Fermat-type equations of signature (13,13,p) via Hilbert cuspforms
aa r X i v : . [ m a t h . N T ] J a n Fermat-type equations of signature (13 , , p ) viaHilbert cuspforms. Luis Dieulefait and Nuno FreitasOctober 15, 2018
Abstract
In this paper we prove that equations of the form x + y = Cz p haveno non-trivial primitive solutions ( a, b, c ) such that 13 ∤ c if p > C . Our method consists in relating asolution ( a, b, c ) to the previous equation to a solution ( a, b, c ) of anotherDiophantine equation with coefficients in Q ( √ a, b, c ) and we prove modularity of them inorder to apply the modular approach via Hilbert cusp forms over Q ( √ Since the proof of Fermat’s Last Theorem by Wiles [27] the modular ap-proach to Diophantine equations has been popularized and achieved great suc-cess in solving equations that previously seemed intractable. In order to attackthe generalized Fermat equation Ax p + By q = Cz r , where 1 /p +1 /q +1 /r < p, q, r ) there exists only a finite number of solutions suchthat ( x, y, z ) = 1. Another important progress was the work of Ellenberg onthe representations attached to Q -curves which allowed to use a special type ofelliptic curves over number fields ( Q -curves) to attack Diophantine equationsover the rationals. In particular, Ellenberg solved the equation A + B = C p (see [13]). For a summary of known results on the equation x p + y q = z r seethe introduction in [6].A particularly important subfamily of the generalized Fermat equation arethe equations of signature ( r, r, p ), that is, Ax r + By r = Cz p with r a fixed prime.In this direction there is work for (3 , , p ) by Kraus [20], Bruin [5], Chen-Siksek[7] and Dahmen [8]; for (5 , , p ) by Billerey [1], Dieulefait-Billerey [2] and fromthe authors [12]; for (7 , , p ) from the second author (currently in preliminaryversion). In this paper we will go further into this family of equations andwe will use a generalized version of the classical modular approach to studyequations of the form x + y = Cz p . (1)1et ( a, b, c ) be a triple of integers such that a + b = Cc p . We say that it isa primitive solution if ( a, b ) = 1 and we will say that it is a trivial solution if abc = 0. Following the terminology introduced by Sophie Germain in her workon the FLT we will divide solutions to (1) into two cases Definition 1.1
A primitive solution ( a, b, c ) of x r + y r = Cz p is called a firstcase solution if r do not divide c , and a second case solution otherwise. The strategy we will use goes as follows: we first relate a possible non-trivialprimitive solution ( a, b, c ) of (1) to a non-trivial primitive solution ( a, b, c ) ofanother Diophantine equation with coefficients in Q ( √ Q ( √
13) that is not a Q -curve. Then using modularity results from Skinner-Wiles and Kisin we provemodularity of our F-H-curves. With modularity established, from the levellowering results for Hilbert modular forms it will follow that the existence ofthe solution ( a, b, c ) implies a congruence between two Galois representations.One of the representations is attached to our Frey-curves and the other to aHilbert newform of a certain level (not depending on c ) and parallel weight(2 , Q ( √
13) can not have non-trivial primitive solutionsand consequently neither (1). The main result in this paper is the followingtheorem:
Theorem 1.2
Let d = 3 , , or and γ be an integer divisible only by primes l (mod 13). If p > is a prime, then:(I) The equation x + y = dγz p has no non-trivial primitive first casesolutions.(II) The equation x + y = 10 γz p has no primitive non-trivial solutions. In what follows we will first prove part (I) of Theorem 1.2 and in the endwe will explain the small tweak needed to conclude part (II). Observe that inpart (II) replacing 10 by twice d for d = 3 , ,
11 the statement is also true buttrivial, because the left-hand side is a sum of two relatively prime squares.We thank John Cremona for providing us a list of elliptic curves that wasuseful to test our strategy. We also want to thank John Voight for computinga list of Hilbert modular forms that was fundamental to finish this work.
In this section we will relate a solution of (1) to a solution of a Diophantineequation with coefficients in Q ( √ x + y over the cyclotomic field Q ( ζ ). Sincethese properties are not exclusive of degree 13 we will prove them in general.Observe that if r is a prime then x r + y r = ( x + y ) φ r ( x, y )2here φ r ( x, y ) = r − Y i =0 ( − i x r − − i y i . Let ζ = ζ r be an r -root of unity and consider the decomposition over thecyclotomic field Q ( ζ ) φ r ( x, y ) = r − Y i =1 ( x + ζ i y ) . (2) Proposition 2.1
