Explicit Modular Approaches to Generalized Fermat Equations David - - PowerPoint PPT Presentation

Explicit Modular Approaches to Generalized Fermat Equations

David Brown

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

Emory University Colloquium February 14, 2011

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
• .

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.

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

Fermat’s Last Theorem

Theorem (Wiles; Taylor-Wiles 1995)

The only integer solutions to the equation xn + yn = zn, n ≥ 3 satisfy xyz = 0.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 3 / 63

Template for the proof of FLT

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 4 / 63

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:

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 4 / 63

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) : y2 = x(x − ap)(x + bp)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 4 / 63

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) : y2 = x(x − ap)(x + bp) j = 28 c2p − apbp3 (abc)2p ∆ = 2−8(abc)2p.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 4 / 63

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) : y2 = x(x − ap)(x + bp) j = 28 c2p − apbp3 (abc)2p ∆ = 2−8(abc)2p. Step 3: (Ribet) Show that the Frey curve E(a,b,c) is not modular.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 4 / 63

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) : y2 = x(x − ap)(x + bp) j = 28 c2p − apbp3 (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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 4 / 63

Modularity is now a theorem

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

Every elliptic curve over Q is modular.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 5 / 63

Elliptic Curves

E : y2 = x3 + ax + b

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 6 / 63

Elliptic Curves - point at infinity

E : zy2 = x3 + axz2 + bz3 ∞ = [0 : 1 : 0]

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Elliptic Curves – addition

E : y2 = x3 + ax + b P = (x0, y0) ∈ Q2 Q = (x1, y1) ∈ Q2

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 8 / 63

Elliptic Curves – addition

E : y2 = x3 + ax + b P = (x0, y0) ∈ Q2 Q = (x1, y1) ∈ Q2 R = (x2, y2) ∈ Q2

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 9 / 63

Elliptic Curves – addition

E : y2 = x3 + ax + b P = (x0, y0) ∈ Q2 Q = (x1, y1) ∈ Q2 R = (x2, y2) ∈ Q2 P + Q = (x2, −y2) ∈ Q2

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 10 / 63

Elliptic Curves – addition

E : y2 = x3 + ax + b E(Q) × E(Q) → E(Q) (P , Q) → P + Q

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 11 / 63

Elliptic Curves - duplication

E : y2 = x3 + ax + b P = (x0, y0) ∈ Q2

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 12 / 63

Elliptic Curves - duplication

E : y2 = x3 + ax + b P = (x0, y0) ∈ Q2 2P = (x3, y3) ∈ Q2

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 12 / 63

Elliptic Curves – identity

E : y2 = x3 + ax + b

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 13 / 63

Elliptic Curves – identity

E : y2 = x3 + ax + b

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

Elliptic Curves – inverses

E : y2 = x3 + ax + b

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 15 / 63

Elliptic Curves – inverses

E : y2 = x3 + ax + b

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 16 / 63

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 = ∞} .

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 17 / 63

Elliptic Curves – two torsion

E : y2 = x3 + ax + b

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 18 / 63

Elliptic Curves – two torsion

E : y2 = x3 + ax + b 2P = 2Q = 2R = ∞

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 18 / 63

Elliptic Curves – structure of torsion

Let E be given by the equation y2 = f (x) = x3 + ax + b. E[n](C) = E[n](Q) ∼ = (Z/nZ)2.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 19 / 63

Elliptic Curves – structure of torsion

Let E be given by the equation y2 = f (x) = x3 + ax + b. E[n](C) = E[n](Q) ∼ = (Z/nZ)2. E[n](Q) may be smaller,

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 19 / 63

Elliptic Curves – structure of torsion

Let E be given by the equation y2 = 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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 19 / 63

Elliptic Curves – torsion

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 20 / 63

Galois Representations associated to an elliptic curve

Let E : y2 = x3 + ax + b be an elliptic curve with a, b ∈ Q.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 21 / 63

Galois Representations associated to an elliptic curve

Let E : y2 = x3 + ax + b be an elliptic curve with a, b ∈ Q. Let GQ := Aut(Q)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 21 / 63

