Explicit Modular Approaches to Generalized Fermat Equations

Explicit Modular Approaches to Generalized FermatEquations

David Brown

University of Wisconsin-MadisonSlides available at http://www.math.wisc.edu/~brownda/slides/

Emory University Colloquium

February 14, 2011

http://www.math.wisc.edu/~brownda/slides/

Basic Problem (Solving Diophantine Equations)

Let f1, . . . , fm ∈ Z[x1, ..., xn] be polynomials and let R be a ring (e.g.,R = Z, Q).

Problem

Describe the set{(a1, . . . , an) ∈ Rn : ∀i , fi (a1, . . . , an) = 0

}.

Fact

Solving diophantine equations is hard.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 2 / 63

Basic Problem (Solving Diophantine Equations)

Let f1, . . . , fm ∈ Z[x1, ..., xn] be polynomials and let R be a ring (e.g.,R = Z, Q).

Problem

Describe the set{(a1, . . . , an) ∈ Rn : ∀i , fi (a1, . . . , an) = 0

}.

Fact

Solving diophantine equations is hard.


Fermat’s Last Theorem

Theorem (Wiles; Taylor-Wiles 1995)

The only integer solutions to the equation

xn + yn = zn, n ≥ 3

satisfy xyz = 0.


Template for the proof of FLT

Step 1: Assume there is a counterexample ap + bp = cp.

Step 2: (Frey) Build an elliptic curve with strange properties:

E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.

Step 3: (Ribet) Show that the Frey curve E(a,b,c) is not modular.

Step 4: Prove that every elliptic curve over Q is modular.





E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.







E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.







E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.







E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.







E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.




Modularity is now a theorem

Theorem (Wiles 1995; Breuil-Conrad-Diamond-Taylor 2002)

Every elliptic curve over Q is modular.


Elliptic Curves

E : y 2 = x3 + ax + b


Elliptic Curves - point at infinity

E : zy 2 = x3 + axz2 + bz3

∞ = [0 : 1 : 0]


Elliptic Curves – addition

E : y 2 = x3 + ax + b

P = (x0, y0) ∈ Q2

Q = (x1, y1) ∈ Q2



E : y 2 = x3 + ax + b

P = (x0, y0) ∈ Q2

Q = (x1, y1) ∈ Q2

R = (x2, y2) ∈ Q2



E : y 2 = x3 + ax + b

P = (x0, y0) ∈ Q2

Q = (x1, y1) ∈ Q2

R = (x2, y2) ∈ Q2

P + Q = (x2,−y2) ∈ Q2



E : y 2 = x3 + ax + b

E (Q)× E (Q)→ E (Q)

(P , Q) 7→ P + Q


Elliptic Curves - duplication

E : y 2 = x3 + ax + b

P = (x0, y0) ∈ Q2

2P = (x3, y3) ∈ Q2


Elliptic Curves - duplication

E : y 2 = x3 + ax + b

P = (x0, y0) ∈ Q2

2P = (x3, y3) ∈ Q2


Elliptic Curves – identity

E : y 2 = x3 + ax + b


Elliptic Curves – identity

E : y 2 = x3 + ax + b


Elliptic Curves – inverses

E : y 2 = x3 + ax + b


Elliptic Curves – inverses

E : y 2 = x3 + ax + b


Elliptic Curves – torsion subgroup

Let n ∈ Z be an integer.

Definition

The n-torsion subgroup E [n] of E is defined to be

ker

(E

[n]−→ E

):= {P ∈ E : nP := P + . . .+ P =∞} .


Elliptic Curves – two torsion

E : y 2 = x3 + ax + b

2P = 2Q = 2R =∞


Elliptic Curves – two torsion

E : y 2 = x3 + ax + b

2P = 2Q = 2R =∞


Elliptic Curves – structure of torsion

Let E be given by the equation y 2 = f (x) = x3 + ax + b.

E [n](C) = E [n](Q) ∼= (Z/nZ)2.

