Several questions about Paillier cryptosystem I have several questions concerning the original Paillier cryptosystem as described in Paillier, Pascal (1999).
"Public-Key Cryptosystems Based on Composite Degree Residuosity Classes".
EUROCRYPT. Springer. pp. 223-238. http://www.springerlink.com/content/kwjvf0k8fqyy2h3d/?MUD=MP

Notation:
$p$ - unencrypted message, plaintext
$c$ - encrypted message, ciphertext
$r$ - random factor
$k_{pub}$ - public key
$k_{priv}$ - private key  

I have a Paillier-encrypted ciphertext ($p$) that is no straight encryption but the result of an arbitrary number of various true or mixed homomorphic operations or re-randomizations.


Assuming I know $p$ and the corresponding private key $k_{priv}$.
Am I able to compute the random factor $r$ from this so that a reencryption of $p$ with $r$ would be identical to $c$, i.e. $E(p, k_{pub}, r) = c$ ?
if positive answer to 1, how do I compute $r$?
if positive answer to 1, is it possible that there exists another $r'$ so that a different plaintext $p'$ encrypted with $r'$ would also result in $c$, i.e.
$Enc(p, k_{pub}, r) = c = E(p', k_{pub}, r')$, $p \neq p'$, $r \neq r'$ ?
if positive answer to 3, could this $r'$ be efficiently computed, i.e. could the owner of $k_{priv}$ be trusted if he would provide $r$ to a given $c$ as (Zero knowledge) proof of correct decryption?
Let us briefly recall the Paillier encryption.
Let $k_{pub} = (N = PQ, g)$ be a public key, where $N$ is the RSA modulus.
The secret key is $\lambda = \mathrm{lcm}(P-1,Q-1)$ (or $P,Q$).
The encryption of $p \in \mathbb{Z}_N$ with randomness $r \in \mathbb{Z}_N^*$ is $C = g^p r^N \bmod{N^2}$.

You can verify $\mathbb{Z}_{N^2}^* \simeq \mathbb{Z}_N \times \mathbb{Z}_N^*$.
As Paillier wrote, there is a subgroup $G = \{z \mid z = y^N \bmod{N^2}\}$ of order $\phi = (P-1)(Q-1)$.
There is a bijective mapping $\tau$ from $\mathbb{Z}_N^*$ to $G$ which maps $y$ to $z = y^N \bmod{N^2}$. The inversion $\tau^{-1}(z)$ is computed by $y = z^{N^{-1} \bmod{\lambda}} \bmod{N}$. Roughly speaking, this is the RSA decryption with $e = N$.

Answers:


Yes.
You can find such $r$. Since you know the plaintext $p$, you can compute $d = g^{-p}C = r^N \bmod{N^2}$. By applying $\tau^{-1}$, you can find $r$. (See Section 5.)
No. There is only one $r \in \mathbb{Z}_N^*$. (See Lemma 3.)\
Makes sense since there may be even a huge number of ciphertexts representing the same plaintext but they all have to decrypt to just one plaintext. Hence r is unique as long as en- and decryption are executed using the same pair of keys.
No.