How to recover $p$ from edwards curve?

IlIIIlllIIllIIl

4/21/24, 7:03 AM

The shape of the edwards curve is equal to $a x^2 + b y^2 = d x^2 y^2 + 1$ in $\mathbb{Z}_p$. Is there a way to know $p$ when we know the five points on the edwards curve without knowing $a, b, d, p,$ and generator $G$?

I tried to find p by using the five points given to me by concatenating the relationship $a x^2 + b y^2 - d x^2 y^2 - 1 ≡ 0 (mod \ p)$, but the expression became too complicated and the result was weird.

Is there a proper way to get p in a coalition using the given 5 points?

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elliptic-curves

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Crypto

poncho

4/21/24, 1:57 PM

This appears to be a homework question, and so I'll just give hints:

Rewrite the equations to the form:

$$ax^2 + by^2 - d(xy)^2 - 1 = kp$$

(where $x^2, y^2, (xy)^2$ are known, $a, b, d, k, p$ are unknowns, and $k$ is some integer)

Given two equations of this form, how can you combine them to come up with a similar form with one less unknown (e.g. where the $a$ unknown does not appear).
How can you repeat the above so that, using four of the equations, we come up with:

$$c = k'p$$

where $c$ is a known constant.

How can you use another set of 4 equations to come up with:

$$c'' = k''p$$

for a (hopefully) different $c''$

How can you use the values $c, c''$ to try to recover $p$?

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Elon Musk

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EN: How to recover $p$ from edwards curve?

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