# Multiplicative order of zeros of the Artin-Schreier Polynomial

This question was asked on NMBRTHRY by Kurt Foster:

If $p$ is a prime number and $\mathbb{F}_p$ the field of $p$ elements, the zeroes of the Artin-Schreier polynomial

$x^p - x - 1 \in \mathbb{F}_p[x]$

obviously have multiplicative order dividing $1 + p + p^2 + \dots + p^{p-1} = (p^p - 1)/(p-1)$ (express the norm as the product of the compositional powers of the Frobenius map)

Once upon a time, long long ago, I read that it had been conjectured (by Shafarevich IIRC) that this is the exact multiplicative order for every prime $p$. Can anyone supply a reference?

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I've never seen it ascribed to Shafarevich, but it is an old question. As a question, it equivalent to determining the period mod p of the sequence of Bell numbers discussed, e.g. in:

Levine, Jack; Dalton, R. E. Minimum periods, modulo p, of first-order Bell exponential integers. Math. Comp. 16 1962 416–423.

But they refer to even older papers. Any conjecture is wishful thinking since we can't get past $p=29$ or so with current technology. (Edit: As Kevin points out in the comments, I am seriously underestimating current technology, so this comment applies only to last century.)

Incidentally, I proved that the order is at least $2^{2.54p}$ in JTNB 16 (2004) 233-239.

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@Felipe: I don't understand what you mean by "we can't get past $p=29$ or so with current technology". I just checked that $(31^{31}-1)/30$ was prime in a gazillionth of a second on a computer, so aren't I done for $p=31$? And if it hadn't been prime then I would have just checked to see if the order of $t$ divided $N/q$ for all prime divisors $q$. Aah---are you saying that we can't get past "the point after which we can't factor random numbers of size about $p^p$"? – Kevin Buzzard Nov 15 '10 at 20:08
@Kevin Your "Aah" is correct :-) – Felipe Voloch Nov 15 '10 at 20:40
I think $p=67$ is getting tough for my laptop. On the other hand I suspect that there are people who do spend their time trying to factor numbers of the form $a^b\pm1$ (the Cunningham Project?) and so one might be able to look up a lot more answers there. – Kevin Buzzard Nov 15 '10 at 21:04
Hmm, seems they like to have $a\leq 12$\ldots – Kevin Buzzard Nov 15 '10 at 22:55

The current computational status is that this is known for all p < 126 and also for p = 137, 149, 157, 163, 167 and 173. See Peter L. Montgomery, Sangil Nahm and Samuel S. Wagstaff Jr., "The period of the Bell numbers modulo a prime", Math. Comp., 79, 271, July 2010, 1793-1800. The method used requires a factorization of (p^p-1)/(p-1) which is hard for larger p :-)

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Thanks for the reference! Here's a link to the paper on Wagstaff's home page: cerias.purdue.edu/assets/pdf/bibtex_archive/2010-01-report.pdf – Victor Miller Nov 16 '10 at 15:52