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Let $n \geq 2$ and we work over the rationals.

Question 0: Is a factorisation into irreducibles of polynomial of the form $p_{n,i}(x)=x^n+x^i+1$ known, where $0<i<n$?

Question 1:Is it true that the $n$ such that all $p_{n,i}(x)$ are irreducible is given by the sequence https://oeis.org/A038754? This is true for $n \leq 55$.

Question 2: In case $p_{n,i}(x)$ is irreducible, what is its Galois group?

Mare
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    You don't say anything about the field you are working over. – Marktmeister Dec 27 '20 at 00:20
  • @Marktmeister Thanks, I changed it and said we work over the rationals (or equivalently $\mathbb{Z}$). – Mare Dec 27 '20 at 00:24
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    I doubt there is a general formula for factoring trinomials, or for telling which ones are irreducible. https://math.stackexchange.com/questions/464715/two-results-about-irreducible-polynomials-over-f-2 discusses the situation over some finite fields. – Gerry Myerson Dec 27 '20 at 00:42
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    Here's a collection of results over finite fields. I know you work over the rationals, but if a monic polynomial is irreducible over a finite field, then it is also irreducible over the rationals. https://math.stackexchange.com/questions/1314257/shortest-irreducible-polynomials-over-bbb-f-p-of-degree-n More of the same: https://math.stackexchange.com/questions/2621288/existence-of-an-irreducible-trinomial-over-finite-fields – Gerry Myerson Dec 27 '20 at 00:51
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    Here's another source that deals specifically with irreducibility over the rationals: https://math.stackexchange.com/questions/3629623/irreducibility-of-a-family-of-polynomials – Gerry Myerson Dec 27 '20 at 01:02
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    Maybe this is even better: https://mathoverflow.net/questions/56579/about-irreducible-trinomials – Gerry Myerson Dec 27 '20 at 01:08
  • Have you had a look at any of those links, Mare? – Gerry Myerson Dec 28 '20 at 12:08
  • @GerryMyerson Yes thanks. It seems question 1 has not been observed yet. – Mare Dec 28 '20 at 12:15
  • I think you can work out an answer to Question 1 from the last link I gave and/or from the papers cited there. – Gerry Myerson Dec 29 '20 at 06:24

1 Answers1

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Concerning Question 1:

Let $n=3^rps$, where $p$ is a prime, $p\ge5$, and $s$ is not a multiple of $3$. Let $m=3^rt$ where $ps>t>0$ and $ps+t$ is a multiple of $3$. Let $\zeta=e^{2\pi i/3^{r+1}}$. Then $\zeta^n+\zeta^m+1$ is the sum of the three cube roots of unity, so it's zero, so $x^n+x^m+1$ is divisible by the minimal polynomial for $\zeta$. That polynomial has degree $2\times3^r$, which is less than $n$, so $x^n+x^m+1$ is reducible.

Now let $n=4t$ for some $t$. Then $$x^n+x^{n/2}+1=x^{4t}+x^{2t}+1=(x^{2t}+x^t+1)(x^{2t}-x^t+1)$$ so $x^n+x^{n/2}+1$ is reducible.

This only leaves $n$ of the form $3^r$ and $2\times3^r$ to consider. Suppose $n$ is of one of these forms, and consider $x^n+x^m+1$, $0<m<n$. At this point, we have to bring in the big result of the papers cited at mathoverflow.net/questions/56579/about-irreducible-trinomials. This says that $x^n+x^m+1$ has at most one non-cyclotomic factor, where by cyclotomic factor I mean a polynomial whose zeros are all at roots of unity. That is, $x^n+x^m+1$ is either $P(x)$ or $Q(x)$ or $P(x)Q(x)$, where $P(x)$ is a cyclotomic factor, and $Q(x)$ is an irreducible non-cyclotomic factor. If it's $Q(x)$, then we're done – we've proved it's irreducible, as requested. So, we assume it has a cyclotomic factor $P(x)$, which has a root $\zeta$, which is a root of unity. Then $\zeta^n+\zeta^m+1=0$, a vanishing sum of three roots of unity, which can only be the sum of the three cube roots of unity. From this, I want to conclude that we must have $n=2\times3^r$, $m=3^r$, and $x^n+x^m+1$ is the minimal polynomial for $\zeta$, hence, irreducible, and we're done, but I'm not seeing it right now. I'll try to come back to finish this up in a day or two.

Gerry Myerson
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