ABC Triples in Families

Transcription

1 Edray Herber Goins Department of Mathematics Purdue University September 30, 2010

2 Abstract Given three positive, relative prime integers A, B, and C such that the first two sum to the third i.e. A+B = C, it is rare to have the product of the primes p dividing them to be smaller than each of the three. In 1985, David Masser and Joseph Osterlé made this precise by defining a quality q(p) for such a triple of integers P = (A,B,C); their celebrated ABC Conjecture asserts that it is rare for this quality q(p) to be greater than 1 even through there are infinitely many examples where this happens. In 1987, Gerhard Frey offered an approach to understanding this conjecture by introducing elliptic curves. In this presentation, we introduce families of triples so that the Frey curve has nontrivial torsion subgroup, and explain how certain triples with large quality appear in these families. We also discuss how these families contain infinitely many examples where the quality q(p) is greater than 1.

3 Mathematical Sciences Research Institute Undergraduate Program MSRI-UP

4 Alexander Barrios Caleb Tillman Charles Watts (Brown University) (Reed College) (Morehouse College) Luis Lomeĺı Jamie Weigandt (Purdue University) (Purdue University)

5 Outline of Talk History of the Conjecture 1 History of the Conjecture The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality 2 Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup 3 of Degree 2 of Degree 4 of Degree 8

6 The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality History of the ABC Conjecture

7 Mason-Stothers Theorem The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality Theorem (W. W. Stothers, 1981; R. C. Mason, 1983) Denote n(ab C) as the number distinct zeroes of the product of relatively prime polynomials A(t), B(t), and C(t) satisfying A+B = C. Then max { deg(a), deg(b), deg(c) } n(ab C) 1. Proof: We follow Lang s Algebra. Explicitly write rad ( ABC ) (t) = a A(t) = A 0 (t α i ) pi deg(a) = i=1 b B(t) = B 0 (t β j ) qj deg(b) = j=1 c C(t) = C 0 (t γ k ) r k deg(c) = k=1 a b (t α i ) (t β j ) i=1 j=1 a i=1 b j=1 c k=1 c (t γ k ) n(abc) = a+b +c k=1 p i q j r k

8 The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality F(t) = A(t) C(t) G(t) = B(t) C(t) = B(t) A(t) = F (t) F(t) G (t) G(t) = a i=1 b j=1 p i t α i q j t β j c i=k c k=1 r k t γ k r k t γ k Clearing, we find polynomials of degrees max { deg(a), deg(b) } n 1: rad ( ABC ) (t) F (t) F(t) = rad ( ABC ) (t) G (t) G(t) = a b p i (t α e ) (t β j ) i=1 e i j=1 c (t γ k ) k=1 c a b r k (t α i ) β j i=1 j=1(t ) (t γ e ) e k b a c q j α i i=1(t ) (t β e ) (t γ k ) e j k=1 c a b r k (t α i ) β j i=1 j=1(t ) (t γ e ) e k k=1 j=1 k=1

9 Examples History of the Conjecture The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality The Mason-Stothers Theorem is sharp. Here is a list of relatively prime polynomials such that A(t)+B(t) = C(t) and max { deg(a), deg(b), deg(c) } = n(ab C) 1. A(t) B(t) C(t) n ( 2t ) 2 ( t 2 1 ) 2 ( t 2 +1 ) 2 8t ( t 2 +1 ) ( t 1 ) 4 ( t +1 ) 4 5 ( ) 3 ( ) ( )( ) 3 16t t +1 t 3 t +3 t 1 5 ( )( ) 16t 3 3 ( ) 3 ( ) t +1 t 3 t +3 t 1 ( 2t ) 4 ( t 4 6t 2 +1 )( t 2 +1 ) 2 ( t 2 1 ) 4 16t ( t 2 1 )( t 2 +1 ) 2 ( t 2 2t 1 ) 4 ( t 2 +2t 1 ) 4 9

10 Restatement of Theorem The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality The polynomial ring Q[t] has an absolute value : Q[t] Z 0 defined by { A(t) 0 if A(t) 0, = 2 deg(a) otherwise. It has the following properties: Multiplicativity: A B = A B Non-degeneracy: A = 0 iff A = 0; A = 1 iff A(t) = A 0 is a unit. Ordering: A B iff deg(a) deg(b). Corollary For each ǫ > 0 there exists a uniform C ǫ > 0 such that the following holds: For any relatively prime polynomials A, B, C Q[t] with A+B = C, max { A, B, C } C ǫ rad(abc) 1+ǫ. Proof: Using Mason-Stothers, we may choose C ǫ = 1/2.

