# Tag Info

1

If you use the raw RSA operation ($M^d \bmod n$ or $M^e \bmod n$), then no, it is unsafe to use the same key, because an attacker could trick the private key holder into signing a message $M$ (i.e. generating $M^d$) which is actually an encrypted message ($M = P^e$), thus allowing the attacker to recover the original plaintext ($(P^e)^d = P$). (The dual ...

3

Short Answer: NO, it is not safe, do NOT do this. Longer Answer: You are true that you can use your RSA keypair for both operations. This approach is used in many applications and scenarios. There are Web Services or Single Sign-On implementations, which enforce you to use the same key pair for both operations. X.509 certificates do not allow you (by ...

1

$Poly1305_{k,r}(N,M)$ is a Carter-Wegman nonce-based MAC, whose security crucially depends on the uniqueness of nonce $N$ for every message $M$. It is defined as $$Poly1305_{k,r}(N,M) = f(M,r) + AES_k(N),$$ where $f(M,r)$ is a polynomial of $r$ with coefficients derived from the binary representation of $M$, and $AES_k(N)$ is the encryption of nonce $N$ ...

1

$Poly1305_{{r,s}}(m)$ is a one-time authenticator - it can be used to authenticate only a single message with any given key $(r,s)$ without violating the security guarantees (the violation is immediate - only two authenticated messages with the same key are required to create a forgery according to the nacl docs). There are two 128 bit key values to this ...

2

My own symmetric cipher also has this property. Here's what I can say about the general meaning of it: It means that the amount of preprocessing of a key is small. So the amount of time from generating/importing a new key to actually starting encrypting is neglegible. It means that the amount of state a cipher uses is small. The state of a cipher generally ...

2

No. You cannot use the same key and IV for more than one vector (with the most AES modes of operation). The only AES mode of operation which is (somewhat) resistant for IV reuse is SIV. For usual modes of operation like CBC, CTR, GCM, etc. reuse of Key+IV pair is a bad mistake. It is important to acknowledge that there are further requirements for ...

2

Speaking in broad strokes, reuse of the key is fine - reuse of the IV: not fine. From wikipedia: "Properties of an IV depend on the cryptographic scheme used. A basic requirement is uniqueness, which means that no IV may be reused under the same key". You also need to decide on a mode of operation, as different modes will dictate different requirements for ...

7

Curve25519 was designed to take advantage of the Montgomery ladder, which combined with Montgomery curves forgoes the $Y$ coordinates, is side-channel resistant, and enables public keys to be any 255-bit string. The ladder looks something like this (pseudocode): Q[0] = P; Q[1] = 2*P; for(int i = log2(exponent) - 2; i >= 0; --i) { Q[ bit(exponent, i)] ...

6

Did you take a look at DjB's paper? One of his design criterias in order to improve performance is "Use a fixed position for the leading 1 in the secret key". The set of secret keys is defined to be $\{\underline{n} : n \in 2^{254} + 8\{0, 1, 2, 3,\ldots, 2^{251}-1\}\}$.

1

Are you trying to prove this for a specific encryption scheme or for any scheme? If you have a specific scheme in mind, you can consider using rejection sampling. In your case, it would be quite straightforward to use : Let's say each key $k\in \{0,1\}^n$ is output by $Gen$ with a probability $p(k)$, and $p_{min} \overset{def}{=} \min\limits_{k} p(k)$. You ...

0

This depends on the degree of non-uniformity and the ability of $\Pi$ to produce uniform key-independent outputs. For instance, a deterministic encryption scheme that always selects $k=k_0$ is just a fixed permutation and can not be used to build a secure scheme without additional tools. However, if $\Pi$ produces a uniform $IV$, simply take it as a key and ...

0

With this cipher, it's pretty easy to retrieve at least 1 key that is consistent with 2 pairs of plaintext,ciphertext . (Other ciphers are better or worse at making it nearly impossible to recover even 1 key consistent with the given plaintext,ciphertext). With this cipher, it is not possible to fully retrieve the key from only 2 known pairs of ...

0

There is a straightforward brute force method. Take for example the lowest 8 bits of everything and check for valid values of $K_1$ and $K_2$, mod $2^8$. You will need about $2^{16}$ checks to get the lower 8 bits of $K_1$ and $K_2$. Proceed then to values mod $2^{16}$, as you know the lower 8 bits of $K_1$ and $K_2$, only bits 8..15 of these must be ...

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