The [Fibonacci polynomials](https://en.wikipedia.org/wiki/Fibonacci_polynomials) are defined recursively by $F\_0(x)=0, F\_1(x)=1$ and $F\_n(x)=xF\_{n-1}(x)+F\_{n-2}(x)$, for $n\geq2$.


While computing certain integrals, I observe the following (numerically) which prompted me to ask:



> 
> **Question.** For $n, k\in\mathbb{N}$, are these always integers?
>  $$\int\_0^1F\_n(k+nz)\,dz$$
> 
> 
> 


To help clarify, here is a list of the first few polynomials:
$$F\_2(x)=x, \qquad F\_3(x)=x^2+1, \qquad F\_4(x)=x^3+2x.$$



integral transform of Fibonacci polynomials is integral
The integral of each individual monomial will be integral. First we have the identity
$$F\_n(x)=\sum\_{i=0}^{\lfloor(n-1)/2\rfloor}{n-i-1\choose i}x^{n-2i-1},$$
so my claim is that
$$\binom{n-i-1}{i}\int\_0^1 (k+nz)^{n-2i-1}dz=\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)} \in \mathbb Z.$$
By the binomial theorem we can write $(k+n)^{n-2i}=k^{n-2i}+nk^{n-2i-1}(n-2i)+n^2d$ for some integer $d$. So we can write
$$\binom{n-i-1}{i}\cdot\frac{(k+n)^{n-2i}-k^{n-2i}}{n(n-2i)}=k^{n-2i-1}\binom{n-i-1}{i}+d\cdot\frac{n}{n-2i}\binom{n-i-1}{i}$$
it suffices to show that
$\frac{n}{n-2i}\binom{n-i-1}{i}$ is an integer.
However we can check that 
$$\frac{n}{n-2i}\binom{n-i-1}{i}=\binom{n-i-1}{i}+2\binom{n-i-1}{i-1}$$ 
and the claim follows.


