Eugene N on Thu, 14 Apr 2011 16:24:01 +0200

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Re: solving

Exactly, thanks!! 

Chosing k=e will do perfectly.
I was looking for a solution, in wich k != e, and this clouded my reason!! 
Sorry for wasting your time!


2011/4/14 Bill Allombert <>
On Thu, Apr 14, 2011 at 07:41:51AM +0300, Eugene N wrote:
> Hi
> What would be the simpliest way to find k and e, such that:
> X(k-e) = nQ = 0 mod Q
> where: k, e, X, n (- GF(Q), and X and Q are known?
> I am sorry if it seems obvious, i have solved equations aX=b mod Q, but i
> dont remember what action to undertake if b =0...

Assuming Q is prime, then obviously either X=0 mod Q or k=e mod Q.