Removing 1 modular exponentiation

[James Muir]

> 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).

-James