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RSA accumulator is an authenticated set built from cryptographic assumptions in hidden-order groups such as $\mathbb{Z}^\times_N$. Any set $S=\{e_1,e_2,\ldots,e_n\}$ can be commited via a commitment $c=g^{\prod_{i\in[n]} \mathsf{H}(e_i)}\in\mathbb{Z}^\times_N$ where $\mathsf{H}:S\rightarrow\mathbb{P}$ is a random oracle that maps the set $S$ to the set of primes $\mathbb{P}$. The memebership proof $\pi_j=g^{\prod_{i\in[n]\setminus \{j\}} \mathsf{H}(e_i)}=c^{1/\mathsf{H}(e_j)}\in\mathbb{Z}^\times_N$ allows anyone to verify $e_j\in S$ if and only if $c=(\pi_j)^{\mathsf{H}(e_j)}$.

My question is what if the commiter commits $c=\pm1$ dishonestly. Then for any odd prime $\mathsf{H}(e_j)$ the membership proof $\pi_j=\pm 1$ (respectively) will fool the verifier. Eliminating $\pm 1$ resolves the issue but does not it eliminate honestly computed commitments too? Or is there any reason to believe that the commitment $c$ can never be $\pm 1$ in the above setting?

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It has been resolved using the discussion of the following question,
Is it possible to have square-free order(s) in $\mathbb{Z}^\times_N$?

In conclusion, the commitment $c=\pm 1$ can be eliminated safely with these three realistic assumptions,

i) $N=(2p+1)(2q+1)$ is a product of two safe primes $p'=2p+1$ and $q'=2q+1$.
ii) $\mathsf{H}$ maps each $e_i$ to odd primes except $p$ and $q$.
iii) The element $g$ is chosen to be $1<g<p'-1$ and $1<g<q'-1$.

Now, we prove that $c\ne\pm 1$ in the above setting.

Denote $d$ as the order of $g$, so $d$ divides the order of $\mathbb{Z}^\times_N$ that is $4pq$. Let, $m=\prod_{i\in[n]} \mathsf{H}(e_i)$ be the exponent. Since, $\mathsf{H}(e_i)$s are odd, $m$ is odd.

If $g^m=1$ then $d$ divides $m$. So, $d$ divides $\mathsf{gcd}(m,4pq)=1$ which implies $d=1$. In $\mathbb{Z}^\times_N$, the only element with order $d=1$ is the identity $1$ itself. But, $g>1$, so, $c\ne 1$.
Similarly, if $g^m=-1$ then $g^{2m}=1$, so $d$ divides $2m$. So, $d$ divides $\mathsf{gcd}(2m,4pq)=2$ which implies $d=2$. It means either $g=-1 \pmod{p'}$ or $g=-1 \pmod{q'}$ or both i.e., $g=-1 \pmod{N}$. Hence, $g\ge p'-1$ or $g\ge q'-1$. But, $g< p'-1$ and $g< q'-1$. Hence, $c\ne-1$.

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