Twisted Edwards curve

A Twisted Edwards curve of equation $10x^2+y^2=1+6x^2y^2$

In algebraic geometry, the Twisted Edwards curves are plane models of elliptic curves, a generalisation of Edwards curves introduced by Bernstein et al. (2007) and named after Harold M. Edwards. Each twisted Edwards curve (as the name suggests) is a twist of an Edwards curve. A twisted Edwards curve E E,a,d over a field K which does not have 2=0 is an affine plane curve defined by the equation:

E E,a,d: $a x^2+y^2= 1+dx^2y^2$

where a, d are distinct non-zero elements of K. An Edwards curve is a twisted Edwards curve with a = 1.

Every twisted Edward curve is birationally equivalent to an elliptic curve in Montgomery form and vice versa.

Group law

As for all elliptic curves, also for the Twisted Edwards curve, it is possible to do some operations between its points, such as adding two of them or doubling (or tripling) one. The results of these operations are always points that belong to the curve itself. In the following sections some formulas are given to obtain the coordinates of a point resulted from an addition between two other points (addition), or the coordinates of point resulted from a doubling of a single point on a curve.

Let K be a field with characteristic different from 2. Let $(x_1,y_1)$ and $(x_2,y_2)$ be points on the Twisted Edward curve. The equation of Twisted Edward curve is written as;

EE,a,d: $ax^2+y^2=1+dx^2y^2$.

The sum of these points $(x_1,y_1) , (x_2,y_2)$ on EE,a,d is:

$(x_1,y_1) + (x_2,y_2) = \left(\frac{x_1y_2+y_1x_2}{1+dx_1x_2y_1y_2} , \frac{y_1y_2-ax_1x_2}{1-dx_1x_2y_1y_2}\right)$

The neutral element is (0,1) and the negative of $(x_1,y_1)$ is ($-x_1,y_1)$

These formulas also work for doubling. If a is a square in K and d is a non-square in K, these formulas are complete: this means that they can be used for all pairs of points without exceptions; so they work for doubling as well, and neutral elements and negatives are accepted as inputs.[1]

Given the following Twisted Edwards curve with a=3 and d=2:

$3x^2 + y^2 = 1 + 2x^2y^2$

it is possible to add the points $P_1=(1,\sqrt{2})$ and $P_2=(1,-\sqrt{2})$ using the formula given above. The result is a point P3 that has coordinates:

$x_3 = \frac{x_1y_2+y_1x_2}{1+dx_1x_2y_1y_2} = 0$

$y_3 = \frac{y_1y_2-ax_1x_2}{1-dx_1x_2y_1y_2} = -1$.

Doubling on Twisted Edward curves

Doubling can be performed with exactly the same formula as addition. Doubling of a point (x1,y1) on the curve EE,a,d is:

[2](x1,y1) = (x3,y3)

where

$x_3= \frac{x_1y_1+y_1x_1}{1+dx_1x_1y_1y_1}=\frac{2x_1y_1}{ax_1^2+y_1^2}$

$y_3= \frac{y_1y_1-ax_1x_1}{1-dx_1x_1y_1y_1}=\frac{y_1^2-ax_1^2}{2-ax_1^2-y_1^2}.$

Example of doubling

Considering the same twisted Edwards curve given in the previous example, with a=3 and d=2, it is possible to double the point $P_1=(1,\sqrt{2})$. The point 2P1 obtained using the formula above has the following coordinates:

$x_3 = \frac{x_1y_1+y_1x_1}{1+dx_1x_1y_1y_1} = \frac{2\sqrt{2}}{5}$

$y_3 = \frac{y_1y_1-ax_1x_1}{1-dx_1x_1y_1y_1} = \frac{1}{3}.$

It is easy to see, with some little computations, that the point $P_3=(\frac{2\sqrt{2}}{5},\frac{1}{3})$ belongs to the curve $3x^2 + y^2 = 1 + 2x^2y^2$.

Extended coordinates

There is another kind of coordinate system with which a point in the Twisted Edwards curves can be represented. A point $(x,y,z)$ on $ax^2+y^2= 1+dx^2y^2$ is represented as X, Y, Z, T satisfying the following equations x=X/Z, y=Y/Z, xy=T/Z.

The coordinates of the point (X:Y:Z:T) are called the extended Twisted Edward coordinates. The identity element is represented by (0:1:1:0). The negative of a point is (-X:Y:Z:-T).

Inverted Twisted Edwards Coordinates

The coordinates of the point $(X_1:Y_1:Z_1)$ are called the inverted twisted Edwards coordinates on the curve $(X^2+aY^2)Z^2= X^2Y^2+dZ^4$with $X_1Y_1Z_1$≠0; this point to the affine one$(Z_1/X_1, Z_1/Y_1)$ on EE,a,d. Bernstein and Lange introduced these inverted coordinates, for the case a=1 and observed that the coordinates save time in addition.

Projective twisted Edward coordinates

The equation for the Projective Twisted Edwards Curve is given as: $(aX^2+Y^2)Z^2= Z^4+dX^2Y^2$ For Z1≠0 the point (X1:Y1:Z1) represents the affine point (x1= X1/Z1, y1= Y1/Z1) on EE,a,d.

Expressing an elliptic curve in twisted Edwards form saves time in arithmetic, even when the same curve can be expressed in the Edwards form. To know more about the speeds of addition and doubling in projective coordinates on Edwards curves, standard coordinates on twisted Edward curves, inverted coordinates on Edwards curves and inverted coordinates on twisted Edwards curves refer to the table in:

http://hyperelliptic.org/EFD/g1p/auto-twisted-extended-1.html

The addition on a projective twisted Edwards curve is given by:

(X3:Y3:Z3)= (X1:Y1:Z1)+(X2:Y2:Z2)

and it costs 10Multiplications + 1Squaring + 2D(multiplication by the curve parameter d) + 7 additions where the 2D are one multiplication by a and one by d.

Algorithm
A= Z1.Z2,
B=A2
C=X1.X2
D=Y1.Y2
E=dC.D
F=B-E
G=B+E
X3= A.F((X1+Y1).(X2+Y2)-C-D)
Y3=A.G.(D-aC)
Z3=F.G[2]

Doubling on projective twisted curves

Doubling on the projective twisted curve is given by:

(X3:Y3:Z3)= 2(X1:Y1:Z1)

This costs 3Multiplications+4Squarings+1D+7additions where 1D is a multiplication by a

Algorithm
B=(X1+Y1)2
C= X12
D=Y12
E=aC
F= E+D
H=Z12
J=F-2H
X3=(B-C-D).J
Y3=F.(E-D)
Z3= F.J[2]