# Tag Info

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I have figured out a way to use TLS with only a EC key by using DSA instead of RSA. I had not realized you could do DSA with a EC key. My mistake was trying to use RSA to sign the certificate. Now I can generate my certificate and self-sign it only with the EC key. The peer will simply check to make sure the certificate was signed by the appropriate node ID ...

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First, one point. Each peer has a well known 256bit peer identifier (the public key of a 256bit elliptic curve keypair). Therefore, you have PKI, even though "there is no PKI 'authority'". Is this a proper use of ECDH anon, and will a simple challenge/response scheme at the application level guarantee peer authentication? Yes, and No. An ...

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If you want $N$ serial numbers, your serial numbers will have to use $n$ bits for uniqueness, where $n = \log_2 N$. So if you have 100 bits to use for the serial, you could use 20 to get about a million serials and have 80 bits to use for a cryptographic MAC or signature. Now there are two approaches, the symmetric and the asymmetric. In the symmetric ...

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It happens that a line usually (not always) cuts three points in a elliptic curve by the Bezout theorem. This is the case for the points and the curve you are asking for. So the sum of two points are defined like the inverse of the third point intercepted by the line that cut $P$ and $Q$ (let's name it $R$). So we need to find $-R$ because $P+Q=-R$ by ...

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The tangent line at point $P(x_1,y_1)$ is: $\frac{\partial f}{\partial x}(x-x_1)+\frac{\partial f}{\partial y}(y-y_1)$ Solving differentials, we obtain: $\frac{\partial f}{\partial x}=3x^2+2x$ $\frac{\partial f}{\partial y}=-2y-1$ For $x_1=0$ and $y_1=0$ we have: $\frac{\partial f}{\partial x}=3x^2+2x = 0$ $\frac{\partial f}{\partial y}=-2y-1 = -1$ ...

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A classical resut is that over a finite field $\mathbb{F}_p$ the group $E(\mathbb{F}_p)$ is either cyclic or isomorphic to the product of two cyclic groups. For the order of the group see this question. For cryptographic purposes, i.e., when you require that the ECDLP is hard, you will firstly have to rule out some weak curves (supersingular or anomalous ...

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The cyclic group over the ECDLP problem is posed is a subset of the set of point of the elliptic curve. That is to say, not all the points in the referred curve will be in the cyclic group. What you've called as $G$ confuses with another notation of the generator of a cyclic group $<G>=\{G,[2]G,\cdots,[n]G=\mathcal{O}\}$. This is the meaning of what ...

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Are there performance, size, or power efficiencies from one curve to another? The larger the curve, the larger the keys and signatures, and likely the slower the computations. There are exceptions to the last one – curve parameters do affect how efficiently they can be implemented, so some curves with good parameters can be faster than slightly smaller ...

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There is no way to generate a secure key-pair from the password "puppies". If you are using a 256-bit elliptic curve and want the full 128-bit security it can offer, any password from which you directly derive a key needs to have over 100 bits of entropy. If you use a key derivation function, like PBKDF2 or scrypt, with parameters that require a few seconds ...

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Given a EC public key, can a different, but plausible and functional private key be derived to match the public key? No, a public key will correspond to only one private key (with one minor exception, which I will explain below). With Elliptic Curve systems, the private key is an integer $d$ between 1 and $q$ (the order the generator point $G$), and ...

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Could Eve capture Bob's public key, then derive some different private key to correspond to it, and then impersonate Bob by saying "I am Bob, and here's my public key." Even though Eve doesn't have Bob's private key, she's able to produce some other private key that's compatible with Bob's public key? No, not unless the public key system is broken. In ...

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I obviously don't understand the mathematics behind ECC, but I am quite certain that the mentioned Multiply is not as simple as regular multiplication, nor as simple as wrapping around a finite field many times over. Because: When Alice and Bob exchange ECDH public keys, and then they each do DeriveKeyMaterial() or CalculateAgreement(), the result is ...

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Please see https://cbcrypt.org specifically https://cbcrypt.org/doku.php#documentation_and_api. There is a method CBCrypt.GenerateKeyPair(string CBCryptHostId, string username, string password) created specifically for this purpose. This does not relate specifically to BitCoin, but if you start with something like a password, salt it with details of where ...

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Such keys are called static keys. Keys that are newly generated each time are called ephemeral keys. Note that you need to trust the public keys of the key pairs to use them for authentication. Please note that there is an issue if you use static keys only for plain Diffie-Hellman: the generated secret will be static as well, as the whole scheme has now ...

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Yes, the same keypairs can be used to derive shared secrets between multiple pairs of parties. If knowing the shared secret between Alice and Bob would help Eve find out the shared secret between Alice and Carol, Eve could just create her own random private key and calculate a "shared" secret between that key and Alice's public key to get the same ...

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As @poncho says, both keys $Q_1=r^{-1}(sR-zG)$ and $Q_2=r^{-1}(sR'-zG)$ will validate the given signature, i.e., $(s^{-1}zG+s^{-1}rQ_i)_x=r\mod{n}$. For some curves, with small but non-zero probability, we have $n\leq(kG)_x<p$, and neither $Q_1$ nor $Q_2$ will validate other signatures made with the original private key $d$. However, by Hasse's theorem, ...

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If you can use something more lightweight than pairings then you should skip pairings. There is a technical property of pairings than is not the case with other crypto primitives. And this is the evaluation of multiplications on the exponent that cannot be achieved in other cases. If you need some sort of summation then you encode in the exponent and then ...

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Pairings in cryptography is a very important tool, the introduction of which has developed a new field, that is pairing-based cryptography. After the independent pioneering work by Joux and by Sakai et al.("Cryptosystems based on pairing"), many pairing-based crypto-systems emerged. In cryptography, pairings are often treated as "black-box", and then we ...

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Actually, it is not possible to uniquely recover the public key from an ECDSA signature $(r,s)$. This remains true even if we also assume you know the curve, the hash function used, and you also have the message that was signed. However, with the signature and the message that was signed, and the knowledge of the curve, it is possible to generate two ...

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You'll get back the same point; however it may be a different representation of the same point. Remember, an elliptic curve point is a solution to a cubic equation in two variables (or the point at infinity); one commonly used equation is $y^2 = x^3 + ax + b$ modulo a large prime $p$. What projective coordinates do is encode the $x$ and $y$ coordinates of ...

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SafeCurves lists some ways to compare the security of elliptic curves. Their security criteria are split to "ECDLP security" and "ECC security". Failing the former basically means "there is no way to use this curve securely in general" while the latter "it is difficult to implement this curve securely". None of the (few) BouncyCastle-supported curves that ...

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You may probably use any curve you like, depending on your special requirements (environment, computational aspects, ...) and the curves implemented by your library (see otus answer refering to some concrete security findings related to specific elliptic curves, and how sensible they are to certain attacks). The reason why the curves are pre-computed, is ...

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You can do it with two machines. https://www.iacr.org/archive/crypto2001/21390136.pdf (this paper is for DSA; it's easy to adapt for ECDSA). Here's an open-source JavaScript implementation of two-party ECDSA signing, using Bitcoin parameters: http://www.jpaulgossip.com/demo/split-key.html Unfortunately the protocol requires at least three rounds of ...

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