Irreducibility of $x^n-x-1$ over $\mathbb Q$ I want to prove that

$p(x):=x^n-x-1 \in \mathbb Q[x]$ for $n\ge 2$ is irreducible.

My attempt.


*

*GCD of coefficients is $1$, $\mathbb Q$ is the field of fractions of $\mathbb Z$, and $\mathbb Z$ is UFD. Hence, $p(x)$ is irreducible over $\mathbb Q$ iff it's irreducible over $\mathbb Z$ (Gauss's lemma).

*Let $m\in \mathbb Z$ such that $\varphi(m)=n$ (Euler's totient). Make reduction of $p(x)$ by modulo $m$. Because of $\overline{x^n}=\overline{x^{\varphi(m)}}=\overline{1}$, we get $\overline{p(x)}=\overline{1-x-1}=\overline{-x}$, which is irreducible. Hence, $p(x)$ is irreducible.
Does this proof is correct?
UPDATE. Thanks to Calvin Lin. My mistake is: not for all $n$ we can find such $m$. OK, but as for the rest, does my proof is correct for such $n$, that $n=\varphi(m)$ for an integer $m$? And can it be some changed for all $n$, i.e. can we find such modulo that $\overline{p(x)}$ is irreducible for every $n\ge 2$?
 A: I will sketch a proof of the irreducibility of $x^n - x - 1$ over ${\mathbf Q}$ for $n \geq 2$ that is simpler than Selmer's argument. I learned of this approach from David Rohrlich, who learned it from Michael Filaseta.
Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$ be nonzero in $F[x]$ for any field $F$. Let $\tilde{f}(x) = x^{\deg f}f(1/x)$ be its reciprocal polynomial:
$\tilde{f}(x) = a_0x^n + a_1x^{n-1} + \cdots + a_{n-1}x + a_n$. If $f(0) \not= 0$ then $f$ and $\tilde f$ have the same degree. Easily if $f = gh$ then $\tilde{f} = \tilde{g}\tilde{h}$, and $\widetilde{cf} = c\tilde{f}$ for any constant $c$.
In terms of factorization over roots, if $f(0) \not= 0$ then
$$
f(x) = c(x-r_1)\cdots(x-r_n) \Longrightarrow \tilde{f}(x) = f(0)c(x - 1/r_1)\cdots(x-1/r_n).
$$
Step 1: Let $f(x) \in {\mathbf Z}[x]$ satisfy (i) $f(0) \not= 0$ and (ii) $f(x)$ and $\tilde{f}(x)$ have no common roots. If $f(x) = g(x)h(x)$ for some nonconstant $g(x)$ and $h(x)$ in $\mathbf Z[x]$, show there is a $k(x)$ in $\mathbf Z[x]$ with $\deg k = \deg f$ such that $f\tilde{f} = k\tilde{k}$ and $k \not= \pm f$ or $\pm\tilde f$. If $f(x)$ is monic and $f(0) = \pm 1$, show you can take $k$ to be monic. (Hint: Use $k = \pm g\tilde{h}$ for a suitable choice of sign.)
Step 2: For $n \geq 2$, show the polynomial $x^n - x - 1$ doesn't have any roots in common with its reciprocal.
Step 3: Let $f(x) = x^n - x - 1$ for $n \geq 2$. Suppose $f\tilde{f} = k\tilde{k}$ for some monic $k \in \mathbf Z[x]$ of degree $n$.  Compare the degree $n$ coefficients on both sides of the equation $f\tilde{f} = k\tilde{k}$ to show that $k$ must be a sum of 3 monomials whose coefficients are all $\pm 1$. Then look at the top 3 nonzero terms on both sides to show $k = f$ or $k = -\tilde{f}$.
Step 4: Combine all the previous steps to deduce irreducibility of $x^n - x - 1$ over $\mathbf Q$.
Exercise: Use the same argument to determine when $x^n + x + 1$, $x^n - x + 1$, and $x^n + x - 1$ are irreducible over $\mathbf Q$. (With a computer you can find in each case that there is a congruence condition on $n$ for it to be reducible, and this turns out to be exactly the condition for $f(x)$ and $\tilde{f}(x)$ to have a common root, which will be either a 3rd or 6th root of unity.)
Update: for another application of this method to prove irreducibility of a family of polynomials, see the MO post here.
A: I doubt an easy proof of the irreducibility exists in general. If $n$ is a prime, then the polynomial is Artin-Schreier and handled easily.
Selmer gave a clever proof in the general case, working explicitly with the roots of the polynomial in $\mathbb{C}$. See E. S. Selmer, On the
irreducibility of certain trinomials, Math. Scand. 4 (1956), 287-302, available here.
