# Tag Info

5

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 ...

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? ...

4

You've mostly pieced it out. This is a DER encoding the the public key, and consists of a sequence of two integers (the first being the modulus, and the second being the exponent). Here is the breakdown of the encoding: 30 The value 30 is used to signify 'sequence'; this is a container that carries a list of DER-encoded objects. 82 01 0a Whenever we ...

3

Okay, I came up with this, it's not a complete answer and the attack presented is pretty weak without a follow-up algorithm for breaking a somewhat unbalanced modulus but let me know if you spot any flaws or have any ideas to improve it... If $n = pq$ with $p, q$ prime and $\gcd(pq, (p - 1)(q - 1)) = r > 1$ then clearly $r = p$ or $r = q$. Now ...

3

A lot of modern cryptography is based on some mathematical assumptions and aims to achieve what is called Computational Security. That means that the adversary (Eve) could get some information about the plaintext with a negligible probability and the adversary is modeled as someone with bounded computational power, storage and bounded time. So all the ...

3

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. ...

3

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.

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

This is called the common modulus attack. Bezout's Identity says that there exists $x$ and $y$ such that $ax + by = gcd(a, b)$. In our case we have $gcd(e_a,e_b) = 1$, so we can find $x$ and $y$ such that $e_{a}x + e_{b}y = 1$ (you can use the extended euclidean algorithm for this). After solving for $x$ and $y$, you compute: $C^xC^y\mod N$ to get $M$. ...

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)

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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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1) Yes. This is the common modulus attack and has actually been answered many times on this forum. 2) Assuming $r$ is prime, yes. $\phi(n)$, (the totient of $n$) can be computed by subtracting 1 from each of $n$'s prime factors and multiplying them together.

1

You actually don't trust the certificate by itself. A certificate is like a diploma that a company (or a domain, e.g. www.crypto.com) obtained from some trusted party, called CA, or Certificate Authority. This diploma states that www.crypto.com is allowed to communicate with you using the public key written in the diploma. But, as you mentioned, ...

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