Example 1.
Let E be the elliptic curve with affine equation x3 + y3 + 1 + 4 x y = 0 and let
O=(0,-1) be the zero element of the group law. Let A=(-1,-2), B=(-1,0). Calculate A+B and 2A.
3
Q
2
1
Q=A+B
B
0
R
O
-1
-2
A
R=2A
-3
-3
-2
-1
0
1
2
3
The line through A, B is clearly x=-1, let's substitute this into the equation of E.
x3 + y3 + 1 + 4 x y . x ® - 1
- 4 y + y3
It can be factorised and we see that the intersection points have y co-ordinates -2, 0 and 2, so the third point of intersection is Q=(-1,2).
x3 + y3 + 1 + 4 x y . x ® - 1  Factor
H- 2 + yL y H2 + yL
The equation of the line through Q and O is y=-3x-1, so let's substitute this
into the equation of E.
x3 + y3 + 1 + 4 x y . y ® - 3 x - 1  Expand
- 13 x - 39 x2 - 26 x3
x3 + y3 + 1 + 4 x y . y ® - 3 x - 1  Factor
- 13 x H1 + xL H1 + 2 xL
The intersection points have x co-ordinates 0, -1 and -1/2, so the third point of
intersection is Q=(-1/2,1/2)=A+B.
2
Ellipticexample.nb
To calculate 2A, we need to calculate the equation of the tangent line to E at
A, for which we need the partial derivatives.
9DAx3 + y3 + 1 + 4 x y, xE, DAx3 + y3 + 1 + 4 x y, yE= . 8x ® - 1, y ® - 2<
8- 5, 8<
Hence the slope is 5/8, and the equation of the tangent line is y=5(x+1)/8-2,
let's substitute this into the equation of E.
x3 + y3 + 1 + 4 x y . :y ®
819
-
512
455 x2
1001 x
-
+
512
512
8
512
H1 + xL2 H- 9 + 7 xL
Hx + 1L - 2>  Expand
637 x3
+
x3 + y3 + 1 + 4 x y . :y ®
91
5
512
5
8
Hx + 1L - 2>  Factor
The intersection points have x co-ordinates -1, -1 and 9/7, so the third intersection point R is (9/7, -4/7).
The equation of the line through R and O is y=x/3-1, so let's substitute this
into the equation of E.
x3 + y3 + 1 + 4 x y . y ® x  3 - 1  Expand
- 3 x + x2 +
28 x3
27
x3 + y3 + 1 + 4 x y . y ® x  3 - 1  Factor
1
27
x H9 + 4 xL H- 9 + 7 xL
The intersection points have x co-ordinates 0, 9/7 and -9/4, so the third intersection point is R=(-9/4, -7/4)=2A.
Ellipticexample.nb
Example 2.
Let E be the elliptic curve with affine equation y 2 = x3 - x2 - 4 x + 4 and let the
point at infinity, O=(0:1:0) be the zero element of the group law. Let A=(1,0),
B=(0,2). Calculate A+B and 2B.
Q=A+B
5
B
A
0
R=R=2B
-5
Q
-2
0
2
4
The line through A, B is y=-2(x-1), let's substitute this into the equation of E.
x3 - x2 - 4 x + 4 - y2 . y ® - 2 Hx - 1L  Expand
4 x - 5 x2 + x3
x3 - x2 - 4 x + 4 - y2 . y ® - 2 Hx - 1L  Factor
H- 4 + xL H- 1 + xL x
The intersection points have x co-ordinates 0, 1 and 4, so the third point of
intersection is Q=(4,-6). Hence A+B=Q=(4,6).
3
4
Ellipticexample.nb
To calculate 2B, we need to calculate the equation of the tangent line to E at
B, for which we need the partial derivatives.
9DAx3 - x2 - 4 x + 4 - y2 , xE, DAx3 - x2 - 4 x + 4 - y2 , yE= . 8x ® 0, y ® 2<
8- 4, - 4<
Hence the slope is -1, and the equation of the tangent line is y=2-x, let's substitute this into the equation of E.
x3 - x2 - 4 x + 4 - y2 . y ® 2 - x  Expand
- 2 x2 + x3
x3 - x2 - 4 x + 4 - y2 . y ® 2 - x  Factor
H- 2 + xL x2
The intersection points have x co-ordinates 0, 0, and 2, so the third intersection point is R=(2,0). Hence 2B=R =(2,0).