E [n](Q) may be smaller, e.g.,

E [2](Q) ∼=

{∞} if f (x) has 0 rational roots

Z/2Z, if f (x) has 1 rational roots

(Z/2Z)2, if f (x) has 3 rational roots




E [n](C) = E [n](Q) ∼= (Z/nZ)2.

E [n](Q) may be smaller,

e.g.,

E [2](Q) ∼=







E [n](C) = E [n](Q) ∼= (Z/nZ)2.

E [n](Q) may be smaller, e.g.,

E [2](Q) ∼=





Elliptic Curves – torsion

E [2](Q) ∼=





Galois Representations associated to an elliptic curve

Let E : y 2 = x3 + ax + b be an elliptic curve with a, b ∈ Q.

Let GQ := Aut(Q)

∼= lim←−KAut(K ).

Let σ ∈ GQ, P = (x , y) ∈ E (Q) Pσ = (xσ, yσ) ∈ E (Q).

If P ∈ E [n], then Pσ ∈ E [n].

Definition

The mod n Galois representation associated to E is the homomorphism

GQ → Aut(E [n]) ∼= GL2(Z/nZ).




Let GQ := Aut(Q)

∼= lim←−KAut(K ).



Definition






Let GQ := Aut(Q) ∼= lim←−KAut(K ).



Definition









Definition









Definition









Definition





E [2](Q) ∼=





Galois Representations: examples

Example

Suppose that E (Q)[2] ∼= (Z/2Z)2. (E.g., E : y 2 = x(x − 1)(x − λ) withλ ∈ Q.) Then

ρE ,2 : GQ → GL2(Z/2Z)

is the trivial homomorphism.

Example

Suppose that E (Q)[2] ∼= Z/2Z. (E.g., E : y 2 = (x2 + D)(x − λ) withD, λ ∈ Q and D > 0.) Then we can choose a basis for E (Q)[2] so thatany σ ∈ GQ acts as a matrix of the form 1 a

0 b

.


Galois Representations: examples

Example

Suppose that E (Q)[2] ∼= (Z/2Z)2. (E.g., E : y 2 = x(x − 1)(x − λ) withλ ∈ Q.) Then

ρE ,2 : GQ → GL2(Z/2Z)

is the trivial homomorphism.

Example

Suppose that E (Q)[2] ∼= Z/2Z. (E.g., E : y 2 = (x2 + D)(x − λ) withD, λ ∈ Q and D > 0.) Then we can choose a basis for E (Q)[2] so thatany σ ∈ GQ acts as a matrix of the form 1 a

0 b

.


Galois Representations from modular forms

Let H := {τ = x + yi ∈ C : y > 0} be the complex upper half plane.

The formula(a bc d

)τ =

aτ + b

cτ + ddefines an action of SL2(Z) on H.

Definition

A modular function is a complex analytic function f : H → C whichis invariant under the action of a congruence subgroup Γ ⊂ SL2(Z)such that f is holomorphic at ∞.

A modular form of weight 2k is a complex analytic functionf : H → C such that f (z)(dz)k is invariant under the action of acongruence subgroup Γ ⊂ SL2(Z) such that f is holomorphic at ∞.

f ///o/o/o/o

��O�O�O

Ef

�� O�O�O

ρf ,n ρEf ,n




The formula(a bc d

)τ =

aτ + b


Definition



f ///o/o/o/o

��O�O�O

Ef

�� O�O�O

ρf ,n ρEf ,n




The formula(a bc d

)τ =

aτ + b


Definition



f ///o/o/o/o

��O�O�O

Ef

�� O�O�O

ρf ,n ρEf ,n




The formula(a bc d

)τ =

aτ + b


Definition



f ///o/o/o/o

��O�O�O

Ef

�� O�O�O

ρf ,n ρEf ,n



Fact

Galois representations associated to modular forms are easy to understandand classify.

Theorem (Modularity)

Every elliptic curve over Q arises from a modular form.

f ///o/o/o/o

��O�O�O

Ef

�� O�O�O

ρf ,n ρEf ,n



Fact




f ///o/o/o/o

��O�O�O

Ef

�� O�O�O

ρf ,n ρEf ,n



Fact




f ///o/o/o/o

��O�O�O

Ef

�� O�O�O

ρf ,n ρEf ,n





E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.







