University of Waterloo November 4th, 2015
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.
From Triangles to Curves Now, we re going to take the information about our triangle and get a new equation which will turn out to represent a curve in the real plane. Let x 2 + y 2 = z 2 and xy = 2N for rationals x, y, z and some congruent number N. Adding and subtracting 2xy = 4N to the first equation gives x 2 + 2xy + y 2 = z 2 + 4N x 2 2xy + y 2 = z 2 4N Factoring and dividing by 4 gives the two equations ( ) x + y 2 = (z/2) 2 + N 2 ( ) x y 2 = (z/2) 2 N 2
From Triangles to Curves With the equations ( ) x + y 2 = (z/2) 2 + N 2 ( ) x y 2 = (z/2) 2 N 2 we multiply these two equations together gives ( ) x + y 2 ( ) x y 2 = ((z/2) 2 + N)((z/2) 2 N) 2 2 (( ) ( )) x + y x y 2 = (z/2) 4 + N(z/2) 2 N(z/2) 2 N 2 2 2 ( ) (x y)(x + y) 2 = (z/2) 4 N 2 4 ( x 2 y 2 ) 2 = (z/2) 4 N 2 4
From Triangles to Curves Letting u = z/2 and v = (x 2 y 2 )/4, the previous equation becomes v 2 = u 4 N 2 Multiplying by u 2 gives (uv) 2 = (u 2 ) 3 N 2 u 2 Finally, we let y = uv and x = u 2 which gives us the equation y 2 = x 3 N 2 x We call such curves where y 2 equals a cubic in x an Elliptic Curve (provided the discriminant is nonzero; this is the case for cubics associated to the Congruent Number Problem).
Examples of an Elliptic Curve Let s look at examples of elliptic curves. What do they look like on the real plane? Let s try to draw y 2 = x 3 x first by drawing y = x 3 x and then trying to draw the elliptic curve.
Drawing y = x 3 x First, note that y = x 3 x = x(x 1)(x + 1) and so the equation has three zeroes at x = 0, ±1. Now let s break this curve into four intervals and see what happens in each interval y = x 3 x = x(x 1)(x + 1). Between and 1, the function is negative. Between 1 and 0, the function is positive. Between 0 and 1, the function is negative. Between 1 and, the function is positive. Lastly, the curve should look smooth with no breaks.
The Cubic Curve y = x 3 x Here is the picture (Using Desmos.com)
The Elliptic Curve y 2 = x 3 x What changes when we make the left hand side y 2 instead of y? For almost all values of x, we will get not 1 but 2 output values (the exceptions are the roots). This means that we no longer have a function, rather a curve. The cubic must be positive to have a real root! So all the areas where the picture is negative are gone. The curve still has no breaks and is symmetric about the x-axis, that is, if I reflect the top half of the picture, it should match the bottom half. The function should still be smooth (even at 1).
The Cubic Curve y = x 3 x Here is the picture (All graphs courtesy of Desmos.com) Notice that the curve has two connected components!
Connected Components Note: In general, not all elliptic curves have two components. Some have one like y 2 = x 3 1: However, the elliptic curves associated to the Congruent Number Problem always have two connect components.
Points on an elliptic curve Elliptic curves have infinitely many real points. As an example, y 2 = x 3 x has infinitely many real points by noticing that the cubic on the right is always positive when x > 1 and hence we can find a y value by taking the square root. So if we take x = 2, then we see that y 2 = 2 3 2 = 6 and so the point P = (2, 6) and Q = (2, 6) are on the curve.
Points on an elliptic curve From the perspective of Diophantine equations, it is interesting to ask: How many integer points are on elliptic curves? For the example y 2 = x 3 x, it turns out that (±1, 0) and (0, 0) are the only integer points, though this is hardly obvious. How many rational points are on elliptic curves? Above, the only rational points are also the integral ones. More on this later.
Group Law of an Elliptic Curve With an elliptic curve, we can actually describe a way to, given two rational points P and Q, create a third rational point R. Let s begin with the elliptic curve y 2 = x 3 x + 1 for illustrative purposes. y 2 = x 3 x + 1
Group Law of an Elliptic Curve y 2 = x 3 x + 1 Let s take the points P = ( 1.324, 0) and Q = (0, 1) (correct to three decimal places).
Group Law of an Elliptic Curve y 2 = x 3 x + 1 Draw the line between P and Q. It intersects the curve in a third point as shown in the picture at coordinates (1.895, 2.43).
