4 added 15 characters in body

My two cents, : Schur proved that the discriminant of $$n!(\frac{x^{n}}{n!} n!\left(\frac{x^{n}}{n!} + \ldots + \frac{x^{2}}{2!} + x+1),$$ x+1\right),$$is equal to (-1)^{n(n-1)/2}(n!)^n, which is not a perfect square as long as n\neq 0\pmod{4}. See this note by K. Conrad for references. On the other hand the discriminant is a polynomial in the coefficients of the given polynomial, so the answer is negative for your general family. This is because a polynomial cannot take only squarefree values at integer tuples unless it is constant. (I'm assuming that you are considering the scaled version of these polynomials, x^n+na_{n-1}x^{n-1}+\cdots+n!a_0)x^n+na_{n-1}x^{n-1}+\cdots+n!a_1x+n!) Explicitly constructing infinitely many polynomials of fixed degree with squarefree discriminant is a very non-trivial task. See the recent article "A construction of polynomials with squarefree discriminants" by K. Kedlaya. 3 added 28 characters in body My two cents, Schur proved that the discriminant of$$n!(\frac{x^{n}}{n!} + \ldots + \frac{x^{2}}{2!} + x+1), 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. (Assuming as in the comments I'm assuming that you scale are considering the coefficients to get a monic polynomial)scaled version of these polynomials, $x^n+na_{n-1}x^{n-1}+\cdots+n!a_0$)

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.

Post Undeleted by Gjergji Zaimi
2 added 92 characters in body
Post Deleted by Gjergji Zaimi
1