that a (E )2 4 (mod l).
p 0
≡ 6≡
Then in particular p ∤ v(3u2 v2) = (rs)l. Define U := u/(3v). From 3v = rl
−
and 3u2 v2 = 3sl, we get U2 = 1/(27)+(s/r2)l. Since p ∤ rs, we obtain for the
−
reduction of U modulo p, denoted U, that
1
(6) U S := α F α2 µ (F ) .
k,p p k p
∈ { ∈ | − 27 ∈ }
For α S consider the elliptic curve over F given by
k,p p
∈
1
(7) E :Y2 =X3+2αX2+ X.
α
27
6 SANDERR.DAHMEN
NotethatE isaquadratictwistofE˜(F ),sothata (E )2 =a (E)2. Ifnowforall
U p p U p
α S we have a (E )2 a (E )2 (mod l), then a (E)2 a (E )2 a (E )2
∈ k,p p α 6≡ p 0 p ≡ p U 6≡ p 0
(mod l), a contradiction which implies that (2) has no solutions for n = l. Note
that we have assumed ρE to be irreducible; one readily obtains (see Section 3.1)
l
from [14] and [15] that this is actually the case if l > 7. This proves the main
theorem of [7].
Theorem 7 ([7, Theorem 1]). Let l >7 be prime and let E be any elliptic curve
0
over Q with conductor 96. If there exists a k Z such that the following three
>0
∈
conditions hold
(1) p:=kl+1 is prime
(2) a (E )2 4 (mod l)
p 0
6≡
(3) a (E )2 a (E )2 (mod l) for all α S ,
p α p 0 k,p
6≡ ∈
where S and E are given by (6) and (7) respectively,
k,p α
then (2) has no solutions for n=l.
Accordingto[7],ithasbeencomputationallyverifiedthatforeveryprimel with
7 < l < 107 and l = 31 there exists a k Z satisfying the three conditions of
>0
6 ∈
Theorem 7, hence (2) has no solutions for n=l if l equals one of these values.
We like to takethis opportunity to pointout a smallomissioninthe proofof[7,
Corollary 3]. This corollary to Theorem 7 states that if l >7 is a Sophie Germain
prime, i.e. p := 2l+1 is prime, p = 1 and p = ( 1)(l+1)/2, then (2) has
7 13 −
no solutions for n = l. As pointed out in [7], the conditions for the Legendre
(cid:0) (cid:1) (cid:0) (cid:1)
symbols are equivalent (by quadratic reciprocity) to S being empty, in which
2,p
case condition 3 of Theorem 7 trivially holds. However, in order to use Theorem
7 to deduce that (2) has no solutions, we need of course prove that condition 2
also holds. This can be done as follows. Note that both isogeny classes of elliptic
curves over Q with conductor 96 contain an elliptic curve with rational torsion
group of order 4. So 4p+1 a (E ), and hence 4a (E ). This implies that if
p 0 p 0
| − |
a (E ) 2 (mod l), then a (E ) 2 2l. That this last inequality cannot
p 0 p 0
≡ ± | ∓ | ≥
hold, follows directly from the Hasse bound ap(E0) 2√p together with l > 3,
| | ≤
which completes the proof of the corollary.
Note that the corollary and its proof are analogously to work in [12], namely
Corollaire 3.2 and its proof.
3. Refinements
In this section we shall prove our main result, Theorem 1.
3.1. Irreducibility. As pointed out in [7], since E has at least one odd prime of
multiplicative reduction and a rationalpoint oforder 2, the irreducibility of ρE for
l
primes l > 7 follows from [14, Corollary 4.4] and [15]. In fact, one does not need
the first mentioned property of E, but only that E has a rational point of order 2
to arrive at the desired conclusion, since it is well-known that the modular curves
X (2l) for primes l > 7 have no noncuspidal rational points. Irreducibility of ρE
0 l
for the remaining primes l>3 can also be obtained.
Theorem 8. Let l > 3 be prime and consider the elliptic curve E given by (5),
where u,v Z with v(3u2 v2)=0. Then ρE is irreducible.
∈ − 6 l
A REFINED MODULAR APPROACH TO x2+y2n=z3 7
Proof. The only cases left to deal with are l=7 and l=5.
The modular curve X (14) is of genus one, it has 2 noncuspidal rationalpoints,
0
whichcorrespondtoellipticcurveswithj-invariantj equalto 33 53or33 53 173;
14
− · · ·
this readily follows from [13, Chapter 5]. Denote the j-invariant of E by j(u,v).
