Twisted Extensions of Fermat s Last Theorem
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1 University of British Columbia December 7th, 2012
2 Why Study Diophantine Equations?
3 Why Study Diophantine Equations? It s a fun challenge.
4 Why Study Diophantine Equations? It s a fun challenge. It gives justification for other studying subjects (for example algebraic number theory or algebraic geometry).
5 Why Study Diophantine Equations? It s a fun challenge. It gives justification for other studying subjects (for example algebraic number theory or algebraic geometry). It leads to other interesting questions.
6 Why Study Diophantine Equations? It s a fun challenge. It gives justification for other studying subjects (for example algebraic number theory or algebraic geometry). It leads to other interesting questions. For example Pell equations, x 2 dy 2 = 1, lead to questions about continued fractions and fundamental units.
7 Why Study Diophantine Equations? It s a fun challenge. It gives justification for other studying subjects (for example algebraic number theory or algebraic geometry). It leads to other interesting questions. For example Pell equations, x 2 dy 2 = 1, lead to questions about continued fractions and fundamental units. Fermat s Last Theorem x n + y n = z n lead to questions about unique factorization domains, cyclotomic fields, elliptic curves and modular forms.
8 The set S and the work of Bennett, Luca and Mulholland Today, I will present known solutions of x 3 + y 3 = p α z n with p a given prime and α 1 an integer.
9 The set S and the work of Bennett, Luca and Mulholland Today, I will present known solutions of x 3 + y 3 = p α z n with p a given prime and α 1 an integer. Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point.
10 The set S and the work of Bennett, Luca and Mulholland Today, I will present known solutions of x 3 + y 3 = p α z n with p a given prime and α 1 an integer. Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point. The set S contains the primes between 5 and 193 (the first exception is 197).
11 The set S and the work of Bennett, Luca and Mulholland Today, I will present known solutions of x 3 + y 3 = p α z n with p a given prime and α 1 an integer. Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point. The set S contains the primes between 5 and 193 (the first exception is 197). It is not known if S is infinite.
12 The set S and the work of Bennett, Luca and Mulholland Today, I will present known solutions of x 3 + y 3 = p α z n with p a given prime and α 1 an integer. Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point. The set S contains the primes between 5 and 193 (the first exception is 197). It is not known if S is infinite. It is known that the complement is infinite. It can be shown that if p 317, 1757 (mod 2040) then p / S.
13 The set S and the work of Bennett, Luca and Mulholland Today, I will present known solutions of x 3 + y 3 = p α z n with p a given prime and α 1 an integer. Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point. The set S contains the primes between 5 and 193 (the first exception is 197). It is not known if S is infinite. It is known that the complement is infinite. It can be shown that if p 317, 1757 (mod 2040) then p / S. In fact, the complement of S forms a set of density one in the primes.
14 The set S and the work of Bennett, Luca and Mulholland Reminder Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point.
15 The set S and the work of Bennett, Luca and Mulholland Reminder Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point. Theorem (Bennett, Luca, Mulholland ) Suppose p 5 is prime and p / S. Let α 1, α Z. Then the equation x 3 + y 3 = p α z n has no solution in coprime nonzero x, y, z Z and prime n with n p 2p.
16 The set S and the work of Bennett, Luca and Mulholland Reminder Definition Let S be the set of primes p 5 for which there exists an elliptic curve E with conductor N E {18p, 36p, 72p} with at least one non-trivial rational 2-torsion point. Theorem (Bennett, Luca, Mulholland ) Suppose p 5 is prime and p / S. Let α 1, α Z. Then the equation x 3 + y 3 = p α z n has no solution in coprime nonzero x, y, z Z and prime n with n p 2p. What about primes in S?
17 An extension of the approach of Bennett, Luca and Mulholland Suppose we have a solution to our Diophantine equation, say a 3 + b 3 = p α c n.
