Pythagoras = $1 million problem. Ken Ono Emory University
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1 Pythagoras = $1 million problem Ken Ono Emory University
2 Pythagoras The Pythagorean Theorem Theorem (Pythagoras) If (a, b, c) is a right triangle, then a 2 + b 2 = c 2.
3 Pythagoras The Pythagorean Theorem Theorem (Pythagoras) If (a, b, c) is a right triangle, then a 2 + b 2 = c 2. Example We have:
4 Pythagoras Proof of the Pythagorean Theorem
5 Pythagoras Proof of the Pythagorean Theorem
6 Pythagoras Proof of the Pythagorean Theorem Four (a, b, c) right triangles and one large c c square.
7 Pythagoras Proof of the Pythagorean Theorem Four (a, b, c) right triangles and one large c c square. This has area: ab + c2 = 2ab + c 2.
8 Pythagoras Proof of the Pythagorean Theorem Four (a, b, c) right triangles and one large c c square. This has area: ab + c2 = 2ab + c 2. As one large square, it has area: (a + b) 2 = a 2 + 2ab + b 2.
9 Pythagoras Proof of the Pythagorean Theorem Four (a, b, c) right triangles and one large c c square. This has area: ab + c2 = 2ab + c 2. As one large square, it has area: (a + b) 2 = a 2 + 2ab + b 2. = c 2 = a 2 + b 2.
10 Pythagoras Definition Integers (a, b, c) form a Pythagorean Triple if a, b, c > 0 and a 2 + b 2 = c 2.
11 Pythagoras Definition Integers (a, b, c) form a Pythagorean Triple if a, b, c > 0 and a 2 + b 2 = c 2. Moreover, it is called primitive if gcd(a, b, c) = 1.
12 Pythagoras Definition Integers (a, b, c) form a Pythagorean Triple if a, b, c > 0 and a 2 + b 2 = c 2. Moreover, it is called primitive if gcd(a, b, c) = 1. Example The first few Pythagorean triples: (3, 4, 5), (5, 12, 13), (2 3,2 4, 2 5), (7, 24, 25), (8, 15, 17), (3 3, 3 4, 3 5)...
13 Pythagoras Definition Integers (a, b, c) form a Pythagorean Triple if a, b, c > 0 and a 2 + b 2 = c 2. Moreover, it is called primitive if gcd(a, b, c) = 1. Example The first few Pythagorean triples: (3, 4, 5), (5, 12, 13), (2 3,2 4, 2 5), (7, 24, 25), (8, 15, 17), (3 3, 3 4, 3 5)... The first few Primitive Pythagorean Triples: (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41),...
14 Pythagoras Natural questions
15 Pythagoras Natural questions Question How many Pythagorean Triples exist?
16 Pythagoras Natural questions Question How many Pythagorean Triples exist? Answer Easy...infinitely many because of scaling.
17 Pythagoras Natural questions Question How many Pythagorean Triples exist? Answer Easy...infinitely many because of scaling. Better Question How many Primitive Pythagorean Triples exist?
18 Classifying Pythagorean Triples Beautiful Theorem
19 Classifying Pythagorean Triples Beautiful Theorem Theorem (Euclid) Every PPT with odd a and even b is of the form (a, b, c) = (st, s2 t 2, s2 + t 2 ) 2 2 where s > t 1 are odd coprime integers.
20 Classifying Pythagorean Triples Beautiful Theorem Theorem (Euclid) Every PPT with odd a and even b is of the form (a, b, c) = (st, s2 t 2, s2 + t 2 ) 2 2 where s > t 1 are odd coprime integers. Example This theorem is easy to use: (s, t) = (17, 5) = (a, b, c) = (85, 132, 157).
21 Rational Points on the unit circle Connection to Unit Circle
22 Rational Points on the unit circle Connection to Unit Circle a 2 + b 2 = c 2 = ( a c ) 2 + ( b c ) 2 = 1.
23 Rational Points on the unit circle Connection to Unit Circle a 2 + b 2 = c 2 = ( a c ) 2 + ( b c ) 2 = 1.
24 Rational Points on the unit circle Connection to Unit Circle a 2 + b 2 = c 2 = ( a c ) 2 + ( b c ) 2 = 1. Question How do we find all the rational points (i.e. x, y rational numbers) on the unit circle?