Let P r be the prime in Q ( ζ ) above the rational prime r andsuppose that ( a, b ) = 1 . Then, any two distinct factors a + ζ i b and a + ζ j b inthe factorization of φ r ( a, b ) are coprime outside P r . Furthermore, if r | a + b then ν P r ( a + ζ i b ) = 1 for all i . Proof:
Suppose that ( a, b ) = 1. Let P be a prime in Q ( ζ ) above p ∈ Q and a common prime factor of a + ζ i b and a + ζ j b , with i > j . Observe that( a + ζ i b ) − ( a + ζ j b ) = bζ j (1 − ζ i − j ) and since P must divide the differenceit can not divide b because in this case it would also divide a , and since a, b are integers p would divide both. So P must be a factor of ζ i (1 − ζ i − j ) but ζ i is a unit for all i then P divides 1 − ζ i − j , that is P = P r . Now for the laststatement in the proposition, suppose that r | a + b . Then, a + ζ i b = a + b − b + ζ i b = ( a + b ) + ( ζ i − b, and since ν P r ( ζ i −
1) = 1 we have ν P r ( a + ζ i b ) = min { r − , } = 1 (cid:4) Corollary 2.2 If ( a, b ) = 1 , then a + b and φ r ( a, b ) are coprime outside r .Furthermore, if r | a + b then ν r ( φ r ( a, b )) = 1 . Proof:
Let p be a prime dividing a + b and φ r ( a, b ) and denote by P a primein Q ( ζ ) above p . P must divide at least one of the factors a + ζ i b . Since a, b are integers P can not divide b then it follows from a + b = a + ζ i b − ζ i b + b = ( a + ζ i b ) + (1 − ζ i ) b that P = P r . Moreover, if r | a + b it follows from the proposition that ν P r ( a + ζ i b ) = 1 for all i then ν P r ( φ r ( a, b )) = r − ν r ( φ r ( a, b )) = 1. (cid:4) Proposition 2.3
Let ( a, b ) = 1 and l (mod r ) be a prime dividing a r + b r .Then l | a + b . Proof:
Since l divides a r + b r , l ∤ ab . Let b be the inverse of − b modulo l . Wehave a r ≡ ( − b ) r (mod l ), hence ( ab ) r ≡ l ). Thus the multiplicativeorder of ab in F l is 1 or r . From the congruence ab ≡ l ) it follows a + b ≡ l ). If l ∤ a + b then the order of ab is r and l ≡ r ). (cid:4) From now on we particularize to r = 13 and we denote φ only by φ . Wehave x + y = ( x + y ) φ ( x, y ), where φ ( x, y ) = x − x y + x y − x y + x y − x y + x y − x y + x y − x y + x y − xy + y . a, b, c ) to (1) with C = dγ , d and γ as in theorem 1.2. Then it follows from corollary 2.2 andproposition 2.3 that there exists a non-trivial primitive solution ( a, b, c ) to φ ( a, b ) = c p , (3)with d | a + b and 13 ∤ a + b or to φ ( a, b ) = 13 c p (4)with d | a + b and 13 | a + b , where in both cases c is only divisible by primescongruent to 1 modulo 13. Consider the factorization of φ in Q ( ζ ) φ ( x, y ) = Y i =1 ( x + ζ i y ) . We have φ = φ φ where φ i are both of degree 6 with coefficients in Q ( √ φ ( x, y ) = ( x + ζy )( x + ζ y )( x + ζ y )( x + ζ y )( x + ζ y )( x + ζ y ) ,φ ( x, y ) = ( x + ζ y )( x + ζ y )( x + ζ y )( x + ζ y )( x + ζ y )( x + ζ y ) . The proof of the following corollary of proposition 2.1 is immediate.
Corollary 2.4
Let P be the prime of Q ( √ above 13. If ( a, b ) = 1 , then φ ( a, b ) and φ ( a, b ) are coprime outside P . Moreover, ν P ( φ i ( a, b )) = 1 or0 for both i if | a + b or ∤ a + b , respectively. Corollary 2.4 and the existence of a solution ( a, b, c ) to (3) or (4) impliesthat for some unit µ there exists a solution ( a, b, c ) (with c an integer in Q ( √ φ ( a, b ) = µc p , (5)with d | a + b and 13 ∤ a + b or to φ ( a, b ) = µ √ c p , (6)with d | a + b and 13 | a + b , respectively.Observe that proposition 2.3 and the form of equation (1) implies that 13 | c is equivalent to 13 | a + b . Moreover, proposition 2.3 also guarantees that whenpassing from the equation in (I) of Theorem 1.2 to equations (5) or (6) the primefactors of γ can be supposed to divide a + b . Since this information will not benecessary to the proof of both parts of theorem 1.2 we can assume that γ = 1.It will also be clear from the proof that the unit µ can be supposed to be 1. Let σ be the generator of G = Gal( Q ( ζ ) / Q ) and K the subfield (of degree6) fixed by σ . Consider the factorization φ = f f f where f ( x, y ) = ( x + ζy )( x + ζ y ) = x + ( ζ + ζ ) xy + y f ( x, y ) = ( x + ζ y )( x + ζ y ) = x + ( ζ + ζ ) xy + y f ( x, y ) = ( x + ζ y )( x + ζ y ) = x + ( ζ + ζ ) xy + y φ with coefficients in K . Now we are interestedin finding a triple ( α, β, γ ) such that αf + βf + γf . Solving a linear system in the coefficients of the f i we find that one of its infinitesolutions in O K is given by α = − ζ + ζ + ζ − ζ β = ζ − ζ − ζ + ζγ = − ζ + ζ + ζ − ζ and verifies ν P ( α ) = ν P ( β ) = ν P ( γ ) = 1.Suppose now that ( a, b, c ) ∈ Z × Z × O Q ( √ is a non-trivial primitive solu-tion to equation (5) or (6) and let A ( a, b ) = αf ( a, b ), B ( a, b ) = βf ( a, b ) and C ( a, b ) = γf ( a, b ). Since A + B + C = 0then we can consider the Frey-curve over K with the classic form E ( a, b ) : y = x ( x − A ( a, b ))( x + B ( a, b )) . In the rest of this section we will denote E ( a, b ) only by E every time it causesno ambiguity. Let P denote the prime of K above 2 (is inert). To the curves E ( a, b ) are associated the following quantities:∆( E ) = 2 ( ABC ) = 2 ( αβγ ) φ ( a, b ) c ( E ) = 2 ( A + AB + B ) = 2 ( AB + BC + AC ) c ( E ) = − ( C + 2 B )( A + 2 B )(2 A + B ) j ( E ) = 2 ( A + AB + B ) ( ABC ) and since ( αβγ ) = P the discriminant of E takes the following values∆( E ) = ( P P c p if 13 ∤ a + b P P c p if 13 | a + b Proposition 3.1
