Consider the following family of polynomials for every integer $d > 0$: $$P_d(X) = X^{d+1} - X^d - 1$$ I was wondering if these were irreducible (over $\mathbb{Q}$) or not. Checking the first few hundred values of $d$ with Mathematica suggests that $P_d(X)$ is reducible iff $d \equiv 4\ (\text{mod}\ 6)$. I checked all the irreducibility criteria I know but didn't find anything that works.
Some more information on the roots: These polynomials each have one positive real root $x_0\in(1,2)$ and, if $d$ is odd, one negative real root on $(-1,0)$. All other roots are non-real with modulus $<x_0$. The polynomial can easily be shown to be squarefree, so all roots are distinct.
Possible generalisation: I also checked for constant terms $a_0$ other than $-1$. For $a_0 = 1$, it seems that it is reducible iff $d > 1$ and $d \equiv 1\ (\text{mod}\ 6)$. For $a_0 = 2$ it seems to be reducible for all even $d$, and if $a_0 = -2$ for all odd $d$. For $a_0$ some other non-zero integer, almost all of them seem to be irreducible except sporadically (e.g. for $a_0 = -6$ and $d = 1$ it factors and for $a_0=-4$ and $d = 2$). I mostly care about the case $a_0 = -1$ though, so I haven't thought much about these.
