Removing 1 modular exponentiation
James Muir
jamuir at scs.carleton.ca
Mon Feb 19 23:23:29 UTC 2007
> 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
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