Division in paillier cryptosystem Is division possible in the Paillier Cryptosystem? i.e. given a the cipher-text $C$ of an integer $M$ the plain-text divisor $D$, and only the public key, can one compute the cipher-text of $M/D$ ?
Yes (and always).

Given $\mathsf{Enc}(a)$ and $b$, you can compute $\mathsf{Enc}(a \cdot b^{-1} \bmod{n})$ by simply computing $\hat{b}=b^{-1} \bmod{n}$ and $Enc(a)^\hat{b} \bmod{n^2}$.

Paillier encryption is built on the bijeective mapping from $(x,y)\in \mathbb{Z}_n \times \mathbb{Z}_n^*$ to:

$E_{g,n}(x,y)=g^x y^n \bmod{n^2}$. 

Generator $g$ is chosen to enforce that $x$ remains in $\mathbb{Z}_n$ (i.e., there exists a $y$ that makes $x$ unique in $\mathbb{Z}_n$). Specifically, it has order $\alpha n$ for a positive $\alpha &lt; \mathrm{LCM}(p-1,q-1)$. A simple optimization is to set $\alpha=1$ by choosing $g=n+1$.

Regardless of variant, in the encryption scheme, $x$ is the message and thus $\mathbb{Z}_n$ is the message space. Contrary to the other answers, this allows $b$ to be generally inverted $\bmod{n}$ (except when 0 or in the negligible case that it is either $p$ or $q$).



Given the confusion, I did a sanity check in Mathematica to make sure (I picked small safe primes but you can set them using RandomPrime): 

p = 23;
q = 83;
n = p*q;
lam = LCM[p - 1, q - 1];

L[u_] := Quotient[u - 1, n];

Trap[y_] := L[PowerMod[y, lam, n^2]];

g = RandomInteger[{1, n}];
While[GCD[Trap[g], n] != 1, g = RandomInteger[{1, n}]]

Enc[m_] := Module[{r, c},
  r = RandomInteger[{1, n - 1}];
  c = Mod[PowerMod[g, m, n^2]*PowerMod[r, n, n^2], n^2]
  ]

Dec[c_] := Mod[Trap[c]*PowerMod[Trap[g], -1, n], n];

m = RandomInteger[{1, n - 1}];
s = RandomInteger[{1, n - 1}];
si = PowerMod[s, -1, n];
c = PowerMod[Enc[m], si, n^2];
Dec[c] == Mod[m*si, n]




This answer is the result of a crash course in Paillier. I typically work in the discrete logarithm setting. Feel free to edit if there are any errors (you probably do know more than me).