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It's not possible because if $B$ knows $b$ and $g^{ab}$, then she can compute $\left(g^{ab}\right)^{b^{-1} \bmod N} = g^a$ (where $N$ is the order of $g$, which is normally prime so that $b^{-1} \bmod N$ exists). EDIT: As noted by poncho in the comments, this assumes we are in a typical discrete logarithm setting, where the order of the group is prime to ...


3

You're right in that there's little chance you can break the logarithm in a well-chosen 512 bit group (using a home computer, in reasonable time — as pointed out by SEJPM, it is possible investing some time and a good amount of money). However, in your case, the parameters are bad: The order of $(\mathbb Z/p\mathbb Z)^\ast$, that is $p-1$, is a smooth ...



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