# $x^m-1 \nmid f(x)$ in $\mathbb{Z}/p\mathbb{Z}[x]$ where $f(x)=(x+1)((x+1)^{2m}+(x+1)^{m}+1)$

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).

• Just for confirmation, by $\Bbb{Z}_p$, do you mean the field with $p$ elements or the $p$-adic integers? Aug 29, 2021 at 9:59
• 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 $m\mid p-1$? Aug 29, 2021 at 10:04
• @ShubhrajitBhattacharya I mean the field with $p$ elements. Aug 29, 2021 at 12:29
• 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. Aug 29, 2021 at 18:08
• @JyrkiLahtonen, exactly! Aug 30, 2021 at 10:41

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 $$p>16$$.

• 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. Aug 31, 2021 at 11:17

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 $$3$$.

• 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). Aug 29, 2021 at 19:27
• @FedericoClerici, I have mentioned that this is an attempt. So, I forgot to write the two cases. Sorry. Aug 29, 2021 at 19:46
• 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 $p=3m+1>13$. Aug 29, 2021 at 20:38