Infact I also expected the same way .. that is, there must be q-1 pairs of solutions .. but I was surprised when I found :

"There are totally p-1 pairs of keys to the equation. so the probablity for finding x & r is : 1/p-1. "

So I asked to help me .. So my problem is still a problem (which I am noting down here again) ..

+++++++++++++++++++++++++++++++++

Let p & q be primes such that q devides p-1.

I want to find r & x when I know y, t in the following equation.

y = r(x+t) mod q,

Here, r,x,t all belong to Zq* { multiplicative group}

I came to know that there are totally p-1 pairs of keys to the equation. so the probablity for finding x & r is : 1/p-1.

How is this true? I want to know the detailed explanation. +++++++++++++++++++++++++++++++++++++++

Thanks,

Bharath

Chris Nash <

chris_nash@...> wrote: Hi there

>I want to find r & x when I know y, t in the following equation.

>y = r(x+t) mod q,

For each value of r in Zq*, there is a unique s such that

rs=1 mod q

Multiply through the equation by s and so

x=ys-t

Hence for each of the q-1 values of r in Zq*, there is a single

solution for x.

Hope that helps

Chris Nash

Lexington KY

UNITED STATES

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