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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

  1. Elliptic Curves – torsion    {∞} if f ( x ) has 0 rational roots      E [2]( Q ) ∼ = Z / 2 Z , if f ( x ) has 1 rational roots       ( Z / 2 Z ) 2 , if f ( x ) has 3 rational roots  David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 22 / 63

  2. Galois Representations: examples Example = ( Z / 2 Z ) 2 . (E.g., E : y 2 = x ( x − 1)( x − λ ) with Suppose that E ( Q )[2] ∼ λ ∈ Q .) Then ρ E , 2 : G Q → GL 2 ( Z / 2 Z ) is the trivial homomorphism. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 23 / 63

  3. Galois Representations: examples Example = ( Z / 2 Z ) 2 . (E.g., E : y 2 = x ( x − 1)( x − λ ) with Suppose that E ( Q )[2] ∼ λ ∈ Q .) Then ρ E , 2 : G Q → GL 2 ( Z / 2 Z ) is the trivial homomorphism. Example = Z / 2 Z . (E.g., E : y 2 = ( x 2 + D )( x − λ ) with Suppose that E ( Q )[2] ∼ D , λ ∈ Q and D > 0.) Then we can choose a basis for E ( Q )[2] so that any σ ∈ G Q acts as a matrix of the form    1 a  .   0 b David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 23 / 63

  4. 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

  5. Galois Representations from modular forms Let H := { τ = x + yi ∈ C : y > 0 } be the complex upper half plane. τ = a τ + b � a b � The formula c τ + d defines an action of SL 2 ( Z ) on H . c d David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 24 / 63

  6. Galois Representations from modular forms Let H := { τ = x + yi ∈ C : y > 0 } be the complex upper half plane. τ = a τ + b � a b � The formula c τ + d defines an action of SL 2 ( Z ) on H . c d Definition A modular function is a complex analytic function f : H → C which is invariant under the action of a congruence subgroup Γ ⊂ SL 2 ( Z ) such that f is holomorphic at ∞ . A modular form of weight 2 k is a complex analytic function f : H → C such that f ( z )( dz ) k is invariant under the action of a congruence subgroup Γ ⊂ SL 2 ( Z ) such that f is holomorphic at ∞ . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 24 / 63

  7. � � � � � � ��� � ��� Galois Representations from modular forms Let H := { τ = x + yi ∈ C : y > 0 } be the complex upper half plane. τ = a τ + b � a b � The formula c τ + d defines an action of SL 2 ( Z ) on H . c d Definition A modular function is a complex analytic function f : H → C which is invariant under the action of a congruence subgroup Γ ⊂ SL 2 ( Z ) such that f is holomorphic at ∞ . A modular form of weight 2 k is a complex analytic function f : H → C such that f ( z )( dz ) k is invariant under the action of a congruence subgroup Γ ⊂ SL 2 ( Z ) such that f is holomorphic at ∞ . E f f ρ f , n ρ E f , n David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 24 / 63

  8. 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

  9. 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

  10. � � � ��� � ��� � � � 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. E f f ρ f , n ρ E f , n David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 25 / 63

  11. Template for the proof of FLT Step 1: Assume there is a counterexample a p + b p = c p . Step 2: (Frey) Build an elliptic curve with strange properties: E ( a , b , c ) : y 2 = x ( x − a p )( x + b p ) c 2 p − a p b p � 3 j = 2 8 � ( abc ) 2 p ∆ = 2 − 8 ( abc ) 2 p . 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

  12. Template for the proof of FLT Step 1: Assume there is a counterexample a p + b p = c p . Step 2: (Frey) Build an elliptic curve with strange properties: E ( a , b , c ) : y 2 = x ( x − a p )( x + b p ) c 2 p − a p b p � 3 j = 2 8 � ( abc ) 2 p ∆ = 2 − 8 ( abc ) 2 p . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 27 / 63

  13. Template for the proof of FLT Step 1: Assume there is a counterexample a p + b p = c p . Step 2: (Frey) Build an elliptic curve with strange properties: E ( a , b , c ) : y 2 = x ( x − a p )( x + b p ) c 2 p − a p b p � 3 j = 2 8 � ( abc ) 2 p ∆ = 2 − 8 ( abc ) 2 p . 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

