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Problem: Suppose that $p$ is a prime. Suppose that there is $m \in \mathbb{N}$ such that $p=1+3m$. Define: $$f(x)=(x+1)((x+1)^{2m}+(x+1)^{m}+1) \in \mathbb{Z}/p\mathbb{Z}[x]$$ I would like to prove that if $p \neq 7$ and $p \neq 13$ then $x^m-1 \nmid f(x)$ in $\mathbb{Z}/p\mathbb{Z}[x]$.

Attempt: I observed that $f(x-1)(x^m-1)=x^p-x$. Then I observed that $x^p-x$ has all its roots in $\mathbb{Z}/p\mathbb{Z}$ and that its roots are all distinct. I would like to prove that for $m \geq 6$ then $x^m-1$ and $((x-1)^m-1)$ share at least a root. This is sufficient because then we would have that $x^p-x$ has at least one multiple roots (I mean that it is not simple).

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  • $\begingroup$ Just for confirmation, by $\Bbb{Z}_p$, do you mean the field with $p$ elements or the $p$-adic integers? $\endgroup$ Aug 29, 2021 at 9:59
  • $\begingroup$ Are you somehow trying to use the fact that for any prime $p$ and $m$ with $p\nmid m$, $\Bbb{Q}_p$ has a primitive $m$-th root of unity iff $m\mid p-1$? $\endgroup$ Aug 29, 2021 at 10:04
  • $\begingroup$ @ShubhrajitBhattacharya I mean the field with $p$ elements. $\endgroup$ Aug 29, 2021 at 12:29
  • $\begingroup$ Ohh, then please use the symbol $\Bbb{Z}/p\Bbb{Z}$ instead of $\Bbb{Z}_p$. The latter is a standard notation for the $p$-adic integers and it might be confusing. $\endgroup$ Aug 29, 2021 at 18:08
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    $\begingroup$ @JyrkiLahtonen, exactly! $\endgroup$ Aug 30, 2021 at 10:41

2 Answers 2

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An element $a\in{\bf Z}/p{\bf Z}$ is a root of $x^m-1$ if and only is it is a non-zero cube. If it is also a root of $(x-1)^m-1$, then $a+1$ is also a non-zero cube. This leads to a ${\bf Z}/p{\bf Z}$-rational point $P$ on the projective genus $1$ curve $x^3=y^3+z^3$ with the property that $x,y,z\not=0$. Since there are nine points with $x,y,z=0$, Hasse's theorem implies that $P$ exists whenever $p+1-2\sqrt{p}>9$. In other words, when $p>16$.

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  • $\begingroup$ I appreciate the answer but unfortunately I do not know the “projective genus 1 curve” neither the Hasse’s theorem. I will give you ‘+1’ but I would like to know if there is an answer which uses only elementary algebra’s arguments. $\endgroup$ Aug 31, 2021 at 11:17
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This is an attempt, please point out if there is an error.

Not a complete answer. Too long for a comment

Consider the polynomial $f(x)=(x+1)((x+1)^{2m}+(x+1)^m+1)$

We get two useful identities, inspired by the proceedings of the OP

$$f(x)((x+1)^m-1)=(x+1)^p-(x+1)\tag{1}$$$$f(x-1)(x^m-1)=x^p-x \tag{2}$$ Let, if possible, $x^m-1\mid f(x)$. Then write $f(x)=(x^m-1)g(x)$ for some $g(x)\in\Bbb{F}_p[x]$. Then $$(x^m-1)((x+1)^m-1)g(x)=(x+1)^p-(x+1)\tag{3}$$ $$(x^m-1)((x-1)^m-1)g(x-1)=x^p-x\tag{4}$$

Therefore, to reach a contradiction, as the OP predicted, it's enough to prove that at least any two among the three $((x+1)^m-1),(x^m-1),((x-1)^m-1)$ share a common root (because for any indeterminate T, the polynomial $T^p-T$ in $\Bbb{F}_p[T]$ has exactly $p$ distinct roots). Well, the elements of $\Bbb{F}_p$ which are not a root of any of the three mentioned, are elements of order three in the multiplicative group $\Bbb{F}_p^{\times}$ of order three. If all these roots of them were distinct, then we would have at most $3m+1-3m=1$ element of order $3$. Indeed, let $$x_1,x_2,\ldots,x_m$$ be the roots of $x^m-1$. Then $$x_1+1,x_2+1,\ldots,x_m+1\\x_1-1,x_2-1,\ldots,x_m-1$$ are the roots of of the rest of the two polynomials $(x+1)^m-1$

This argument works in two cases

Case 1 when $(x+1)^m-1$ and $x^m-1$ share a root

Case 2 When $x^m-1$ and $(x-1)^m-1$ share a common root

Case 3 What to do when $(x-1)^m-1$ and $(x+1)^m-1$ share a common root?

and $(x-1)^m-1$ and all these $3m$ elements are in $\Bbb{F}_p$. But this is a contradiction to Sylow's theorem which says that there exists a subgroup of order $3$ and hence at least $2$ elements of order $3$.

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  • $\begingroup$ Unfortunately I think that this doesn't really work: you have to prove that either $(x+1)^m-1$ and $x^m-1$ share a root or that $x^m-1$ and $(x-1)^m-1$ share a root, you can't mix up $(x+1)^m-1$ and $(x-1)^m-1$, since they belong to different identities (i.e. 3 and 4). $\endgroup$ Aug 29, 2021 at 19:27
  • $\begingroup$ @FedericoClerici, I have mentioned that this is an attempt. So, I forgot to write the two cases. Sorry. $\endgroup$ Aug 29, 2021 at 19:46
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    $\begingroup$ Well, now it's fine; it's missing the "hard" part of the problem, i.e. proving that $(x+1)^m-1$ and $x^m-1$ always share a common factor for all $p=3m+1>13$. $\endgroup$ Aug 29, 2021 at 20:38

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