For both values of j it is completely straightforward to check that the equation
14
j(u,v)=j has no solutions with u,v Z and (3u2 v2)v =0. This proves that
14
∈ − 6
ρE is irreducible for l=7.
l
Since the modular curve X (10) is of genus zero and has a rational cusp (which
0
is nonsingular of course), it has infinitely many noncuspidal rational points. So we
have to work a little harder to obtain the irreducibility of ρE for l = 5. Up to
l
quadratic twist, the Frey curve E is 2-isogenous to
E′ :Y2 =X3 6uX2+3(3u2 v2)X
− −
(replacing X by X+2u, gives us back the model for E given in Section 2). So ρE
′ 5
is irreducible if and only if ρE′ is irreducible. The j-invariant j (u,v) of E (which
5 ′ ′
is a twist of the j-map from X(2) to X(1)) is given by
1728(u2+v2)3
j (u,v) =
′
v2(3u2 v2)2
−
1728u2(u2 3v2)2
= − +1728.
v2(3u2 v2)2
−
The j-map from the modular curve X (5) to X(1), denoted j , is given by
0 5
(t2+10st+5s2)3
j (s,t) =
5 s5t
(t2+4st s2)2(t2+22st+125s2)
= − +1728.
s5t
Bycomparingj 1728,weseethatevery[u:v] P1(Q)suchthatρE′ isirreducible
− ∈ 5
gives rise to a rational point on the curve C in P1 P1 give by
×
1728u2(u2 3v2)2 (t2+4st s2)2(t2+22st+125s2)
C : − = − .
v2(3u2 v2)2 s5t
−
Letting
t u(u2 3v2) st
X := , Y := −
s v(3u2 v2) · t2+4st s2
− −
defines a covering by C of the elliptic curve over Q given by
1728Y2 =(X2+22X+125)X.
By checking that this elliptic curve has rank 0 and torsion subgroup of order 2,
we get that the only rational points on C are those with [s : t] = [1 : 0] or
[s : t]= [0:1]. For these values of [s : t] we have j (s,t)= , so they correspond
5
to cusps on X (5). We conclude that ρE′, and hence ρE, is∞irreducible for l = 5.
0 l l
This completes the proof. (cid:3)
This irreducibility result obviously leads to the following strengthening of The-
orem 7.
Theorem 9. Theorem 7 holds true with the condition l>7 replaced by l>3.
8 SANDERR.DAHMEN
3.1.1. l = 5. In order to show that (2) has no solutions for n = l := 5, it suffices
by Theorem 9 to show that there exists a k Z satisfying the three conditions
>0
∈
ofTheorem7. Let k:=2, then p:=kl+1=11is prime, a (E )2 1 4 (mod l)
p 0
≡ 6≡
and S is empty because 1/27 1 are not squares in F . So the conditions are
2,p p
±
satisfied for k =2 and we conclude that this proves Theorem 1 in case n=l =5.
Remark 10. By our irreducibility result, Theorem 8, the corollary mentioned at
the end of Section 2 also holds with the condition l > 7 replaced by l > 3 (the
case l = 7 is in fact trivial, because 7 is not a Sophie Germain prime). So for
obtaining Theorem 1 in case n = l := 5 it sufficed to check that p = 1 and
7
p =( 1)(l+1)/2 for p=11.
13 − (cid:0) (cid:1)
3(cid:0).2.(cid:1)Using more local information. For l := 31 there is no k Z known
>0
∈
which satisfies the three conditions of Theorem 7 (it seems in fact very unlikely
thatsuchak exists), sothatwe cannotusethis theoremto provethe nonexistence
ofsolutionsof(2)forn=l=31. Thisisnottoosurprising,since,looselyspeaking,
l=31isveryfarfrombeingaSophieGermainprimeinthe sensethatthe smallest
k Z such that kl+1 is prime, which is k = 10, is not so small compared to
>0
∈
l=31.
Nowletl >3beprime,supposethat(x,y,z)isasolutionto(2)forn=landlet
u,v,r,sbeasinLemma4. Theideaistofactortheleft-handsideof3u2 v2 =3sl
−
over the ring of integers R := Z[√3] in order to obtain more local information on
u,v, which, together with Lemma 6 should lead to a contradiction. We claim that
(8) 3v=rl, √3u v =√3xlǫ and √3u+v =√3xlǫ 1
− 1 2 −
for some nonzero x ,x R and unit ǫ R . Note that R has class number one.
