# Finite fields: Is multiplicative order of $x^p - x - 1$ equal to $\frac{p^p-1}{p-1}$ over GF(p)? [duplicate]

Possible Duplicate:
Multiplicative order of zeros of the Artin-Schreier Polynomial

I will be grateful for any reference to some literature on the following question (to the best of my knowledge the answer is not known):

Is multiplicative order of roots of $x^p - x - 1$ equal to $\frac{p^p-1}{p-1}$ over $GF(p)$?

Brief Note: $P(x) = x^p - x - 1$ is irreducible. (Proof: Observe that composition with $x+1$ does not change $P$, so if polynomial Q(x) is a factor than Q(x+a) is also a factor.) Norm($x$) = 1, so the maximum possible order for $x$ is $\frac{p^p-1}{p-1}$.

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## marked as duplicate by Qiaochu Yuan, Todd Trimble♦, Felipe Voloch, Andres Caicedo, Anton GeraschenkoDec 26 '10 at 17:52

I believe you mean the multiplicative order of a root of $x^p-x-1$... –  Zev Chonoles Dec 26 '10 at 6:51
Sure! (just saving space) –  Alexey Lvov Dec 26 '10 at 6:56
By the way: To the best of my knowledge currently there are no known nontrivial series of irreducible polynomials over finite fields with known mult. order at all. I do not expect a direct answer to this question, rather a reference to some relevant literature. –  Alexey Lvov Dec 26 '10 at 7:03
Duplicate: mathoverflow.net/questions/46133/… –  Qiaochu Yuan Dec 26 '10 at 9:47
It's better to close it as a duplicate than to delete it. That way, if somebody has the same question in the future, they have a better chance of finding the right thread (i.e. different wordings will effectively point to the same place). –  Anton Geraschenko Dec 26 '10 at 17:55