11 The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality Multiplicative Dedekind-Hasse Norms Let R be a Principal Ideal Domain with quotient field K: Q[t] in Q(t) with primes t α. Z in Q with primes p. Define rad(a) of an ideal a is the intersection of primes p containing it: rad(a) = i (t α i) for A(t) = A 0 i (t α i) ei in Q[t]. rad(a) = i p i for A = i pei i in Z. Theorem (Richard Dedekind; Helmut Hasse) R is a Principal Ideal Domain if and only if there exists an absolute value : R Z 0 with the following properties: Multiplicativity: A B = A B Non-degeneracy: A = 0 iff A = 0; A = 1 iff A is a unit. Ordering: A B if A divides B. We may define a = A as a = AR. Hence rad(a) a as rad(a) a.

12 ABC Conjecture History of the Conjecture The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality Conjecture (David Masser, 1985; Joseph Oesterlé, 1985) For each ǫ > 0 there exists a uniform C ǫ > 0 such that the following holds: For any relatively prime integers A, B, C Z with A+B = C, max { A, B, C } C ǫ rad(abc) 1+ǫ. Lemma The symmetric group on three letters acts on the set of ABC Triples: σ : A B B C, τ : A B B σ 3 = 1 A where τ 2 = 1 C A C C τ σ τ = σ 2 In particular, we may assume 0 < A B < C.

13 Quality of ABC Triples The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality Corollary If the ABC Conjecture holds, then limsupq(a,b,c) 1 for the quality q(a,b,c) = max{ ln A, ln B, ln C } ln rad(abc). Proof: Say ǫ = ( limsupq(p) 1 ) /3 is positive. Choose a sequence P k = (A k,b k,c k ) with q(p k ) 1+2ǫ. But this must be finite because max { A k, B k, C k } C ǫ rad(a k B k C k ) [ ] 1+ǫ q(p k ) exp q(p k ) 1 ǫ lnc ǫ. Question For each ǫ > 0, there are only finitely many ABC Triples P = (A,B,C) with q(p) 1+ǫ. What is the largest q(p) can be?

14 Exceptional Quality History of the Conjecture The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality Proposition (Bart de Smit, 2010) There are only 233 known ABC Triples P = (A,B,C) with q(p) 1.4. Rank A B C q(a,b,c)

15 The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality ABC Conjecture Home Page

16 The Mason-Stothers Theorem ABC Conjecture Examples with Exceptional Quality ABC at Home

17 Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup The ABC Conjecture and Elliptic Curves

18 Frey s Observation History of the Conjecture Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Theorem (Gerhard Frey, 1989) Let P = (A,B,C) be an ABC Triple, that is, a triple of relatively prime integers such that A+B = C. Then the corresponding curve E A,B,C : y 2 = x(x A)(x +B) has remarkable properties. Question How do you explain this to undergraduates? Answer: You don t!

19 Elliptic Curves History of the Conjecture Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Definition An elliptic curve E over a field K is a nonsingular projective curve of genus 1 possessing a K-rational point O. Proposition An elliptic curve E over K is birationally equivalent to a cubic equation E : y 2 = x 3 3c 4 x 2c 6 having nonzero discriminant (E) = 12 3( c 2 6 c3 4). Define the set of K-rational points on E as the projective variety { } E(K) = (x 1 : x 2 : x 0 ) P 2 (K) x2 2 x 0 = x1 3 3c 4x 1 x0 2 2c 6x0 3 where O = (0 : 1 : 0) is on the line x 0 = 0. In practice, K = Q(t) or Q.

20 Group Law History of the Conjecture Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Definition Let E be an elliptic curve over a field K. Define the operation as P Q = (P Q) O.

21 Mordell-Weil Group History of the Conjecture Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Theorem (Henri Poincaré, 1901) Let E be an elliptic curve defined over a field K. Then E(K) is an abelian group under with identity O. Theorem (Louis Mordell, 1922) Let E be an elliptic curve over Q. Then E(Q) is finitely generated: there exists a finite group E(Q) tors and a nonnegative integer r such that E(Q) E(Q) tors Z r. Definition E(Q) is called the Mordell-Weil group of E. The finite set E(Q) tors is called the torsion subgroup of E, and the nonnegative integer r is called the rank of E.

22 Classification of Torsion Subgroups Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Theorem (Barry Mazur, 1977) Let E is an elliptic curve over Q. Then { Z N where 1 N 10 or N = 12; E(Q) tors Z 2 Z 2N where 1 N 4. Corollary (Gerhard Frey, 1989) For each ABC Triple, the elliptic curve E A,B,C : y 2 = x(x A)(x +B) has discriminant (E A,B,C ) = 16A 2 B 2 C 2 and E A,B,C (Q) tors Z 2 Z 2N. Question For each ABC Triple, which torsion subgroups do occur?

