Rational Points on an Elliptic Curve Dr. Carmen Bruni University of - PowerPoint PPT Presentation

Rational Points on an Elliptic Curve Dr. Carmen Bruni University of Waterloo November 11th, 2015 Lest We Forget Dr. Carmen Bruni Rational Points on an Elliptic Curve

Revisit the Congruent Number Problem Congruent Number Problem Determine which positive integers N can be expressed as the area of a right angled triangle with side lengths all rational. For example 6 is a congruent number since it is the area of the 3 − 4 − 5 right triangle. Dr. Carmen Bruni Rational Points on an Elliptic Curve

Enter Elliptic Curves The associated equations with the congruent number problem, namely x 2 + y 2 = z 2 xy = 2 N can be converted to an elliptic curve of the form Y 2 = X 3 − N 2 X . We also saw that we can reduce our problem to considering only squarefree numbers N . Dr. Carmen Bruni Rational Points on an Elliptic Curve

Going Backwards The belief now is that solving problems related to elliptic curves might be easier than the originally stated problem. The question now that occurs is can we go from an elliptic curve of the form y 2 = x 3 − N 2 x to a rational right triangle with area N ? Dr. Carmen Bruni Rational Points on an Elliptic Curve

Key Theorem 1 Theorem 1. Let ( x , y ) be a point with rational coordinates on the elliptic curve y 2 = x 3 − N 2 x where N is a positive squarefree integer. Suppose that x satisfies three conditions: 1 x is the square of a rational number 2 x has an even denominator 3 x has a numerator that shares no common factor with N Then there exists a right angle triangle with rational sides and area N, that is, N is congruent. Dr. Carmen Bruni Rational Points on an Elliptic Curve

Key Theorem 2 Theorem 2. A number N is congruent if and only if the elliptic curve y 2 = x 3 − N 2 x has a rational point P = ( x , y ) distinct from (0 , 0) and ( ± N , 0) . Thus, determining congruent numbers can be reduced to finding rational points on elliptic curves! Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 1 Let ( x , y ) be a point with rational coordinates on the elliptic curve y 2 = x 3 − N 2 x where N is a positive squarefree integer where x is a rational square, has even denominator (in lowest terms) and has a numerator that shares no common factor with N . Our goal is to trace backwards the proof from last week. Let u = √ x which is given to be rational. Set v = y u giving v 2 = y 2 = x 2 − N 2 . u 2 = x 3 − N 2 x x Let d be the smallest integer such that du ∈ Z (namely the denominator of u in lowest terms). Note that d is even by assumption and that d 4 is the denominator for u 2 = x . Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 1 Since v 2 = x 2 − N 2 and N 2 is an integer, then d 4 is also the denominator of v 2 . Multiplying everything by d 4 gives ( d 2 v ) 2 = ( d 2 x ) 2 − ( d 2 N ) 2 . Since ( d 2 v ) 2 = ( d 2 x ) 2 − ( d 2 N ) 2 , the triple ( d 2 v , d 2 x , d 2 N ) forms a Pythagorean triple. Since the numerator of x shares no common factor with N , we have that this is a primitive triple and thus, by problem set 1, there exist integers a and b of opposite parity such that d 2 v = a 2 − b 2 d 2 x = a 2 + b 2 d 2 N = 2 ab Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 1 Create the triangle with sides 2 a / d , 2 b / d and 2 u . This satisfies the Pythagorean Theorem since (2 a / d ) 2 + (2 b / d ) 2 = 4 a 2 / d 2 + 4 b 2 / d 2 = 4 / d 2 ( a 2 + b 2 ) = 4 / d 2 ( d 2 x ) = 4 x = (2 u ) 2 and it has area N since A = 1 2 · 2 a d · 2 b d = 2 ab d 2 = N . Dr. Carmen Bruni Rational Points on an Elliptic Curve

Summary of Key Theorem 1 From the triple, d 2 v = a 2 − b 2 d 2 x = a 2 + b 2 , d 2 N = 2 ab we can add and subtract twice the first to the last equation to get d 2 ( x + N ) = a 2 + 2 ab + b 2 = ( a + b ) 2 d 2 ( x − N ) = a 2 − 2 ab + b 2 = ( a − b ) 2 Taking square roots yields √ √ d x + N = a + b d x − N = a − b (where above we’ve assumed that 0 < b < a ). Adding and subtracting and dividing by 2. gives expressions for a and b , namely √ √ √ √ a = d / 2( x + N + x − N ) b = d / 2( x + N − x − N ) Dr. Carmen Bruni Rational Points on an Elliptic Curve

Example of Key Theorem 1 Let’s find the triangle for N = 7. On the elliptic curve y 2 = x 3 − 7 2 x , we close our eyes and pray we can find a triple that consists of integers. After some trying we see that ( x , y ) = (25 , 120) gives a solution. Adding the point to itself gives 2 P = ( x 2 P , − y 2 P ) where (using the formulas from last time) m = 3 x 2 − 7 2 � 913 � = 913 = − 1685 b = 120 − 25 2 y 120 120 24 � 2 x 2 P = m 2 − 2 x = 113569 � 337 14400 = 120 y 2 P = mx 2 P + b = − 17631503 1728000 Hence 2 P = (113569 / 14400 , 17631503 / 1728000). Now, d is the denominator of √ x 2 P = 337 / 120 and so d = 120. Dr. Carmen Bruni Rational Points on an Elliptic Curve

