Re: Breaking RSA: Totient indirect factorization
On Wed, Nov 14, 2007 at 10:59:42PM +0100, gandlf wrote:
..
> Algorithm
> ---------
>
> - Repeat "a = a^n mod m" with n from 2 to m, saving all the results in
> a table until a == 1 (Statement 4).
Do I understand correctly that this step of your proposed algorithm
can identify the private key corresponding to (e.g.) a 1024 bit public
key, but only by doing on the order of Sum(2..2^1024) = ~ 2^1025
modular exponentiations and storing the results? If so, that would
come to approximately 1E307 modular exponentiation operations.
Divide that out by (for example) teraflops and the expected lifetime
of the universe, and I don't think you will get a pleasing result.
-- Clifton
--
Clifton Royston -- cliftonr@xxxxxxxxxxxxxxxxxx / cliftonr@xxxxxxxx
President - I and I Computing * http://www.iandicomputing.com/
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