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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 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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 The discriminant of $g(x)$ is not an integer but is the reciprocal of an large integer all of whose prime factors are $n$ or smaller. [The exact formula can be obtained form $g'(x) = g(x) - x^n/n!$.] What would it mean for such a number to be "squarefree"? – Noam D. Elkies Oct 21 2011 at 4:29 Embarassing moment: Sorry, I meant the discriminants the polynomials obtained after multiplying by $n!$, as Gjergji point out, and the answer in general is indeed straightforward. I just didn't see that. – Cosmin Pohoata Oct 21 2011 at 4:59 Concerning the discriminant of $x^n-x-1$, David Boyd asked whether it was squarefree for prime $n$ at the Western Number Theory meeting of December, 2000. The following February, Greg Martin proved that there are an infinity of primes $n$ for which the discriminant is not squarefree. In particular, he found the example $n=257$, which entails divisibility by $59^2$. See ma.utexas.edu/users/goddardb/WCNT11/2000.PDF problem 9. – Gerry Myerson Oct 21 2011 at 5:00 A small comment: if $p$ exactly divides the discriminant of a monic $f(x)\in\mathbf{Z}[x]$, then there are precisely $n-1$ distinct homomorphism from $\mathbf{Z}[x]/(f(x))$ to $\overline\mathbf{F}_p$, where $n$ is the degree of $f(x)$. If $p>n$ then this condition suffices to ensure that $p$ exactly divides the discriminant of $f(x)$. May be this is of no use for you, or you might already be perfectly aware of this. – Tommaso Centeleghe Oct 21 2011 at 8:26

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, $x^n+na_{n-1}x^{n-1}+\cdots+n!a_1x+n!$)
 What does it mean for a polynomial in several variables to be monic? – Gerry Myerson Oct 21 2011 at 4:44 Fixed.  – Gjergji Zaimi Oct 21 2011 at 4:52 Nice. Thanks. By the way, is there any elementary way to prove the irreducibility of the truncanted exponential? //elementary in the sense of the proof for xN−x−1 (by Selmer?) which just exploited some inequalities between the absolute values of the roots. I just realized that the only way I know to do this is with Newton polygons. – Cosmin Pohoata Oct 21 2011 at 5:07