If I know an element $x$ in a group, and it's inverse $x^{-1}$, can I guess the modulo, or with a probability?

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    $\begingroup$ The question makes sense in the group $\Bbb Z_n^*$ for unknown $n$. Hint: what do you know about the ordinary product $x^{-1}\,x$ ? $\endgroup$ – fgrieu Jan 15 '19 at 7:49
  • $\begingroup$ @fgrieu The product $x^{-1}x= nk+1$ for an integer $k$, with the bits of $n$ increase and for larger $x$, it will be more complex? But even more complex, it will always get the modulo in polynomial time? $\endgroup$ – w.qi Jan 15 '19 at 13:04

As said in the comments above, you have that $x \cdot x^{-1} = 1 \mod n $. Another way to say that is $k \times n=(x \cdot x^{-1}) -1$

So no, you can't always find the exact value $n$ from $x$ and $x^{-1}$.

Take $x=12 \ $ and $x^{-1}=17$. You have $x \cdot x^{-1} = 204$. So $n$ is a divisor of $204-1=203$, meaning, $ n \in \{29,203\}$ (other divisors are smaller than $\max(x,x^{-1})$ so they can't be $n$)

The only cases where you can find $n$ is when the only divisor of $kn$ greater than $x$ and $x^{-1}$ is $kn$ itself. In this case, $n=kn=(x \cdot x^{-1}) -1 $

  • $\begingroup$ You are right, so can I think it is a difficult factoring problem as the $n$ is large enough? $\endgroup$ – w.qi Jan 15 '19 at 14:39
  • $\begingroup$ Can you please give some details about your problem ? What I understand is, given $x$ and $x^{-1}$, can an adversary retrieve $p$ and $q$ such that $pq=n$ ? Is it correct ? $\endgroup$ – Bissi Jan 15 '19 at 14:49
  • $\begingroup$ I'm sorry, I mean that while $x^{-1}x$ is large, so the $kn$ is large enough, so it can be seen as a hard problem: from $kn$ to find $k$ and $n$? $\endgroup$ – w.qi Jan 15 '19 at 15:00
  • $\begingroup$ What would be the use of the $x$, $x^{-1}$ and $k$ values ? The public key would be $(x,x^{-1})$ and the secret one $(k,n)$ ? Your question need a bit more of contextualization. Where did you get this idea from ? $\endgroup$ – Bissi Jan 15 '19 at 15:22
  • $\begingroup$ Note: you can add: if we know more than one pair we can find more about the $n$. $\endgroup$ – kelalaka Jan 15 '19 at 18:39

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