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One would put a requirement for the group first, rather than one for just prime $p$. That is, require that taking a logarithm (calculate an index) to be hard in this group. For a multiplicative modulo-$p$ group, this would also require choosing a generator of an order $q$ such that $q$ is a prime, is large enough, and divides $p-1$. Alternatively, it would ...


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If you mean a prime in the neighborhood of $2^{80}$, well, that is incorrect. A prime that small will allow someone to commit to a value, and then reveal another one. A Pederson commitment is a value $g^x h^r \bmod p$, where $g$, $h$ and $p$ are public values, $x$ is the value being committed to, and $r$ is a random value. To reveal the commitment, you ...


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Give a zero-knowledge proof that $y_1 \times y_2$ is a Quadratic Residue. [Extra verbage included because a one line answer feels too brief] If we have $y_1 = x_1^2 t^{b_1}$ and $y_2 = x_2^2 t^{b_2}$, then $y_1 y_2 = (x_1x_2)^2 t^{b_1 + b_2}$. If $b_1 = b_2$, this product is either $(x_1x_2)^2$ (if $b_1 = b_2 = 0$), or $(x_1x_2t)^2$ (if $b_1 = b_2 = 1$), ...



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