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).