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6

As mikeazo notes in the comments, RSA operates on the ring $\mathbb Z / n\mathbb Z$ of integers modulo $n$, for a given modulus $n = pq$. In this ring, $$E(m) \cdot t^e \equiv m^e \cdot t^e \equiv (mt)^e \equiv E(mt)\ \pmod n.$$ In particular, for $n = 35$, $e = 23$, $$17^{23} \cdot 2^{23} \equiv 33 \cdot 18 \equiv 594 \equiv 34 \equiv 34^{23}\ ...


4

Yes. Assume that the attacker knows the ciphertext $c = c_1 \mathbin\| c_2$, the initialization vector $v$ and the plaintext $m = m_1 \mathbin\| m_2$. This tells them that $D_k(c_1) = m_1 \oplus v$ and $D_k(c_2) = m_2 \oplus c_1$, where $D_k(\cdot)$ denotes block cipher decryption under the (unknown) key $k$. In particular, this implies that, if the ...


3

I just want to add some additional information to the answer of Ilmari. As Ilmari has already described in his answer, when using RSA you work in the ring of integers ${\mathbb Z}/{\mathbb Z}_n$, which is also called a residue class ring. This means that it consists of the set of residue classes $[i]$, where the $i$'th class is defined as the set $\{a ...


2

The answer to your question is contained in the Authenticity bound (Theorem 5.1). This is because Authenticity implies non-malleability (see e.g. http://eprint.iacr.org/2011/092.pdf). Note that only one term in the bound refers to the length of the tag (referred to by the variable $\tau$): $$\mathbf{Adv}_{OCB}^{auth}[\mathrm{Perm}(n), \tau] (A) \leq ...


1

Ed25519 in the default implementation is malleable. It includes the public key $A$ in the hashed message, so it cannot be modified It includes $R$ in the hashed message, so it cannot be modified $S$ is encoded as a 256 bit. But since it's a scalar, $S^\prime = S + k \cdot l$ is equivalent to $S$ for any integral $k$ (where $l$ is the order of the subgroup, ...


1

Depending on how malleability is defined, the question actually has some merit. Given to the Wikipedia definition of malleability, a cipher is malleable if there exists at least one function $g$ over the set of possible cipher texts, and one function $f$ over the set of possible plain texts, such that given any cipher text $c_0$, the cipher text $c_1 = ...


1

With the usual definition of AES-CTR or Xsalsa20, no, it is not possible to reliably perform some arithmetic operation $\odot$ (such as addition or multiplication) with a known constant $a$ on ciphertext, without some additional knowledge or restriction. In these stream ciphers, the ciphertext is the exclusive-OR of plaintext with some unknown keystream $K$, ...



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