Tag Info

New answers tagged

2

Curve25519 makes use of a special x-coordinate only form to achieve faster multiplication. Ed25519 uses Edwards curve for similar speedups, but includes a sign bit. While it could have been done differently, doing it this way simplifies implementations that only need one of encryption or signing.


6

If I understand your question correctly, you are essentially asking if points in Edwards and Montgomery curves can be represented in Weierstrass coordinates. This is true; in fact, any elliptic curve over a prime field can be represented in Weierstrass form $\mathcal{E}_{w}^{a, b} : y^2 = x^3 + ax + b$, and by extension its points can too. The question, ...


4

No, Curve25519 signature is not vulnerable to bad RNG during signature generation; that's because Curve25519 signature needs no random number during signature generation. By contrast, in ECDSA, a fresh random number is needed for each signature, and if it gets known, that allows to recover the private key from the signature and public key; same if the same ...


5

Any key generation algorithm for any cryptosystem is going to be weak if the attacker can predict what seed was used to generate the key. They can just generate the same key. However, assuming the the random number generator is not that bad, different algorithms start to look different. If you are just using the output of the random number generator as a ...


3

1. The equation $-x^2+y^2=1-(121665/121666)x^2y^2$ defining the curve $E$ is quadratic in $x$, hence for any given $y\in\mathbb F_q$, there are at most two points on $E$ which have $y$ as their second coordinate. In this case, the two possible $x$-coordinates for a point on $E$ with $y$-coordinate $4/5\in\mathbb F_q$ are the solutions to the equation $$ ...


1

The goal of this method is to achieve collision-resilience (resistance against collision attacks). The second hash can be viewed as $H(R || M)$ for message M and some randomness R that is unknown to an attacker. Now, even if an attacker could efficiently find collisions for $H$, he cannot use this ability to run the standard forgery attack that works as ...



Top 50 recent answers are included