I was reading the paper One-way Accumulators: A Decentralized Alternative to Digital Signatures by Benaloh and de Mare [link], and in section 4.2, they say that given $z\in (\mathbb{Z}/n\mathbb{Z})^*$ and a set $$\{(z^{1/m_1}, m_1),(z^{1/m_2}),\dots, (z^{1/m_k}, m_k)\}$$ with $m_i \in \mathbb{Z}$, computing $z^{1/r}$ for $r \in \mathbb{Z}$ is hard, which I get since the order of $(\mathbb{Z}/n\mathbb{Z})^*$ is unknown.

However, they say that the problem becomes easy if $r \mid m_1m_2\cdots m_k$, and I do not get how it becomes easy.

For the case $k=1$, we have $z, (z^{1/m_1}, m_1) \text{ and } r \mid m_1$, this is easy since $m_1 = k\cdot r$ for $k\in \mathbb{Z}$ and $(z^{1/m_1})^k = z^{1/r}$

For the case $k=2$, when we have $ m_1 m_2 = k^\prime \cdot r$, we can compute $z^{1/m_1}z^{1/m_2} = z^{(m_1+m_2)/m_1m_2}$ and get $z^{(m_1+m_2)/r}$. However, I do not know how to get rid of the value $m_1+m_2$ to obtain $z^{1/r}$

The case $k>2$ seems even harder.

Any hint on how to move forward will be welcomed

  • $\begingroup$ HINT: Bezout's identity $\endgroup$
    – Daniel S
    Aug 9 at 7:51
  • $\begingroup$ Would it be better to have the title say "r-th root" in $\mathbb Z/n\mathbb Z$? For consistency with the body of your Question. $\endgroup$
    – hardmath
    Aug 9 at 12:40

1 Answer 1


The key is to use extended euclidean algorithm. Here's a numerical example, see if you can figure it out yourself.

Given $z, a := z^{1 / 3}, b := z^{1 / 5}$, we can compute $z^{1 / 15} = a^{-1} \cdot b^2 = z^{-5 / 15 + 6 / 15}$.


  • $\begingroup$ but this only works if for $z^{1/m_1}, z^{1/m_2}$, $m_1$ and $m_2$ are coprime. However, in the paper, they didn't mention something about the gcd of the $m_i$ $\endgroup$
    – vxek
    Aug 9 at 14:24
  • $\begingroup$ @vxek nope, read about extended euclidean algorithm again $\endgroup$
    – Gareth Ma
    Aug 9 at 15:08
  • $\begingroup$ it seems like using extended euclidean algorithm/ Bezout's identity, I can find $z^{1/r}$ if $r = \dfrac{m_1m_2}{\gcd({m_1,m_2})}$ But I am still struggling to see how the $r$-th root is easy if $r$ is a random integer that divides $m_1m_2$ $\endgroup$
    – vxek
    Aug 9 at 15:59
  • $\begingroup$ @vxek Ah, so you are asking how is computing $z^{1 / 24}$ easy when given $z^{1 / 4}$ and $z^{1 / 6}$. Yes, you are correct that it is hard here. The paper assumes that $m_i$ are mutually coprime e.g. when they're all primes. Stop using symbols :) $\endgroup$
    – Gareth Ma
    Aug 9 at 17:22

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