[21710] in cryptography@c2.net mail archive
Re: is breaking RSA at least as hard as factoring or vice-versa?
daemon@ATHENA.MIT.EDU (Taral)
Mon Apr 3 00:31:00 2006
X-Original-To: cryptography@metzdowd.com
X-Original-To: cryptography@metzdowd.com
Date: Sun, 2 Apr 2006 19:04:01 -0500
From: Taral <taralx@gmail.com>
To: "Travis H." <solinym@gmail.com>
Cc: Cryptography <cryptography@metzdowd.com>
In-Reply-To: <d4f1333a0604012341s72dbf45sde8dfbd7a5d8750f@mail.gmail.com>
On 4/2/06, Travis H. <solinym@gmail.com> 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 two.
This implication runs both ways. Given d and e (and pq), one can
compute p and q. Proving this is an exercise left to the reader.
--
Taral <taralx@gmail.com>
"You can't prove anything."
-- G=F6del's Incompetence Theorem
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