Let P be a prime of K distinct from P and P . The curves E ( a, b ) have good or multiplicative reduction at P . Moreover, the curves havegood ( ν P ( N E ) = 0 ) or bad additive ( ν P ( N E ) = 2 ) reduction at P if | a + b or ∤ a + b , respectively. In particular, E has multiplicative reduction atprimes dividing c . Proof:
To the results used in this proof we followed [22]. Let P be has inthe hypothesis and observe that υ P (∆( E )) = 2 pυ P ( c ). Then if P ∤ c we have υ P (∆) = 0 and the curve has good reduction. It follows from proposition 2.1that A ( a, b ), B ( a, b ) and C ( a, b ) are pairwise coprime outside P and recallthat the three are divisible by P . If P | c then P must divide only one among A, B or C . From the form of c it can be seen that υ P ( c ) = 0. Also, υ P (∆) > E has multiplicative reduction at P . Moreover, we see from ∆( E ) that if13 | a + b the equation is not minimal and E has good reduction at P . On theother hand if 13 ∤ a + b then ν P (∆) = 6 and ν P ( c ) > P . 5 Proposition 3.2
The short Weierstrass model of the curves E ( a, b ) is definedover Q ( √ . Proof:
First observe that σ (mod σ ) generates Gal( K/ Q ( √ E are defined over K they are invariant under σ and in particular j ( E )is invariant by σ by definition. We also have that σ ( A ) = B, σ ( B ) = C, σ ( C ) = A, and from j ( E ) = 2 ( AB + BC + CA ) ( ABC ) it is clear that j is also invariant under σ . Then the j -invariant that a priori belonged to K of degree 6, in reality is in Q ( √ E ( a, b ) in theshort Weierstrass form to get a model E : y = x + a x + a , where a = − AB + BC + CA ) a = − A + 3 A B − AB − B )Since a is clearly invariant under σ and a = − A + 3 A B − AB − B ) == − − B − C ) + 3( − B − C ) B − − B − C ) B − B ) == − B + 3 B C − BC − C ) = σ ( a )we conclude that the short Weierstrass model is already defined over Q ( √ (cid:4) Writing E in the short Weierstrass form we get an elliptic curve E definedover Q ( √
13) given by E : y = x + a ( a, b ) x + a ( a, b ) ,a ( a, b ) = (216 w − a + ( − w + 5616) a b +(1728 w − a b + ( − w + 5616) ab +(216 w − b ,a ( a, b ) = ( − w + 44928) a + (49248 w − a b +( − w + 471744) a b + (152928 w − a b ++( − w + 471744) a b + (49248 w − ab ++( − w + 44928) b + (50193 w + 182520) b , where w = 13. Writing a curve in short Weierstrass form changes the values of∆, c and c according to ∆( E ) = 6 ∆( E ), c ( E ) = 6 c ( E ) and c ( E ) =6 c ( E ). Observe that 2 is inert in Q ( √
13) (and in K ) and denote by 2 and w the ideals in Q ( √
13) above 2 and 13, respectively.6 roposition 3.3
The possible values for the conductors of E ( a, b ) are N E = 2 s w rad ( c ) , where s = 3 , and rad ( c ) is the product of the prime factors of c . Moreover, if | a + b then s = 3 if | a + b and s = 4 if ∤ a + b . Proof:
As in proposition 3.1 we followed the results in [22] to compute theconductor. Since the primes dividing 6 do not ramify in K/ Q ( √
13) and do notdivide c the conductor of E and E is same at these primes.Since ( w ) = P in K we see from Proposition 3.1 that ν w (∆( E )) = 4 or2 if 13 | a + b or 13 ∤ a + b , respectively. Also, ν w ( c ( E )) > ≥ ν w ( N E ) = 2.It easily can be seen that ν (∆( E )) = 16, ν ( c ( E )) = 11 and ν ( c ( E )) ≥
8. Table IV in [22] tell us that the equation is not minimal and after a changeof variables we have ν (∆( E )) = 4, ν ( c ( E )) = 5 and ν ( c ( E )) ≥
4. Wecheck that in the columns where ν ( c ( E )) = 5 and ν (∆( E )) = 4 we have ν ( N E ) = 2 , ,
4. The conductors at 2 of the curves E (1 , −
1) and E (1 ,
1) are 2 and 2 , respectively. The case s = 2 never happens. This is a direct consequenceof Proposition 2 in [22] by taking r = w/ / a, b ) ≡ (1 , , (0 ,
1) (mod 2)or r = 0 if ( a, b ) ≡ (1 ,
1) (mod 2). Since the same proposition 2 guaranteesthat all the possible conductors at 2 will occur for the pairs ( a, b ) modulo 4 weuse SAGE to compute the conductor for all these pairs and easily verify thestatement by inspection. (cid:4)
From now on we will write E to denote E . E ( a, b ) Let ρ E,p : G Q ( √ → GL ( Q p ) be the p -adic representation associated with E and ¯ ρ E,p its reduction modulo p . In this section we will prove that ρ E,p ismodular and that ¯ ρ E,p is irreducible for p >
