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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