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2

If now the prime value $P$ gets increased for constant $k$ (order of $g$ still $k$) does this decrease the security? No, for random choice of $g$ among $g$ with large constant order $k$. Why this? If $P \rightarrow \infty$ it would be the normal logarithm .. When $P$ grows, $g$ also grows. So we never reach the threshold where $g^a<P$. Thus we can ...


1

The MAC under attack is built from a hash $h$ (assumed to process a hashed message by splitting it in blocks processed only once, as most hashes do), with a key split into $K_1$ and $K_2$ of equal size. It computes the MAC of a message $x$ as $$\operatorname{MAC}_{K_1\mathbin\|K_2}(x)=h(K_1\mathbin\|x\mathbin\|K_2)$$ The attack in the paper recovers $K_1$ ...


5

This scheme suffers from a classic problem of textbook RSA which is mitigated e.g. by RSA-KEM (as outlined by kelalaka) or RSA-OAEP. When you compute $k^3\bmod N$, you'll experience that $$c=k^3\stackrel{k< 2^{256}}{\leq}\left(2^{256}\right)^3=2^{768}\ll 2^{2000}<N$$ Now remember how $x\bmod N$ works: If $x\geq N$, then you recursively compute and ...


4

What you describe is a little away from the RSA-KEM (KEM : Key Encapsulation Mechanism). As pointed out by SEjPM, in the comments, an AES-128 key when encrypted with the public modulus has almost 768 bits and this can be recovered by the cube-root attack. Here is the RSA-KEM; RSA-KEM mitigates the attack that you have. RSA-KEM for a single recipient with ...


3

meet-in-the-middle attacks rely on storing the results in a map Only the naive version of MitM does. The impractical memory requirement (size and accesses) of naive MitM can be greatly lowered with a relatively modest increase in other computation costs, and the calculation distributed to independent devices. The reference paper is Paul C. van Oorschot and ...


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