  14. Template for the proof of FLT Step 1: Assume there is a counterexample a p + b p = c p . Step 2: (Frey) Build an elliptic curve with strange properties: E ( a , b , c ) : y 2 = x ( x − a p )( x + b p ) c 2 p − a p b p � 3 j = 2 8 � ( abc ) 2 p ∆ = 2 − 8 ( abc ) 2 p . 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

  15. 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

  16. 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 F 0 = 0 , F 1 = F 2 = 1 , F 6 = 8 , F 12 = 144 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 28 / 63

  17. More applications of the modular method Theorem (Darmon, Merel 1997) Any pairwise coprime integer solution to the equation x n + y n = z 2 , n ≥ 4 satisfies xyz = 0 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 29 / 63

  18. Template for the proof of Darmon-Merel Step 1: Assume there is a counterexample a p + b p = c 2 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 30 / 63

  19. Template for the proof of Darmon-Merel Step 1: Assume there is a counterexample a p + b p = c 2 . Step 2: (Frey) Build an elliptic curve with strange properties: x 2 + a p E ( a , b , c ) : y 2 + xy = x 3 + c − 1 2 6 x 4 ∆ = 1 2 12 ( a 2 b ) p j = − 2 6 (3 a p − 4 c 2 ) 3 ( a 2 b ) p if ab is even. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 30 / 63

  20. Template for the proof of Darmon-Merel Step 1: Assume there is a counterexample a p + b p = c 2 . Step 2: (Frey) Build an elliptic curve with strange properties: E ( a , b , c ) : y 2 = x 3 + 2 cx 2 + a p x ∆ = 2 6 ( a 2 b ) p j = − 2 6 (3 a p − 4 c 2 ) 3 ( a 2 b ) p if ab is odd. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 31 / 63

  21. Template for the proof of Darmon-Merel Step 1: Assume there is a counterexample a p + b p = c 2 . Step 2: (Frey) Build an elliptic curve with strange properties:   y 2 + xy = x 3 + c − 1 4 x 2 + a p  if ab is even 2 6 x  E ( a , b , c ) :  y 2 = x 3 + 2 cx 2 + a p x  if ab is odd  David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 32 / 63

  22. Template for the proof of Darmon-Merel Step 1: Assume there is a counterexample a p + b p = c 2 . Step 2: (Frey) Build an elliptic curve with strange properties:   y 2 + xy = x 3 + c − 1 4 x 2 + a p  if ab is even 2 6 x  E ( a , b , c ) :  y 2 = x 3 + 2 cx 2 + a p x  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

  23. Template for the proof of Darmon-Merel Step 1: Assume there is a counterexample a p + b p = c 2 . Step 2: (Frey) Build an elliptic curve with strange properties:   y 2 + xy = x 3 + c − 1 4 x 2 + a p  if ab is even 2 6 x  E ( a , b , c ) :  y 2 = x 3 + 2 cx 2 + a p x  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

  24. Template for the proof of Darmon-Merel Step 1: Assume there is a counterexample a p + b p = c 2 . Step 2: (Frey) Build an elliptic curve with strange properties:   y 2 + xy = x 3 + c − 1 4 x 2 + a p  if ab is even 2 6 x  E ( a , b , c ) :  y 2 = x 3 + 2 cx 2 + a p x  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

  25. Generalized Fermat Equations Fix p , q , r ∈ N such that χ = 1 p + 1 q + 1 r − 1 < 0. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 33 / 63

  26. 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 x p + y q = z r has only finitely many coprime solutions with xyz � = 0 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 33 / 63

  27. 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

  28. 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 x 2 + y 3 = z 7 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

  29. Generalized Fermat Equations – Known Solutions The ‘known’ solutions to the equation x p + y q = z r 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

