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I recently discovered the benaloh cryptosystem. I am working with the system as it is discribed in the following link: https://en.wikipedia.org/wiki/Benaloh_cryptosystem

However I need some help in order to understand why we need \begin{equation} {gcd(r,(p-1)/r)} \end{equation}

As far as I understand the condition \begin{equation} r \mid (p-1) \end{equation} guarantees the existence of the subgroup of order (p-1)/r which contains the r-residues.

The third condition

\begin{equation} {gcd(r,(q-1))} \end{equation}

allows us to say that there are

\begin{equation} \mid \mathbb{Z}_n^* \mid /r \end{equation} r-residues mod n.

what does the other condition add?

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  • $\begingroup$ You need those conditions to be sure that $r^2 \not | \phi(n)$ $\endgroup$ – ddddavidee Jan 14 '16 at 21:40
  • $\begingroup$ why is this important here? $\endgroup$ – user28082 Jan 14 '16 at 22:10
  • $\begingroup$ otherwise your secret key $x$ would be a $r$-residue. and the decryption would not work. I'll try to write down the whole in a complete answer during next weekend. $\endgroup$ – ddddavidee Jan 14 '16 at 22:16
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$gcd(r,(p-1)/r)$ needs to be $1$ in Benaloh cryptosystem that mean that $(r)$ and $(p-1)$ are preliminary among them.

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    $\begingroup$ sorry i dont understand your answer. Could you please explain what do you mean by preliminary among them? $\endgroup$ – user28082 Jan 15 '16 at 22:57
  • $\begingroup$ it's mean they have to be prime between them e.g 9 and 26 are prime between them. $\endgroup$ – Rafik Hamza Jan 17 '16 at 21:31
  • $\begingroup$ then what you say is not true. $gcd(r,(p−1)/r)$ only says that $r$ and $(p-1)/r$ are prime between them. But surely not that $r$ and $p-1$ are prime between them (which is not the case by the way). This doesn't seem to be an answer to my initial question why we need this condition. $\endgroup$ – user28082 Jan 17 '16 at 22:23
  • $\begingroup$ yep, i got you , u want to know why this condition is set up, the authors of this algorithm set this condition which is gcd(r,(p−1)/r)=gcd(r,p−1)=1 maybe u have just to read the exect article. for all what i know, it's necessary for the algorithm to work correctly, u may want to see the work of "Benaloh's Dense Probabilistic Encryption Revisited" arxiv.org/pdf/1008.2991.pdf $\endgroup$ – Rafik Hamza Jan 19 '16 at 20:58

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