OSEC

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Re: Breaking RSA: Totient indirect factorization

From: Alexander Klimov (alserkliinbox.ru)
Date: Thu Nov 15 2007 - 02:29:19 CST


On Wed, 14 Nov 2007, gandlf wrote:
> 1) m = p*q -> RSA modulus
>
> [...]
>
> 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).

:-)

So what is the expected running time of your algorithm? For example,
how long it will take on average to factor a 1024-bit modulus?

> Impact
> ------
>
> PKI vendors must change modulus generator algorithms to discard
> totients with lower factors.

You may be interested in ``Are 'Strong' Primes Needed for RSA?'' by
Ron Rivest and Robert Silverman.

--
Regards,
ASK