5

Note. In this question I am using the following definition of irreducible polynomial: "Polynomial is irreducible if it cannot be factored into polynomials with coefficients in the same domain that have both a positive degree." (See this Wikipedia article)

I am trying to prove the following claim:

Let $P_n(x)=D_n(x,1)-x^n$ , where $D_n(x,1)$ is a Dickson polynomial of the first kind with parameter $\alpha=1$ . Then, $P_n(x)$ is irreducible over $\mathbb{Z}$ for all even $n$ greater than two.

I have verified this claim for all $n$ up to $2000$. Here is the link to the SageMathCell.

Some of my observations:

  1. In case when $n$ is even $P_n(x)$ is even , i.e. $P_n(-x)=P_n(x)$.
  2. In case when $n=2p$, where $p$ is a prime number, vector of the slopes of the Newton polygon of the polynomial $P_n(x)$ with respect to $p$ is of the form: $[-1/(2p-2),-1/(2p-2),\ldots,-1/(2p-2)]$ .
  3. In case when $n$ is even Eisenstein's criterion, Cohn's irreducibility criterion and Perron's irreducibility criterion cannot be applied to polynomial $P_n(x)$ .

Any hints are welcomed.

Pedja
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  • Do you have some idea of the location of the roots of $P_n$, at least for small values of $n$? I'm wondering if some kind of root-location type result might be useful. – Sarvesh Ravichandran Iyer Jan 20 '22 at 04:23
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    Eisenstein's criterion, applied to the reciprocal, settles the case $n=2p$. At least with $p=2,3,5,7$. All the coefficients except the constant term are divisible by $p$. And the leading term is not divisible by $p^2$. – Jyrki Lahtonen Jan 20 '22 at 06:34
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    Forget the "at least". Follows easily from the formula for Dickson polynomials. – Jyrki Lahtonen Jan 20 '22 at 06:41
  • @JyrkiLahtonen But we cannot apply Eisenstein's criterion to $-10x^8 + 35x^6 - 50x^4 + 25x^2 - 2$ for example because $5$ divides $-10$ and $5$ doesn't divide $-2$. Correct me if I am wrong. – Pedja Jan 20 '22 at 07:22
  • @SarveshRavichandranIyer Thanks, but I don't see how that information could be helpful. – Pedja Jan 20 '22 at 07:57
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    Apply Eisenstein to the reciprocal. The sequence of coefficients can be reversed without affecting irreducibility. $5\nmid(-2)$, $5\mid25,-50,35,-10$, $5^2\nmid(-10)$ is enough. – Jyrki Lahtonen Jan 20 '22 at 08:14
  • @JyrkiLahtonen Oh I see. Thanks. – Pedja Jan 20 '22 at 08:15
  • @Peđa I see you figured out your answer , but I was wondering if something like the general principle mentioned here can be used. – Sarvesh Ravichandran Iyer Jan 20 '22 at 10:22
  • @SarveshRavichandranIyer Thank you for the link. – Pedja Jan 20 '22 at 12:29
  • $P_4(x)$ and $P_8(x)$ are reducible over the integers as both can be factored into $2\cdot Q(x)$ where $Q(x)\in\Bbb Z[x]$, and $2$ is not a unit. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Jan 22 '22 at 12:17
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    @TheSimpliFire FWIW: In theory, you are right. For this very reason the definition of an irreducible element of $D[x]$, $D$ an integral domain, is sometimes modified to read: A polynomial in $D[x]$ is irreducible if it is not a constant, and cannot be written as a product of two lower degree polynomials. My source uses such a convention and I copy that to my own coming lecture notes :-) Should $D$ be a field, this modification won't change anything. – Jyrki Lahtonen Jan 22 '22 at 17:14
  • Perhaps, we can take advantage of the fact that $D_n(t+1/t,1)=t^n+1/t^n$, i.e. $P_n(t+1/t)=t^n+1/t^n-(t+1/t)^n$. Not sure how exactly one could use that, though. – richrow Jan 22 '22 at 20:24

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