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September 20th, 2017, 10:24 AM   #1
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Order of an element in multiplicative group


I am looking for the answer how many possible values has $\displaystyle Ord_{p} (2)$
I suspect that there is only two solutions : $\displaystyle 0.5*(p-1)$ and $\displaystyle p-1$. Could someone prove it?
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September 20th, 2017, 01:17 PM   #2
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I'm afraid your suspicion is wrong: $Ord_{31}(2) = 5 < 0.5 * (31-1) < 31 - 1$. The subsequent mersenne primes are all also counterexamples.

It's easy to see that $Ord_{p}(2) > \log_2(p)$, but I can't think of a straightforward way to improve upon this bound.
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September 20th, 2017, 04:20 PM   #3
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Found this.

They mentioned your observation that the order is greater than $\log_2(p)$. Evidently not much is known about this.
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