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I'm lacking quite some mathematical knowledge here, but could anyone please explain to me why the Paillier cryptosystem is still (additive/multiplicative) homomorphic despite introducing a random value?

$[\![w_1w_2]\!]_g$ = $[\![w_1]\!]_g + [\![w_2]\!]_g\ \bmod\ n$

Does it have to do with the value being sampled from $\mathbb{Z}^*_n$? Is the only requirement for the above statement to hold that $w_1, w_2 \in \mathbb{Z}^*_{n^2}$?

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  • $\begingroup$ Welcome to crypto.SE! Note that in the question's equation, the product is modulo $n^2$, and the sum is modulo $n$. Another way to define the homomorphism in Paillier's cryptosystem is $\forall (m_1,m_2,r_1,r_2),\quad D(E(m_1,r_1)\cdot E(m_2,r_2) \bmod n^2)\;=\;m_1+m_2\bmod n$. where $m_i$ are the messages, and $r_i$ the randomness used in encryption. This is not an answer. $\endgroup$ – fgrieu♦ Dec 16 '19 at 13:40
  • $\begingroup$ Thanks, in the paper only the sum contains a modulo though? $\endgroup$ – chibi03 Dec 16 '19 at 14:15
  • $\begingroup$ [updated] In this paper, ciphertexts are in $\Bbb Z_{n^2}^*$ (the subset of ,$\Bbb Z_{n^2}$ with elements are coprime with $n$), thus multiplication of ciphertexts implicitly reduce modulo $n^2$. Definitely, one can reduce modulo $n^2$ after multiplication of ciphertexts. That's a good idea since it reduces size, and a practical implementation could/should enforce ciphertext in $[0,n^2)$. Plus, in some protocols, the reduction could help prevent traceability. This is not an answer either. $\endgroup$ – fgrieu♦ Dec 16 '19 at 22:11
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The difficult part about understanding the Paillier cryptosystem is to understand what the $L$ function in the cryption actually does and why it works. The good news is: To understand the homomorphism, that detail can be put on hold.

The best way to understand homomorphism is to have a close look at the encryption function. Here it is: $$ E(m) = r^n g^m \mod n^2$$

If we take this apart, we can see:

  • The modulus is $n^2$. That means we operate on a multiplicative group with order $n \cdot \lambda$ (with $\lambda=lcm(p-1,q-1)$).
  • This means, there exist subgroups of order $p,q,n,pa, qa,$ and $na$, where $a$ is any proper divisor of $\lambda$.
  • The random number $r$ has an order equal to one of those subgroup orders. It could be divisible by $p,q$ (actually, it's overwhelmingly likely to be the order $n\lambda$). However, if we take that to the $n$-th potency, then the order of $r^n$ is not divisible by either $p$ or $q$, it can only be a divisor of $\lambda$. That is a property that can be used.

Now for the decryption, it is enough that somehow the decryption function can:

  • Remove any masking factor which is such an $r^n$, without knowing which $r$ it is.
  • Get $m$ back from the remaining number $g^m$, which is basically a discrete logarithm in this special kind of group.

Now for the homorphism, just encrypt two messages, build their product and do some very basic transoformations:

$$ E(m_1) = c_1 = g^{m_1}{r_1}^n$$ $$ E(m_2) = c_2 = g^{m_2}{r_2}^n$$ $$ c_1 c_2 = g^{m_1}{r_1}^n g^{m_2}{r_2}^n = g^{m_1+m_2} (r_1r_2)^n$$

Clearly, this is just the same as using the encryption method with the message $m_1+m_2$ and the random number $(r_1r_2)$. And for that the decryption works just like for a single ciphertext.

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