E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.

Step 3: (Ribet) Classify possibilities for E(a,b,c),p (modularity is onetool used in this classification).

Step 4: The output of step 3 turns out to be empty!





E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.







E(a,b,c) : y 2 = x(x − ap)(x + bp)

j =28(c2p − apbp

)3

(abc)2p

∆ = 2−8(abc)2p.




Other applications of the modular method

The ideas behind the proof of FLT now permeate the study of diophantineproblems.

Theorem (Bugeaud, Mignotte, Siksek 2006)

The only Fibonacci numbers that are perfect powers are

F0 = 0, F1 = F2 = 1, F6 = 8, F12 = 144.


Other applications of the modular method

The ideas behind the proof of FLT now permeate the study of diophantineproblems.

Theorem (Bugeaud, Mignotte, Siksek 2006)

The only Fibonacci numbers that are perfect powers are

F0 = 0, F1 = F2 = 1, F6 = 8, F12 = 144.


More applications of the modular method

Theorem (Darmon, Merel 1997)

Any pairwise coprime integer solution to the equation

xn + yn = z2, n ≥ 4

satisfies xyz = 0.


Template for the proof of Darmon-Merel

Step 1: Assume there is a counterexample ap + bp = c2.


E(a,b,c) : y 2 + xy = x3 +c − 1

4x2 +

ap

26x

∆ =1

212(a2b)p

j = − 26(3ap − 4c2)3

(a2b)p

if ab is even.





E(a,b,c) : y 2 + xy = x3 +c − 1

4x2 +

ap

26x

∆ =1

212(a2b)p

j = − 26(3ap − 4c2)3

(a2b)p

if ab is even.





E(a,b,c) : y 2 = x3 + 2cx2 + apx

∆ = 26(a2b)p

j = −26(3ap − 4c2)3

(a2b)p

if ab is odd.





E(a,b,c) :

y 2 + xy = x3 + c−1

4 x2 + ap

26 x if ab is even

y 2 = x3 + 2cx2 + apx if ab is odd

Step 3: E(a,b,c) has CM (complex multiplication).Step 4: There are only finitely many E/Q with CM (up to twists).Step 5: (Easy) Find all triples (a, b, c) such that E(a,b,c) has CM.





E(a,b,c) :

y 2 + xy = x3 + c−1

4 x2 + ap

26 x if ab is even


Step 3: E(a,b,c) has CM (complex multiplication).

Step 4: There are only finitely many E/Q with CM (up to twists).Step 5: (Easy) Find all triples (a, b, c) such that E(a,b,c) has CM.





E(a,b,c) :

y 2 + xy = x3 + c−1

4 x2 + ap

26 x if ab is even


Step 3: E(a,b,c) has CM (complex multiplication).Step 4: There are only finitely many E/Q with CM (up to twists).

Step 5: (Easy) Find all triples (a, b, c) such that E(a,b,c) has CM.





E(a,b,c) :

y 2 + xy = x3 + c−1

4 x2 + ap

26 x if ab is even


Step 3: E(a,b,c) has CM (complex multiplication).Step 4: There are only finitely many E/Q with CM (up to twists).Step 5: (Easy) Find all triples (a, b, c) such that E(a,b,c) has CM.


Generalized Fermat Equations

Fix p, q, r ∈ N such that χ = 1p + 1

q + 1r − 1 < 0.

Theorem (Darmon, Granville 1995)

The equationxp + yq = z r

has only finitely many coprime solutions with xyz 6= 0.


Generalized Fermat Equations

Fix p, q, r ∈ N such that χ = 1p + 1

q + 1r − 1 < 0.

Theorem (Darmon, Granville 1995)

The equationxp + yq = z r

has only finitely many coprime solutions with xyz 6= 0.


