Question

In connection with the Galois theoretic results surrounding the irreducibility of $f(x)= x^{N}-x-1$ over $\mathbb{Q}$, I've been trying to prove for a while that the discriminant of $f$ is actually squarefree as it sounded plausible. After failed attemps I started believing that this might not be actually true so I did a computer verification which showed that for $N=257$ this is actually wrong (however not too wrong; there was just a square of a prime that appeared). So now I was wondering whether the discriminat of the truncanted exponential, i.e. $$g(x) = \frac{x^{n}}{n!} + \ldots + \frac{x^{2}}{2!} + x+1,$$ is squarefree or if in general the discriminants of polynomials of type $$\pm \frac{x^{n}}{n!} + a_{n-1} \frac{x^{n-1}}{(n-1)!} + \ldots + a_{2} \frac{x^{2}}{2!} + 1,$$ where the $a_{i}$'s are integers, are squarefree (question motivated of course by the fact that they are irreducible over $\mathbb{Q}$ with Galois group $S_{n}$ - Schur).

//Of course this is harder to verify since we no longer have a "nice" formula for polynomials of this form.

Answer 1

My two cents: Schur proved that the discriminant of $$n!\left(\frac{x^{n}}{n!} + \ldots + \frac{x^{2}}{2!} + x+1\right),$$ is equal to $(-1)^{n(n-1)/2}(n!)^n$, which is not a perfect square as long as $n\neq 0\pmod{4}$. See this note by K. Conrad for references. On the other hand the discriminant is a polynomial in the coefficients of the given polynomial, so the answer is negative for your general family. This is because a polynomial cannot take only squarefree values at integer tuples unless it is constant. (I'm assuming that you are considering the scaled version of these polynomials, $x^n+na_{n-1}x^{n-1}+\cdots+n!a_1x+n!$)

Explicitly constructing infinitely many polynomials of fixed degree with squarefree discriminant is a very non-trivial task. See the recent article "A construction of polynomials with squarefree discriminants" by K. Kedlaya.