Group Law of an Elliptic Curve Draw the vertical line through the point which must intersect the curve in a third point, in our case, R = (1.895, 2.43) (this is the same as reflecting about the x-axis). Define P + Q = R for points on an elliptic curve (note that this isn t just adding the coordinates!) y 2 = x 3 x + 1
Group Law of an Elliptic Curve If P = Q, then we can still add points. Here, we use the tangent line to find a third point of intersection. To the right, we start with the point P = ( 1, 1) on the same elliptic curve. y 2 = x 3 x + 1
Group Law of an Elliptic Curve Using calculus, we can calculate the tangent line at P to be y = x + 2. This intersects the elliptic curve at the point (3, 5). y 2 = x 3 x + 1
Group Law of an Elliptic Curve y 2 = x 3 x + 1 Reflecting as before gives us that 2P = P + P = (3, 5).
Group Law of an Elliptic Curve What about if the line between P and Q is vertical? We define a point at infinity and call it R = O. This point intersects all vertical lines. In this case, we also call Q = P (this is the reflection of P about the x-axis). Thus P P = P + Q = R = O y 2 = x 3 x + 1
Your Turn! y 2 = x 3 x + 1 Try an example. Add the points P = (0, 1) and Q = (3, 5).
Your Turn! The slope of the line between P and Q is y 2 = x 3 x + 1 m = 5 1 3 0 = 4 3 and the y intercept is b = 1 since P = (0, 1) is on the line y = 4 3 x + 1. Thus the equation of the line between P and Q is y = 4 3 x + 1.
Your Turn! Where does the line y = 4 3 x + 1 intersect y 2 = x 3 x + 1? Plug the equation of the line into the elliptic curve to get: ( 4 3 x + 1)2 = x 3 x + 1 16 9 x 2 + 8 3 x + 1 = x 3 x + 1 x 3 16 9 x 2 11 3 x = 0 x(x 2 16 9 x 11 3 ) = 0 The last quadratic must have x = 3 as a root since we know the line intersects at the points P = (0, 1) and Q = (3, 5). So factoring the above gives x(x 3)(x + 11 9 ) = 0 Thus the other point of intersection occurs when x = 11 9. The corresponding y value is y = 4 3 ( 11 9 ) + 1 = 44 27 + 1 = 17 27
Your Turn! y 2 = x 3 x + 1 This line intersects the elliptic curve at the point ( 11 9, 17 27 ). Then finally, reflecting (negating the y-coordinate) gives the point R = ( 11 9, 17 27 )
Formulas For Adding Points Let s summarize the above for adding two points P = (x 1, y 1 ) and Q = (x 2, y 2 ) on the elliptic curve y 2 = x 3 + Cx + D. Let l be the line connecting P and Q and suppose l is defined by y = mx + b We can describe the slope m and the y-intercept b via m = { y2 y 1 x 2 x 1 3x1 2+C 2y 1 If P Q If P = Q and b = y 1 mx 1 where again we used calculus to compute the tangent line in the case when P = Q.
Formulas For Adding Points As in our example, we can find the intersection of y 2 = x 3 + Cx + D and y = mx + b by solving (mx + b) 2 = x 3 + Cx + D m 2 x 2 + 2mxb + b 2 = x 3 + Cx + D 0 = x 3 m 2 x 2 + (C 2mb)x + D b 2
Formulas For Adding Points This new polynomial has x 1 and x 2 as solutions since P and Q are on both the line and the curve. Hence, 0 = x 3 m 2 x 2 + (C 2mb)x + D b 2 = (x x 1 )(x x 2 )(x x 3 ) = x 3 (x 1 + x 2 + x 3 )x 2 + (x 1 x 2 + x 1 x 3 + x 2 x 3 )x x 1 x 2 x 3 which must hold for all values of x. Hence the coefficients on either side match up. Thus, comparing the x 2 coefficients on either side gives m 2 = (x 1 + x 2 + x 3 ) x 3 = m 2 x 1 x 2 and y 3 = mx 3 + b. Hence reflecting gives P + Q = (x 3, y 3 )
Formulas For Adding Points on y 2 = x 3 N 2 x When we add P = (x, y) = Q on the elliptic curve y 2 = x 3 N 2 x with N squarefree, the formula for the x-coordinate of P + P becomes: (x 2 N 2 ) 2 (2y) 2 (see the problem set). Notice here that the x-coordinate is a square, has an even denominator and the numerator shares no common factor with N provided P (0, 0) or (±N, 0) (see the problem set).
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.
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.
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!
Next Time We prove these theorems. We figure out how to go from a rational point on an elliptic curve to a rational right triangle with area N. We revisit Don Zagier s example. We discuss some tricks for finding rational points on elliptic curves.