Galois Representations associated to an elliptic curve

Let E : y2 = x3 + ax + b be an elliptic curve with a, b ∈ Q. Let GQ := Aut(Q) ∼ = lim ← −K Aut(K).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 21 / 63

Galois Representations associated to an elliptic curve

Let E : y2 = x3 + ax + b be an elliptic curve with a, b ∈ Q. Let GQ := Aut(Q) ∼ = lim ← −K Aut(K). Let σ ∈ GQ, P = (x, y) ∈ E(Q) Pσ = (xσ, yσ) ∈ E(Q).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 21 / 63

Galois Representations associated to an elliptic curve

Let E : y2 = x3 + ax + b be an elliptic curve with a, b ∈ Q. Let GQ := Aut(Q) ∼ = lim ← −K Aut(K). Let σ ∈ GQ, P = (x, y) ∈ E(Q) Pσ = (xσ, yσ) ∈ E(Q). If P ∈ E[n], then Pσ ∈ E[n].

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 21 / 63

Galois Representations associated to an elliptic curve

Let E : y2 = x3 + ax + b be an elliptic curve with a, b ∈ Q. Let GQ := Aut(Q) ∼ = lim ← −K Aut(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).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 21 / 63

Elliptic Curves – torsion

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 22 / 63

Galois Representations: examples

Example

Suppose that E(Q)[2] ∼ = (Z/2Z)2. (E.g., E : y2 = x(x − 1)(x − λ) with λ ∈ Q.) Then ρE,2 : GQ → GL2(Z/2Z) is the trivial homomorphism.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 23 / 63

Galois Representations: examples

Example

Suppose that E(Q)[2] ∼ = (Z/2Z)2. (E.g., E : y2 = x(x − 1)(x − λ) with λ ∈ Q.) Then ρE,2 : GQ → GL2(Z/2Z) is the trivial homomorphism.

Example

Suppose that E(Q)[2] ∼ = Z/2Z. (E.g., E : y2 = (x2 + D)(x − λ) with D, λ ∈ Q and D > 0.) Then we can choose a basis for E(Q)[2] so that any σ ∈ GQ acts as a matrix of the form    1 a b    .

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 23 / 63

Galois Representations from modular forms

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 24 / 63

Galois Representations from modular forms

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

c d

• τ = aτ + b

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 24 / 63

Galois Representations from modular forms

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

c d

• τ = aτ + b

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

Definition

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 24 / 63

Galois Representations from modular forms

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

c d

• τ = aτ + b

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

Definition

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

• Ef
• ρf ,n

ρEf ,n

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 24 / 63

Galois Representations from modular forms

Fact

Galois representations associated to modular forms are easy to understand and classify.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 25 / 63

Galois Representations from modular forms

Fact

Galois representations associated to modular forms are easy to understand and classify.

Theorem (Modularity)

Every elliptic curve over Q arises from a modular form.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 25 / 63

Galois Representations from modular forms

Fact

Galois representations associated to modular forms are easy to understand and classify.

Theorem (Modularity)

Every elliptic curve over Q arises from a modular form. f

• Ef
• ρf ,n

ρEf ,n

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 25 / 63

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) : y2 = x(x − ap)(x + bp) j = 28 c2p − apbp3 (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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 26 / 63

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) : y2 = x(x − ap)(x + bp) j = 28 c2p − apbp3 (abc)2p ∆ = 2−8(abc)2p.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 27 / 63

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) : y2 = x(x − ap)(x + bp) j = 28 c2p − apbp3 (abc)2p ∆ = 2−8(abc)2p. Step 3: (Ribet) Classify possibilities for E(a,b,c),p (modularity is one tool used in this classification).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 27 / 63

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) : y2 = x(x − ap)(x + bp) j = 28 c2p − apbp3 (abc)2p ∆ = 2−8(abc)2p. Step 3: (Ribet) Classify possibilities for E(a,b,c),p (modularity is one tool used in this classification). Step 4: The output of step 3 turns out to be empty!