  30. Generalized Fermat Equations – Known Solutions The ‘known’ solutions to the equation x p + y q = z r with χ = 1 p + 1 q + 1 r − 1 < 0 and xyz � = 0 are the following: 1 p + 2 3 = 3 2 ( − 1) 2 p + 2 3 = 3 2 2 5 + 7 2 = 3 4 7 3 + 13 2 = 2 9 2 7 + 17 3 = 71 2 3 5 + 11 4 = 122 2 17 7 + 76271 3 = 21063928 2 1414 3 + 2213459 2 = 65 7 9262 3 + 15312283 2 = 113 7 43 8 + 96222 3 = 30042907 2 33 8 + 1549034 2 = 15613 3 David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 35 / 63

  31. 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

  32. 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

  33. 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

  34. ( p , q , r ) such that χ < 0 and the solutions to x p + y q = 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 , 2 n , 2 n } 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 ℓ, 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 , 2 n , 5) Chen n ≥ 3 ∗ (4 , 2 n , 3) Bennett-Chen n ≥ 3 (6 , 2 n , 2) Bennett-Chen n ≥ 3 (2 , 6 , n ) Bennett-Chen n ≥ 3 David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 37 / 63

  35. 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

  36. 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 x 2 + y 3 = z 10 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

  37. 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 x 2 + y 3 = z 10 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 x 2 + y 3 = z n conjectured to have only trivial solutions. (3 2 + ( − 2) 3 = 1 n is considered to be trivial.) David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 38 / 63

  38. Framework for solving x 2 + y 3 = z n Step 1: Assume there is a counterexample a 2 + b 3 = c n . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

  39. Framework for solving x 2 + y 3 = z n Step 1: Assume there is a counterexample a 2 + b 3 = c n . Step 2: Study the elliptic curve E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a ∆ = − 12 3 c n j = 12 3 b 3 / c n . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

  40. Framework for solving x 2 + y 3 = z n Step 1: Assume there is a counterexample a 2 + b 3 = c n . Step 2: Study the elliptic curve E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a ∆ = − 12 3 c n j = 12 3 b 3 / c n . Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , n . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

  41. Framework for solving x 2 + y 3 = z n Step 1: Assume there is a counterexample a 2 + b 3 = c n . Step 2: Study the elliptic curve E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a ∆ = − 12 3 c n j = 12 3 b 3 / c n . Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , n . Step 4: For such ρ , classify all elliptic curves E for which ρ E , n ∼ = ρ . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 39 / 63

  42. Framework for solving x 2 + y 3 = z n Step 1: Assume there is a counterexample a 2 + b 3 = c n . Step 2: Study the elliptic curve E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a ∆ = − 12 3 c n j = 12 3 b 3 / c n . Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , n . Step 4: For such ρ , classify all elliptic curves E for which ρ E , 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

  43. Framework for solving x 2 + y 3 = z n For large n , this template (conjecturally) works! Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , n . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 40 / 63

  44. Framework for solving x 2 + y 3 = z n For large n , this template (conjecturally) works! Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , n . For large n , there are 13 possibilities for ρ E ( a , b , c ) , n , which are ‘independent of n ’. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 40 / 63

  45. Framework for solving x 2 + y 3 = z n For large n , this template (conjecturally) works! Step 4: For a fixed ρ , classify all elliptic curves E for which ρ E , n ∼ = ρ . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 41 / 63

  46. Framework for solving x 2 + y 3 = z n For large n , this template (conjecturally) works! Step 4: For a fixed ρ , classify all elliptic curves E for which ρ E , n ∼ = ρ . This would follow from a standard conjecture: David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 41 / 63

  47. Framework for solving x 2 + y 3 = z n For large n , this template (conjecturally) works! Step 4: For a fixed ρ , classify all elliptic curves E for which ρ E , 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 ρ E , p ∼ = ρ E ′ , p . Then E is isogenous to E ′ . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 41 / 63

  48. Template breaks down for x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , 10 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

  49. Template breaks down for x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , 10 . Known tools for classifying ρ E ( a , b , c ) , 10 fail. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

  50. Template breaks down for x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , 10 . Known tools for classifying ρ E ( 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