23 Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Proposition (EHG and Jamie Weigandt, 2009) All possible subgroups do occur and infinitely often. Proof: Choose relatively prime integers m and n. We have the following N-isogeneous curves: A B C E A,B,C (Q) tors ( 2mn ) 2 ( m 2 n 2) 2 ( m 2 + n 2) 2 Z 2 Z 4 8mn ( m 2 + n 2) ( m n ) 4 ( m + n ) 4 Z 2 Z 2 16mn 3 ( m + n ) 3 ( m 3n ) ( m + 3n )( m n ) 3 Z 2 Z 6 16m 3 n ( m + n )( m 3n ) 3 ( m + 3n ) 3 ( m n ) Z 2 Z 2 ( ) 2mn 4 ( m 4 6m 2 n 2 + n 4) (m 2 + n 2) 2 ( m 2 n 2) 4 16mn ( m 2 n 2) (m 2 + n 2) 2 ( m 2 2mn n 2) 4 ( m 2 + 2mm n 2) 4 Z 2 Z 8 Z 2 Z 2

24 Examples History of the Conjecture Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Rank of Quality E A,B,C (Q) tors m n Quality q(a,b,c) Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z Z 2 Z

25 Exceptional Quality Revisited Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Proposition There are infinitely many ABC Triples P = (A,B,C) with q(p) > 1. Proof: For each positive integer k, define the relatively prime integers A k = 1, B k = 2 k+2( 2 k 1 ), and C k = ( 2 k+1 1 ) 2. Then P k = (A k,b k,c k ) is an ABC Triple. Moreover, rad ( A k B k C k ) = rad ( (2 k+1 2)(2 k+1 1) ) (2 k+1 2)(2 k+1 1) < C k. Hence q(p k ) = max{ ln A k, ln B k, ln C k } ln rad(a k B k C k ) = ln C k ln rad(a k B k C k ) > 1. Corollary If the ABC Conjecture holds, then limsupq(a,b,c) = 1.

26 Quality by Torsion Subgroup Elliptic Curves Torsion Subgroups Quality by Torsion Subgroup Question Fix N = 1, 2, 3, 4. Let F(N) denote the those ABC Triples (A,B,C) such that E A,B,C (Q) tors Z 2 Z 2N. What can we say about lim sup q(a,b,c)? (A,B,C) F(N) Theorem (Alexander Barrios, Caleb Tillman and Charles Watts, 2010) Fix N = 1, 2, 4. There are infinitely many ABC Triples with E A,B,C (Q) tors Z 2 Z 2N and q(a,b,c) > 1. In particular, if the ABC Conjecture holds, then limsup P F(N) q(p) = 1. Proof: Use the formulas above to create a dynamical system!

27 of Degree 2 of Degree 4 of Degree 8 Can we use elliptic curves to find ABC Triples with exceptional quality q(a, B, C)?

28 of Degree 2 of Degree 4 of Degree 8 In what follows, we will substitute A k = m, B k = n, and C k = m+n. A B C E A,B,C (Q) tors ( 2mn ) 2 ( m 2 n 2) 2 ( m 2 + n 2) 2 Z 2 Z 4 8mn ( m 2 + n 2) ( m n ) 4 ( m + n ) 4 Z 2 Z 2 16mn 3 ( m + n ) 3 ( m 3n ) ( m + 3n )( m n ) 3 Z 2 Z 6 16m 3 n ( m + n )( m 3n ) 3 ( m + 3n ) 3 ( m n ) Z 2 Z 2 ( ) 2mn 4 ( m 4 6m 2 n 2 + n 4) (m 2 + n 2) 2 ( m 2 n 2) 4 16mn ( m 2 n 2) (m 2 + n 2) 2 ( m 2 2mn n 2) 4 ( m 2 + 2mm n 2) 4 Z 2 Z 8 Z 2 Z 2

29 Motivation History of the Conjecture of Degree 2 of Degree 4 of Degree 8 Consider a sequence {P 0,..., P k, P k+1,...} defined recursively by A k+1 B k+1 = A2 4A k k B Bk 2 ( k ) A2 k or 2 Ak B k. C k+1 B 2 k Proposition (EHG and Jamie Weigandt, 2009) If the following properties hold for k = 0, they hold for all k 0: i. A k, B k, and C k are relatively prime, positive integers. ii. A k +B k = C k. iii. A k 0 (mod 16) and C k 1 (mod 4). Corollary For ǫ > 0, there exists δ such that max { ln A k, ln B k, ln C k } > ǫ when k δ. Hence q(p k ) > 1 for all k 0 if and only if q(p 0 ) > 1. There exists an infinite sequence with q(p k ) > 1. C 2 k