Examples of Generalized Fermat Equations

χ =1

2+

1

3+

1

7− 1 = − 1

42< 0

Theorem (Poonen, Schaefer, Stoll 2008)

The coprime integer solutions to x2 + y 3 = z7 are the 16 triples

(±1,−1, 0), (±1, 0, 1), ±(0, 1, 1), (±3,−2, 1),

(±71,−17, 2), (±2213459, 1414, 65), (±15312283, 9262, 113),

(±21063928,−76271, 17) .


Examples of Generalized Fermat Equations

χ =1

2+

1

3+

1

7− 1 = − 1

42< 0

Theorem (Poonen, Schaefer, Stoll 2008)

The coprime integer solutions to x2 + y 3 = z7 are the 16 triples

(±1,−1, 0), (±1, 0, 1), ±(0, 1, 1), (±3,−2, 1),

(±71,−17, 2), (±2213459, 1414, 65), (±15312283, 9262, 113),

(±21063928,−76271, 17) .


Generalized Fermat Equations – Known Solutions

The ‘known’ solutions to the equation xp + yq = z r withχ = 1

p + 1q + 1

r − 1 < 0 and xyz 6= 0 are the following:

1p + 23 = 32 (−1)2p + 23 = 32 25 + 72 = 34

73 + 132 = 29 27 + 173 = 712 35 + 114 = 1222

177 + 762713 = 210639282 14143 + 22134592 = 657

92623 + 153122832 = 1137 438 + 962223 = 300429072

338 + 15490342 = 156133



The ‘known’ solutions to the equation xp + yq = z r withχ = 1

p + 1q + 1

r − 1 < 0 and xyz 6= 0 are the following:

1p + 23 = 32 (−1)2p + 23 = 32 25 + 72 = 34

73 + 132 = 29 27 + 173 = 712 35 + 114 = 1222

177 + 762713 = 210639282 14143 + 22134592 = 657

92623 + 153122832 = 1137 438 + 962223 = 300429072

338 + 15490342 = 156133



Conjecture (Beal, Granville, Tijdeman-Zagier)

This is a complete list of coprime non-zero solutions such that1p + 1

q + 1r − 1 < 0.

$100,000 prize for proof of conjecture......or even for a counterexample.





q + 1r − 1 < 0.

$100,000 prize for proof of conjecture...

...or even for a counterexample.





q + 1r − 1 < 0.

$100,000 prize for proof of conjecture......or even for a counterexample.


(p, q, r) such that χ < 0 and the solutions to xp + yq = z r have been determined.

{n, n, n} Wiles,Taylor-Wiles, building on work of many others{2, n, n} Darmon-Merel, others for small n{3, n, n} Darmon-Merel, others for small n{5, 2n, 2n} Bennett(2, 4, n) Ellenberg, Bruin, Ghioca n ≥ 4(2, n, 4) Bennett-Skinner; n ≥ 4{2, 3, n} Poonen-Shaefer-Stoll, Bruin. 6 ≤ n ≤ 9{2, 2`, 3} Chen, Dahmen, Siksek; primes 7 < ` < 1000 with ` 6= 31{3, 3, n} Bruin; n = 4, 5{3, 3, `} Kraus; primes 17 ≤ ` ≤ 10000(2, 2n, 5) Chen n ≥ 3∗

(4, 2n, 3) Bennett-Chen n ≥ 3(6, 2n, 2) Bennett-Chen n ≥ 3(2, 6, n) Bennett-Chen n ≥ 3


Main Theorem

χ = 12 + 1

3 + 110 − 1 = − 1

15 is maximal among unsolved Fermat equations.

Theorem (B., 2011)

The only coprime integer solutions to the equation

x2 + y 3 = z10

are the 12 triples

(±1,−1, 0), (±1, 0,±1), (0, 1,±1), (±3,−2,±1).

It is the first generalized Fermat equation of the form x2 + y 3 = zn

conjectured to have only trivial solutions.

(32 + (−2)3 = 1n is considered to be trivial.)


Main Theorem

χ = 12 + 1

3 + 110 − 1 = − 1


Theorem (B., 2011)


x2 + y 3 = z10

are the 12 triples

(±1,−1, 0), (±1, 0,±1), (0, 1,±1), (±3,−2,±1).





