Re: Removing 1 modular exponentiation

[Mike Perry <mikepery@xxxxxxxxxx>]

Thus spake James Muir (jamuir@xxxxxxxxxxxxxxx):

> >Problem is: (g^X)^k = g for some given k. Find X equivalent to 1/k.
> >
> >Rewrite as (g^k)^X = g
> >
> >Seems like you need to take the Discrete Log of both sides to get your
> >X=1/k value. This is hard.
> 
> Since we are working modulo p and we know that g is a generator of ZZ_p 
> its order is p-1.  So, to find X above you just need to solve:
> 
> 	k*X == 1 (mod p-1)
> 
> This can be done efficiently using the extended Euclidean Algorithm 
> (provided that gcd(k,p-1)=1).

Doh! You're right. The kittens are saved! (For now)

-- 
Mike Perry
Mad Computer Scientist
fscked.org evil labs