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Is it true that for every $n \in \mathbb{N}$, $x^{n}-x-1$ is irreducible in $\mathbb{Z}[x]$?

The standard irreducibility criteria seem to fail.

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    $\begingroup$ Yes, this is true; it is due to Selmer. The complete list of reducible trinomials with $\pm 1$ coefficients has been obtained by Ljunggren, see the reference in Gerry Myerson's answer here: mathoverflow.net/questions/56579/… $\endgroup$ Aug 4, 2014 at 17:42
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    $\begingroup$ @VesselinDimitrov: I think your comment is worth to be posted as an answer. $\endgroup$
    – GH from MO
    Aug 4, 2014 at 18:04
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    $\begingroup$ Keith Conrad gave shorter and more elementary proof of irreducibility of $x^{n}-x-1$ at math.stackexchange.com/a/800835/5814 $\endgroup$ Sep 17, 2015 at 13:54

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This is true; it is due to Selmer. Ljunggren (On the irreducibility of certain trinomials and quadrinomials, Math. Scand. 1960) has obtained the complete list of reducible trinomials with $\pm 1$ coefficients. For more details see Gerry Myerson's answer to this question, which contains a review of Ljunggren's paper: About irreducible trinomials . To this I may add Prasolov's monograph Polynomials as a (probably) more accessible reference; there, you will find a complete treatment of Ljunggren's result.

Added. As Greg Martin notes, the trinomial $x^n-x-1$ in fact has Galois group $S_n$. Osada's paper to which he refers is The Galois groups of the polynomials $X^n + aX^l+b$ (J. Number Theory 25, pp. 230-238, 1987), although the result already appears in an earlier paper by E. Nart and N. Vila, quoted [7] in loc.cit. However, the proof of the maximality of the Galois group uses Selmer's result that the trinomial is irreducible, and does not yield a new proof of irreducibility.

The argument [that Selmer's result implies full Galois group] is presented in section 4.4 of Serre's 1988 course Topics in Galois Theory, see Remark 2 on page 42. You may find this text on the web:

http://www.msc.uky.edu/sohum/ma561/notes/workspace/books/serre_galois_theory.pdf

To summarize the argument, which also works for a wider class of irreducible trinomials, one checks directly that $x^n-x-1$ has at most one double root at each ramified prime. Consequently, all non-trivial inertia subgroups are generated by a transposition. Since $\mathbb{Q}$ has no unramified extensions, the Galois group of any polynomial over $\mathbb{Q}$ is generated by its inertia subgroups, and hence we see that our $G \subset S_n$ is generated by transpositions of $S_n$. Selmer's result implies that $G$ is also transitive, and it is a standard fact (proved in every course in Galois theory) that a transitive subgroup of $S_n$ generated by transpositions must be the full $S_n$.

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    $\begingroup$ For that matter, H. Osada proved that the Galois group of $x^n-x-1$ is always $S_n$. $\endgroup$ Aug 5, 2014 at 7:11
  • $\begingroup$ Sounds awesome! Can you give a reference? $\endgroup$
    – Pablo
    Aug 6, 2014 at 5:15

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