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 Apr 22 revised Trapdoor and RSA (Schneier) Use y consistently with the first citation Apr 22 revised Trapdoor and RSA (Schneier) Polish Apr 22 revised Trapdoor and RSA (Schneier) Polish Apr 22 revised Trapdoor and RSA (Schneier) Polish Apr 21 revised Trapdoor and RSA (Schneier) Polish Apr 21 revised Trapdoor and RSA (Schneier) Polish Apr 21 revised Trapdoor and RSA (Schneier) Polish Apr 21 revised Trapdoor and RSA (Schneier) Polish Apr 21 revised Trapdoor and RSA (Schneier) Use the question's notation Apr 21 answered Trapdoor and RSA (Schneier) Apr 21 revised Trapdoor and RSA (Schneier) Give second citation Apr 21 comment Calculating the discrete logarithm Hint: observe that $p-1$ is the product of $u=64$ and $v=19$, and $\gcd(u,v)=1$, and you know the unknowns ($\log2$, $\log3$, and $\log5$) modulo $u$ and modulo $v$; now apply the Chinese Remainder Theorem. Apr 20 comment Is the 'nonce' in bitcoin really a nonce? In cryptography, we use 'nonce' including when in legitimate use, the likelihood of reuse is low, which can be achieved by making a random choice in a large set. I do not know if that's how bitcoin uses the term. The question would be much better with a link to the authoritative bitcoin specification with the use of 'nonce' considered. Apr 20 revised Trapdoor and RSA (Schneier) proper citation, in preparation of re-opening Apr 20 awarded Nice Question Apr 20 comment Possibility for same Private Key Generation in identity-based encryption The simpler approximation $p(n)\approx{n(n-1)\over2r}$ also gives $\approx3\cdot 10^{-42}$. This is negligibly low compared to the risk of destruction of human life on earth by a comet during the forthcoming year (which is considerably more than $10^{-9}$). It would follow that the short answer is: no, there is no practical risk that well-behaving IBE servers accidentally generate the same master secret key components. Apr 19 awarded Talkative Apr 19 revised Turing's (still?) classified inference engine algorithm? retag Apr 19 comment How to crack unknown 8-bit encryption algorithm? I'm assuming we know the operating mode (CBC), but nothing about the 8-bit block cipher (this might be what the question is about, or not). Then, the coupon collector problem tells us that we build the full dictionary with an average 1563 bytes of known plaintext (we have a useful dictionary much before that). As pointed by @mikeaso (I guess), we can reduce this with iteratively chosen plaintext (chosing the next byte of plaintext from the previous byte of ciphertext). Apr 19 comment How to crack unknown 8-bit encryption algorithm? Assuming the question is about an 8-bit unknown block cipher in CBC mode, hint: build a dictionary of input/output pairs for the block cipher.