97. These results together allow usto apply the lowering the level theorems for Hilbert modular forms.
Theorem 4.1
Let F be a totally real cyclic number field and E and ellipticcurve defined over F . Suppose that 3 splits in F and E has good reduction atthe primes above 3. Then E is modular. Let N E denote the conductor of E and put ρ = ρ E, . Since the representation¯ ρ = ¯ ρ E, : Gal( ¯ Q / Q ( √ → GL ( F ) is odd it is absolutely irreducible if andonly if it is irreducible. The following key lemma is a known result and aconsequence of the work of Langlands-Tunnell. For references where it is usedsee [21], [26], [14] and [17]. Lemma 4.2 If ¯ ρ is irreducible then it is modular arising from an Hilbert new-form over F of parallel weight (2 , . Moreover, if ρ is ordinary we can supposethat f is ordinary at 3. roof: Let t be a prime in F above 3.For the second statement of the theorem, by hypothesis we know that¯ ρ | I t = (cid:18) ¯ χ ∗ (cid:19) , where χ is the 3-adic cyclotomic character. It is know from the work of Jarvis,Rajaei and Fujiwara that since ¯ ρ f, | D t ≡ ¯ ρ | D t and ρ | D t is a Barsotti-Taterepresentation for all t (because E has good reduction at 3) we can choose f to have parallel weight (2 ,
2) and level coprime with 3, hence ρ f, | D t is alsoBarsotti-Tate. Since ¯ ρ f, | D t is ordinary we can apply a result of Breuil (see [4],chapter 9) to conclude that ρ f, is ordinary. (cid:4) We will now prove the theorem. A similar argument but over Q was givenby the first author in [11]. Proof (of theorem 4.1): We divide the proof into three cases:(1) Suppose that ¯ ρ and ¯ ρ | G Q ( √− are both abs. irreducible. Here we applycorollary 2.1.3 in [19]. Condition (1) holds because E has good reductionat the primes above 3 and 3 splits in F . Lemma 4.2 guarantees condition(2) and (3) is obvious. Then ρ is modular.(2) Suppose that ¯ ρ is abs. irreducible and ¯ ρ | G F ( √− abs. reducible. Thismeans that the image of P (¯ ρ ) is Dihedral. Namely, that the image of ¯ ρ iscontained in the normalizer N of a Cartan subgroup C of GL (¯ F ) but notcontained in C . Moreover, the restriction to Q ( √−
3) of our representationhas its image inside C . Thus, the composition of ¯ ρ with the quotient N/C ,Gal( ¯ Q /F ) → N → N/C, (7)gives the quadratic character of F ( √− /F which ramifies at 3 because 3is unramified in F .Let t be a prime in F above 3. Since E has good reduction at 3, therestriction of the residual representation ¯ ρ to the inertia subgroup I t hasonly two possibilities¯ ρ | I t = (cid:18) ¯ χ ∗ (cid:19) or (cid:18) ψ ψ (cid:19) , where χ is the 3-adic cyclotomic character and ψ is the fundamentalcharacter of level 2.If we suppose that ¯ ρ | I t acts through level 2 fundamental characters, theimage of I t by P (¯ ρ ) gives a cyclic group of order 4 >
2, thus it has to becontained in P ( C ) (if it not contained in P ( C ) and has order 4 it mustbe isomorphic to C × C ). But this implies that the quadratic characterdefined by composition (7) should be unramified at 3, contradicting thefact that this character corresponds to F ( √− ρ | I t is reducible. Since ρ is crystallinewith Hodge-Tate weights 0 and 1 and 3 splits in F we apply a resultof Breuil (see [4], chapter 9) to conclude that ρ | D t is reducible henceordinary. Indeed, we can apply this result of classification of crystalline8epresentations at 3 because the highest Hodge-Tate weight is w = 1 andso 3 > w + 1. Since t is arbitrary the previous holds for all primes t above3 hence ρ is ordinary. Therefore, we can suppose that the form f given byLemma 4.2 is ordinary thus all conditions are satisfied to apply Theorem5.1 in [25] to conclude that ρ is modular.(3) Suppose that ¯ ρ is abs. reducible. We exclude again the case of the fun-damental characters of level 2, but this time this is automatic because ofreducibility. Then ¯ ρ ss = ǫ ⊕ ǫ − ¯ χ, where ǫ ramifies only at the primes of N E . Also, since the representation isodd and F is real, reducibility must take place over F . Again by the resultof Breuil on crystalline representations we conclude that ρ | D t is reducible.Observe now that ǫ must be quadratic because F ∗ has two elements then ǫ/ ( ǫ − ¯ χ ) = ¯ χ and the extension F ( √− / Q is abelian because F is cyclicand disjoint from Q ( √− ρ is modular. (cid:4) Proposition 4.3
Let p > be a prime. The representation ¯ ρ E,p is absolutelyirreducible.
