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seen Apr 11 at 11:40

Feb
28
awarded  Teacher
Feb
27
awarded  Commentator
Feb
27
comment Is sharing the modulus for multiple RSA key pairs secure?
@Ricky: What a stupid remark. The original question was if sharing modulus was secure. I show that it isn't because you can easily recover the primes given $e$, $d$ and $n$, and I show that in a setting that applies to all standard RSA out there. Hence I believe it's a highly satisfactory answer to the original question. Note that the other answer makes the same assumption but use a much more involved algorithm.
Feb
27
comment Is sharing the modulus for multiple RSA key pairs secure?
@Ricky: Not sure what you mean? In my comment, $\phi$ is the size of the multiplicative group modulo $n$, i.e. $\phi = (p-1)(q-1)$ for the standard RSA setup with two primes. In almost all cases in practice, $e$ is fixed at $65537$ and $d$ is taken to be the multiplicative inverse of $e$ modulo $\phi$, that is $ed = 1 \, (\textrm{mod} \, \phi)$ (if it exists - otherwise new primes are generated).
Feb
27
answered Is sharing the modulus for multiple RSA key pairs secure?
Feb
26
asked Sextic twist of BN pairing parameters vs security
Feb
25
comment Sextic twist optimization of BN pairing - cubic root extraction required?
Great thanks... didn't think of f(x) = 0 (mod f(x)), so now it all makes sense :)
Feb
25
accepted Sextic twist optimization of BN pairing - cubic root extraction required?
Feb
24
revised Sextic twist optimization of BN pairing - cubic root extraction required?
Prettified the math
Feb
24
asked Sextic twist optimization of BN pairing - cubic root extraction required?
Feb
5
comment Generating bilinear pairing parameters - running time of finding member of p-torsion group
Thanks just the answer I was looking for :-) I already know the number of points on the curve over the base field so it is easy to calculate the number of points over the extended field.
Feb
5
comment Generating bilinear pairing parameters - running time of finding member of p-torsion group
gmoktop: Thanks!
Feb
5
accepted Generating bilinear pairing parameters - running time of finding member of p-torsion group
Feb
4
awarded  Scholar
Feb
4
awarded  Supporter
Feb
4
comment Generating bilinear pairing parameters - running time of finding member of p-torsion group
Thanks, but does this algorithm perform (much) better than the original? Because it seems improbable to get a success in step 1-3 (probability ~2^512 / 2^3072)? And steps 4-5 are the original.
Feb
4
comment Generating bilinear pairing parameters - running time of finding member of p-torsion group
Yes, replacing p by p^2 immediately solved the problem (it found a point on the first iteration). Yes I read I can use twist curves but wanted to keep it as simple as possible in the beginning. By the way can you recommend a way to calculate the number of points on the curve (over the q^12 field)? Right now I use Sage and it's very fast but I need to implement my own point counting. I was wondering if this type of curve, due to its construction, has characteristics that makes it easier than in the general case?
Feb
4
comment Generating bilinear pairing parameters - running time of finding member of p-torsion group
gmoktop: Actually what I'm looking for is a point in the p-torsion group, i.e. a point that satisfies p*P = O so it must have either order p, or the order must be a divisor in p. But since p is prime, this should mean the point must have order p. I'm not sure I understand what you mean that the algorithm gives points of order is divided by p^s? What is this algorithm (the one in my post!) called and is there a reference where I could read more about it? I only saw it mentioned for use for this specific problem but I don't know how it works!
Feb
3
awarded  Editor
Feb
3
revised Generating bilinear pairing parameters - running time of finding member of p-torsion group
Rephrased question due to better understanding of the problem.