# Tag Info

7

If $p$ and $q$ are 1024-bit primes, then by definition of the bit size of an integer (at least, a prime in a cryptographic context with glimpses of RSA), $2^{1023}\le p<2^{1024}$ and $2^{1023}\le q<2^{1024}$. Thus their product $n=pq$ verifies $2^{2046}\le n<2^{2048}$, and $n$ is a 2047-bit or 2048-bit integer. We show by exhibition that both cases ...

5

Yes, you encrypt the file with a symmetric key, then encrypt that symmetric key with each of the recipients public keys. gpg can do this by adding multiple --recipient options.

4

This exists. It is called Broadcast Encryption http://en.wikipedia.org/wiki/Broadcast_encryption . Latest research even allows for Traitor tracing http://en.wikipedia.org/wiki/Traitor_tracing , meaning that even if two people give a part of their secret keys to form a "pirate decryptor", there is an algorithm which will find one of the users that colluded. ...

4

You got what a "semiprime" number is; it's a number which is the product of two primes. When people talk about "multi-prime RSA", what they mean is something which is pretty much the standard RSA algorithm; however the modulus is the product of at least 3 prime numbers (as opposed to standard RSA, which has only 2 prime factors). Why would anyone do this? ...

3

Any integer $n$ can be represented in binary form in $\left \lfloor{log_2{n}}\right \rfloor + 1$ bits. Now coming to your problem where $n = pq$. Number of bits to represent $n$ is $\left \lfloor{log_2{n}}\right \rfloor + 1 = \left \lfloor{log_2{pq}}\right \rfloor + 1 = \left \lfloor{log_2{p}+log_2{q}}\right \rfloor + 1$. By the properties of the floor ...

2

For CPA security it is actually enough that the first scheme, i.e., $\pi = (gen, enc, dec)$ is CPA secure. Lets define the CPA game of a general scheme $\pi = (Gen, Enc, Dec)$ against an adversary $A$ as follows: We sample $(pk, sk) \leftarrow Gen(1^\lambda)$, and send $pk$ to $A$. $A$ outputs messages $m_0$ and $m_1$. We sample $b \leftarrow \{0,1\}$ (a ...

2

I don't believe he is answering the right question. You essentially asked "why are public keys so much larger than symmetric keys", and after his first sentence (which started to address the question, but was a bit vague), he tried to answer the distinct question "why are public key operations so much slower" (not that he got the details of that correct; ...

2

One real problem is that lack of authentication between the two sides. Here's one possible problem: Alice generates an RSA keypair (we assume Alice is using proper random numbers) Alice sends the public key as plain text to Bob. Eve intercepts this message, and forwards on a message to Bob with her public key Bob generates a 3DES session key: ...

2

DrLecter's comment should really be an answer, but one point which bears emphasis is that it is not sufficient for crypto purposes to say "a (cyclic) group of prime order $p$", because even though mathematically all groups of order $p$ are essentially the same group (i.e., they are isomorphic), we can't really say the same thing in crypto. When doing crypto ...

1

I recently came across a paper that may interest you that I think answers your question. To quote from the abstract: Unfortunately, in all existing HD wallets---including BIP32 wallets---an attacker can easily recover the master private key given the master public key and any child private key. This vulnerability precludes use cases such as a combined ...

1

Asymmetric keys have to be much larger than symmetric keys because 1) there are less asymmetric keys for a given number of bits (key space), and 2) there are patterns within the asymmetric keys themselves. To compare, consider that the ECRYPT II recommendations on key length suggest a 128-bit symmetric key is as strong as a 3,248-bit asymmetric key, and ...

1

I don't think it is right. The reason why RSA in particular uses such a high bit count, is that RSA's security is based on factorization of integers and integers with up to 100 digits (roughly 300 bits) can be "easily" factorized with the Quadratic Sieve. In general, there are asymmetric ciphers like those based on elliptic curve cryptography that also use ...

1

Regarding public key: In two prime RSA you often compute the secret key from the public as $d=e^{-1} \pmod {(p-1)(q-1)},\;$ but using $d=e^{-1} \pmod {\mathrm{lcm}(p-1,q-1)}$ may have some advantages.

1

No, the total number of bits after multiplication will be 2*1024. In binary form take for eg. 3 (2 bits = 11) * 3 (2 bits = 11) = 9 (4 bits = 1001 in binary)

1

Guess the catch in the video is in how the participants exchange details 'publicly'. If the Man-In-The-Middle can intercept and manipulate what is being 'publicly' shared, then the attempt to eavesdrop would still be successful.

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