TITLE: Show that $u_1^3+u_2^3+\cdots+u_n^3$ is a multiple of $u_1+u_2+\cdots+u_n$
QUESTION [12 upvotes]: Let $k$ be a positive integer.
Define $u_0 = 0\,,\ u_1 = 1\ $ and $\ u_n = k\,u_{n-1}\ -\ u_{n-2}\,,\
n \geq 2$.
Show that for each integer $n$, the number
  $u_{1}^{3} + u_{2}^{3} + \cdots + u_{n}^{3}\ $ is a multiple of
  $\ u_{1} + u_{2} + \cdots + u_{n}$.


Computing a few terms I found 
\begin{align*}u_0 &= 0\\u_1 &= 1\\u_2 &= k\\u_3 &= k^2-1\\u_4 &= k(k^2-1)-k = k^3-2k\\u_5 &= k(k^3-2k)-(k^2-1) = k^4-3k^2+1\\u_6 &= k(k^4-3k^2+1)-(k^3-2k) = k^5-4k^3+3k.\end{align*}
I am not sure how we can use this to solve the question, but I think it may help. Cubing these expressions seems very computational so there must be an easier way.

REPLY [2 votes]: Given the equation
$$
u_n=ku_{n-1}-u_{n-2}\tag{1}
$$
where $u_0=0$ and $u_1=1$, we get the solution
$$
u_n=\frac{\alpha^n-\alpha^{-n}}{\alpha-\alpha^{-1}}\tag{2}
$$
where
$$
\alpha=\frac{k+\sqrt{k^2-4}}2\tag{3}
$$
except when $k=2$ where the solution is
$$
u_n=n\tag{4}
$$
and the result for $k=2$ follows from the fact that the sum of the cubes of the first $n$ consecutive integers is the square of the sum of the first $n$ consecutive integers. A proof without words is given in this answer.
For the solution $(2)$, we get $\alpha^2+\alpha+1=(k+1)\alpha$ and
$$
\begin{align}
&\sum_{j=0}^{n-1}\left(\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\right)^3\\
&=\sum_{j=0}^{n-1}\frac{\alpha^{3j}-3\alpha^j+3\alpha^{-j}-\alpha^{-3j}}{\alpha^3-3\alpha+3\alpha^{-1}-\alpha^{-3}}\\
&=\frac1{\alpha^3-3\alpha+3\alpha^{-1}-\alpha^{-3}}\left(\frac{\alpha^{3n}-1}{\alpha^3-1}-3\frac{\alpha^n-1}{\alpha-1}+3\frac{\alpha^{-n}-1}{\alpha^{-1}-1}-\frac{\alpha^{-3n}-1}{\alpha^{-3}-1}\right)\\
&=\frac1{\alpha^3-3\alpha+3\alpha^{-1}-\alpha^{-3}}\left(\frac{\left(\alpha^{3n}-1\right)\left(1-\alpha^{3-3n}\right)}{\alpha^3-1}-\frac{3\left(\alpha^n-1\right)\left(1-\alpha^{1-n}\right)}{\alpha-1}\right)\tag{5}
\end{align}
$$
and
$$
\begin{align}
&\sum_{j=0}^{n-1}\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\\
&=\frac1{\alpha-\alpha^{-1}}\left(\frac{\alpha^{n}-1}{\alpha-1}-\frac{\alpha^{-n}-1}{\alpha^{-1}-1}\right)\\
&=\frac1{\alpha-\alpha^{-1}}\left(\frac{\left(\alpha^{n}-1\right)\left(1-\alpha^{1-n}\right)}{\alpha-1}\right)\tag{6}
\end{align}
$$
Therefore, we can compute the ratios
$$
\begin{align}
r_{n-1}
&=\left.\sum_{j=0}^{n-1}u_j^3\middle/\sum_{j=0}^{n-1}u_j\right.\\
&=\left.\sum_{j=0}^{n-1}\left(\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\right)^3\middle/\sum_{j=0}^{n-1}\frac{\alpha^j-\alpha^{-j}}{\alpha-\alpha^{-1}}\right.\\
&=\frac1{\alpha^2-2+\alpha^{-2}}\left(\frac{\left(\alpha^{2n}+\alpha^n+1\right)\left(\alpha^{2-2n}+\alpha^{1-n}+1\right)}{(k+1)\alpha}-3\right)\\
&=\frac1{k^2-4}\left(\frac{\left(\alpha^{2n}+\alpha^n+1\right)\left(\alpha^{2n-2}+\alpha^{n-1}+1\right)}{(k+1)\alpha^{2n-1}}-3\right)\\
&=\frac1{k^2-4}\left(\frac{\alpha^{2n-1}+\alpha^{n}+\alpha^{n-1}+\alpha^{1}+1+\alpha^{-1}+\alpha^{1-n}+\alpha^{-n}+\alpha^{1-2n}}{k+1}-3\right)\\
&=\frac1{k^2-4}\left(\frac{\alpha^{2n-1}+\alpha^{n}+\alpha^{n-1}+\alpha^{1-n}+\alpha^{-n}+\alpha^{1-2n}}{k+1}-2\right)\tag{7}
\end{align}
$$
Due to the equation
$$
\begin{align}
&(x-1)\left(x-\alpha\right)\left(x-\alpha^{-1}\right)\left(x-\alpha^2\right)\left(x-\alpha^{-2}\right)\\
&=(x-1)\left(x^2-kx+1\right)\left(x^2-\left(k^2-2\right)x+1\right)\\
&=x^5-mx^4+kmx^3-kmx^2+mx-1\tag{8}
\end{align}
$$
where $m=k^2+k-1$, the ratios $r_n$ in $(7)$ satisfy the relation
$$
r_n=mr_{n-1}-kmr_{n-2}+kmr_{n-3}-mr_{n-4}+r_{n-5}\tag{9}
$$
Computing the first few values of $r_{n-1}$ yields
$$
\begin{align}
r_{-2}
&=\frac1{k^2-4}\left(\frac{\alpha^{-3}+\alpha^{-1}+\alpha^{-2}+\alpha^2+\alpha^1+\alpha^3}{k+1}-2\right)\\
&=1\\
r_{-1}
&=\frac1{k^2-4}\left(\frac{\alpha^{-1}+\alpha^0+\alpha^{-1}+\alpha^1+\alpha^0+\alpha^1}{k+1}-2\right)\\
&=0\\
r_{0}
&=\frac1{k^2-4}\left(\frac{\alpha^1+\alpha^1+\alpha^0+\alpha^0+\alpha^{-1}+\alpha^{-1}}{k+1}-2\right)\\
&=0\\
r_{1}
&=\frac1{k^2-4}\left(\frac{\alpha^3+\alpha^2+\alpha^1+\alpha^{-1}+\alpha^{-2}+\alpha^{-3}}{k+1}-2\right)\\
&=1\\
r_{2}
&=\frac1{k^2-4}\left(\frac{\alpha^5+\alpha^3+\alpha^2+\alpha^{-2}+\alpha^{-3}+\alpha^{-5}}{k+1}-2\right)\\
&=k^2-k+1
\end{align}\tag{10}
$$
The recurrence $(9)$ and the computations $(10)$ ensure that for all $n$, $r_n\in\mathbb{Z}$.