Proof:
Since ¯ ρ E,p is odd and Q ( √
13) is totally real it is known that ¯ ρ E,p isabsolutely irreducible if and only if it is irreducible then we only need to ruleout the case where ¯ ρ E,p is reducible and has the form¯ ρ E,p = (cid:18) ǫ − χ p ∗ ǫ (cid:19) , (8)where χ p is the mod p cyclotomic character and ǫ is a character of G Q ( √ with values in F p . Since the image of inertia at semistable primes is of theform (cid:18) ∗ (cid:19) the conductor of ǫ only contains additive primes. By the workof Carayol the conductor at bad additive primes of ¯ ρ E,p is the same as that of ρ E,p . Since the conductors of ǫ and ǫ − are equal it follows from proposition 3.3that the cond( ǫ ) = P P or P P . The characters of G Q ( √ with conductordividing P P are in correspondence with the characters of the finite group H = ( O Q ( √ / P P ) ∗ ∼ = Z / Z × Z / Z × Z / Z . The group of characters of H is dual of H then all the characters have orderdividing 12. In particular ǫ is a root of the polynomial q := x − p )for any p . Let P be a prime above 3. By taking traces on equality (8) we get a P ≡ ǫ (Frob P ) + 3 ǫ − (Frob P ) (mod p) , ǫ (Frob P ) satisfies for any p the polynomial q := x − a P x +3 (mod p ). Let ζ = ζ , then the resultant of q and q is given byres( q , q ) = Y i =1 ( a P + q a P − − ζ i )( a P − q a P − − ζ i )= Y i =1 ( ζ i − a P ζ i + 3)Since | a P | ≤ ǫ (Frob P ) is a common root of the q i (mod p ) then res( q , q ) ≡ p ), which is impossible if p >
97. Thus ¯ ρ E,p is absolutely irreducible if p > (cid:4)
With the theorems above we are now able to lower the level. See Jarvis-Meekin [18] for an application of the level lowering results for Hilbert modularforms in [23], [15] and [16]. In the present case we apply these results alongthe same lines. Denote by S ( N ) the set of Hilbert modular cusp forms ofparallel weight (2 ,
2) and level N . It follows from the modularity that thereexists a newform f in S ( P i P rad( c )) where i = 3 or 4 such that ρ E,p isisomorphic to the p -adic representation associated with f , which we denoteby ρ f ,p . Since the semistable primes of E , i.e. those dividing c , appear to a p -th power in the discriminant ∆( E ) we know by an argument of Hellegouarchthat the representation ¯ ρ E,p will not ramify at these primes. Furthermore, whenreducing to the residual representation the conductor at the bad additive primescan not decrease hence ¯ ρ E,p has conductor equal to that of ρ E,p without thefactor rad( c ), that is P i P with i = 3 or 4. Since ρ E,p is modular then ¯ ρ E,p ,when irreducible, is modular and by the results on level lowering for Hilbertmodular forms we know that there exists a newform f in S ( P i P ) such thatits associated mod p Galois representation satisfies ρ E,p ≡ ρ f ,p ≡ ρ f,p (mod P ) . (9) In this section we will find a contradiction to congruence (9). This showsthat the Frey-curves associated with primitive non-trivial first case solutions( a, b, c ) to equation (5) or (6) can not exist and ends the proof of part (I) inTheorem 1.2. To find the desired contradiction we use the trace values a L ( ρ E,p )and a L ( ρ f,p ) for some primes L of Q ( √
13) and the Hilbert modular newforms f in the spaces predicted in the previous section. Let w ∈ Q ( √
13) be such that w = 13 and consider the following prime ideals in Q ( √ L = h i , L = h w i L = h ( w + 1) i , L = h ( − w + 1) i ,L = h ( w + 9) i , L = h ( − w + 9) i ,L = h ( − w − i , L = h ( − w + 5) i ,L = h (3 w + 1) i , L = h (3 w − i ,L = h i , L = h i , L = h i .
10n one hand, to obtain the values of a L ( ρ f,p ), with the aid of John Voight weused algorithms to compute Hilbert modular forms implemented in MAGMA[3] (an expository account can be found in [10]). John Voight gave us two listscorresponding to all forms with integer coefficients such that a L = a L = 0and of levels P s P for s = 3 ,
4. With MAGMA we have done the same toall dividing levels and by putting together both informations we obtained allnewforms in the spaces S ( P s P ) for s = 3 , Q f = Q . A list of co-efficients corresponding to the newforms obtained this way can be found in theappendix A. Moreover, a consequence of the method used is that any newformin the two previous spaces with Q f strictly containing Q must have a Fouriercoefficient outside Q at the prime L above 3. John Voight also computed thefactorization of the characteristic polynomial of the Hecke operator T L in bothspaces (see appendix B).On the other hand, for every prime L in Q ( √
13) of good reduction for E ,such that L is above a rational prime l ≤