25 Rational Points on the unit circle Sample points...
26 Rational Points on the unit circle Sample points... Obvious rational points on the unit circle: (±1, 0) and (0, ±1).
27 Rational Points on the unit circle Sample points... Obvious rational points on the unit circle: (±1, 0) and (0, ±1). Some much less obvious points: ( 4 5, 3 ) ( 45, 5 53, 28 ),..., 53 ( , ),
28 Rational Points on the unit circle
29 Rational Points on the unit circle
30 Rational Points on the unit circle Rational pts P ( 1, 0) have lines with rational slopes m.
31 Rational Points on the unit circle Rational pts P ( 1, 0) have lines with rational slopes m. By substituting y = mx + m into x 2 + y 2 = 1 = x 2 + (mx + m) 2 = 1.
32 Rational Points on the unit circle Rational pts P ( 1, 0) have lines with rational slopes m. By substituting y = mx + m into x 2 + y 2 = 1 = x 2 + (mx + m) 2 = 1. One root is x = 1
33 Rational Points on the unit circle Rational pts P ( 1, 0) have lines with rational slopes m. By substituting y = mx + m into x 2 + y 2 = 1 = x 2 + (mx + m) 2 = 1. ( ) One root is x = 1 and the other gives P = 1 m 2 m 2 +1, 2m. m 2 +1
34 Rational Points on the unit circle Rational Points Theorem (Chord Method) The rational points on the unit circle are: {( ) } 1 m 2 ( 1, 0) m 2 + 1, 2m m 2 : m rational. + 1
35 Rational Points on the unit circle Rational Points Theorem (Chord Method) The rational points on the unit circle are: {( ) } 1 m 2 ( 1, 0) m 2 + 1, 2m m 2 : m rational. + 1 Remark By drawing and intersecting lines, we determined all the rational points from a single point ( 1, 0).
36 Rational Points on the unit circle Natural Questions
37 Rational Points on the unit circle Natural Questions Can one solve other Diophantine equations from a finite seed set of points by intersecting lines?
38 Rational Points on the unit circle Natural Questions Can one solve other Diophantine equations from a finite seed set of points by intersecting lines? How many points are needed for starters?
39 Rational Points on the unit circle Natural Questions Can one solve other Diophantine equations from a finite seed set of points by intersecting lines? How many points are needed for starters? What if one cannot find any points to start with?
40 Congruent Numbers An ancient problem Definition An integer is congruent if it is the area of a right triangle with rational sidelengths.
41 Congruent Numbers An ancient problem Definition An integer is congruent if it is the area of a right triangle with rational sidelengths. Problem (Arab Scholars) Classify all of the congruent numbers.
42 Congruent Numbers Is this an easy problem?
43 Congruent Numbers Is this an easy problem? Example Here are some facts:
44 Congruent Numbers Is this an easy problem? Example Here are some facts: 6 is congruent thanks to (3, 4, 5).
45 Congruent Numbers Is this an easy problem? Example Here are some facts: 6 is congruent thanks to (3, 4, 5). 5 is congruent since
46 Congruent Numbers Is this an easy problem? Example Here are some facts: 6 is congruent thanks to (3, 4, 5). 5 is congruent since ( ) ( ) 20 2 = 3 ( ) 41 2 and = 5.
47 Congruent Numbers Is this an easy problem? Example Here are some facts: 6 is congruent thanks to (3, 4, 5). 5 is congruent since ( ) ( ) 20 2 = 3 ( ) 41 2 and = 5. 1 is not congruent because???.
48 Congruent Numbers Zagier s Example
49 Congruent Numbers Zagier s Example Example The number 157 is congruent, since it is the area of ( ,, ).
50 Congruent Numbers Zagier s Example Example The number 157 is congruent, since it is the area of ( ,, ). Remark The problem of classifying congruent numbers is probably hard.
