Solution
= Solution
Let $x_1,\ldots,x_k$ be the distinct zeros in $(-1,1)$ at which $P_n$ changes sign, and put $Q(x)=\prod_{j=1}^k(x-x_j)$. Then $P_nQ$ has constant sign and is nonzero except at finitely many points, so
$$
\int_{-1}^1P_nQr\ne0.
$$
If $k<n$, this contradicts the orthogonality of $P_n$ to the lower-degree polynomial $Q$. Hence $k\geq n$; degree $n$ forces exactly $n$ distinct, simple roots, all in $(-1,1)$.