> > Although the Carmichael numbers fool the Fermat test
> > (that is, $a^{n-1} = 1 (n)$) for *all* a,
I thought it would work properly if a shares a factor with n.
> Yes I guess the difference is that with MR you are trying to find a
> number that is *likely* a prime, whereas with Fermat you are
On Fri, 11 Nov 2005, Joseph Ashwood wrote:
> From: "Charlie Kaufman" <[EMAIL PROTECTED]>
>
> >I've heard but not confirmed a figure of one failure in 20 million. I've
> >never heard an estimate of the probability that two runs would fail to
> >detect the composite. It couldn't be better than one fa
> Don't ever encrypt the same message twice that way, or you're likely to
> fall to a common modulus attack, I believe.
Looks like it (common modulus attack involves same n, different (e,d) pairs).
However, you're likely to be picking a random symmetric key as the
"message", and Schneier even sug
> > Don't ever encrypt the same message twice that way, or you're likely to
> > fall to a common modulus attack, I believe.
>
> Looks like it (common modulus attack involves same n,
> different (e,d) pairs).
>
> However, you're likely to be picking a random symmetric key as the
> "message", and