  51. Template breaks down for x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , 10 . Known tools for classifying ρ E ( a , b , c ) , 10 fail. E.g., Ribet’s level lowering theorem fails for n = 2. ρ E ( 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

  52. Template breaks down for x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 3: Explicitly classify possibilities for ρ E ( a , b , c ) , 10 . Known tools for classifying ρ E ( a , b , c ) , 10 fail. E.g., Ribet’s level lowering theorem fails for n = 2. ρ E ( a , b , c ) , n may be reducible for both n = 2 and 5. Definition We say that ρ : G Q → GL 2 ( F ℓ ) is reducible if there is some subspace ℓ such that for every P ∈ W , P ρ ( σ ) ∈ W . W ⊂ F 2 David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 42 / 63

  53. Template breaks down forth x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 4: For a fixed ρ , classify all elliptic curves E for which ρ E , n ∼ = ρ . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 43 / 63

  54. Template breaks down forth x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 4: For a fixed ρ , classify all elliptic curves E for which ρ E , n ∼ = ρ . Using one prime is not enough. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 43 / 63

  55. Template breaks down forth x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 4: For a fixed ρ , classify all elliptic curves E for which ρ E , 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

  56. Template breaks down forth x 2 + y 3 = z 10 E ( a , b , c ) : y 2 = x 3 + 3 bx − 2 a Step 4: For a fixed ρ , classify all elliptic curves E for which ρ E , 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

  57. Elliptic Curves – torsion    {∞} if f ( x ) has 0 rational roots      E [2]( Q ) ∼ = Z / 2 Z , if f ( x ) has 1 rational roots       ( Z / 2 Z ) 2 , if f ( x ) has 3 rational roots  David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 44 / 63

  58. Step 3: Classifying mod 2 Galois representations Let E be given by the equation y 2 = f ( x ) := x 3 + 3 bx − 2 a Fact The splitting field of the polynomial f ( x ) completely determines ρ E , 2 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 45 / 63

  59. Step 3: Classifying mod 2 Galois representations Let E be given by the equation y 2 = f ( x ) := x 3 + 3 bx − 2 a Fact The splitting field of the polynomial f ( x ) completely determines ρ E , 2 . The splitting field K of x 3 + 3 bx − 2 a is unramified outside of { 2 , 3 } and of degree at most 6. David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 45 / 63

  60. Step 3: Classifying mod 2 Galois representations Let E be given by the equation y 2 = f ( x ) := x 3 + 3 bx − 2 a Fact The splitting field of the polynomial f ( x ) completely determines ρ E , 2 . The splitting field K of x 3 + 3 bx − 2 a 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

  61. Step 3: Classifying mod 2 Galois representations Let E be given by the equation y 2 = f ( x ) := x 3 + 3 bx − 2 a Fact The splitting field of the polynomial f ( x ) completely determines ρ E , 2 . The splitting field K of x 3 + 3 bx − 2 a 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

  62. Step 3: Progress for ℓ = 2 Lemma There are elliptic curves { E 1 , . . . , E n } such that for every ( a , b , c ) such that a 2 + b 3 = c 10 , there is an i such that ρ E ( a , b , c ) , 2 ∼ = ρ E i , 2 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 46 / 63

  63. Step 3: Progress for ℓ = 2 Lemma There are elliptic curves { E 1 , . . . , E n } such that for every ( a , b , c ) such that a 2 + b 3 = c 10 , there is an i such that ρ E ( a , b , c ) , 2 ∼ = ρ E i , 2 . Wanted : a similar lemma for ρ E ( a , b , c ) , 5 . David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 46 / 63

  64. Step 3: Progress for ℓ = 2 Lemma There are elliptic curves { E 1 , . . . , E n } such that for every ( a , b , c ) such that a 2 + b 3 = c 10 , there is an i such that ρ E ( a , b , c ) , 2 ∼ = ρ E i , 2 . Wanted : a similar lemma for ρ E ( a , b , c ) , 5 . Problem : ρ E ( a , b , c ) , 5 may be reducible, thus modularity won’t help! David Brown (UW-Madison) Generalized Fermat Equations February 14, 2011 46 / 63

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