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seen Jun 29 '12 at 19:42

Jun
21
awarded  Scholar
Jun
21
comment How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
Thanks, that clears it all up.
Jun
21
accepted How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
Jun
20
awarded  Student
Jun
20
comment How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
That makes sense since the order of $G_q$ is q. Nonetheless, I'm still not sure why I can apply the DLP oracle in $G_q$ to $t^2 = (s^2)^z)$. However some research teaches me that that's a property when $p$ is a safe prime, I'll try to prove that. Thanks for all your help :)!
Jun
20
comment How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
This is the best I can do so far : if I can find $z$ in $(t^2) = (s^2)^z mod p$ by using my oracle in $G_q$ (I don't know why I can do this, but let's assume...). Once I find $z$ it should simply equal $y$, since $t = \sqrt{(s^2)^z} = s^z$, and $t = s^y$, hence $z = y$. But that just doesn't feel right to me :/
Jun
20
comment How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
Then how can I use my oracle in the subgroup? I think I'm missing some algebra to comprehend what you're trying to say
Jun
20
comment How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
I'm not seeing it yet, but just for clarity, you mean $(t^2) = (s^2)^z\,mod\,q$, not $mod\,p$, right?
Jun
20
comment How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
I see how to use your hint to get to the first solution (more or less), but I already had that :). I still don't see how to use it to calculate mod (2q+1)
Jun
20
awarded  Editor
Jun
20
revised How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
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Jun
20
asked How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?