51 Elliptic curves Another Chord Law Group Law E : y 2 = x 3 + Ax + B
52 Elliptic curves Example E : y 2 = x(x 3)(x + 32)
53 Elliptic curves Example E : y 2 = x(x 3)(x + 32) We find that P + Q = ( , )
54 Elliptic curves Big theorems
55 Elliptic curves Big theorems Theorem (Classical Fact) The rational points on an elliptic curve form an abelian group.
56 Elliptic curves Big theorems Theorem (Classical Fact) The rational points on an elliptic curve form an abelian group. Theorem (Mordell) The rational points of an elliptic curve form a finitely generated abelian group.
57 Elliptic curves Big theorems Theorem (Classical Fact) The rational points on an elliptic curve form an abelian group. Theorem (Mordell) The rational points of an elliptic curve form a finitely generated abelian group. Question What kind of groups arise?
58 Elliptic curves Examples of Groups of Rational Points
59 Elliptic curves Examples of Groups of Rational Points E Group # of Finite Pts y 2 = x(x 1)(x + 1) Z/2Z Z/2Z 3 y 2 = x Z/6Z 5 y 2 = x Z Z y 2 = x x + 10 Z/1Z 0
60 Elliptic curves A Classical Diophantine criterion
61 Elliptic curves A Classical Diophantine criterion Theorem An integer D is congruent if and only if the elliptic curve has infinitely many points. E D : y 2 = x(x + D)(x D)
62 Elliptic curves Some data
63 Elliptic curves Some data Example The first few congruent numbers: 5, 6, 7, 13, 14, 15, 20, 21, 22, 23,....
64 Elliptic curves Some data Example The first few congruent numbers: 5, 6, 7, 13, 14, 15, 20, 21, 22, 23,.... The first few non-congruent numbers: 1, 2, 3, 4, 8, 9, 10, 11, 12, 16, 17, 18, 19,....
65 Elliptic curves Some data Example The first few congruent numbers: 5, 6, 7, 13, 14, 15, 20, 21, 22, 23,.... The first few non-congruent numbers: 1, 2, 3, 4, 8, 9, 10, 11, 12, 16, 17, 18, 19,.... Conjecture Half of the integers are congruent.
66 Elliptic curves How do we make use of this criterion?
67 Elliptic curves How do we make use of this criterion? Good question...
68 Elliptic curves How do we make use of this criterion? Good question...a $1 million question!
69 $1 million bounty
70 $1 million bounty Definition (Trace mod p) For primes p, let a(p) := p #{(x, y) (mod p) : y 2 x 3 x (mod p)}.
71 $1 million bounty Definition (Trace mod p) For primes p, let a(p) := p #{(x, y) (mod p) : y 2 x 3 x (mod p)}. Example For p = 7 we have the 7 points mod 7: {(0, 0), (1, 0), (4, 2), (4, 5), (5, 1), (5, 6), (6, 0)}.
72 $1 million bounty Definition (Trace mod p) For primes p, let a(p) := p #{(x, y) (mod p) : y 2 x 3 x (mod p)}. Example For p = 7 we have the 7 points mod 7: {(0, 0), (1, 0), (4, 2), (4, 5), (5, 1), (5, 6), (6, 0)}. = a(7) = 7 7 = 0.
73 $1 million bounty A very strange phenomenon Define integers A(n) by A(n)x n := x (1 x 4n ) 2 (1 x 8n ) 2 = x 2x 5 3x n=1 n=1
74 $1 million bounty A very strange phenomenon Define integers A(n) by A(n)x n := x (1 x 4n ) 2 (1 x 8n ) 2 = x 2x 5 3x n=1 n=1 Then for primes p we have:
75 $1 million bounty A very strange phenomenon Define integers A(n) by A(n)x n := x (1 x 4n ) 2 (1 x 8n ) 2 = x 2x 5 3x n=1 n=1 Then for primes p we have: p a(p) A(p)
76 $1 million bounty A very strange phenomenon Define integers A(n) by A(n)x n := x (1 x 4n ) 2 (1 x 8n ) 2 = x 2x 5 3x n=1 n=1 Then for primes p we have: p a(p) A(p) Theorem (Modularity) If p is prime, then A(p) = a(p).
77 $1 million bounty The Hasse-Weil Function
78 $1 million bounty The Hasse-Weil Function For D, define the function L(D, s) := n=1 ( D ) n A(n) n s. Example For D = 1, we find that L(1, s) =
79 $1 million bounty So what?
