TITLE: are these polynomials or rationals functions?
QUESTION [15 upvotes]: Let $x$ be a variable. Define the following family of sequences (reminiscent of Lucas polynomials) according to the rule: $P_0(x):=0, P_1(x):=1$ and for $n\geq2$ by
$$P_n(x)=xP_{n-1}(x)-P_{n-2}(x).$$
Notice that $P_n(2)=n$ for every $n\in\mathbb{Z}_{\geq0}$. Here are a few examples:
$$P_2=x, \qquad P_3=x^2-1, \qquad P_4=x^3-2x, \qquad P_5=x^4-3x^2+1.$$

QUESTION 1. Empirical evidence suggest that, for each fixed integers $n, k\geq1$,
  $$Q_{n,k}(x):=\frac{P_1(x)^{2k-1}+P_2(x)^{2k-1}+\cdots+P_n(x)^{2k-1}}{P_1(x)+P_2(x)+\cdots+P_n(x)} \tag1$$
  is a polynomial in $x$.
This is trivial for $k=1$. Is it true for other odd powers $2k-1$? 

REMARKS. 
(1) Specialized values $2k-1=3$ or $5$, etc are still interesting to me.
(2) Even the case of special valuations for $x\in\mathbb{Z}$ are appealing as well, which means (1) becomes a claim on integrality of sequences.

QUESTION 2. Encouraged by the success with QUESTION 1, how about this?
  $$R_n(x)=\prod_{j=1}^n\frac{P_j(x)^{2k-1}+\cdots+P_n(x)^{2k-1}}{P_j(x)} \tag2$$
  is a polynomial in $x$.

REPLY [5 votes]: Question 2.
Since $P_n(2\cos t)=\sin(nt)/\sin t$, we get that the roots of $P_n$ are $2\cos(\pi k/n)$,
$k=1,\dots,n-1$. That is, a number $\kappa=2\cos(\pi a/b)$, $0<a<b$, gcd$(a,b)=1$, is a root
of $P_1\dots P_n$ of multiplicity $[n/b]$. We have to prove that $\kappa$ is a root of the numerator
of your fraction of multiplicity at least $[n/b]$. Fix odd integer $r$
and consider the sum $$S_{rmn}(t)=\sum_{j=m}^n \sin(rjt)=-\frac{\sin\frac{n+m}2 rt 
\sin\frac{n-m-1}2 rt}{\sin \frac{rt}2}.$$
As Christian explained, the $m$-th multiple in the numerator is a linear combination
of $(\sin t)^{-1}S_{rmn}(t)$ for different odd $r$. We see that if $n+m$ or $n-m-1$ is divisible by
$2b$ (note that both divisibilities are not possible for the same $m$), 
then $S_{rmn}(\pi a/b)=0$. The total number of such $m$ 
is the total number of multiples of $2b$ in the set $\{-1,0,1,\dots,2n\}\setminus \{n-1,n\}$.
It equals $1+[2n/(2b)]-\delta\geqslant [n/b]$ as desired (here $\delta$ equals 1 
if $n-1$ or $n$ is divisible by $2b$,
else $\delta=0$.)