is breaking RSA at least as hard as factoring or vice-versa?
Taral
taralx at gmail.com
Sun Apr 2 20:04:01 EDT 2006
On 4/2/06, Travis H. <solinym at 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 at gmail.com>
"You can't prove anything."
-- Gödel's Incompetence Theorem
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