Tag Info

New answers tagged

0

For anyones who's interested - From this paper http://scholar.lib.vt.edu/theses/public/etd-32298-93111/materials/etd.pdf I found that it's best to find the degree of the polynomial you're going to be using based on the number of bits in $N$. Let k = $log_2N$ Briggs gives experimental bounds $d= 5$ for $k \geq 110$ $d= 4$ for $80 > k > 110$ $d= 3$ ...


0

Your Question is related to the well know RABIN Cryptosystem which is similary like RSA, except the public exponenent is 2. As fgrieu mentionned it, decipherment can be easilly processed by the CRT algorithm, but some precautions must beforehand be observed during the key generation. In fact the resolution the equation gives 4 roots, which mean that the ...


0

In this case, take a look to the GQ identication scheme, from Guillou and Quisquater.


-2

Generally speaking, the bit length of the exponents are the same as the modulus. Depending on the implementation, there may also be some extra data in the public/private key structure. Here is a good discussion on the concept of key length as it relates to the public/private values: Private key length bytes


1

The term you are looking for is Modular Arithmetic. In the case of 1535, if it is indeed a combination of 3 values ranging 0 to 25, you do the following: 1535 mod 26 = 1 (1535 - 1) / 26 = 59 59 mod 26 = 7 (59 - 7) / 26 = 2 The set of values that generated 1535 is 2,7,1 (c,h,b?), which can easily be verified.


-1

Google Quantum A.I. Lab Team Post, Jan 20th 2014 Just FYI https://plus.google.com/+QuantumAILab/posts/DymNo8DzAYi


0

Here you have an overkill of information, really. We know that $e_A$ and $d_A$ are related by the fact that $e_A d_A \equiv 1 \pmod{\phi(N)}$, normally, and then this answer gives an algorithm. But here we have more info, as both $e_A d_A - 1$ and $e_B d_B - 1$ are multiples of $\phi(N) = (p-1)(q-1)$. We need some multiple of their gcd, that is close to $N$ ...


0

Brievly, when you multiply a n-bits integer by a m-bit integer, the product is $(m+n-\epsilon)$-bit integer, with $\epsilon \leq 1$. For RSA the security recommandations specify that secret factors of the public modulus must be of approximatively the same size. if the difference in bit is modest (less than a machine granularity) it could be approved. There ...


0

if the adversary knows $m^{e_1}$, $m^{e_2}$ and $d_1e_2$, is it able to guess m? Obviously not; if he could, then they could break RSA. Here's the RSA problem, given $m^e$, $e$ (and the modulus $N$), recover $m$. Suppose that we can solve your problem, that is, we had a black box that, given $m^{e_1}$, $m^{e_2}$ and $d_1e_2$, could recover $m$. Then, ...


0

Outline of a elementary attack: if $\mid p_i-q_i \leq 2^s$, we can write for each $p_i=a.2^s+\tau_i$, with $\tau_i \leq 2^s$, and we can imagine that if we determine the integer a, we can factor $N_i$ by a brute force attack. We know that $p_i$ have exactly the same bit size and if $N_i$ have exactly the same bit size (n=1024, 2048, ... s=64), the sizes of ...


0

Do not use raw RSA at all. First simple attack starts when (plain ^ public_exponent) < n , i.e. when the message to be encrypted is small. Then there are more sophisticated attacks and forgeries as soon as 1 plain text and the corresponding cipher text are known, or even can be guessed. And then you have to review all possible attacks on RSA for the ...


1

I think the confusion lies in mixing two distinct hard problems. RSA is related to the difficulty of factoring large numbers. The scheme you link to in the paper is based on the difficulty of solving discrete logarithms. Since they are very different hard problems, you can't really compare the two in terms of what must be kept secret. For RSA, ...



Top 50 recent answers are included