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 27 / 63

Other applications of the modular method

The ideas behind the proof of FLT now permeate the study of diophantine problems.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 28 / 63

Other applications of the modular method

The ideas behind the proof of FLT now permeate the study of diophantine problems.

Theorem (Bugeaud, Mignotte, Siksek 2006)

The only Fibonacci numbers that are perfect powers are F0 = 0, F1 = F2 = 1, F6 = 8, F12 = 144.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 28 / 63

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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 29 / 63

Template for the proof of Darmon-Merel

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 30 / 63

Template for the proof of Darmon-Merel

Step 1: Assume there is a counterexample ap + bp = c2. Step 2: (Frey) Build an elliptic curve with strange properties: E(a,b,c) : y2 + xy = x3 + c − 1 4 x2 + ap 26 x ∆ = 1 212 (a2b)p j = − 26(3ap − 4c2)3 (a2b)p if ab is even.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 30 / 63

Template for the proof of Darmon-Merel

Step 1: Assume there is a counterexample ap + bp = c2. Step 2: (Frey) Build an elliptic curve with strange properties: E(a,b,c) : y2 = x3 + 2cx2 + apx ∆ = 26(a2b)p j = −26(3ap − 4c2)3 (a2b)p if ab is odd.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 31 / 63

Template for the proof of Darmon-Merel

Step 1: Assume there is a counterexample ap + bp = c2. Step 2: (Frey) Build an elliptic curve with strange properties: E(a,b,c) :        y2 + xy = x3 + c−1

4 x2 + ap 26 x

if ab is even y2 = x3 + 2cx2 + apx if ab is odd

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 32 / 63

Template for the proof of Darmon-Merel

Step 1: Assume there is a counterexample ap + bp = c2. Step 2: (Frey) Build an elliptic curve with strange properties: E(a,b,c) :        y2 + xy = x3 + c−1

4 x2 + ap 26 x

if ab is even y2 = x3 + 2cx2 + apx if ab is odd Step 3: E(a,b,c) has CM (complex multiplication).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 32 / 63

Template for the proof of Darmon-Merel

Step 1: Assume there is a counterexample ap + bp = c2. Step 2: (Frey) Build an elliptic curve with strange properties: E(a,b,c) :        y2 + xy = x3 + c−1

4 x2 + ap 26 x

if ab is even y2 = 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).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 32 / 63

Template for the proof of Darmon-Merel

Step 1: Assume there is a counterexample ap + bp = c2. Step 2: (Frey) Build an elliptic curve with strange properties: E(a,b,c) :        y2 + xy = x3 + c−1

4 x2 + ap 26 x

if ab is even y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 32 / 63

Generalized Fermat Equations

Fix p, q, r ∈ N such that χ = 1

p + 1 q + 1 r − 1 < 0.

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Generalized Fermat Equations

Fix p, q, r ∈ N such that χ = 1

p + 1 q + 1 r − 1 < 0.

Theorem (Darmon, Granville 1995)

The equation xp + yq = zr has only finitely many coprime solutions with xyz = 0.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 33 / 63

Examples of Generalized Fermat Equations

χ = 1 2 + 1 3 + 1 7 − 1 = − 1 42 < 0

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 34 / 63

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 + y3 = 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) .

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 34 / 63

Generalized Fermat Equations – Known Solutions

The ‘known’ solutions to the equation xp + yq = zr with χ = 1

p + 1 q + 1 r − 1 < 0 and xyz = 0 are the following:

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 35 / 63

Generalized Fermat Equations – Known Solutions

The ‘known’ solutions to the equation xp + yq = zr with χ = 1

p + 1 q + 1 r − 1 < 0 and xyz = 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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 35 / 63

Generalized Fermat Equations – Known Solutions

Conjecture (Beal, Granville, Tijdeman-Zagier)

This is a complete list of coprime non-zero solutions such that

1 p + 1 q + 1 r − 1 < 0.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 36 / 63

Generalized Fermat Equations – Known Solutions

Conjecture (Beal, Granville, Tijdeman-Zagier)

This is a complete list of coprime non-zero solutions such that

1 p + 1 q + 1 r − 1 < 0.