1 2 ∗
∈ ∈
Using gcd(u,v) = 1, 2uv and 3v we readily get (√3u v,√3u+v) = (√3) and
| | −
√3 √3u v. From (√3u v)(√3u+v)=3sl the claim now follows.
|| ± −
As before, let E be given by (5), let E be any elliptic curves over Q with
0
conductor 96, let k Z be such that p := kl +1 is prime and suppose that
>0
∈
a (E )2 4 (mod l). Note that by Theorem 8 we get that ρE is irreducible.
p 0 6≡ l
Again, by Lemma 6 we get p ∤ v(3u2 v2)=(rs)l and we see that if we can show
−
that a (E )2 a (E)2 (mod l), then we reach a contradiction which shows that
p 0 p
6≡
(2) hasnosolutions forn=l. Nowsuppose furthermorethatpsplits in R. Denote
bypanyofthe2primesofRlyingabovep,foranyx Rdenotebyxthereduction
∈
of x modulo p in R/p F and set r :=√3. By reducing (8) modulo p we get
p 3
≃
3v =ζ , r u v =r ζ ǫ, and r u+v =r ζ ǫ 1
0 3 3 1 3 3 2 −
−
for some ζ ,ζ ,ζ µ (F ). Let U :=u/(3v) as before, set ζ :=ζ /ζ ,ζ :=ζ /ζ
0 1 2 ∈ k p 1′ 1 0 2′ 2 0
and divide by 3r v =r ζ to obtain
3 3 0
1 1
(9) U =ζ ǫ, U + =ζ ǫ 1.
− 3r 1′ 3r 2′ −
3 3
By Dirichlet’s unit theorem R = 1,ǫ for some fundamental unit ǫ R (we
∗ f f
h− i ∈
can take for example ǫ = 2+√3). So ζ ǫ,ζ ǫ 1 belong to the subgroup of F
f 1′ 2′ − ∗p
generated by µ (F ) and ǫ , which we denote by G . If ǫ µ (F ) one easily
k p f k,p f k p
6∈
obtains that G = F . Together with (9) this only gives us back the original
k,p ∗p
information U S . If however ǫ µ (F ), then of course G = µ (F ) and
k,p f k p k,p k p
∈ ∈
A REFINED MODULAR APPROACH TO x2+y2n=z3 9
we get
1 1
(10) U ∈Sk′,p :=(µk(Fp)+ 3r )∩(µk(Fp)− 3r ).
3 3
Thismightbemuchstrongerinformation,becausepossibly(andheuristicallyspeak-
ing, for large k even very likely) S is strictly contained in S . So suppose
k′,p k,p
ǫ k = 1. Since a (E)2 = a (E )2, we reach a contradiction if for all α S we
f p p U ∈ k′,p
have a (E )2 a (E )2 (mod l). We arrive at the following.
p α p 0
6≡
Theorem 11. Let l > 3 be prime and let E be any elliptic curve over Q with
0
conductor 96. If there exists a k Z such that the following five conditions hold
>0
∈
(1) p:=kl+1 is prime
(2) p splits in Z[√3]
(3) pNorm (2+√3)k 1
| Q(√3)/Q −
(4) a (E )2 4 (mod l)
p 0 6≡ (cid:0) (cid:1)
(5) a (E )2 a (E )2 (mod l) for all α S ,
p α 6≡ p 0 ∈ k′,p
where S and E are given by (10) and (7) respectively,
k′,p α
then (2) has no solutions for n=l.
Remark 12. Suppose that k Z satisfies the first two conditions of Theorem
>0
∈
7. Heuristically speaking, by considering the expected size of S , it seems highly
k,p
unlikelythatifk&l,thelastconditionissatisfied. Nowcondition3ofTheorem11
is very restrictive. But if k Z does satisfy the first four conditions of Theorem
>0
∈
11, then the expected size of S is much smaller then that of S and it only
k′,p k,p
becomes highly unlikely that the last condition is satisfied when k & l2. This is
of course all very rough, but the main idea about the benefits of Theorem 11 are
hopefully clear.
3.2.1. l = 7. Let l := 7. There are no problems with irreducibility. However, a
large computer search did not reveal a k Z satisfying the three conditions of
>0
∈
Theorem 7 or the five conditions of Theorem 11 (and we believe that it is very
unlikely that such a k Z exists).