Main Theorem

χ = 12 + 1

3 + 110 − 1 = − 1


Theorem (B., 2011)


x2 + y 3 = z10

are the 12 triples

(±1,−1, 0), (±1, 0,±1), (0, 1,±1), (±3,−2,±1).





Framework for solving x2 + y 3 = zn

Step 1: Assume there is a counterexample a2 + b3 = cn.

Step 2: Study the elliptic curve

E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123cn

j = 123b3/cn.

Step 3: Explicitly classify possibilities for ρE(a,b,c),n.

Step 4: For such ρ, classify all elliptic curves E for which ρE ,n ∼= ρ.

Step 5: For each such E , find all (a, b, c) such that E ∼= E(a,b,c).





E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123cn

j = 123b3/cn.








E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123cn

j = 123b3/cn.








E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123cn

j = 123b3/cn.








E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123cn

j = 123b3/cn.






For large n, this template (conjecturally) works!

Step 3:

Explicitly classify possibilities for ρE(a,b,c),n.

For large n, there are 13 possibilities for ρE(a,b,c),n, which are‘independent of n’.




Step 3:

Explicitly classify possibilities for ρE(a,b,c),n.

For large n, there are 13 possibilities for ρE(a,b,c),n, which are‘independent of n’.




Step 4:

For a fixed ρ, classify all elliptic curves E for which ρE ,n ∼= ρ.

This would follow from a standard conjecture:

Conjecture (Frey-Mazur)

Let p > 23 be a prime and E and E ′ be elliptic curves such thatρE ,p ∼= ρE ′,p. Then E is isogenous to E ′.




Step 4:








Step 4:






Template breaks down for x2 + y 3 = z10

E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 3:

Explicitly classify possibilities for ρE(a,b,c),10.

Known tools for classifying ρE(a,b,c),10 fail.

E.g., Ribet’s level lowering theorem fails for n = 2.ρE(a,b,c),n may be reducible for both n = 2 and 5.

Definition

We say that ρ : GQ → GL2(F`) is reducible if there is some subspaceW ⊂ F2

` such that for every P ∈W , Pρ(σ) ∈W .



E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 3:




Definition





E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 3:



E.g., Ribet’s level lowering theorem fails for n = 2.

ρE(a,b,c),n may be reducible for both n = 2 and 5.

Definition





E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 3:




Definition





E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 3:




Definition




Template breaks down forth x2 + y 3 = z10

E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 4:


Using one prime is not enough.

E.g., there are infinitely many elliptic curves over Q with trivial mod 2representation (E : y 2 = x(x − 1)(x − λ)).

Multiprime approaches seem to be computationally infeasible.



E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 4:







E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 4:







E(a,b,c) : y 2 = x3 + 3bx − 2a

Step 4:







E [2](Q) ∼=





Step 3: Classifying mod 2 Galois representations

Let E be given by the equation y 2 = f (x) := x3 + 3bx − 2a

Fact

The splitting field of the polynomial f (x) completely determines ρE ,2.

The splitting field K of x3 + 3bx − 2a is unramified outside of {2, 3}and of degree at most 6.

(Hermite) There are only finitely many such fields.

These days there are sophisticated algorithms for enumerating such K .




Fact








Fact








Fact






Step 3: Progress for ` = 2

Lemma

There are elliptic curves {E1, . . . ,En} such that for every (a, b, c) suchthat a2 + b3 = c10, there is an i such that

ρE(a,b,c),2∼= ρEi ,2.

Wanted: a similar lemma for ρE(a,b,c),5.

Problem: ρE(a,b,c),5 may be reducible, thus modularity won’t help!



Lemma


ρE(a,b,c),2∼= ρEi ,2.





Lemma


ρE(a,b,c),2∼= ρEi ,2.




Parameter spaces for Galois representations

Definition

XE (n) is the parameter space for pairs (E ′, ψ), where E ′ is an elliptic curveand ψ : ρE ,n → ρE ′,n is a symplectic isomorphism of mod n Galoisrepresentations.



