Show that $\sum_{i=0}^{\frac{p-1}{2}} {{\frac{p-1}{2}}\choose {i}}^2 x^{\frac{p-1}{2}-i}$ is separable This is a problem occurs in my research. For any algebraically closed field $k$ of characteristic $p$. I want to show that $\sum_{i=0}^{\frac{p-1}{2}} {{\frac{p-1}{2}}\choose {i}}^2 x^{\frac{p-1}{2}-i}$ is a separable polynomial over this field. I am trying to prove that $p$ does not divide the discriminant of the polynomial. It seems like all the prime factors of the discriminant are less than $p$.
 A: Let me elaborate on Noam D. Elkies' comment. If we denote $n=(p-1)/2$, the discriminant of this polynomial $g(x)$ is non-zero modulo $p$ if and only if the discriminant of Legendre's polynomial $f(x)=2^{-n}\sum_{k=0}^n \binom{n}{k}^2(x-1)^{n-k}(x+1)^k=2^{-n}(x-1)^ng((x+1)/(x-1))$ is non-zero modulo $p$ (the roots of $f$ and $g$ are obtained from each other by fractional linear functions, thus if $g$ has only simple roots, so does $f$ and vice versa). The discriminant of Legendre's polynomial and even of Jacobi's polynomial is known, see, for example, the formula on page 5 here. Indeed the prime divisors do not exceed $2n<p$. 
A: This is not an answer, but rather a reformulation. Write $\equiv_p$ for congruence mod $p$. Here $p$ is odd.
Start with $\binom{\frac{p-1}2}i\equiv_p\binom{2i}i4^{-i} \mod p$. If we drop $x^{\frac{p-1}2}$ and then replace $x^{-1}\rightarrow x$, and letting
$$f_p(x):=\sum_{i=0}^{\frac{p-1}2}\binom{\frac{p-1}2}i^2x^i,$$
the problem concerns $f_p(x)$ being "separable". Now, this is equivalent to showing that
$$f_p(x)\equiv_p\sum_{i=0}^{\frac{p-1}2}\binom{2i}i^216^{-i}x^i$$
is "separable". Of course, the same goes to
$$g_p(y):=\sum_{i=0}^{\frac{p-1}2}\binom{2i}i^2y^i.$$
