26 2. Elliptic curves L E(Q). Thus, P + Q is the reflection of R across the x-axis, i.e., P + Q = (− 5 9 , 100 27 ). Using Proposition 1.1.3, we may try to use the point P + Q = (− 5 9 , 100 27 ) to find a (new) right triangle with rational sides and area equal to 5, but this point corresponds to the triangle ( 20 3 , 3 2 , 41 6 ), the same triangle that corresponds to Q = (−4,6). In order to find a new triangle, let us find Q + Q = 2Q. The line L in this case is the tangent line to E at Q. The slope of L can be found using implicit differentiation on y2 = x3 25x: 2y dy dx = 3x2 25, so dy dx = 3x2 25 2y . Hence, the slope of L is m = 23 12 and L : y = 23 12 (x + 4) + 6. In order to find R we need to solve y = 23 12 (x + 4) + 6 y2 = x3 25x. Simplifying yields x3 529 144 x2 1393 18 x 1681 9 = 0, which factors as (x + 4)2(144x 1681) = 0. Once again, two factors were expected: x = −4 needs to be a double root because L is tangent to E at Q = (−4,6). The third factor tells us that the x coordinate of R is x = 1681 144 , and y = 23 12 (x+4)+6 = 62279 1728 . Thus, Q + Q = 2Q = ( 1681 144 , 62279 1728 ). This point corresponds to the right triangle (a, b, c) = 1519 492 , 4920 1519 , 3344161 747348 . Example 2.4.2. Let E : y2 = x3 +1 and put P = (2,3). Let us find P , 2P , 3P , etc. In order to find 2P , first we need to find the tangent line to E at P , which is y 3 = 2(x 2) or y = 2x 1. The third point of intersection is R = (0, −1), so 2P = (0,1). To find 3P , we add P and 2P . The third point of inter- section of E with the line that goes through P and 2P is R = (−1,0) hence, 3P = (−1,0).
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