Let's say we have a point $nP = (x,y)$ on a curve $E$ over a prime $p$. The corresponding Edwards curve coordinates are $(u,v)$. I want to construct the point corresponding to $(u,-v)$ on the Edwards curve.

I can construct $(-u,-v)$ easily, that just amounts to $-nP$, flipping the $y$ coordinate. But I don't know about $(u,-v)$.

Is anything known about this construction?

See also, my related question here:


I think this amounts to a similar thing.

  • $\begingroup$ The question is not clear. Why do you start with a generic point (x,y) but then you move to Edwards. Can't we just start with an Edwards curve. It's not clear whether you want to find a scalar that you get you to (u,-v) or you want something else. Also, -nP should get you to (-u,v) not (-u,-v) $\endgroup$
    – Ruggero
    Sep 6 at 10:10
  • $\begingroup$ Yes, I want the scalar that gets me to $(u,-v)$ on the Edwards. See also this: mathoverflow.net/questions/454076/… I believe these are all equivalent problems. $\endgroup$ Sep 6 at 10:13

1 Answer 1


If $P=(u,v)$ then $-P=(-u,v)$

Now it follows from addition law here that if we denote $Q=(0,-1)$(Q is some point of order $2$) then

$P+Q=(-u,-v)$. It follows that $-(P+Q)=-P+Q=(u,-v).$

Is this what are you looking for?


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