29 and l = 19, we use SAGE to gothrough all the possible residual elliptic curves for all pairs ( a, b ) ∈ F l × F l andcompute all the possible values for a L ( ρ E,p ) = a L ( E ): a L ∈ {− , − } ,a L ∈ {− , − , } ,a L ∈ {− , − , } ,a L ∈ { , − , − , − } ,a L ∈ {− , , , − , , − , } ,a L ∈ { , , , , − , − } ,a L ∈ { , , , − , − , − } ,a L ∈ { , , , , − , − , − , − } ,a L ∈ { , , , − , − , − } ,a L ∈ { , , , − , − , − , − , − } ,a L ∈ { , , , , − , − , − , − , − } Before proceeding to eliminate the newforms we divide them into two sets: • S1: The newforms in S ( P i P ) for i = 3 , Q f = Q . • S2: The newforms in the same levels with Q f strictly containing Q .Note that equations (5) and (6) have trivial solutions (1 , , ± (0 , , ± (1 , , , − , − , , E (1 , E (0 ,
1) and E (1 , −
1) that indeed exist and so there must benewforms associated with them in S1 which a priori will not be possible toeliminate only by comparing the a L .Going through all the forms in S a L ′ s with the possibilities for our Frey-curves we immediately eliminate allexcept 4 newforms. Here we have eliminated a newform if one of its coefficients a L is not on the corresponding list above. This can be done because the valueof p in the statement of Theorem 1.2 is very large hence congruence (9), whenspecified at a trace at L for a prime L of small norm, does not hold modulosuch large prime p unless a L ( f ) = a L ( E ). For example, the first form in the11ppendix satisfies a L ( f ) = − a L ( E ) ∈ {− , − , } it is clear that − ≡ − , − , p ) can not hold for p >
11. The four remaining newformscorrespond to the trivial solutions above plus the twist by − E (1 , E (1 ,
1) has level P P and the other three P P . Intable 1 we list their first eigenvalues. a L a L a L a L a L a L a L a L a L a L a L f -1 1 7 3 1 7 2 -7 -3 -1 3 f -1 1 3 7 -7 -1 2 -3 -7 -1 3 f -1 -3 -1 -5 5 -9 -6 -3 1 -5 15 f -3 -1 1 -3 -3 -9 -2 -7 5 -11 -15Table 1: Values of a L To be able to eliminate these newforms we need to use the extra conditionson d and a + b . Recall that the solutions ( a, b, c ) to equation (5) or (6) satisfy d | a + b . Recomputing the possibilities for some a L but with this extra con-dition we find that a L = − a L = − d = 3) , a L = − d = 5) , a L = −
11 (if d = 7) or a L = −
15 (if d = 11). By checking in table 1 we seethat any of the previous conditions is enough to eliminate all f i except for f .Actually, f is the newform associated with the trivial solution (1 , − ,
0) andcan not be eliminated this way as expected. Finally, if we assume that the so-lution is first case, Proposition 3.1 together with condition 13 ∤ a + b guaranteesthat when restricted to Gal( ¯ Q /K ) the representations ρ f ,p and ρ E,p will havedifferent inertia at P and thus can not be isomorphic modulo P .To finish the argument we have to eliminate also the newforms in S
2. Recallthat we know the factorization (appendix B) of the characteristic polynomial of T L which we denote by p . If for f in S a L ( E ) ≡ c L ( f ) (mod P ) . Let p c ( x ) be the minimal polynomial of c L ( f ) which must be a non-linear factorof p . Thus, p c ( a L ( E )) ≡ p c ( c L ( f )) ≡ P ) (10)and p c ( a L ( E )) = 0 because c L ( f ) Z . Since a L ( E ) ∈ { , − } by computing p c (3) and p c ( −
1) for all p c a non-linear factors of p we have all the possibilitiesfor p c ( a L ( E )) and we can see that congruence (10) can not hold if p > p > (cid:4) Remark 5.1
It is also possible to eliminate the newforms in S without know-ing the factorization of p but this would result in the bound p > for theexponent. Indeed, let p c = P r n x n be the minimal polynomial of a non integer c L ( f ) . All the roots c σL satisfy the Weil bound since they are coefficients ofthe conjugated form f σ . Moreover, by knowing the dimension of S ( P s P ) wecan bound all | r n | using the binomial coefficients. Putting these bounds togetherwe find only a finite number of possibilities for the non-zero value p c ( a L ( E )) .The details for this argument can be found in the first version of this work athttp://arxiv.org/abs/1112.4521 x + y = ( x + y ) φ ( x , y ) = 10 z p , (11)Proposition 2.3 and Corollary 2.2, that a solution ( a, b, c ) must verify 10 | a + b .To a primitive solution ( a, b, c ) we now attach the Frey-curve E ( a , b ). Observealso that 4 ∤ a + b by looking modulo 4. It now follows from Proposition 3.3,modularity and lowering the level that the set S1 will only have newforms oflevel P P . This means that after comparing the values a L as before, weeliminate all newforms except for the one corresponding to the curve E (1 , | a + b is enough to deal with thisnewform. In fact, recall that in this case the Frey curve has a = −
