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How can I transform a complete twisted Edwards curve $ax^2+y^2 = 1+dx^2y^2$ with not square $d$ and square $a$ into an isomorphic Edwards curve $X^2+Y^2 = 1+DX^2Y^2$ with a square $-D$ i.e. $D = -r^2$?

I tried to set $X = \frac{x}{\sqrt{a}}; Y=y$, but $-\frac{d}{a}$ is also a non square (at least for Edwards25519). This answer is not working as well (i.e. $-1/d$ is not a square), because $-1$ is square.

Is it even possible to do such a transformation?

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    $\begingroup$ If your number $d$ is not a square, you can always work in the extension field, where the square does exist. E.g. work in $F[u]/(u^2-d)$, where $F$ is your original field. $\endgroup$
    – Fractalice
    Nov 17, 2021 at 17:12
  • $\begingroup$ @Fractalice, Thanks! However, I'm not sure it will work for the injective encoding I'm implementing (eprint.iacr.org/2013/373.pdf). Is there any other way? Btw, how to find $u$? Is it a square root of something? $\endgroup$
    – pintor
    Nov 17, 2021 at 17:47

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