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About the Naccache-Stern cryptosystem, I have found two different encryption algorithms:

  • In the original paper from Naccache and Stern, the encryption step is performed by calculating $c = g^m \hspace{2mm} (mod \hspace{2mm} n)$, with $m$ being the plaintext, with $\sigma$ being optionally kept private. Then decryption is performed by applying some series of congruences mod n, which are easy to understand.
  • However, on Wikipedia (https://en.wikipedia.org/wiki/Naccache%E2%80%93Stern_cryptosystem), and many other sites, such as Crypto Wiki, the encryption step is performed differently: previously, an element $x \in \mathbb{Z}_{n}$ is randomly selected, and the plaintext $m$ is encrypted as $c = x^{\sigma}g^{m} \hspace{2mm} mod \hspace{2mm} n$, forcing $ \sigma$ to be part of the public key. Now, as for the explanation on why decryption works, they apply the following congruence: $$ c_{i} = c^{\phi(n)/p_{i}} \equiv x^{\sigma\phi(n)/p_{i}}g^{m\phi(n)/p_{i}} \hspace{2mm} mod \hspace{2mm} n \equiv g^{m\phi(n)/p_{i}} \hspace{2mm} mod \hspace{2mm} n $$ so basically, the same process as in the paper is applied, but with this different step, which I don't understand why is correct. Eventually, we get $c_{i}$ is an element of the cyclic subgroup generated by $g$.

So, my questions here are:

  • Is this congruence true? If so, why, if $x$ was randomly chosen?
  • Does this mean that $x$ is also an element of the cyclic subgroup generated by $g$?
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1 Answer 1

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Yes, the congruence is true it follows because $p_i$ divides $\sigma$ and so $\sigma\phi(n)/p_i\equiv 0\pmod{\phi(n)}$. The blinding value $x$ is chosen in order to provide ciphertext indistinguishability. By multiplying by a random $x^\sigma$, any given plaintext can encrypt in multiple ways. Showing that two ciphertext correspond to the same plaintext is equivalent to showing that their quotient mod $n$ is a $\sigma$th power (and hence solving the $\sigma$th power residuacity problem).

It is not necessary for $x$ to be in the same subgroup as that generated by $g$, in particular $x$ could be a non-residue for both $p$ and $q$ and still work as a blinding factor. This is important as it should be hard to distinguish elements of $\langle g\rangle$ from elements $-\langle g\rangle$ by quadratic residuacity whereas the sender can cheaply pick a blinding factors without this information.

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  • $\begingroup$ Thank you so much for your answer! Just to see if I got it right, since $\sigma\phi(n)/p_{i} \equiv 0 \hspace{2mm} mod \hspace{1mm} \phi(n)$, then that power of $x$ is congruent to 1 modulo $n$? $\endgroup$
    – hectorvr14
    Commented Dec 19, 2023 at 15:49
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    $\begingroup$ @hectorvr14 that's right and it will be true for any $x$ coprime to $n$ rather than just the powers of $g$. $\endgroup$
    – Daniel S
    Commented Dec 19, 2023 at 15:52

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