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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 appearead)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.

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# Searching for polynomials with squarefree discriminant

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 appearead). 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.