is breaking RSA at least as hard as factoring or vice-versa?

So I'm reading up on unconditionally secure authentication in Simmon's Contemporary Cryptology, and he points out that with RSA, given d, you could calculate e (remember, this is authentication not encryption) if you could factor n, which relates the two. However, the implication is in the less

Re: Unforgeable Blinded Credentials

I came across the same problem a couple of years ago (and indeed iterated through private/public key solutions with a colleague). The problem is that you can still give your private key to somebody else. There's no real deterrent unless that private key is used for many other purposes,

Re: Unforgeable Blinded Credentials

On Sat, Apr 01, 2006 at 12:35:12PM +0100, Ben Laurie wrote: However, anyone I show this proof to can then masquerade as a silver member, using my signed nonce. So, it occurred to me that an easy way to prevent this is to create a private/public key pair and instead of the nonce use the hash of

Re: is breaking RSA at least as hard as factoring or vice-versa?

On 4/2/06, Travis H. [EMAIL PROTECTED] wrote: So I'm reading up on unconditionally secure authentication in Simmon's Contemporary Cryptology, and he points out that with RSA, given d, you could calculate e (remember, this is authentication not encryption) if you could factor n, which relates

Re: is breaking RSA at least as hard as factoring or vice-versa?

At 1:41 -0600 2006/04/02, Travis H. wrote: So I'm reading up on unconditionally secure authentication in Simmon's Contemporary Cryptology, and he points out that with RSA, given d, you could calculate e (remember, this is authentication not encryption) if you could factor n, which relates the