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6

A is acting as a square-root oracle in that protocol. We can use that oracle to factor $n$ and break the scheme. Suppose you are an attacker that wants to impersonate A. You: Pick a random $m$; Send $m^2$ to A; Compute $p = \gcd(m_1 - m, n)$, thus factoring $n$. This works with probability $1/2$ for each attempt.


4

Nightcracker's method works fine. There also are deterministic solutions to select the correct ciphertext that require very few additional bits. One very useful ingredient is the use of the Jacobi symbol. For example, you might look at The Rabin cryptosystem revisited by M. Elia, M. Piva and D. Schipani (http://arxiv.org/pdf/1108.5935.pdf).


4

This is a solution that should work with very high probability, but possibly can fail. As a bonus it also resists tampering with the ciphertext. As encrypter generate a random key (say a 128-bit key for AES128-CTR) and encrypt the plaintext using that key. Then compute a MAC over the ciphertext (for example using HMAC-SHA1) using the same key. Finally you ...



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