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 36 / 63

Generalized Fermat Equations – Known Solutions

Conjecture (Beal, Granville, Tijdeman-Zagier)

This is a complete list of coprime non-zero solutions such that

1 p + 1 q + 1 r − 1 < 0.

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 36 / 63

(p, q, r) such that χ < 0 and the solutions to xp + y q = zr 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, 2ℓ, 3} Chen, Dahmen, Siksek; primes 7 < ℓ < 1000 with ℓ = 31 {3, 3, n} Bruin; n = 4, 5 {3, 3, ℓ} Kraus; primes 17 ≤ ℓ ≤ 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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 37 / 63

Main Theorem

χ = 1

2 + 1 3 + 1 10 − 1 = − 1 15 is maximal among unsolved Fermat equations.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 38 / 63

Main Theorem

χ = 1

2 + 1 3 + 1 10 − 1 = − 1 15 is maximal among unsolved Fermat equations.

Theorem (B., 2011)

The only coprime integer solutions to the equation x2 + y3 = z10 are the 12 triples (±1, −1, 0), (±1, 0, ±1), (0, 1, ±1), (±3, −2, ±1).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 38 / 63

Main Theorem

χ = 1

2 + 1 3 + 1 10 − 1 = − 1 15 is maximal among unsolved Fermat equations.

Theorem (B., 2011)

The only coprime integer solutions to the equation x2 + y3 = 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 + y3 = zn conjectured to have only trivial solutions. (32 + (−2)3 = 1n is considered to be trivial.)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 38 / 63

Framework for solving x2 + y 3 = zn

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

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) : y2 = x3 + 3bx − 2a ∆ = −123cn j = 123b3/cn.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

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) : y2 = x3 + 3bx − 2a ∆ = −123cn j = 123b3/cn. Step 3: Explicitly classify possibilities for ρE(a,b,c),n.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

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) : y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

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) : y2 = 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).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

Framework for solving x2 + y 3 = zn

For large n, this template (conjecturally) works!

Step 3:

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 40 / 63

Framework for solving x2 + y 3 = zn

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’.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 40 / 63

Framework for solving x2 + y 3 = zn

For large n, this template (conjecturally) works!

Step 4:

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 41 / 63

Framework for solving x2 + y 3 = zn

For large n, this template (conjecturally) works!

Step 4:

For a fixed ρ, classify all elliptic curves E for which ρE,n ∼ = ρ. This would follow from a standard conjecture:

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 41 / 63

Framework for solving x2 + y 3 = zn

For large n, this template (conjecturally) works!

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 that ρE,p ∼ = ρE ′,p. Then E is isogenous to E ′.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 41 / 63

Template breaks down for x2 + y 3 = z10

E(a,b,c) : y2 = x3 + 3bx − 2a

Step 3:

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

Template breaks down for x2 + y 3 = z10

E(a,b,c) : y2 = x3 + 3bx − 2a

Step 3:

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

Template breaks down for x2 + y 3 = z10

E(a,b,c) : y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

Template breaks down for x2 + y 3 = z10

E(a,b,c) : y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

Template breaks down for x2 + y 3 = z10

E(a,b,c) : y2 = 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(Fℓ) is reducible if there is some subspace W ⊂ F2

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

Template breaks down forth x2 + y 3 = z10

E(a,b,c) : y2 = x3 + 3bx − 2a

Step 4:

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 43 / 63

Template breaks down forth x2 + y 3 = z10

E(a,b,c) : y2 = x3 + 3bx − 2a

Step 4:

For a fixed ρ, classify all elliptic curves E for which ρE,n ∼ = ρ. Using one prime is not enough.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 43 / 63

Template breaks down forth x2 + y 3 = z10

E(a,b,c) : y2 = x3 + 3bx − 2a

Step 4:

For a fixed ρ, classify all elliptic curves E for which ρE,n ∼ = ρ. Using one prime is not enough.

