I am trying to prove that $f(x)=x^n+nx+n$ has Galois group $S_n$ over the rationals. The discriminant of this polynomial is $\Delta= (-1)^{n(n-1)/2}n^n(-(n-1)^{n-1}+(-1)^n n^{n-1})$. The Newton polygon of this polynomial to a prime $p|n$ is given by a line with slope $-v_p(n)/n$, hence this polynomial is irreducible. It seems that the factor $d=n^{n-1}+(-1)^{n+1}(n-1)^{n-1}$ of the discriminant is always a squarefree number, hence the question. I have read in Serre's Topics in Galois Theory the following argument, but I am not so confident to interpret if it means, that $d$ is always a squarefree number: Let $p$ be a prime $p|\Delta$, $p$ does not divide $n$, hence $p|d$. This will happen if $f(x)$ and $f'(x)$ have a common root mod $p$. We have $f'(x) = nx^{n-1}+n = 0 \text{ mod } p$, hence since $p$ does not divide $n$, we have $x^{n-1} = -1 \text{ mod } p $. Now we plug in $x^{n-1} = -1 \text{ mod } p $ in $f(x) = 0 \text{ mod } p $ and get $x = \frac{n}{1-n} \text{ mod } p $. Now the agument of Serre goes like this: "Hence there can be at most one double root $\text{mod} p$ for each ramified prime $p$ which does not divide $n$. This shows that the inertia subgroup at $p$ is either trivial, or is of order two, generated by transpositions. But $G=Gal(f)$ is generated by its inertia subgroups, because $\mathbb{Q}$ has no nontrivial unramified extension. But $G$ is transitive, since $f(x)$ is irreducible."

Does this argument prove, that $d$ is a squarefree number? I am not so confident with inertia subgroups and ramified primes.

**Edit:**

As it turns out when asking questions, I found a counterexample for $n=29$ and $n=47$. In this cases $d$ is nearly a squarefree number, meaning that there is a prime $q$ with $v_q(d) = 2$. Hence my modified questions are:

1) Does the argument of Serre imply that $v_q(d) \le 2$ for each prime $q$ dividing $d$?

2) How does one prove, that the $Gal(f) = S_n$ over the rationals?

Squarefree values of trinomial discriminants, by David Boyd, Greg Martin and Mark Thom. See here: arxiv.org/pdf/1402.5148.pdf . The conjectured density of the values $n$ having $d$ square-free is about $0.9934466\ldots$ - so the vast majority. But even proving infinitely many square-free values appears beyond reach of current methods. $\endgroup$ – Vesselin Dimitrov Feb 6 '17 at 9:24