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I have a question.

How can we the quickest to test whether $x$ is a primitive root modulo $n$?

On the Wikipedia page I found information about a possible algorithm.

This algorithm, however, must know all the divisors.

I found the question: Here, but I did not understand much (I need test number $x$).

Can you think of something?

Can this be done in a polynomial time?

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Aurelio
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  • $\mathbb{Z}/(n\mathbb{Z})^$ is not always a cyclic group, so what does your question really* mean? – Jack D'Aurizio Mar 20 '17 at 21:31
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    $x\in\mathbb{Z}/(p\mathbb{Z})^*$ is a generator iff $$x^{\frac{p-1}{q}}\neq 1\pmod{p}$$ for every prime divisor $q$ of $p-1$. – Jack D'Aurizio Mar 20 '17 at 21:32
  • And assuming GRH, there is a generator for $\mathbb{Z}/(p\mathbb{Z})^*$ in the first $C\log^2(p)$ elements for any prime $p$ large enough. – Jack D'Aurizio Mar 20 '17 at 21:33
  • There is no faster way? – Aurelio Mar 20 '17 at 21:35
  • Not that I am aware of. A random element (better if not that random, i.e. better avoiding perfect powers) has a positive probability to be a generator, hence you may just test them till getting a generator. GRH ensures this approach gives a generator pretty soon. – Jack D'Aurizio Mar 20 '17 at 21:42
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    And you have to perform a test for every prime divisor, not every divisor. – Jack D'Aurizio Mar 20 '17 at 21:43
  • Without the factorization of $p-1$ , it is probably impossible to determine the order, unless we use a quantum computer. – Peter Mar 20 '17 at 21:44
  • How large is the prime you have to deal with ? – Peter Mar 20 '17 at 21:46
  • This is a theoretical question. Even the Shor algorithm will be sufficient. Thanks for your attention! – Aurelio Mar 20 '17 at 22:12

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