18 An extension of the approach of Bennett, Luca and Mulholland Suppose we have a solution to our Diophantine equation, say a 3 + b 3 = p α c n. Associate to this solution a Frey curve E a,b : y 2 = f (x) where f (x) := (x b + a)(x 2 + (a b)x + (a 2 + ab + b 2 ))
19 An extension of the approach of Bennett, Luca and Mulholland Suppose we have a solution to our Diophantine equation, say a 3 + b 3 = p α c n. Associate to this solution a Frey curve E a,b : y 2 = f (x) where f (x) := (x b + a)(x 2 + (a b)x + (a 2 + ab + b 2 )) This last quadratic has discriminant 3(a + b) 2 and hence splits completely over F l when ( ) 3 l = 1, that is when l 1 (mod 6).
20 An extension of the approach of Bennett, Luca and Mulholland Suppose we have a solution to our Diophantine equation, say a 3 + b 3 = p α c n. Associate to this solution a Frey curve E a,b : y 2 = f (x) where f (x) := (x b + a)(x 2 + (a b)x + (a 2 + ab + b 2 )) This last quadratic has discriminant 3(a + b) 2 and hence splits completely over F l when ( ) 3 l = 1, that is when l 1 (mod 6). Hence, 4 #E a,b (F l ) and thus a l (E a,b ) := l + 1 #E a,b (F l ) l + 1 (mod 4).
21 An extension of the approach of Bennett, Luca and Mulholland Suppose we have a solution to our Diophantine equation, say a 3 + b 3 = p α c n. Associate to this solution a Frey curve E a,b : y 2 = f (x) where f (x) := (x b + a)(x 2 + (a b)x + (a 2 + ab + b 2 )) This last quadratic has discriminant 3(a + b) 2 and hence splits completely over F l when ( ) 3 l = 1, that is when l 1 (mod 6). Hence, 4 #E a,b (F l ) and thus a l (E a,b ) := l + 1 #E a,b (F l ) l + 1 (mod 4). Ribet s level lowering applied to E a,b gives us a newform f of level 18p, 36p or 72p. When the newform is irrational or if the newform is rational and does not have two torsion, we can show that n p 2p.
22 An extension of the approach of Bennett, Luca and Mulholland For rational newforms with two torsion, suppose that the associated elliptic curve F has the property that a l (F ) l + 1 (mod 4) for some prime l 1 (mod 6).
23 An extension of the approach of Bennett, Luca and Mulholland For rational newforms with two torsion, suppose that the associated elliptic curve F has the property that a l (F ) l + 1 (mod 4) for some prime l 1 (mod 6). Ribet s level lowering gives us that n (a l (E a,b ) a l (F )) for all but finitely many primes l.
24 An extension of the approach of Bennett, Luca and Mulholland For rational newforms with two torsion, suppose that the associated elliptic curve F has the property that a l (F ) l + 1 (mod 4) for some prime l 1 (mod 6). Ribet s level lowering gives us that n (a l (E a,b ) a l (F )) for all but finitely many primes l. We already know that a l (E a,b ) l + 1 a l (F ) (mod 4). A result of Kraus states that a prime where they differ must occur at some value of l p 2 and thus the Hasse bound says that this difference at l is small compared to p 2p.
25 An extension of the approach of Bennett, Luca and Mulholland For rational newforms with two torsion, suppose that the associated elliptic curve F has the property that a l (F ) l + 1 (mod 4) for some prime l 1 (mod 6). Ribet s level lowering gives us that n (a l (E a,b ) a l (F )) for all but finitely many primes l. We already know that a l (E a,b ) l + 1 a l (F ) (mod 4). A result of Kraus states that a prime where they differ must occur at some value of l p 2 and thus the Hasse bound says that this difference at l is small compared to p 2p. Hence in this case we get an additional restriction on n. Our goal is thus to classify the following set.
26 The set P b Definition Let P b be the set of primes p 5 such that for every elliptic curve E with conductor N E {18p, 36p, 72p} we have that 4 #E tor (Q) and at least one curve having a non-trivial rational 2-torsion point. Also, at all curves F with non-trivial rational 2-torsion, we require that there exists a prime l 1 (mod 6) such that a l (F ) l + 1 (mod 4). Note P b S.
