# Homomorphic properties of Paillier

I'm curious about the homomorphic properties of Paillier. So, basically if I have $$\textsf{Dec}(\textsf{sk}, \textsf{Enc}(\textsf{pk}, \alpha) \cdot \textsf{Enc}(\textsf{pk}, \alpha^{-1}))$$, I will get as result $$\alpha + \alpha^{-1}$$. But, does it also mean that if I have $$\textsf{Dec}(\textsf{sk}, \textsf{Enc}(\textsf{pk}, \alpha)^{\textsf{Enc}(\textsf{pk}, \alpha^{-1})})$$, then the result will be $$\alpha \cdot \alpha^{-1}$$, which will basically cancel each other, and will be left with 1?

No, there is no reason that $$\textsf{Dec}(\textsf{sk}, \textsf{Enc}(\textsf{pk},\alpha)^{\textsf{Enc}(\textsf{pk}, \alpha^{-1})})$$ would be $$\alpha\cdot\alpha^{-1}$$, including when we spread $$\bmod N$$ or $$\bmod N^2$$ here and there.
What does apply is: for overwhelmingly most $$\alpha$$ and $$k$$ in $$\Bbb Z$$, it holds that $$\textsf{Dec}(\textsf{sk},\textsf{Enc}(\textsf{pk}, \alpha)^k\bmod N^2)=k\cdot\alpha\bmod N$$. We could take $$k=\alpha^{-1}\bmod N$$ and get $$\textsf{Dec}(\textsf{sk},\textsf{Enc}(\textsf{pk},\alpha)^{(\alpha^{-1}\bmod N)}\bmod N^2)=1$$, but that's not useful anyway, since $$\alpha^{-1}\bmod N$$ reveals $$\alpha\bmod N$$.
• It is not defined in which group it is computed $$\alpha^{-1}$$, and that matters.
• $$\textsf{Dec}(\textsf{sk}, \textsf{Enc}(\textsf{pk}, \alpha) \cdot \textsf{Enc}(\textsf{pk}, \alpha^{-1}))$$ will be $$\alpha+\alpha^{-1}\bmod N$$, which may or may not be $$\alpha+\alpha^{-1}$$.
Given two plaintexts $$\alpha$$ and $$\beta$$, Pailler cryptosystem $$\mathcal{E}$$ homomotphic property is: $$\mathcal{E}(\alpha)\times \mathcal{E}(\beta)=\mathcal{E}(\alpha+\beta)$$. So, $$\mathcal{E}(\alpha)^n=\mathcal{E}(n\alpha)$$. In your example, $$n=\mathcal{E}(\alpha^{-1})$$ and thus after decryption you will have $$\mathcal{E}(\alpha^{-1})\times \alpha$$ ans not $$\alpha^{-1} \times \alpha$$. This is a high level answer, but you need to define the modulo $$N$$ of your system.