# Montgomery curve

In mathematics the Montgomery curve is a form of elliptic curve introduced by Peter L. Montgomery in 1987,[1] different from the usual Weierstrass form. It is used for certain computations, and in particular in different cryptography applications.

## Definition

A Montgomery curve of equation ${\textstyle {\frac {1}{4}}y^{2}=x^{3}+{\frac {5}{2}}x^{2}+x}$

A Montgomery curve over a field K is defined by the equation

${\displaystyle M_{A,B}:By^{2}=x^{3}+Ax^{2}+x}$

for certain A, BK 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 ${\displaystyle P,Q}$  consists of finding a third one ${\displaystyle R}$  such that ${\displaystyle R=P+Q}$ ; "doubling" a point consists of computing ${\displaystyle [2]P=P+P}$  (For more information about operations see The group law) and below.

A point ${\displaystyle P=(x,y)}$  on the elliptic curve in the Montgomery form ${\displaystyle By^{2}=x^{3}+Ax^{2}+x}$  can be represented in Montgomery coordinates ${\displaystyle P=(X:Z)}$ , where ${\displaystyle P=(X:Z)}$  are projective coordinates and ${\displaystyle x=X/Z}$  for ${\displaystyle Z\neq 0}$ .

Notice that this kind of representation for a point loses information: indeed, in this case, there is no distinction between the affine points ${\displaystyle (x,y)}$  and ${\displaystyle (x,-y)}$  because they are both given by the point ${\displaystyle (X:Z)}$ . However, with this representation it is possible to obtain multiples of points, that is, given ${\displaystyle P=(X:Z)}$ , to compute ${\displaystyle [n]P=(X_{n}:Z_{n})}$ .

Now, considering the two points ${\displaystyle P_{n}=[n]P=(X_{n}:Z_{n})}$  and ${\displaystyle P_{m}=[m]P=(X_{m}:Z_{m})}$ : their sum is given by the point ${\displaystyle P_{m+n}=P_{m}+P_{n}=(X_{m+n}:Z_{m+n})}$  whose coordinates are:

${\displaystyle X_{m+n}=Z_{m-n}((X_{m}-Z_{m})(X_{n}+Z_{n})+(X_{m}+Z_{m})(X_{n}-Z_{n}))^{2}}$
${\displaystyle Z_{m+n}=X_{m-n}((X_{m}-Z_{m})(X_{n}+Z_{n})-(X_{m}+Z_{m})(X_{n}-Z_{n}))^{2}}$

If ${\displaystyle m=n}$ , then the operation becomes a "doubling"; the coordinates of ${\displaystyle [2]P_{n}=P_{n}+P_{n}=P_{2n}=(X_{2n}:Z_{2n})}$  are given by the following equations:

${\displaystyle 4X_{n}Z_{n}=(X_{n}+Z_{n})^{2}-(X_{n}-Z_{n})^{2}}$
${\displaystyle X_{2n}=(X_{n}+Z_{n})^{2}(X_{n}-Z_{n})^{2}}$
${\displaystyle Z_{2n}=(4X_{n}Z_{n})((X_{n}-Z_{n})^{2}+((A+2)/4)(4X_{n}Z_{n}))}$

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 ${\displaystyle (A+2)/4}$ , so ${\displaystyle A}$  can be chosen in order to have a small D.

## Algorithm and example

The following algorithm represents a doubling of a point ${\displaystyle P_{1}=(X_{1}:Z_{1})}$  on an elliptic curve in the Montgomery form.

It is assumed that ${\displaystyle Z_{1}=1}$ . The cost of this implementation is 1M + 2S + 1*A + 3add + 1*4. Here M denotes the multiplications required, S indicates the squarings, and a refers to the multiplication by A.

${\displaystyle XX_{1}=X_{1}^{2}\,}$
${\displaystyle X_{2}=(XX_{1}-1)^{2}\,}$
${\displaystyle Z_{2}=4X_{1}(XX_{1}+aX_{1}+1)\,}$

### Example

Let ${\displaystyle P_{1}=(2,{\sqrt {3}})}$  be a point on the curve ${\displaystyle 2y^{2}=x^{3}-x^{2}+x}$ . In coordinates ${\displaystyle (X_{1}:Z_{1})}$ , with ${\displaystyle x_{1}=X_{1}/Z_{1}}$ , ${\displaystyle P_{1}=(2:1)}$ .