27 The set P b Definition Let P b be the set of primes p 5 such that for every elliptic curve E with conductor N E {18p, 36p, 72p} we have that 4 #E tor (Q) and at least one curve having a non-trivial rational 2-torsion point. Also, at all curves F with non-trivial rational 2-torsion, we require that there exists a prime l 1 (mod 6) such that a l (F ) l + 1 (mod 4). Note P b S. Primes in P b include 53, 83, 149, 167, 173, 199,... (sequence A in OEIS).
28 The set P b Definition Let P b be the set of primes p 5 such that for every elliptic curve E with conductor N E {18p, 36p, 72p} we have that 4 #E tor (Q) and at least one curve having a non-trivial rational 2-torsion point. Also, at all curves F with non-trivial rational 2-torsion, we require that there exists a prime l 1 (mod 6) such that a l (F ) l + 1 (mod 4). Note P b S. Primes in P b include 53, 83, 149, 167, 173, 199,... (sequence A in OEIS). Classifying these points gives the following theorem.
29 New Result Theorem (B. 2013?) Suppose that p S and that for each curve of conductor {18p, 36p, 72p}, we have that the rational torsion subgroup is not divisible by 4. Then p P b if and only if every elliptic curve with conductor in {18p, 36p, 72p} and non-trivial rational two torsion has discriminant not of the form 3m 2 for any integer m.
30 New Result Theorem (B. 2013?) Suppose that p S and that for each curve of conductor {18p, 36p, 72p}, we have that the rational torsion subgroup is not divisible by 4. Then p P b if and only if every elliptic curve with conductor in {18p, 36p, 72p} and non-trivial rational two torsion has discriminant not of the form 3m 2 for any integer m. In fact, we can also give the type of such primes that are in S but not in P b.
31 New Result Theorem (B. 2013?) Suppose that p S and that for each curve of conductor {18p, 36p, 72p}, we have that the rational torsion subgroup is not divisible by 4. Then p P b if and only if every elliptic curve with conductor in {18p, 36p, 72p} and non-trivial rational two torsion has discriminant not of the form 3m 2 for any integer m. In fact, we can also give the type of such primes that are in S but not in P b. Hence, we have Theorem (B. 2013?) Suppose p 5 is prime and p P b S. Let α 1, α Z. Then the equation x 3 + y 3 = p α z n has no solution in coprime nonzero x, y, z Z and prime n with n p 2p.
32 Primes in S
33 Primes in S p = 5, 7, 11, 13, 17, 19, 23, 29, 31, 47, 67, 73, 193, 1153 p = 2 a 3 b ± 1 p = 3 b ± 2 a p n = t 2 ± 2 a 3 b, n = 1 or the least prime divisor of n is 7. 3 b p = t a p = 3t 2 ± 2 a p = t b p = t b+1 4p = t b+1 4p n = 3t and p 1 (mod 4), n = 1, 2 p = 3t 2 2 a with a = 2, 4, 5 3 b p n = t , n = 1 or the least prime divisor of n is 7. p = 3t a and a = 2, 4, 5
34 Primes in P b p = 5, 7, 11, 13, 17, 19, 23, 29, 31, 47, 67, 73, 193, 1153 p = 2 a 3 b ± 1 p = 3 b ± 2 a p n = ±(t 2 2 a 3 b ), n = 1, and a, b not both even, a 4. 3 b p = t a, t ±1 p = 3t 2 ± 2 a p = t b, b even p = t b+1 4p = t b+1 4p n = 3t and p 1 (mod 4), n = 1, 2 p = 3t 2 2 a with a = 2, 4, 5 3 b p n = t , n = 1 or the least prime divisor of n is 7. p = 3t a and a = 2, 4, 5
35 Proof Outline First, suppose that p P b S and assume towards a contradiction that E = 3m 2.
36 Proof Outline First, suppose that p P b S and assume towards a contradiction that E = 3m 2. We may assume that any E with conductor in {18p, 36p, 72p} and with a non-trivial 2 torsion point has the form E : y 2 = x(x 2 + a 2 x + a 4 )
37 Proof Outline First, suppose that p P b S and assume towards a contradiction that E = 3m 2. We may assume that any E with conductor in {18p, 36p, 72p} and with a non-trivial 2 torsion point has the form E : y 2 = x(x 2 + a 2 x + a 4 ) By definition, there exists a prime l 1 (mod 6) such that 4 #E(F l ).
