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 failure is 20
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
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
