TITLE: True or false: If you square the coefficients in the expansion of $(x+1)^n$, the resulting polynomial has $n$ distinct real roots.
QUESTION [6 upvotes]: It seems that if you square the coefficients in the expansion of $(x+1)^n$, the resulting polynomial has $n$ distinct real roots. (I experimented using desmos.com.)
In other words, I am looking for a proof (or counter-example) for the proposition that polynomial $\displaystyle p(x) = \sum_{k=0}^n\binom{n}{k}^2x^k$ has exactly $n$ distinct real roots.
But I have not found a proof. I tried induction, to no avail.
(Context: I am looking for an expanded polynomial with exactly, say, $100$ distinct real roots.)

REPLY [3 votes]: We have $p(x) = (1-x)^n P_n\left(\frac{1+x}{1-x}\right)$ where $P_n$ is the Legendre polynomial of degree $n$. This follows from the Rodrigues formula, among other ways. (Mathematica spotted it for me.) The Legendre polynomials are orthogonal polynomials, so by standard theory their roots interlace each other, and in particular $P_n(x)$ has $n$ distinct real roots. Moreover it is easy to see $-1$ is not a root.
The function $x \mapsto \frac{1+x}{1-x}$ is bijective from $\mathbb{R} \setminus \{1\}$ to $\mathbb{R} \setminus \{-1\}$. Hence the $n$ distinct roots of $P_n$ are hit by $n$ distinct values in $\mathbb{R} \setminus \{1\}$, so $p(x)$ has at least, hence exactly, $n$ distinct real roots.