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 43 / 63

Template breaks down forth x2 + y 3 = z10

E(a,b,c) : y2 = x3 + 3bx − 2a

Step 4:

For a fixed ρ, classify all elliptic curves E for which ρE,n ∼ = ρ. Using one prime is not enough.

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

Multiprime approaches seem to be computationally infeasible.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 43 / 63

Elliptic Curves – torsion

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 44 / 63

Step 3: Classifying mod 2 Galois representations

Let E be given by the equation y2 = f (x) := x3 + 3bx − 2a

Fact

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 45 / 63

Step 3: Classifying mod 2 Galois representations

Let E be given by the equation y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 45 / 63

Step 3: Classifying mod 2 Galois representations

Let E be given by the equation y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 45 / 63

Step 3: Classifying mod 2 Galois representations

Let E be given by the equation y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 45 / 63

Step 3: Progress for ℓ = 2

Lemma

There are elliptic curves {E1, . . . , En} such that for every (a, b, c) such that a2 + b3 = c10, there is an i such that ρE(a,b,c),2 ∼ = ρEi,2.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 46 / 63

Step 3: Progress for ℓ = 2

Lemma

There are elliptic curves {E1, . . . , En} such that for every (a, b, c) such that 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 46 / 63

Step 3: Progress for ℓ = 2

Lemma

There are elliptic curves {E1, . . . , En} such that for every (a, b, c) such that 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!

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 46 / 63

Parameter spaces for Galois representations

Definition

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 47 / 63

Elliptic Curves – torsion

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 48 / 63

Parameter spaces for Galois representations

Definition

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

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 : y2 = x(x − 1)(x − λ).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 49 / 63

Other parameter spaces

Recall that ρ : GQ → GL2(Fℓ) is reducible if there is some subspace W ⊂ F2

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 50 / 63

Other parameter spaces

Recall that ρ : GQ → GL2(Fℓ) is reducible if there is some subspace W ⊂ F2

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

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 W is an invariant subgroup of size p).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 50 / 63

Other parameter spaces

Recall that ρ : GQ → GL2(Fℓ) is reducible if there is some subspace W ⊂ F2

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

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 W is an invariant subgroup of size p).

Example (X0(5))

Let E : y2 = x3 + 3bx − 2a, and suppose ρE,5 is reducible. Then there exists a t ∈ Z such that 123 b3 a2 + b3 = (t2 + 250t + 3125)3 t5 .

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 50 / 63

Step 3: Intermediate Modular curves

Goal

Explicitly classify possibilities for ρE(a,b,c),5.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 51 / 63

Step 3: Intermediate Modular curves

Goal

Explicitly classify possibilities for ρE(a,b,c),5. X(10)

• X(5)
• X0(5)
• X(2)

π

X0(2) X(1)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 51 / 63

Step 3: Intermediate Modular curves

Goal

Explicitly classify possibilities for ρE(a,b,c),5. X(10)

• X(5)
• X0(5)
• X(2)

π

X0(2) X(1)

π: (E, ψ: ρtriv ∼ = ρE,2) → (E, W ).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 51 / 63

Step 3: Intermediate Modular curves: p = 2

Aut(X(2)/X(1)) ∼ = GL2(F2) ∼ = S3. X0(2)

• X(2)
• X(1)

X∆

• David Brown (UW-Madison)

Generalized Fermat Equations February 14, 2011 52 / 63

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)

• X(2)
• X(1)

X∆

• David Brown (UW-Madison)

Generalized Fermat Equations February 14, 2011 52 / 63

Step 3: Intermediate Modular curves: p = 2

Define X∆ to be the quotient of X(2) by the normal subgroup A3. X0(2)

• X(2)
• X(1)

X∆

• David Brown (UW-Madison)

Generalized Fermat Equations February 14, 2011 53 / 63

Step 3: Intermediate Modular curves: p = 2

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)

• X(2)
• X(1)

X∆

• David Brown (UW-Madison)

Generalized Fermat Equations February 14, 2011 53 / 63

Step 3: Intermediate Modular curves

Goal

Explicitly classify possibilities for ρE(a,b,c),5. X(10)