Then:

${\displaystyle XX_{1}=X_{1}^{2}=4\,}$
${\displaystyle X_{2}=(XX_{1}-1)^{2}=9\,}$
${\displaystyle Z_{2}=4X_{1}(XX_{1}+AX_{1}+1)=24\,}$

The result is the point ${\displaystyle P_{2}=(X_{2}:Z_{2})=(9:24)}$  such that ${\displaystyle P_{2}=2P_{1}}$ .

Given two points ${\displaystyle P_{1}=(x_{1},y_{1})}$ , ${\displaystyle P_{2}=(x_{2},y_{2})}$  on the Montgomery curve ${\displaystyle M_{A,B}}$  in affine coordinates, the point ${\displaystyle P_{3}=P_{1}+P_{2}}$  represents, geometrically the third point of intersection between ${\displaystyle M_{A,B}}$  and the line passing through ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$ . It is possible to find the coordinates ${\displaystyle (x_{3},y_{3})}$  of ${\displaystyle P_{3}}$ , in the following way:

1) consider a generic line ${\displaystyle ~y=lx+m}$  in the affine plane and let it pass through ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$  (impose the condition), in this way, one obtains ${\displaystyle l={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}$  and ${\displaystyle m=y_{1}-\left({\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}\right)x_{1}}$ ;

2) intersect the line with the curve ${\displaystyle M_{A,B}}$ , substituting the ${\displaystyle ~y}$  variable in the curve equation with ${\displaystyle ~y=lx+m}$ ; the following equation of third degree is obtained:

${\displaystyle x^{3}+(A-Bl^{2})x^{2}+(1-2Blm)x-Bm^{2}=0.}$

As it has been observed before, this equation has three solutions that correspond to the ${\displaystyle ~x}$  coordinates of ${\displaystyle P_{1}}$ , ${\displaystyle P_{2}}$  and ${\displaystyle P_{3}}$ . In particular this equation can be re-written as:

${\displaystyle (x-x_{1})(x-x_{2})(x-x_{3})=0}$

3) Comparing the coefficients of the two identical equations given above, in particular the coefficients of the terms of second degree, one gets:

${\displaystyle -x_{1}-x_{2}-x_{3}=A-Bl^{2}}$ .

So, ${\displaystyle x_{3}}$  can be written in terms of ${\displaystyle x_{1}}$ , ${\displaystyle y_{1}}$ , ${\displaystyle x_{2}}$ , ${\displaystyle y_{2}}$ , as:

${\displaystyle x_{3}=B\left({\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}\right)^{2}-A-x_{1}-x_{2}.}$

4) To find the ${\displaystyle ~y}$  coordinate of the point ${\displaystyle P_{3}}$  it is sufficient to substitute the value ${\displaystyle x_{3}}$  in the line ${\displaystyle ~y=lx+m}$ . Notice that this will not give the point ${\displaystyle P_{3}}$  directly. Indeed, with this method one find the coordinates of the point ${\displaystyle ~R}$  such that ${\displaystyle R+P_{1}+P_{2}=P_{\infty }}$ , but if one needs the resulting point of the sum between ${\displaystyle P_{1}}$  and ${\displaystyle P_{2}}$ , then it is necessary to observe that: ${\displaystyle R+P_{1}+P_{2}=P_{\infty }}$  if and only if ${\displaystyle -R=P_{1}+P_{2}}$ . So, given the point ${\displaystyle ~R}$ , it is necessary to find ${\displaystyle ~-R}$ , but this can be done easily by changing the sign to the ${\displaystyle ~y}$  coordinate of ${\displaystyle ~R}$ . In other words, it will be necessary to change the sign of the ${\displaystyle ~y}$  coordinate obtained by substituting the value ${\displaystyle x_{3}}$  in the equation of the line.

