\documentclass{article} \pagestyle{empty} \usepackage{amsmath} \begin{document} \newcommand{\C}{{\bf C}} \title{How to convert an elliptic curve from Montgomery to Weierstra\ss{} coordinates} \maketitle Curve in Montgomery coordinates: \begin{displaymath} E: B Y^2 = X^3 + A X^2 + X \end{displaymath} With a given point $(X, ?)$ we can set $Y=1$ and choose the isomorphic curve % \begin{displaymath} G Y^2 = X^3 + A X^2 + X \end{displaymath} % with $G = X^3 + A X^2 + X$. Now (X, 1) is a valid point on this curve. Dividing the equation by $G^3$ yields % \begin{displaymath} \left(\frac{Y}{G}\right)^2 = \left(\frac{X}{G}\right)^3 + \frac{A}{G} \left(\frac{X}{G}\right)^2 + \frac{1}{G^2} \left(\frac{X}{G}\right) \end{displaymath} % so we can set $x = X/G$, $y = Y/G$ and receive the curve \begin{displaymath} y^2 = x^3 + \frac{A}{G} x^2 + \frac{1}{G^2} x \end{displaymath} This is not in short Weierstra\ss{} form yet. By choosing instead $x = (X+A/3)/G$, $y = Y/G$, $a = (1-A^2/3)/G^2$, $b = A (2 A^2 - 9)/(27 G^3)$, we get \begin{displaymath} y^2 = x^3 + a x + b \end{displaymath} This is the desired curve isomorphic to $E$, with $(x,y)$ corresponding to $(X,?)$ on $E$. \end{document}