38 Proof Outline First, suppose that p P b S and assume towards a contradiction that E = 3m 2. We may assume that any E with conductor in {18p, 36p, 72p} and with a non-trivial 2 torsion point has the form E : y 2 = x(x 2 + a 2 x + a 4 ) By definition, there exists a prime l 1 (mod 6) such that 4 #E(F l ). The discriminant of the quadratic above has the form q = 3k 2 for some k m.
39 Proof Outline First, suppose that p P b S and assume towards a contradiction that E = 3m 2. We may assume that any E with conductor in {18p, 36p, 72p} and with a non-trivial 2 torsion point has the form E : y 2 = x(x 2 + a 2 x + a 4 ) By definition, there exists a prime l 1 (mod 6) such that 4 #E(F l ). The discriminant of the quadratic above has the form q = 3k 2 for some k m. Since l 1 (mod 6), this quadratic splits in F l and thus 4 #E(F l ), a contradiction.
40 Proof Outline For the opposite direction, suppose that E = 3m 2. We want that p P b.
41 Proof Outline For the opposite direction, suppose that E = 3m 2. We want that p P b. To show this, we are looking for a prime l 1 (mod 6) so that the curve does not split and does not contain a point of order 4 in F l.
42 Proof Outline For the opposite direction, suppose that E = 3m 2. We want that p P b. To show this, we are looking for a prime l 1 (mod 6) so that the curve does not split and does not contain a point of order 4 in F l. Look at the numerator of x(2p), which is (x 2 a 4 ) 2. We want an l where a 4 is a non-quadratic residue modulo l (this gives the 2 torsion criteria), E is a non-quadratic residue modulo l (this prevents the original elliptic curve from splitting) and 3 is a quadratic residue modulo l (this is l 1 (mod 6)).
43 Proof Outline For the opposite direction, suppose that E = 3m 2. We want that p P b. To show this, we are looking for a prime l 1 (mod 6) so that the curve does not split and does not contain a point of order 4 in F l. Look at the numerator of x(2p), which is (x 2 a 4 ) 2. We want an l where a 4 is a non-quadratic residue modulo l (this gives the 2 torsion criteria), E is a non-quadratic residue modulo l (this prevents the original elliptic curve from splitting) and 3 is a quadratic residue modulo l (this is l 1 (mod 6)). This can be accomplished.
44 The Truth How far can this technique extend?
45 The Truth How far can this technique extend? So far we can show x 3 + y 3 = p α z n has no solution if n p 2p provided that p / S or when p P b S. Can we extend this to all of S?
46 The Truth How far can this technique extend? So far we can show x 3 + y 3 = p α z n has no solution if n p 2p provided that p / S or when p P b S. Can we extend this to all of S? Suppose that p = x 3 + y 3. Factoring (in one case) gives p = 3x 2 3x + 1, where the right hand side is irreducible. Hence, Schinzel s hypothesis H suggests that this should be prime infinitely often.
47 The Truth How far can this technique extend? So far we can show x 3 + y 3 = p α z n has no solution if n p 2p provided that p / S or when p P b S. Can we extend this to all of S? Suppose that p = x 3 + y 3. Factoring (in one case) gives p = 3x 2 3x + 1, where the right hand side is irreducible. Hence, Schinzel s hypothesis H suggests that this should be prime infinitely often. Hence, whenever p = x 3 + y 3, we should always have a solution, namely when α = 1, z = 1 and n is any natural number. Thus any primes in S that are a sum of two cubes will not have a theorem statement like the above.
48 Thank you.
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