The Paillier CryptoSystem has a public key that $(g,n)$ and the private key which can be exclusive to $\lambda$, where the decryption scheme is:

$m = L(c^\lambda \bmod n^2)/L(g^\lambda \bmod n^2) \bmod n$

Since $1/L(g^\lambda \bmod n^2)$ is fixed and always needed for decryption, it is usually computed once and denoted as $\mu$.

What information does $\mu$ leaks about $\lambda$? Because at the end of the day, even if I have $\mu$, I cannot decrypt. i.e. Can I get $\lambda$ from $\mu$?

A Side Note on the way $\mu$ is constructed, that I think proves the correctness of the assumption:

\begin{align} g &= (1+n)^\alpha \cdot \mathcal{B}^n \pmod{n^2} & & \text{$g$ in the $n^{\text{th}}$ root form} \\ g^\lambda &= (1+n)^{\alpha\lambda} \cdot \mathcal{B}^{n\lambda} \pmod{n^2} & &\text{so base on carmichael's theorem} \\ g^\lambda &= (1+n)^{\alpha\lambda} \pmod{n^2} & & \text{again, based on $n^{\text{th}}$ root rule}\\ g^\lambda &= 1+n\alpha\lambda \pmod{n^2}& & \\ L(g^\lambda) &= \alpha\lambda \pmod{n^2}& &\\ \mu &= 1/\alpha\lambda \pmod{n^2} & \end{align} So, since it is impossible to get $\alpha$ given $g$, the main complexity of the encryption scheme itself, and the last equation is a function of two variable, and there is no way to find either variable.

  • $\begingroup$ Could you give your reference? The original paper only replaces $\lambda$ with $\alpha$ on page 10. From the article; Note that this time, the encryption function's trapdoorness relies on the knowledge of $\alpha$ (instead of $\lambda$) as secret key. $\endgroup$
    – kelalaka
    Oct 6, 2019 at 21:10
  • $\begingroup$ Aside from the technical details, The question still remains, given $1/L(g^{\alpha} \ mod \ n^2)$, which is fixed, can I get $\alpha$ $\endgroup$ Oct 6, 2019 at 21:20
  • 1
    $\begingroup$ Huh? You were asking about $\lambda$ in the question, not $\alpha$, right? $\endgroup$
    – Maarten Bodewes
    Oct 7, 2019 at 9:02
  • $\begingroup$ yes, $\lambda$ (The decryption key). $\endgroup$ Oct 7, 2019 at 9:15

1 Answer 1


What information does $\mu$ leak about $\lambda$?

The safe assumption is: all. It must be assumed that knowledge of $\mu$, together with the public key, allows computing $\lambda$ (which allows decryption and factorization of $n$).

At least, that holds in Paillier's scheme as described in Jonathan Katz and Yehuda Lindell's Introduction to Modern Cryptography (section 13.2.2). In this we have $p$ and $q$ of equal size, $g=n+1$, $\lambda=(p-1)(q-1)$, and $\mu=\lambda^{-1}\bmod n$. It follows that $\lambda=\mu^{-1}\bmod n$, allowing computation of $\lambda$ from $\mu$ and $n$ (using e.g. the extended Euclidean algorithm, which is inexpensive).

While that does not immediately tell how to compute $\lambda$ from $\mu$ and $n$ in Paillier's scheme as in the question, that's enough to show that we can't safely reveal $\mu$.

  • $\begingroup$ Ok, but as far as I know, The fast method is using extended Euler's method that needs either $\phi(n)$ or $\lambda(n)$ which are very hard to compute. Based on this the problem remains hard. $\endgroup$ Oct 7, 2019 at 18:07
  • $\begingroup$ Another hard way is to use Carmichael's Theorem since $\mu$ is relatively prime with n, $\mu^\lambda = 1\ mod\ n$, which is again not feasible! $\endgroup$ Oct 7, 2019 at 18:10
  • $\begingroup$ Hum, it is rather $\mu\,\lambda\equiv1\pmod n$, at least in the variant that I present. $\endgroup$
    – fgrieu
    Oct 7, 2019 at 18:49
  • $\begingroup$ Why do you need $\phi(n)$ or $\lambda(n)$ in order to compute the extended Euclidean algorithm on $(\mu, n)$ to find the Bézout coefficients $(\lambda, k)$ satisfying $\lambda \mu + k n = 1$? (Except, of course, insofar as $\lambda$ means $\lambda(n)$ here and revealing $\mu$ is a way to leak $\lambda(n)$.) $\endgroup$ Oct 7, 2019 at 19:21
  • $\begingroup$ It was a response to the OP, yes. @WalidAshraf See above. $\endgroup$ Oct 8, 2019 at 13:41

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