E [2](Q) ∼=





Parameter spaces for Galois representations

Definition

XE (n) is the parameter space for pairs (E ′, ψ), where E ′ is an elliptic curveand ψ : ρE ,n → ρE ′,n is a symplectic isomorphism of mod n Galoisrepresentations.

Example

Let E be an elliptic curve with E (Q)[2] ∼= (Z/2Z)2 (so that ρE ,2 is trivial).Then E is of the form

E : y 2 = x(x − 1)(x − λ).


Other parameter spaces

Recall that ρ : GQ → GL2(F`) is reducible if there is some subspaceW ⊂ F2


Definition

X0(p) is the parameter space for elliptic curves such that ρE ,p is reducible(more precisely – pairs (E ,W ⊂ E [p]), where E is an elliptic curve and Wis an invariant subgroup of size p).

Example (X0(5))

Let E : y 2 = x3 + 3bx − 2a, and suppose ρE ,5 is reducible. Then thereexists a t ∈ Z such that

123 b3

a2 + b3=

(t2 + 250t + 3125)3

t5.





Definition


Example (X0(5))


123 b3

a2 + b3=

(t2 + 250t + 3125)3

t5.





Definition


Example (X0(5))


123 b3

a2 + b3=

(t2 + 250t + 3125)3

t5.


Step 3: Intermediate Modular curves

Goal


X (10) //

��

X (5)

��X0(5)

��X (2)

π // X0(2) // X (1)

π : (E , ψ : ρtriv∼= ρE ,2) 7→ (E ,W ).



Goal


X (10) //

��

X (5)

��X0(5)

��X (2)

π // X0(2) // X (1)

π : (E , ψ : ρtriv∼= ρE ,2) 7→ (E ,W ).



Goal


X (10) //

��

X (5)

��X0(5)

��X (2)

π // X0(2) // X (1)

π : (E , ψ : ρtriv∼= ρE ,2) 7→ (E ,W ).


Step 3: Intermediate Modular curves: p = 2

Aut(X (2)/X (1)) ∼= GL2(F2) ∼= S3.

X0(2) is the quotient of X (2) by a transposition.

X0(2)

##GGGGGGGG

X (2)

;;wwwwwwww

##GGGG

GGGG

GX (1)

X∆

;;wwwwwwwww



Aut(X (2)/X (1)) ∼= GL2(F2) ∼= S3.

X0(2) is the quotient of X (2) by a transposition.

X0(2)

##GGGGGGGG

X (2)

;;wwwwwwww

##GGGG

GGGG

GX (1)

X∆

;;wwwwwwwww



Define X∆ to be the quotient of X (2) by the normal subgroup A3.

X∆ classifies pairs (E , z) such that z2 = j(E )− 123 = c6(E )2/∆E .

X0(2)

##GGGGGGGG

X (2)

;;wwwwwwww

##GGGG

GGGG

GX (1)

X∆

;;wwwwwwwww



Define X∆ to be the quotient of X (2) by the normal subgroup A3.

X∆ classifies pairs (E , z) such that z2 = j(E )− 123 = c6(E )2/∆E .

X0(2)

##GGGGGGGG

X (2)

;;wwwwwwww

##GGGG

GGGG

GX (1)

X∆

;;wwwwwwwww



Goal


X (10) //

��

X (5)

��X0(5)

��X (2) // X∆

// X (1)



X (10) //

��

""FF

FF

FX (5)

��X //

��

X0(5)

��X (2) // X∆

// X (1)

X classifies triples (E ,W , z) such that

z2 = j(E )− 122 = c4(E )2/∆E ,W is an invariant subspace of E [5] of order 5.



X (10) //

��

""FF

FF

FX (5)

��X //

��

X0(5)

��X (2) // X∆

// X (1)

X classifies triples (E ,W , z) such that

z2 = j(E )− 122 = c4(E )2/∆E ,W is an invariant subspace of E [5] of order 5.



