Re: solving

[Eugene N <neverov.biks.07.1@gmail.com>]

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!

Eugene 

2011/4/14 Bill Allombert <Bill.Allombert@math.u-bordeaux1.fr>

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.

Cheers
Bill