In mathematics, the Montgomery curve is a form of elliptic curve introduced by Peter L. Montgomery in 1987, different from the usual Weierstrass form. It is used for certain computations, and in particular in different cryptography applications.

Definition

A Montgomery curve over a field K is defined by the equation for certain A, B ∈ K and with B(A2 − 4) ≠ 0 . Generally this curve is considered over a finite field K (for example, over a finite field of q elements, K = Fq ) with characteristic different from 2 and with A ≠ ±2 and B ≠ 0 , but they are also considered over the rationals with the same restrictions for A and B .

Montgomery arithmetic

It is possible to do some "operations" between the points of an elliptic curve: "adding" two points P, Q consists of finding a third one R such that R=P+Q; "doubling" a point consists of computing [2]P=P+P (For more information about operations see The group law) and below. A point P=(x,y) on the elliptic curve in the Montgomery form can be represented in Montgomery coordinates P=(X:Z), where P=(X:Z) are projective coordinates and x=X/Z for Z\ne 0. Notice that this kind of representation for a point loses information: indeed, in this case, there is no distinction between the affine points (x,y) and (x,-y) because they are both given by the point (X:Z). However, with this representation it is possible to obtain multiples of points, that is, given P=(X:Z), to compute. Now, considering the two points and : their sum is given by the point whose coordinates are: If m=n, then the operation becomes a "doubling"; the coordinates of are given by the following equations: The first operation considered above (addition) has a time-cost of 3M+2S, where M denotes the multiplication between two general elements of the field on which the elliptic curve is defined, while S denotes squaring of a general element of the field. The second operation (doubling) has a time-cost of 2M + 2S + 1D, where D denotes the multiplication of a general element by a constant; notice that the constant is (A+2)/4, so A can be chosen in order to have a small D.

Algorithm and example

The following algorithm represents a doubling of a point on an elliptic curve in the Montgomery form. It is assumed that Z_1=1. The cost of this implementation is 1M + 2S + 1A + 3add + 14. Here M denotes the multiplications required, S indicates the squarings, and a refers to the multiplication by A.

Example

Let be a point on the curve. In coordinates (X_1:Z_1), with x_1=X_1/Z_1, P_1=(2:1). Then: The result is the point such that P_2=2P_1.

Addition

Given two points, on the Montgomery curve M_{A,B} in affine coordinates, the point P_3=P_1+P_2 represents, geometrically the third point of intersection between M_{A,B} and the line passing through P_1 and P_2. It is possible to find the coordinates (x_3,y_3) of P_3, in the following way:

1. consider a generic line ~y=lx+m in the affine plane and let it pass through P_1 and P_2 (impose the condition), in this way, one obtains and ;
2. intersect the line with the curve M_{A,B}, substituting the ~y variable in the curve equation with ~y=lx+m; the following equation of third degree is obtained: As it has been observed before, this equation has three solutions that correspond to the ~x coordinates of P_1, P_2 and P_3. In particular this equation can be re-written as:
3. Comparing the coefficients of the two identical equations given above, in particular the coefficients of the terms of second degree, one gets: So, x_3 can be written in terms of x_1, y_1, x_2, y_2, as:
4. To find the ~y coordinate of the point P_3 it is sufficient to substitute the value x_3 in the line ~y=lx+m. Notice that this will not give the point P_3 directly. Indeed, with this method one find the coordinates of the point ~R such that, but if one needs the resulting point of the sum between P_1 and P_2, then it is necessary to observe that: if and only if -R=P_1+P_2. So, given the point ~R, it is necessary to find ~-R, but this can be done easily by changing the sign to the ~y coordinate of ~R. In other words, it will be necessary to change the sign of the ~y coordinate obtained by substituting the value x_3 in the equation of the line. Resuming, the coordinates of the point, P_3=P_1+P_2 are:

Doubling

Given a point P_1 on the Montgomery curve M_{A,B}, the point [2]P_1 represents geometrically the third point of intersection between the curve and the line tangent to P_1; so, to find the coordinates of the point P_3=2P_1 it is sufficient to follow the same method given in the addition formula; however, in this case, the line y = lx + m has to be tangent to the curve at P_1, so, if with then the value of l, which represents the slope of the line, is given by: by the implicit function theorem. So and the coordinates of the point P_3, P_3=2P_1 are:

Equivalence with twisted Edwards curves

Let K be a field with characteristic different from 2. Let M_{A,B} be an elliptic curve in the Montgomery form: with , and let E_{a,d} be an elliptic curve in the twisted Edwards form: with The following theorem shows the birational equivalence between Montgomery curves and twisted Edwards curve: Theorem (i) Every twisted Edwards curve is birationally equivalent to a Montgomery curve over K. In particular, the twisted Edwards curve E_{a,d} is birationally equivalent to the Montgomery curve M_{A,B} where, and. The map: is a birational equivalence from E_{a,d} to M_{A,B}, with inverse: Notice that this equivalence between the two curves is not valid everywhere: indeed the map \psi is not defined at the points v = 0 or u + 1 = 0 of the M_{A,B}.

Equivalence with Weierstrass curves

Any elliptic curve can be written in Weierstrass form. In particular, the elliptic curve in the Montgomery form can be transformed in the following way: divide each term of the equation for M_{A,B} by B^3, and substitute the variables x and y, with and respectively, to get the equation To obtain a short Weierstrass form from here, it is sufficient to replace u with the variable : finally, this gives the equation: Hence the mapping is given as In contrast, an elliptic curve over base field \mathbb{F} in Weierstrass form can be converted to Montgomery form if and only if E_{a,b} has order divisible by four and satisfies the following conditions: When these conditions are satisfied, then for we have the mapping