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How can I transform an elliptic curve over the real numbers in Weierstrass form

$y^2=x^3+ax+b$

into a cubic of the form

$y^2=x(x-c)(x-d)$?

anna90
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    Put another way, a pure translation in $x$ to move a real root (there is one) from where it is to $0.$ – Will Jagy Oct 30 '13 at 20:12
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    See also here: http://math.stackexchange.com/questions/477834/making-an-elliptic-curve-out-of-a-cubic-polynomial-made-a-cube-or-ax3bx2cx. – Dietrich Burde Oct 30 '13 at 20:14

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