is breaking RSA at least as hard as factoring or vice-versa?
Travis H.
solinym at gmail.com
Sun Apr 2 03:41:59 EDT 2006
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 useful direction; namely, that
factoring n is at least as hard as breaking RSA. As of the books
publication in 1992, it was not known whether the decryption of almost
all ciphers for arbitrary exponents e is as hard as factoring.
This is news to me! What's the current state of knowledge?
--
Security Guru for Hire http://www.lightconsulting.com/~travis/ -><-
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