Example of Key Theorem 1 Finding the a and b values gives... √ √ a = d / 2( x + N + x − N ) � � = 120 / 2( 113569 / 14400 + 7 + 113569 / 14400 , − 7) = 288 and √ √ b = d / 2( x + N − x − N ) � � = 120 / 2( 113569 / 14400 + 7 − 113569 / 14400 , − 7) = 175 This gives the triangle 2 √ x = 337 2 a d = 2 · 288 = 24 2 b d = 2 · 175 = 35 120 5 120 12 60 which indeed has area 7 and is a right angle triangle (the side lengths satisfy the Pythagorean Identity) Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 2 Theorem 3. A number N is congruent if and only if the elliptic curve y 2 = x 3 − N 2 x has a rational point P = ( x , y ) distinct from (0 , 0) and ( ± N , 0) . We have already seen that if N is congruent, then we can find a rational point on the elliptic curve. Now, suppose our elliptic curve has a rational point P = ( x , y ) where P is not one of (0 , 0) and ( ± N , 0). Our goal will be to show that 2 P satisfies the conditions of the previous theorem. Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 2 Using the results of adding a point to itself from last time, we see that the x -coordinate of P + P on the elliptic curve y 2 = x 3 + Ax + B is given by � 3 x 2 + A − 2 x = 9 x 4 + 6 Ax 2 + A 2 � 2 − 2 x 2 y 4 y 2 = 9 x 4 + 6 Ax 2 + A 2 4( x 3 + Ax + B ) − 2 x = 9 x 4 + 6 Ax 2 + A 2 4( x 3 + Ax + B ) + − 2 x · 4( x 3 + Ax + B ) 4( x 3 + Ax + B ) = 9 x 4 + 6 Ax 2 + A 2 4( x 3 + Ax + B ) + − 8 x 4 − 8 Ax 2 − 8 Bx 4( x 3 + Ax + B ) = x 4 − 2 Ax 2 − 8 Bx + A 2 4( x 3 + Ax + B ) Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 2 Using the results of adding a point to itself from last time, we see that the x -coordinate of P + P on the elliptic curve y 2 = x 3 + Ax + B is given by � 3 x 2 + A − 2 x = x 4 − 2 Ax 2 − 8 Bx + A 2 � 2 4( x 3 + Ax + B ) 2 y Specializing to when A = − N 2 and B = 0 (that is, on the elliptic curve y 2 = x 3 − N 2 x ) gives us the formula for the x -coordinate of P + P as x 4 + 2 N 2 x 2 + N 4 = ( x 2 + N 2 ) 2 . 4( x 3 − N 2 x ) (2 y ) 2 Notice that by our restriction on the rational point P , the denominator is nonzero and the numerator is nonzero. Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 2 The x -coordinate of P + P , namely x 4 + 2 N 2 x 2 + N 4 = ( x 2 + N 2 ) 2 . 4( x 3 − N 2 x ) (2 y ) 2 satisfies that it is the square of a rational number. It is also true that the numerator shares no common factor with N . Suppose p divides x 2 + N 2 and p divides N for some prime p . Then p | x and hence p 3 divides x 3 − N 2 x = y 2 . Hence p 3 divides y 2 . Thus, in the x -coordinate above, we can factor out a p 2 in the numerator and cancel it with a p 2 in the denominator. By repeating this, the numerator can be reduced so that it shares no common factor with N . So it suffices to show that the number has an even denominator. Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 2 The x -coordinate of P + P , namely x 4 + 2 N 2 x 2 + N 4 = ( x 2 + N 2 ) 2 . 4( x 3 − N 2 x ) (2 y ) 2 immediately appears to have an even denominator but we need to be careful. What happens if the factor of 4 in the denominator cancels with the numerator? In what cases is this possible? Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 2 The x -coordinate of P + P , namely x 4 + 2 N 2 x 2 + N 4 = ( x 2 + N 2 ) 2 . 4( x 3 − N 2 x ) (2 y ) 2 Case 1: x and N are even . Then 2 divides both x and N which means that the numerator and N share a common factor. Applying the previous argument shows that we can reduce the fraction. Dr. Carmen Bruni Rational Points on an Elliptic Curve

Proof of Key Theorem 2 The x -coordinate of P + P , namely x 4 + 2 N 2 x 2 + N 4 = ( x 2 + N 2 ) 2 . 4( x 3 − N 2 x ) (2 y ) 2 Case 2: x and N are odd . In this case, write x = 2 a + 1 and N = 2 b + 1. Plugging in and simplifying gives ( x 2 + N 2 ) 2 = 16( a 2 + a + b 2 + b ) 2 + 16( a 2 + a + b 2 + b ) + 4 Hence 4 exactly divides the numerator. Since y 2 = x 3 − N 2 x , we have that y is even and so at least 16 divides the denominator. Hence the denominator is even. Thus, this point P satisfies the conditions of the previous theorem and so the number N is congruent. Dr. Carmen Bruni Rational Points on an Elliptic Curve