# Reflecting a point on the Edwards curve

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 $$(u,-v)$$.

https://math.stackexchange.com/questions/4764173/can-we-ever-scale-elliptic-curve-points?noredirect=1#comment10116801_4764173

I think this amounts to a similar thing.

• 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) Commented Sep 6, 2023 at 10:10
• 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. Commented Sep 6, 2023 at 10:13

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