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 $x$.
> 
> 
> 



are these polynomials or rationals functions?
This is response to **QUESTION 1**.


As Fedor pointed out, we're dealing with the [Chebyshev polynomials](https://en.wikipedia.org/wiki/Chebyshev_polynomials#Trigonometric_definition) $P\_n(2\cos t)=\sin nt/\sin t$. So we must show that if
$$
\sum\_{n=1}^N \sin nt = 0 , \quad\quad\quad\quad (1)
$$
then also $\sum\_{n=1}^N \sin^m nt = 0$ for any odd exponent $m\ge 1$.


We may take $0<t<\pi/2$. Also, the sum in (1) can of course be [evaluated,](https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Lagrange.27s_trigonometric_identities) and we find that (1) is equivalent to
$$
\cos t/2 = \cos (N+1/2) t . \quad\quad\quad\quad (2)
$$
I now claim that if (2) holds for $t$, then it also holds for any multiple of $t$. To see this, we just notice that (2) means that $s=t/2$ satisfies $(2N+1)s = 2\pi M\pm s$, for some $M\ge 1$ and a choice of sign (recall that $0<s<\pi/4$). In other words, (2) requires $s$ to be a rational multiple of $\pi$ with denominator $N$ or $N+1$, and clearly this property is preserved under taking integer multiples.


Now everything is clear: $\sin^m\alpha$ can be written as a linear combination of $\sin j\alpha$, with $j$ odd, and we have just seen that (1) implies that also $\sum\_{n=1}^N \sin njt = 0$ for any odd $j$.


