If an Elliptic Curve is of the form E:y^2=x^3+ax+b then how do we convert this Legendre Form y^2=x(x-1)(x-L)?

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    $\begingroup$ Factor $x^3 + ax + b = (x-x_1) (x-x_2) (x-x_3)$. Translate $x$ by $x_1$ to make the first factor $x$. Then multiply $x$ by $x_2-x_1$ and $y$ by $(x_2-x_1)^{3/2}$ and divide by $(x_2-x_1)^3$ to get $y^2 = x (x-1) (x-L)$ with $L = (x_3-x_1)/(x_2-x_1)$. If $x^3+ax+b$ does not factor completely then $L$ must be in a field larger than the one used to define the curve. $\endgroup$ – Noam D. Elkies Jun 19 '12 at 4:28
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    $\begingroup$ @Noam: This is an answer, not a comment. $\endgroup$ – Martin Brandenburg Jun 19 '12 at 9:39


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