0

When constructing irreducible polynomials in $\mathbb{F}_2[x]$, one notices that $x^2+x+1$, $x^3+x+1$, and $x^4+x+1$ are irreducible, but $x^5+x+1=(x^3+x^2+1)(x^2+x+1)$ is reducible in $\mathbb{F}_2[x]$. It is natural to ask for which positive integers $n$, the trinomial $x^n+x+1$ is irreducible in $\mathbb{F}_2[x]$. The list of such $n$ is tabulated in OEIS as the sequence A002475.

I understand that there is no reasonable pattern we can predict in this sequence of values of $n$. However, I wonder whether the following questions have been studied in the literature:

Problem 1. Are there infinitely many positive integers $n$ such that $x^n+x+1$ is an irreducible polynomial in $\mathbb{F}_2[x]$?

Problem 2. Are there infinitely many positive integers $n$ such that $x^n+x+1$ is a reducible polynomial in $\mathbb{F}_2[x]$?

Thank you!

user26857
  • 53,190
Prism
  • 11,782
  • @user26857 Wow, that is nice! Thank you. Can you say a few words about the proof of this result? Feel free to expand on your comment in an answer box. – Prism Oct 13 '23 at 16:50
  • 2
    No need to expand, this has been asked very often, see here for example and the duplicates linked to to. – Dietrich Burde Oct 13 '23 at 16:52
  • 2
    Cf. https://www.sciencedirect.com/science/article/pii/S0019995870902640 – Travis Willse Oct 13 '23 at 16:56
  • @Dietrich Thank you! I see that Problem 2 has indeed been discussed at great length on the website, although I have yet to see Problem 1 being addressed (but I will look more closely). – Prism Oct 13 '23 at 17:03
  • 2
    Well, even more when $n\equiv 2\pmod{3}$ we have $x^n+x+1$ is divisible by $x^2+x+1$ (over $\mathbb{Z}$!) because $x^3-1=(x^2+x+1)(x-1)$ allows you to "subtract 3 from $n$". So you can only look for $n\equiv 0,1\pmod{3}$ in 1 as a start. – user10354138 Oct 13 '23 at 17:08
  • 1
    For more examples, $x^6+x+1$ and $x^7+x+1$ are irreducible. Here I explain why $x^{127}+x+1$ and $x^{2^{127}-1}+x+1$ are irreducible. My guess would be that it is not known whether there are infinitely many irreducible polynomials of this form. A statistical argument suggest that to be the case though. After all, a random degree $n$ polynomial is irreducible with roughly the probability $1/n$. Here we have a single polynomial of degree $n$ for all $n$, and the series $\sum_{n=1}^\infty1/n$ diverges, so.... – Jyrki Lahtonen Oct 13 '23 at 18:21