80 $1 million bounty So what? D Congruent (Y/N) L(D, 1) 5 Y 0 6 Y 0 7 Y 0 8 N N N N N Y 0 14 Y 0 15 Y 0
81 $1 million bounty Birch and Swinnerton-Dyer Conjecture Conjecture If E/Q is an elliptic curve and L(E, s) is its L-function, then L(E, 1) = 0 if and only if #E(Q) = +.
82 $1 million bounty Birch and Swinnerton-Dyer Conjecture Conjecture If E/Q is an elliptic curve and L(E, s) is its L-function, then L(E, 1) = 0 if and only if #E(Q) = +. Corollary Assuming BSD, D is congruent iff L(D, 1) = 0.
83 $1 million bounty Kolyvagin s Theorem Theorem (Kolyvagin) If L(E, 1) 0, then #E(Q) < +.
84 $1 million bounty Kolyvagin s Theorem Theorem (Kolyvagin) If L(E, 1) 0, then #E(Q) < +. Remark If ord s=1 (L(E, s)) {0, 1}, then he proves that this order is the number of generators.
85 $1 million bounty A strange criterion using modularity
86 $1 million bounty A strange criterion using modularity Theorem (Tunnell, 1983) If D is odd and square-free, then L(D, 1) = 0 if and only if #{2x 2 + y z 2 = D} = 1 2 #{2x 2 + y 2 + 8z 2 = D}.
87 $1 million bounty A strange criterion using modularity Theorem (Tunnell, 1983) If D is odd and square-free, then L(D, 1) = 0 if and only if #{2x 2 + y z 2 = D} = 1 2 #{2x 2 + y 2 + 8z 2 = D}. In particular, BSD implies that D is congruent iff we have equality.
88 $1 million bounty A strange criterion using modularity Theorem (Tunnell, 1983) If D is odd and square-free, then L(D, 1) = 0 if and only if #{2x 2 + y z 2 = D} = 1 2 #{2x 2 + y 2 + 8z 2 = D}. In particular, BSD implies that D is congruent iff we have equality. Remark
89 $1 million bounty A strange criterion using modularity Theorem (Tunnell, 1983) If D is odd and square-free, then L(D, 1) = 0 if and only if #{2x 2 + y z 2 = D} = 1 2 #{2x 2 + y 2 + 8z 2 = D}. In particular, BSD implies that D is congruent iff we have equality. Remark (1) There is a similar criterion for even square-free D.
90 $1 million bounty A strange criterion using modularity Theorem (Tunnell, 1983) If D is odd and square-free, then L(D, 1) = 0 if and only if #{2x 2 + y z 2 = D} = 1 2 #{2x 2 + y 2 + 8z 2 = D}. In particular, BSD implies that D is congruent iff we have equality. Remark (1) There is a similar criterion for even square-free D. (2) By Kolyvagin, no equality = D is not congruent.
91 $1 million bounty A strange criterion using modularity Theorem (Tunnell, 1983) If D is odd and square-free, then L(D, 1) = 0 if and only if #{2x 2 + y z 2 = D} = 1 2 #{2x 2 + y 2 + 8z 2 = D}. In particular, BSD implies that D is congruent iff we have equality. Remark (1) There is a similar criterion for even square-free D. (2) By Kolyvagin, no equality = D is not congruent. (3) The converse may require solving the $1 million problem.
92 In closing... Some facts...
93 In closing... Some facts... It is easy to classify Pythagorean Triples.
94 In closing... Some facts... It is easy to classify Pythagorean Triples....motivates using chords to study rational points.
95 In closing... Some facts... It is easy to classify Pythagorean Triples....motivates using chords to study rational points....morphs into the chord law for elliptic curves.
96 In closing... Some facts... It is easy to classify Pythagorean Triples....motivates using chords to study rational points....morphs into the chord law for elliptic curves....hard to classify congruent numbers.
97 In closing... Some facts... It is easy to classify Pythagorean Triples....motivates using chords to study rational points....morphs into the chord law for elliptic curves....hard to classify congruent numbers. If we could... maybe we d win $1 million!
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