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Jun
2
comment Precomputation attacks on RSA
The precomputation portion is harder than factoring the modulus, but the size of that portion's output and the amount of per-key computation (aka, the important parts of a precomputation attack) are both trivial. $\;$
Jun
2
comment Precomputation attacks on RSA
It works by precomputing a collision anyway. $\:$ (Surely that part is much easier $\hspace{1.6 in}$ than creating "about $2^{950}$ keys".) $\;\;\;\;$
Jun
2
comment Precomputation attacks on RSA
@poncho : $\;\;\;$ Where does that signature scheme use anything about the message other than it's length and it's hash? $\:$ If it doesn't, then there is trivially a precomputation attack based on finding a hash collision whose messages have the same short length, submitting one of them to the signing oracle, and then outputting the other along with the signature. $\;\;\;\;\;\;\;$
Jun
1
comment Creating a license system based on asymmetric encryption (RSA or ECDSA)
@marstato : $\;\;\;$ What signed data? $\:$ Since the license-holder has the private key for their key-pair, the license-holder could sign any ID with that. $\;\;\;\;\;\;\;$
Jun
1
revised How can I make a message into a blocks for NTRU?
fixed TeX error that was introduced by e-sushi
May
31
answered How to prove identity without revealing identity
May
31
revised How can I make a message into a blocks for NTRU?
improved grammar
May
31
comment Creating a license system based on asymmetric encryption (RSA or ECDSA)
@marstato : $\:$ Anyone could just change the ID, since it's not tied to anything else they have. $\hspace{.85 in}$
May
31
comment Creating a license system based on asymmetric encryption (RSA or ECDSA)
If you "create asymmetric key pairs and provide the private key along with an ID as the license code" $\hspace{.17 in}$ "what keeps hundreds of users from using the same license" code? $\;$
May
31
comment RSA: Fermat's Little Theorem and the multiplicative inverse relationship between mod n and mod phi(n)
Even if there was such a symmetry, I don't see how that would help, since the encryption algorithm is not doing any computations mod phi(n). $\:$ Furthermore, powering is different from multiplying. $\:$ (p.s., What is "a multilicative inverse"?) $\;\;\;\;$
May
30
comment RSA: Fermat's Little Theorem and the multiplicative inverse relationship between mod n and mod phi(n)
What you're missing is that "if we had made" $ed = 1\ (mod\ n)$ then $\hspace{2.21 in}$ "raising $C$ to $d$ will" usually not "get us $M$ again". $\;$
May
30
comment Update to “Cryptographic Right Answers”
@otus : $\;\;\;$ You need that since you need some authentication essentially everywhere. $\:$ The point is $\hspace{.41 in}$ to make as much of the authentication as possible (including "actual software updates") internal. $\hspace{.58 in}$
May
30
comment Secure Secret sharing
So ... does this differ from homomorphic encryption? $\;$
May
30
comment Update to “Cryptographic Right Answers”
@otus : $\:$ Also, even though "revocations and trust chains are much more complicated than simply replacing the trusted key", they also provide more security "than simply replacing the trusted key", $\hspace{.28 in}$ since they make the "updates" usually not need any external authentication mechanism. $\hspace{1.07 in}$
May
30
comment Update to “Cryptographic Right Answers”
His advice for Client-server application security is not valid. $\:$ One should instead distribute the signature verification key for one's own offline certificate authority with the client code, and not trust any other certificate authority as a root. $\;\;\;\;$
May
29
revised How can one parallelize tasks in CTR-AES for maximum performance?
fixed capitalization and grammar
May
28
comment Need to generate new DH keys for every message in OTR
One could alternatively use forward-secure symmetric cryptography. $\;$
May
28
revised Why is plain-hash-then-encrypt not a secure MAC?
fixed typo
May
28
answered Why is plain-hash-then-encrypt not a secure MAC?
May
27
comment For enciphering messages with AES in CTR mode, do I need a different key for each message?
@hunter ... where "used" refers to "by the block cipher" rather than "by the encryption scheme". $\hspace{.58 in}$