poncho
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 3h comment typical block size in RSA You mean that this is the second block of some message (that someone else has encrypted)? This can be covered by making the tag (that identifies the message that this block is a member of) be a 16 byte random number that the encryptor picks. No one else (other than the holder of the decryption key) can know what that is, so no one else can generate such a block. 3h comment typical block size in RSA Actually, you don't really need to use distinct RSA keys; instead, you use a few of the 86 or 214 bytes you have in include a tag that declares that these blocks are the part of the same message, and a 'fragment number' to give the order. However, we'll all agree that encrypting a symmetric key is far cheaper... 3h answered typical block size in RSA 5h revised RSA private key finding method Fixed math typo 10h comment Public key exponent coprime with totient proof @fgrieu: If that's what he's asking, then a related answer (with, a bit of work, does answer the question) is crypto.stackexchange.com/questions/31980/… 1d comment What is base of $log$ in Subexponential Algo for DLP? @RickyDemer: nope, they don't, because they're either in a $O(\textit{formula})$ notation, or multiplied by an unspecified constant; either form eats constant factors 1d comment Breaking the Rabin Signature system with chosen message attack And, by selecting a $y$ with Jacobi symbol -1, I really mean selecting one with $\Bigl(\frac{y}{n}\Bigr) = -1$ 1d comment Breaking the Rabin Signature system with chosen message attack Actually, we would consider a cryptosystem broken if there is an attack which succeeds with probability $10^{-6}$. However, since you asked: en.wikipedia.org/wiki/Jacobi_symbol 1d answered Breaking the Rabin Signature system with chosen message attack 2d comment Why is AES-128 considered secure while RSA needs a 1024-bit size to be considered secure Actually, to answer the question: while factoring the modulus may be difficult, it is significantly easier than trying to brute force the private exponent, or even trying possible primes $p$ to look for a factor... 2d comment Do most TLS 1.2 implementations express curves in a canonical form when performing EC arithmetic? Yes, that is correct. As you yourself stated, "all EC curves are isomorphic to a curve in Weierstrass form", however not all EC curves are isomorphic to a curve in Montgomery/Edwards form. 2d comment Do most TLS 1.2 implementations express curves in a canonical form when performing EC arithmetic? The curves used most often in TLS 1.2 are the "NIST curves", which cannot be expressed in Montgomery or Edwards form. 2d comment I have a challenge for anyone willing You're right; this isn't a "let us solve your cryptogram" site. In addition, it is expected that answers are posted here (after all, this is a Q/A site). Perhaps sci.crypt on usenet might be more suitable? 2d answered Vulnerabilities in specifying the hash algorithm used for signature verification on a per-message basis 2d revised RSA private key finding method At least, lets get the math right... 2d comment Caculated time One Point Multiplication with double and method How certain are you on the timings of the double and add routines? If they are moderately faster than you think, that'd explain the performance difference. 2d comment How to design a deliberately weak PRNG for experimentation? @pg1989: actually, the weaknesses in MT would be far too subtle for an ML to pick up on. It would appear that someone would need to select the biases, and then create a PRNG that deliberately achieves those biases. 2d answered Determining collision resistance of a function? 2d comment RSA private key finding method Maybe she tries the various possibilities of $d$ until she hits one where $(M^e)^d = M$. With these toy parameters, that is eminently doable... Feb 9 revised OpenSSL & TLS1.2 with aes-cbc and initialization vector added 426 characters in body