>0
∈
Although the modular methods can give us many congruence relations for a
hypothetical solution (x,y,z) of (2) with n = l = 7, such as 2x and 29y, we are
| |
not able to show that (2) has no solutions for n=l=7 alongthese lines. Anyway,
in [17] all finitely many nonzero coprime a,b,c Z satisfying a2 +b3 = c7 are
∈
determinedandonereadilychecksthatfornoneofthe solutions cis asquare. So
−
we conclude that (2) has no solutions for n=l=7, as mentioned before.
3.2.2. l = 31. Now let l := 31, k := 718 and p := kl +1 = 22259. Then one
quickly verifies that p is a prime that splits in Z[√3] and that ǫ k = 1. We
f
calculate a (E ) = 140, so a (E )2 8 4 (mod l). Finally, the set S is
p 0 ± p 0 ≡ 6≡ k′,p
easilydeterminedexplicitly. Allthe elementsα S aregiveninthe firstcolumn
∈ k′,p
of Table 1 together with the corresponding values of a (E )2 (mod l).
p α
Recall that a (E )2 8 (mod l), so from the second column of Table 1 we see
p 0
≡
that a (E )2 a (E )2 (mod l) for all α S . By Theorem 11 we can now
p α 6≡ p 0 ∈ k′,p
conclude that (2) has no solutions for n = l = 31. This concludes the proof of
Theorem 1.
10 SANDERR.DAHMEN
α a (E )2 (mod l)
p α
±
127 28
1852 19
2818 1
3146 10
3615 9
3764 16
4419 18
5889 25
7994 20
8058 0
8330 19
10171 18
10561 5
Table 1. elements of S with corresponding values of a (E )2 (mod l)
k′,p p α
Remark 13. For l = 31 the smallest k Z satisfying the five conditions of
>0
∈
Theorem 11 is k = 718. Another k Z satisfying these conditions is k = 2542,
>0
∈
and it seems very likely that there are no other such k.
4. Extra information from quadratic reciprocity
In this section we shall prove Theorem 3.
In [8, Section 4], quadratic reciprocity (over Q) is used to obtain that there
are no nonzero coprime integers a,b,c satisfying a3 +b3 = cn for n Z with
>0
∈
n 51,103 or 105 (mod 106). The same method can be applied to (2). The key
≡
information obtained is the following.
Lemma 14. Let l >3 be prime, suppose that (x,y,z) is a solution to (2) for n=l
and let r,s be as in Lemma 4. Then s r2 is a square modulo 7.
−
Proof. Let u,v be as in Lemma 4. Using (4) we get
(11) s r2 3(sl r2l)=3u2 28v2.
− | − −
Supposethatqisanoddprimethatdividess r2. Then(3u)2 3 7(2v)2 (mod q).
− ≡ ·
From 3v = rl and gcd(r,s) = 1 we obtain q ∤ 2v, so 3 7 is a square modulo q,
·
i.e. 37 0,1 . By quadratic reciprocity we obtain q 0,1 . We have
q· ∈ { } 37 ∈ { }
·
−317 (cid:16)=1(cid:17). However 327 =−1,butweclaimthatord2(s−(cid:0)r2)i(cid:1)seven,sothatweget
(cid:0)s·−3r(cid:1)72 ∈{0,1}. We(cid:0)c·la(cid:1)imfurthermorethat s−3r2 =1,fromwhichitnowfollows
·
t(cid:16)hats(cid:17) r2 isa squaremodulo 7. Itremainto(cid:16)prove(cid:17)ourtwoclaims. ByTheorem8
−
and the discussionin Section 2, we know that 2u, so 2∤v. Suppose first that 2 u,
| ||
then3(u/2)2 7v2 4 (mod 8),henceord (sl r2l)=ord (3u2 28v2)=4. Next
2 2
− ≡ − −
suppose that 4u, then we immediately get ord (sl r2l) = ord(3u2 28v2) = 2.
2
| − −
So in any case, ord (sl r2l) is even. From the fact that r,s are both odd and
2
−
by counting terms in (sl r2l)/(s r2), we get that the latter quantity is odd.
− −
We conclude that ord (s r2) = ord (sl r2l) is even, which proves our first
2 2
− −
claim. For our second claim, note that 3r,3 ∤ s and l is odd, so it remains to
|
prove that sl is a square modulo 3. From 3v we get 3v2/3, so from (4) we get
| |