E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123c10 = −3(23 · 3 · c5)2

X

��

//

��

X0(5)

��X∆

// X (1)



E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123c10 = −3(23 · 3 · c5)2

X(−3)

��

//

��

X0(5)

��X(−3∆) // X (1)

X(−3) classifies triples (E ,W , z) such that

−3z2 = c4(E )2/∆E ,W is an invariant subspace of E [5] of order 5.



E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123c10 = −3(23 · 3 · c5)2

X(−3)

��

//

��

X0(5)

��X(−3∆) // X (1)

X(−3) classifies triples (E ,W , z) such that

−3z2 = c4(E )2/∆E ,W is an invariant subspace of E [5] of order 5.



E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123c10 = −3(23 · 3 · c5)2

X(−3)

��

//

��

X0(5)

��X(−3∆) // X (1)

Reducible ρE(a,b,c),5 thus give rise to a point on X(−3)(Q).

X(−3) turns out to be an elliptic curve, with X(−3)(Q) ∼= Z/5Z.



E(a,b,c) : y 2 = x3 + 3bx − 2a

∆ = −123c10 = −3(23 · 3 · c5)2

X(−3)

��

//

��

X0(5)

��X(−3∆) // X (1)

Reducible ρE(a,b,c),5 thus give rise to a point on X(−3)(Q).

X(−3) turns out to be an elliptic curve, with X(−3)(Q) ∼= Z/5Z.


Step 3: Classifying ρE(a,b,c),`

Lemma

There are elliptic curves {E1, . . . ,En} and {E ′1, . . . ,E ′n′} such that forevery (a, b, c) such that a2 + b3 = c10, there exists an i and j such that

ρE(a,b,c),2∼= ρEi ,2

andρE(a,b,c),5

∼= ρE ′j ,5.

Thus, E(a,b,c) gives rise to a point on XEi(2)(Q) and a point on XE ′

j(5)(Q).


Step 3: Classifying ρE(a,b,c),`

Lemma

There are elliptic curves {E1, . . . ,En} and {E ′1, . . . ,E ′n′} such that forevery (a, b, c) such that a2 + b3 = c10, there exists an i and j such that

ρE(a,b,c),2∼= ρEi ,2

andρE(a,b,c),5

∼= ρE ′j ,5.


j(5)(Q).


Step 4: Classify elliptic curves with a given pair of Galoisrepresentations


j(5)(Q).

Step 4:

For a fixed i , j , classify all elliptic curves E for which ρE ,2 ∼= ρEi ,2 andρE ,5 ∼= ρE ′

j ,2.


Step 4: Elliptic Chabauty

XEi ,E′j(10) //

��

/KE ′j $$J

JJ

JJ

XE ′j(5)

/KE ′j

��XEi

//

��

X0(5)

��XEi

(2) // X∆Ei// X (1)

For every coprime (a, b, c) such that a2 + b3 = c10, we can find someEi , E ′j and a point on P ∈ XEi

(KE ′j) such that j(P) ∈ X (1)(Q).

This latter set is finite, and in fact computable (via p-adic integrationand other methods).



XEi ,E′j(10) //

��

/KE ′j $$J

JJ

JJ

XE ′j(5)

/KE ′j

��XEi

//

��

X0(5)

��XEi

(2) // X∆Ei// X (1)






XEi ,E′j(10) //

��

/KE ′j $$J

JJ

JJ

XE ′j(5)

/KE ′j

��XEi

//

��

X0(5)

��XEi

(2) // X∆Ei// X (1)





Conclusion: New ideas for x2 + y 3 = z10

The template fails for x2 + y 3 = z10.

1) Known tools for classifying ρE(a,b,c),` fail for ` = 2, 5.

New idea: supplement classical classification techniques with numberfield enumeration and non-traditional parameter spaces.

2) Its not enough to classify only the mod 2 or the mod 5representation.

3) Classifying both the mod 2 and mod 5 at the same time leads to‘high genus parameter spaces’.

New idea: translate the work to low genus parameter spaces, but overlarger number fields than Q.


































Explicit Modular Approaches to Generalized Fermat Equations

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