On Thu, Oct 11, 2007 at 04:48:33PM -0000, Cyptmon wrote:
>
> Thanks for the replies, I am sorry I made a mistake, I get the private
> exponent as 1024 bits, the public exponent is the smaller one and it
> is 17 always. Why is it always 17 irrespective of the key size that I
> specify?
There's nothing wrong using this value;
it's prime so would fit any \phi(n), and
it results in (rather) small number of operations while calculating exponent
... InvertibleRSAFunction {
void Initialize(RandomNumberGenerator &rng,
unsigned int modulusBits,
const Integer &e = 17);
...
};
> Isnt the public key chosen as something that is relatively prime to
> Totient function of n (Phi (n ) where n is p * q)
17 is a prime, so it's relatively prime to anything
> Thanks
>
> PS: I hope GenerateRandomWithKeySize is the right function to use to
> generate different sized keys for RSA as directed by the second
> argument passed to it
>
>
>
>
> On Oct 10, 3:12 am, Vadym Fedyukovych <[EMAIL PROTECTED]> wrote:
> > On Tue, Oct 09, 2007 at 10:23:38PM -0000, Cyptmon wrote:
> >
> > > I am using RSA 1024 bit keys for RSA signing with SHA-256. I had a
> > > question, I get a public key which is indeed 1024 bits in size but the
> > > private key is always not more than 2 digits. I use
> > > GenerateRandomWithKeySize with 1024 as the second argument to get the
> > > keys. What is the difference between using this and the rsa1024.dat
> > > file provided
> >
> > > Thanks
> >
> > 1024 is modulus size.
> > 17 and 2^16+1 are public exponent values that are used often by default
> >
> >
>
>
>
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