30 of Degree 2 of Degree 4 of Degree 8 Z 2 Z 2 and Z 2 Z 4 Consider a sequence {P 0,..., P k, P k+1,...} defined recursively by A k+1 8A k B ( k A 2 +B ) 2 ( ) 2 k B k+1 = ( ) k 2Ak B 4 Ak B k ( k or A 2 k B 2 2 k). C ( k+1 A 2 k +Bk) 2 2 C 4 k Proposition (Alexander Barrios, Caleb Tillman and Charles Watts, 2010) If the following properties hold for k = 0, they hold for all k 0: i. A k, B k, and C k are relatively prime, positive integers. ii. A k +B k = C k. iii. A k 0 (mod 16) and C k 1 (mod 4). Corollary For ǫ > 0, there exists δ such that max { ln A k, ln B k, ln C k } > ǫ when k δ. Hence q(p k ) > 1 for all k 0 if and only if q(p 0 ) > 1. There exist infinitely E Pk (Q) tors Z 2 Z 2N for N = 1, 2; q(p k ) > 1.

31 Z 2 Z 6? History of the Conjecture of Degree 2 of Degree 4 of Degree 8 Consider a sequence {P 0,..., P k, P k+1,...} defined recursively by A 16A k+1 k B B k+1 ( ) k 3 3 ( ) = Ak +B k Ak 3B ( k. )( ) 3 Ak +3B k Ak B k C k+1 Proposition (Alexander Barrios, Caleb Tillman and Charles Watts, 2010) If the following properties hold for k = 0, they hold for all k 0: i. A k, B k, and C k are relatively prime integers. ii. A k +B k = C k. iii. A k 0 (mod 16) and C k 1 (mod 4). Question What condition do we need to guarantee that these are positive integers?

32 Z 2 Z 6? History of the Conjecture of Degree 2 of Degree 4 of Degree 8 We sketch why perhaps 3.214B k > A k. Define the rational number 1 x k = A 1 ( ) 3 ( ) Ak +B k Ak 3B k = k x = k+1 x k 16A k B 3 B k k A k B k = x4 k 6x2 k 8x k x k The largest root is x 0 = , so 0 < x k+1 < x k < x 0 is decreasing.

33 of Degree 2 of Degree 4 of Degree 8 Z 2 Z 8 Consider a sequence {P 0,..., P k, P k+1,...} defined recursively by A ( ) 4 k+1 2Ak B B k+1 ( k )( = A 4 k 6A 2 k B2 k +B4 k A 2 k +B 2 2 k). C ( k+1 A 2 k Bk) 2 4 Proposition (Alexander Barrios, Caleb Tillman and Charles Watts, 2010) If the following properties hold for k = 0, they hold for all k 0: i. A k, B k, and C k are relatively prime, positive integers. ii. A k +B k = C k and B k > 3.174A k. iii. A k 0 (mod 16) and C k 1 (mod 4). Corollary For ǫ > 0, there exists δ such that max { ln A k, ln B k, ln C k } > ǫ when k δ. Hence q(p k ) > 1 for all k 0 if and only if q(p 0 ) > 1. There exists a sequence E Pk (Q) tors Z 2 Z 8 and q(p k ) > 1.

34 of Degree 2 of Degree 4 of Degree 8 Z 2 Z 8 We sketch why B k > 3.174A k. Define the rational number x k = B x k+1 x k = k = A k ( )( A 4 k 6A 2 k B2 k +B4 k A 2 k +Bk) 2 2 ( ) 4 B k 2Ak B k A k = x8 k 4x6 k 16x5 k 10x4 k 4x2 k +1 16xk The largest root is x 0 = , so x 0 < x k < x k+1 is increasing.

35 Example History of the Conjecture of Degree 2 of Degree 4 of Degree 8 We can generate many examples of ABC Triples P = (A,B,C) with E A,B,C (Q) tors Z 2 Z 8 and q(a,b,c) > 1. We consider the recursive sequence defined by A ( ) 4 k+1 2Ak B k B k+1 ( )( = A 4 k 6A 2 k B2 k +B4 k A 2 k +Bk) 2 2. ( A 2 k Bk) 2 4 C k+1 Initialize with P 0 = ( 16 2, 63 2, 65 2) so that we have i. A k, B k, and C k are relatively prime, positive integers. ii. A k +B k = C k and B k > 3.174A k. iii. A k 0 (mod 16) and C k 1 (mod 4). k A k B k C k q(p k )

36 of Degree 2 of Degree 4 of Degree 8 Questions?