2, and thisis different from the corresponding coefficient a of E (1 , (cid:4) Appendix a L a L a L a L a L a L a L a L a L a L a L a L -3 -3 -3 -3 -4 -4 -9 2 2 -13 -18-3 -1 -5 -1 -9 5 -6 1 -3 -5 15-3 -1 -1 3 3 9 2 -7 5 11 15-3 -1 1 -3 -3 -9 -2 -7 5 -11 -15-3 -1 5 1 9 -5 6 1 -3 5 -15-3 1 -7 -7 3 -1 6 7 -9 1 9-3 1 7 7 -3 1 -6 7 -9 -1 -9-1 -3 -3 1 -9 -3 -2 5 -7 -11 -15-1 -3 -1 -5 5 -9 -6 -3 1 -5 15-1 -3 1 5 -5 9 6 -3 1 5 -15-1 -3 3 -1 9 3 2 5 -7 11 15-1 -1 3 3 -6 -6 1 0 0 5 22-1 1 -7 -3 -1 -7 -2 -7 -3 1 -3-1 1 -5 7 5 3 -6 1 5 -1 3-1 1 5 -7 -5 -3 6 1 5 1 -3-1 1 7 3 1 7 2 -7 -3 -1 3-1 3 -1 7 2 2 -7 -8 0 1 6-1 3 1 -7 -2 -2 7 -8 0 -1 -60 0 6 6 8 8 -6 2 2 -10 -181 -3 -7 -7 -1 3 6 -9 7 1 91 -3 7 7 1 -3 -6 -9 7 -1 -91 -1 -7 5 -3 -5 6 5 1 1 -31 -1 -3 -7 -7 -1 -2 -3 -7 1 -31 -1 3 7 7 1 2 -3 -7 -1 31 -1 7 -5 3 5 -6 5 1 -1 31 1 -3 -3 -1 -1 6 3 3 13 211 1 -3 -3 0 0 -1 6 6 -13 141 1 -3 -3 4 4 -9 -6 -6 11 -181 1 3 3 -4 -4 9 -6 -6 -11 18Table 2: a L values for newforms of level P P L a L a L a L a L a L a L a L a L a L a L a L values for newforms of level P P L a L a L a L a L a L a L a L a L a L a L -1 -1 -3 -3 1 1 6 3 3 13 21-1 -1 3 3 -1 -1 -6 3 3 -13 -21-1 -1 3 3 0 0 1 6 6 13 -14-1 -1 3 3 4 4 9 -6 -6 -11 18-1 -1 3 7 7 -1 -2 3 7 -1 3-1 -1 5 -7 5 -3 -6 -1 -5 1 -3-1 -1 7 -5 3 -5 6 -5 -1 -1 3-1 -1 7 3 -1 7 -2 7 3 -1 3-1 1 -7 5 3 5 6 5 1 1 -3-1 1 -3 -7 7 1 -2 -3 -7 1 -3-1 1 -3 -3 -6 6 1 0 0 -5 -22-1 1 -3 -3 -1 1 -6 -3 -3 13 21-1 1 -3 -3 0 0 1 -6 -6 -13 14-1 1 -3 -3 3 -3 10 9 9 -11 5-1 1 -3 -3 4 -4 9 6 6 11 -18-1 1 3 3 -4 4 -9 6 6 -11 18-1 1 3 3 -3 3 -10 9 9 11 -5-1 1 3 3 0 0 -1 -6 -6 13 -14-1 1 3 3 1 -1 6 -3 -3 -13 -21-1 1 3 3 6 -6 -1 0 0 5 22-1 1 3 7 -7 -1 2 -3 -7 -1 3-1 1 7 -5 -3 -5 -6 5 1 -1 3-1 3 -7 -7 1 -3 6 -9 7 1 9-1 3 -3 1 9 -3 2 -5 7 -11 -15-1 3 -1 -5 -5 -9 6 3 -1 -5 15-1 3 1 5 5 9 -6 3 -1 5 -15-1 3 3 -1 -9 3 -2 -5 7 11 15-1 3 7 7 -1 3 -6 -9 7 -1 -90 0 -6 -6 -8 8 -6 -2 -2 10 180 0 -6 -6 8 -8 -6 -2 -2 10 180 0 -6 -6 8 8 6 2 2 10 180 0 -3 -3 -4 -4 9 -1 -1 -2 6Table 4: a L values for newforms of level P P (cont.)16 L a L a L a L a L a L a L a L a L a L a L a L values for newforms of level P P (cont.)17 L a L a L a L a L a L a L a L a L a L a L a L values for newforms of level P P (cont.)18 L a L a L a L a L a L a L a L a L a L a L a L values for newforms of level P P (cont.) p The polynomial p on S ( P P ) has the following factors:[ x − , , [ x − , , [ x − , , [ x, , [ x + 1 , , [ x + 2 , , [ x + 3 , x − x +2 , , [ x − x − , , [ x − x +1 , , [ x − x − , , [ x − x − , , [ x − x − , , [ x − x − , , [ x − x − , , [ x − x − , , [ x − , , [ x − , , [ x − , , [ x − , , [ x + x − , , [ x + x − , , [ x + x − , , [ x + x − , , [ x +2 x − , , [ x + 2 x − , , [ x + 3 x − , , [ x + 3 x + 1 , , [ x + 4 x + 2 , , [ x − x +3 x +1 , , [ x − x − x +13 , , [ x − x − x +7 , , [ x − x − x +1 , , [ x − x − , , [ x − x +7 , , [ x − x − , , [ x − x +1 , , [ x +2 x − x − , , [ x +2 x − x − , , [ x + 3 x − x − , , [ x + 4 x + 3 x − , , [ x − x − x + 10 x +2 , , [ x − x + 8 x − , , [ x − x − x + 23 x − , , [ x − x − x + 22 x − , , [ x − x − x +8 x − , , [ x − x − x +8 x +4 , , [ x − x − x +10 x +1 , , [ x − x − x + 8 x − , , [ x + 2 x − x − x − , , [ x + 2 x − x − x − , , [ x +2 x − x − x +4 , , [ x +2 x − x − x +1 , , [ x +2 x − x − x − , , [ x +3 x − x − x − , , [ x +4 x − x − x +2 , , [ x +4 x − x − , , [ x − x − x +36 x +18 x − x +29 , , [ x − x − x +31 x +15 x − x − , , [ x − x − x +23 x +15 x − x − , , [ x − x − x +16 x +35 x − x − , , [ x − x − x + 11 x + 49 x − x − , , [ x + x − x − x +49 x +27 x − , , [ x +2 x − x − x +35 x +26 x − , , [ x +3 x − x − x + 18 x + 68 x + 29 , , [ x + 3 x − x − x + 15 x + 25 x − , , [ x + 3 x − x − x +15 x +13 x − , , [ x − x + x +48 x − x − x +115 x − x +4 , , [ x − x − x + 57 x + 124 x − x − x + 588 x + 16 , , [ x + 3 x − x − x +124 x +327 x − x − x +16 , , [ x +6 x + x − x − x +54 x + 115 x + 48 x + 4 , , [ x − x − x + 50 x − x − x + 125 x + 50 x − x +4 , , [ x − x − x +32 x +38 x − x − x +75 x +37 x +4 , , [ x − x − x +34 x +75 x − x − x +106 x − x − , , [ x − x − x +16 x +114 x − x − x + 165 x + 181 x − , , [ x + x − x − x + 114 x +76 x − x − x + 181 x + 128 , , [ x + 2 x − x − x + 75 x + 158 x − x − x − x + 4 , , [ x + 3 x − x − x + 38 x + 100 x − x − x +197 x − , , [ x +4 x − x − x − x +156 x +125 x − x − x − , , [ x − x − x +30 x +153 x − x − x +863 x − x − x +242 x +22 x − , , [ x + x − x − x + 153 x + 276 x − x − x − x + 513 x +242 x − x − , , [ x − x − x + 48 