Resuming, the coordinates of the point ${\displaystyle P_{3}=(x_{3},y_{3})}$ , ${\displaystyle P_{3}=P_{1}+P_{2}}$  are:

${\displaystyle x_{3}={\frac {B(y_{2}-y_{1})^{2}}{(x_{2}-x_{1})^{2}}}-A-x_{1}-x_{2}}$
${\displaystyle y_{3}={\frac {(2x_{1}+x_{2}+A)(y_{2}-y_{1})}{x_{2}-x_{1}}}-{\frac {B(y_{2}-y_{1})^{3}}{(x_{2}-x_{1})^{3}}}-y_{1}}$

## Doubling

Given a point ${\displaystyle P_{1}}$  on the Montgomery curve ${\displaystyle M_{A,B}}$ , the point ${\displaystyle [2]P_{1}}$  represents geometrically the third point of intersection between the curve and the line tangent to ${\displaystyle P_{1}}$ ; so, to find the coordinates of the point ${\displaystyle P_{3}=2P_{1}}$  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 ${\displaystyle P_{1}}$ , so, if ${\displaystyle M_{A,B}:f(x,y)=0}$  with

${\displaystyle f(x,y)=x^{3}+Ax^{2}+x-By^{2},}$

then the value of l, which represents the slope of the line, is given by:

${\displaystyle l=-\left.{\frac {\partial f}{\partial x}}\right/{\frac {\partial f}{\partial y}}}$

by the implicit function theorem.

So ${\displaystyle l={\frac {3x_{1}^{2}+2Ax_{1}+1}{2By_{1}}}}$  and the coordinates of the point ${\displaystyle P_{3}}$ , ${\displaystyle P_{3}=2P_{1}}$  are:

{\displaystyle {\begin{aligned}x_{3}&=Bl^{2}-A-x_{1}-x_{1}={\frac {B(3x_{1}^{2}+2Ax_{1}+1)^{2}}{(2By_{1})^{2}}}-A-x_{1}-x_{1}\\y_{3}&=(2x_{1}+x_{1}+A)l-Bl^{3}-y_{1}\\&={\frac {(2x_{1}+x_{1}+A)(3{x_{1}}^{2}+2Ax_{1}+1)}{2By_{1}}}-{\frac {B(3{x_{1}}^{2}+2Ax_{1}+1)^{3}}{(2By_{1})^{3}}}-y_{1}.\end{aligned}}}

## Equivalence with twisted Edwards curves

Let ${\displaystyle K}$  be a field with characteristic different from 2.

Let ${\displaystyle M_{A,B}}$  be an elliptic curve in the Montgomery form:

${\displaystyle M_{A,B}:Bv^{2}=u^{3}+Au^{2}+u}$

with ${\displaystyle A\in K\smallsetminus \{-2,2\}}$ , ${\displaystyle B\in K\smallsetminus \{0\}}$

and let ${\displaystyle E_{a,d}}$  be an elliptic curve in the twisted Edwards form:

${\displaystyle E_{a,d}\ :\ ax^{2}+y^{2}=1+dx^{2}y^{2},\,}$

with ${\displaystyle a,d\in K\smallsetminus \{0\},\quad a\neq d.}$

The following theorem shows the birational equivalence between Montgomery curves and twisted Edwards curve:[2]

Theorem (i) Every twisted Edwards curve is birationally equivalent to a Montgomery curve over ${\displaystyle K}$ . In particular, the twisted Edwards curve ${\displaystyle E_{a,d}}$  is birationally equivalent to the Montgomery curve ${\displaystyle M_{A,B}}$  where ${\displaystyle A={\frac {2(a+d)}{a-d}}}$ , and ${\displaystyle B={\frac {4}{a-d}}}$ .

The map:

${\displaystyle \psi \,:\,E_{a,d}\rightarrow M_{A,B}}$
${\displaystyle (x,y)\mapsto (u,v)=\left({\frac {1+y}{1-y}},{\frac {1+y}{(1-y)x}}\right)}$

is a birational equivalence from ${\displaystyle E_{a,d}}$  to ${\displaystyle M_{A,B}}$ , with inverse:

${\displaystyle \psi ^{-1}}$ : ${\displaystyle M_{A,B}\rightarrow E_{a,d}}$
${\displaystyle (u,v)\mapsto (x,y)=\left({\frac {u}{v}},{\frac {u-1}{u+1}}\right),a={\frac {A+2}{B}},d={\frac {A-2}{B}}}$

Notice that this equivalence between the two curves is not valid everywhere: indeed the map ${\displaystyle \psi }$  is not defined at the points ${\displaystyle v=0}$  or ${\displaystyle u+1=0}$  of the ${\displaystyle M_{A,B}}$ .

