Removing 1 modular exponentiation

Watson Ladd watsonbladd at
Tue Feb 20 02:08:27 UTC 2007

Mike Perry wrote:
> Thus spake Mike Perry (mikepery at
>> Thus spake James Muir (jamuir at
>>>> 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)
> Oh wait, heh, this is an extra modular exponentiation hidden in the
> f^(b/k) step then. The kittens have been put in jeopardy for nothing! ;)
But f^(b/k) requires 1 exponentiation and 1 multiplication. So this does
save some work. The issue is that it is insecure.

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