New answers tagged implementation
0
Here's a guess: when you ask to compute h ** (a * b), it actually computes h ** (a * b mod q), where $q$ is the order of $g$ in $G1$.
If that is the case, well, a random element from $G2$ is not going to be a divisor of $q$ - any element that is the result of a pairing operation will be; however most elements of $G2$ cannot be the result of a pairing ...
Top 50 recent answers are included