x + 261 x − x − x +2024 x +3449 x − x − x +6520 x +3355 x − x − x +1104 x − , , [ x +2 x − x − x +261 x +442 x − x − x +3449 x +4958 x − x − x +3355 x +4294 x − x − x − , , [ x − x − x + 184 x + 101 x − x + 1048 x + 13080 x − x − x +52906 x + 54352 x − x − x + 56412 x + 5600 x − x − x +568 , , [ x − x − x + 150 x + 387 x − x − x + 16799 x − x − x + 59322 x + 117310 x − x − x + 190984 x − x + 32604 x − x − , , [ x − x − x + 82 x + 318 x − x − x + 5426 x + 5939 x − x − x + 37297 x + 2528 x − x + 10028 x + 20669 x − x − x + 1259 , , [ x − x − x +84 x + 308 x − x − x + 4994 x + 4927 x − x − x +19011 x +6382 x − x − x +2003 x +627 x − x − , , [ x − x − x +78 x +270 x − x − x +4416 x +4153 x − x − x +22693 x +3772 x − x +782 x +7699 x − x − x +13 , , [ x − x − x +80 x +264 x − x − x +4500 x +3737 x − x − x +22123 x − x − x + 5878 x + 5249 x − x + 688 x − , , [ x − x − x + 64 x + 461 x − x − x + 4968 x + 12385 x − x − x + 24760 x + 31587 x − x − x − x + 910 x +56 x − , , [ x + 2 x − x − x + 461 x + 802 x − x − x +12385 x +15862 x − x − x +31587 x +14558 x − x +640 x +910 x − x − , , [ x + 3 x − x − x + 308 x + 921 x − x − x + 4927 x + 13943 x − x − x + 6382 x + 10700 x − x − x + 627 x + 20 x − , , [ x + 3 x − x − x + 318 x + 915 x − x − x + 5939 x + 18603 x − x − x + 2528 x + 41228 x +10028 x − x − x + 1970 x + 1259 , , [ x + 3 x − x − x +264 x + 837 x − x − x + 3737 x + 13433 x − x − x − x + 18304 x + 5878 x − x − x − x − , , [ x + 3 x − x − x + 270 x + 815 x − x − x + 4153 x + 13377 x − x − x + 3772 x + 20184 x + 782 x − x − x + 514 x + 13 , , [ x +4 x − x − x + 387 x + 2233 x − x − x − x +65971 x + 59322 x − x − x + 21264 x + 190984 x + 132221 x +32604 x + 1200 x − , , [ x + 6 x − x − x + 101 x + 2206 x +1048 x − x − x + 39490 x + 52906 x − x − x +23122 x + 56412 x − x − x + 1480 x + 568 , , [ x − x − x +164 x + 1304 x − x − x + 51323 x + 147499 x − x − x +2322310 x +2957083 x − x − x +19892990 x +10323950 x − x − x +28202264 x +2963409 x − x +220229 x + 3889959 x − x − x + 72460 x − , , [ x − x − x + 160 x + 1210 x − x − x + 47937 x + 126405 x − x − x + 2085994 x + 2381831 x − x − x +17541450 x +9224648 x − x − x +27067090 x +6187223 x − x − x +5264635 x +99800 x − x +47264 x +14987 , , [ x +3 x − x − x + 1304 x + 3858 x − x − x + 147499 x +427049 x − x − x + 2957083 x + 8374150 x − x − x +10323950 x +30317349 x − x − x +2963409 x +14792310 x +220229 x − x − x +380960 x +72460 x +2647 , , [ x +20 x − x − x + 1210 x + 3680 x − x − x + 126405 x +390929 x − x − x + 2381831 x + 7410738 x − x − x +9224648 x +27292133 x − x − x +6187223 x +16180684 x − x − x + 99800 x + 738836 x + 47264 x − , p on S ( P P ) has the following factors:[ x − , , [ x − , , [ x − , , [ x, , [ x + 1 , , [ x + 2 , , [ x + 3 , , [ x − x +2 , , [ x − x − , , [ x − x + 1 , , [ x − x − , , [ x − x − , , [ x − x − , , [ x − x − , , [ x − x − , , [ x − , , [ x − , , [ x − , , [ x − , , [ x + x − , , [ x + x − , , [ x + x − , , [ x + 2 x − , , [ x + 3 x − , , [ x + 3 x +1 , , [ x − x − x +13 , , [ x − x +7 , , [ x − x − , , [ x +2 x − x − , , [ x +2 x − x − , , [ x + 4 x + 3 x − , , [ x − x − x + 10 x + 2 , , [ x − x − x +22 x − , , [ x − x − x +10 x +1 , , [ x +2 x − x − x − , , [ x +2 x − x − x +4 , , [ x +2 x − x − x − , , [ x +3 x − x − x − , , [ x +4 x − x − , , [ x − x − x +36 x +18 x − x +29 , , [ x − x − x +31 x +15 x − x − , , [ x − x − x +23 x +15 x − x − , , [ x − x − x +11 x +49 x − x − , , [ x + 2 x − x − x + 35 x + 26 x − , , [ x − x + x + 48 x − x − x +115 x − x +4 , , [ x − x − x +57 x +124 x − x − x +588 x +16 , , [ x − x − x +50 x − x − x +125 x +50 x − x +4 , , [ x − x − x +16 x +114 x − x − x +165 x +181 x − , , [ x +2 x − x − x + 75 x + 158 x − x − x − x + 4 , , [ x + 3 x − x − x + 38 x +100 x − x − x +37 x − , , [ x + x − x − x +153 x +276 x − x − x − x +513 x +242 x − x − , , [ x − x − x +48 x +261 x − x − x + 2024 x + 3449 x − x − x + 6520 x + 3355 x − x − x +1104 x − , , [ x − x − x +150 x +387 x − x − x + 16799 x − x − x + 59322 x + 117310 x − x − x + 190984 x − x + 32604 x − x − , , [ x − x − x +64 x + 461 x − x − x + 4968 x + 12385 x − x − x +24760 x + 31587 x − x − x − x + 910 x + 56 x − , , [ x +3 x − x − x +308 x +921 x − x − x +4927 x +13943 x − x − x +6382 x +10700 x − x − x +627 x +20 x − , , [ x +3 x − x − x +318 x +915 x − x − x +5939 x +18603 x − x − x + 2528 x + 41228 x + 10028 x − x − x + 1970 x +1259 , , [ x + 3 x − x − x + 264 x + 837 x − x − x +3737 x + 13433 x − x − x − x + 18304 x + 5878 x − x − x − x − , , [ x + 3 x − x − x + 270 x + 815 x − x − x + 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