## Equivalence with Weierstrass curves

Any elliptic curve can be written in Weierstrass form. In particular, the elliptic curve in the Montgomery form

${\displaystyle M_{A,B}}$ : ${\displaystyle By^{2}=x^{3}+Ax^{2}+x,}$

can be transformed in the following way: divide each term of the equation for ${\displaystyle M_{A,B}}$  by ${\displaystyle B^{3}}$ , and substitute the variables x and y, with ${\displaystyle u={\frac {x}{B}}}$  and ${\displaystyle v={\frac {y}{B}}}$  respectively, to get the equation

${\displaystyle v^{2}=u^{3}+{\frac {A}{B}}u^{2}+{\frac {1}{B^{2}}}u.}$

To obtain a short Weierstrass form from here, it is sufficient to replace u with the variable ${\displaystyle t-{\frac {A}{3B}}}$ :

${\displaystyle v^{2}=\left(t-{\frac {A}{3B}}\right)^{3}+{\frac {A}{B}}\left(t-{\frac {A}{3B}}\right)^{2}+{\frac {1}{B^{2}}}\left(t-{\frac {A}{3B}}\right);}$

finally, this gives the equation:

${\displaystyle v^{2}=t^{3}+\left({\frac {3-A^{2}}{3B^{2}}}\right)t+\left({\frac {2A^{3}-9A}{27B^{3}}}\right).}$

Hence the mapping is given as

${\displaystyle \psi }$ : ${\displaystyle M_{A,B}\rightarrow E_{a,b}}$
${\displaystyle (x,y)\mapsto (t,v)=\left({\frac {x}{B}}+{\frac {A}{3B}},{\frac {y}{B}}\right),a={\frac {3-A^{2}}{3B^{2}}},b={\frac {2A^{3}-9A}{27B^{3}}}}$

In contrast, an elliptic curve over base field ${\displaystyle \mathbb {F} }$  in Weierstrass form

${\displaystyle E_{a,b}}$ : ${\displaystyle v^{2}=t^{3}+at+b}$

can be converted to Montgomery form if and only if ${\displaystyle E_{a,b}}$  has order divisible by four and satisfies the following conditions:[3]

1. ${\displaystyle z^{3}+az+b=0}$  has at least one root ${\displaystyle \alpha \in \mathbb {F} }$ ; and
2. ${\displaystyle 3\alpha ^{2}+a}$  is a quadratic residue in ${\displaystyle \mathbb {F} }$ .

When these conditions are satisfied, then for ${\displaystyle s=({\sqrt {3\alpha ^{2}+a}})^{-1}}$  we have the mapping

${\displaystyle \psi ^{-1}}$ : ${\displaystyle E_{a,b}\rightarrow M_{A,B}}$
${\displaystyle (t,v)\mapsto (x,y)=\left(s(t-\alpha ),sv\right),A=3\alpha s,B=s}$ .

2. ^ Daniel J. Bernstein, Peter Birkner, Marc Joye, Tanja Lange and Christiane Peters (2008). "Twisted Edwards Curves". Progress in Cryptology – AFRICACRYPT 2008. Lecture Notes in Computer Science. Vol. 5023. Springer-Verlag Berlin Heidelberg. pp. 389–405. doi:10.1007/978-3-540-68164-9_26. ISBN 978-3-540-68159-5.{{cite book}}: CS1 maint: multiple names: authors list (link)
3. ^ Katsuyuki Okeya, Hiroyuki Kurumatani, and Kouichi Sakurai (2000). Elliptic Curves with the Montgomery-Form and Their Cryptographic Applications. Public Key Cryptography (PKC2000). doi:10.1007/978-3-540-46588-1_17.{{cite conference}}: CS1 maint: multiple names: authors list (link)