Let $p$ be a prime number and denote $Q=(p^p-1)/(p-1)$. Is it true that $x^Q-x-1$ irreducible over $GF(p)[x]$?
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3$\begingroup$ This reads like a homework question, how did you come across it, if not? $\endgroup$David Roberts– David Roberts ♦2025-04-27 03:29:58 +00:00Commented Apr 27 at 3:29
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2$\begingroup$ @DavidRoberts To see that there is no root in $GF(p)$ is very easy. But is proving irreducibility really such an easy homework like assignment? $\endgroup$Peter Mueller– Peter Mueller2025-04-27 07:10:26 +00:00Commented Apr 27 at 7:10
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1$\begingroup$ @PeterMueller I don't know, I'm saying it looks to me like a homework question, but I'm not an algebraist, so I'm not an authority. That's why if I left open the option that it might have arisen in the course of research and asked the OP to explain more. $\endgroup$David Roberts– David Roberts ♦2025-04-27 07:52:19 +00:00Commented Apr 27 at 7:52
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$\begingroup$ I should say that this is not an easy question, if the answer is positive, then we can get a lot of interesting results.@everyone $\endgroup$xwcao– xwcao2025-04-27 11:44:31 +00:00Commented Apr 27 at 11:44
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1$\begingroup$ @xwcao I believe that your question would get more attention and interest if you tell more about the background and the "lot of interesting results" one would get from a positive answer. $\endgroup$Peter Mueller– Peter Mueller2025-04-28 07:48:39 +00:00Commented Apr 28 at 7:48
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