• X(5)
• X0(5)
• X(2)

X∆ X(1)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 54 / 63

Step 3: Intermediate Modular curves

X(10)

• X(5)
• X
• X0(5)
• X(2)

X∆ X(1)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 55 / 63

Step 3: Intermediate Modular curves

X(10)

• X(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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 55 / 63

Step 3: Intermediate Modular curves

E(a,b,c) : y2 = x3 + 3bx − 2a ∆ = −123c10 = −3(23 · 3 · c5)2 X

• X0(5)
• X∆

X(1)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 56 / 63

Step 3: Intermediate Modular curves

E(a,b,c) : y2 = x3 + 3bx − 2a ∆ = −123c10 = −3(23 · 3 · c5)2 X(−3)

• X0(5)
• X(−3∆)

X(1)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 57 / 63

Step 3: Intermediate Modular curves

E(a,b,c) : y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 57 / 63

Step 3: Intermediate Modular curves

E(a,b,c) : y2 = 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).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 58 / 63

Step 3: Intermediate Modular curves

E(a,b,c) : y2 = 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.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 58 / 63

Step 3: Classifying ρE(a,b,c),ℓ

Lemma

There are elliptic curves {E1, . . . , En} and {E ′

1, . . . , E ′ n′} such that for

every (a, b, c) such that a2 + b3 = c10, there exists an i and j such that ρE(a,b,c),2 ∼ = ρEi,2 and ρE(a,b,c),5 ∼ = ρE ′

j ,5. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 59 / 63

Step 3: Classifying ρE(a,b,c),ℓ

Lemma

There are elliptic curves {E1, . . . , En} and {E ′

1, . . . , E ′ n′} such that for

every (a, b, c) such that a2 + b3 = c10, there exists an i and j such that ρE(a,b,c),2 ∼ = ρEi,2 and ρE(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). David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 59 / 63

Step 4: Classify elliptic curves with a given pair of Galois representations

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

j (5)(Q).

Step 4:

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

j ,2. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 60 / 63

Step 4: Elliptic Chabauty

XEi,E ′

j (10)

• /KE′

j

• XE ′

j (5)

/KE′

j

• XEi
• X0(5)
• XEi(2)

X∆Ei X(1)

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 61 / 63

Step 4: Elliptic Chabauty

XEi,E ′

j (10)

• /KE′

j

• 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 some Ei, E ′

j and a point on P ∈ XEi(KE ′

j ) such that j(P) ∈ X(1)(Q). David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 61 / 63

Step 4: Elliptic Chabauty

XEi,E ′

j (10)

• /KE′

j

• 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 some Ei, 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 integration and other methods).

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 61 / 63

Conclusion: New ideas for x2 + y 3 = z10

The template fails for x2 + y3 = z10. 1) Known tools for classifying ρE(a,b,c),ℓ fail for ℓ = 2, 5.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 62 / 63

Conclusion: New ideas for x2 + y 3 = z10

The template fails for x2 + y3 = z10. 1) Known tools for classifying ρE(a,b,c),ℓ fail for ℓ = 2, 5.

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 62 / 63

Conclusion: New ideas for x2 + y 3 = z10

The template fails for x2 + y3 = z10. 1) Known tools for classifying ρE(a,b,c),ℓ fail for ℓ = 2, 5.

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

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

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 62 / 63

Conclusion: New ideas for x2 + y 3 = z10

The template fails for x2 + y3 = z10. 1) Known tools for classifying ρE(a,b,c),ℓ fail for ℓ = 2, 5.

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

2) Its not enough to classify only the mod 2 or the mod 5 representation. 3) Classifying both the mod 2 and mod 5 at the same time leads to ‘high genus parameter spaces’.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 62 / 63

Conclusion: New ideas for x2 + y 3 = z10

The template fails for x2 + y3 = z10. 1) Known tools for classifying ρE(a,b,c),ℓ fail for ℓ = 2, 5.

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

2) Its not enough to classify only the mod 2 or the mod 5 representation. 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 over larger number fields than Q.

David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 62 / 63