# Show that $u_1^3+u_2^3+\cdots+u_n^3$ is a multiple of $u_1+u_2+\cdots+u_n$

1. Let $k$ be a positive integer.
2. Define $u_0 = 0\,,\ u_1 = 1\$ and $\ u_n = k\,u_{n-1}\ -\ u_{n-2}\,,\ n \geq 2$.
3. 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.

• $u_3 = k^2 - 1$, not $k^2 - k$. – Robert Israel Jul 15 '16 at 17:55
• @RobertIsrael Thanks, fixed. – user19405892 Jul 15 '16 at 18:04

It appears that $$y_n = \dfrac{u_1^3 + \ldots u_n^3}{u_1 + \ldots + u_n}$$ satisfies the recurrence relation $$y_n = (k^2+k-1)(y_{n-1} - k y_{n-2} + k y_{n-3} - y_{n-4}) + y_{n-5} \ \text{for}\ n \ge 6$$ Given that $y_1, \ldots, y_5$ are integers, this would imply that all $y_n$ are integers.

EDIT: Writing $\cos(\theta) = k/2$, we have $$u_n = \frac{\sin(n\theta)}{\sin(\theta)}$$ which can be verified by induction. Don't worry about $\theta$ being real only for $|k|\le 2$.

Using this we can obtain closed-form formulas for $u_1 + \ldots + u_n$ and $u_n^3 + \ldots + u_n$, and $y_n$ (it's rather tedious if working by hand, but elementary)

$$y_n = \frac{-\cos((2n+1)\theta) + 2 \cos(\theta) - \cos((n+1)\theta) - \cos(n\theta) + 1}{\cos(\theta) - \cos(3\theta) - \cos(2\theta)+1}$$

and it can be verified directly that this satisfies the recurrence above.

• How did you find out $y_n$ satisfied the recurrence? – user19405892 Jul 15 '16 at 18:12
• Using the gfun package in Maple. – Robert Israel Jul 15 '16 at 18:12
• I am looking for an elementary way of solving this question. – user19405892 Jul 15 '16 at 18:13
• @lhf: Sorry, that was $u_n$ not $a_n$. – Robert Israel Jul 15 '16 at 20:59

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 $n$, $r_n\in\mathbb{Z}$.

fiddling with small $k.$ Take $x = \Sigma u_j, \; \; y = \Sigma u_j^3.$

CONCLUSION: for $k \geq 2,$ $$\color{blue}{ y = \frac{x^2 ((k-2) x + 3)}{k+1},}$$ while $\color{blue}{x \equiv 0,1 \pmod {k+1}}.$

When $k=2,$ $$y = x^2.$$ This comes up pretty often, the sum of the consecutive cubes (starting with $1$) is the square of the sum of the consecutive numbers.

When $k=3,$ $$y = \frac{x^2 (x + 3)}{4}.$$ For this one, you need to know that $x \equiv 0,1 \pmod 4.$

This already suggests that $k=4$ gives $y = a x^4 + b x^3 + c x^2 + d x,$ with rational coefficients, and some restrictions on $x$ that make $y$ an integer. If true, the coefficients can be found by taking four $x$ points, then making and inverting a certain four by four rational matrix. NOPE, not that much effort required. Stays cubic, a three by three matrix would have been enough...

Not quite what I expected: for $k=4,$ $$y = \frac{x^2 (2 x + 3)}{5},$$ while $x \equiv 0,1 \pmod 5.$

For $k=5,$ $$y = \frac{x^2 (3 x + 3)}{6},$$ while $x \equiv 0,1 \pmod 6.$

For $k=6,$ $$y = \frac{x^2 (4 x + 3)}{7},$$ while $x \equiv 0,1 \pmod 7.$

Apparently, for $k \geq 2,$ $$\color{red}{ y = \frac{x^2 ((k-2) x + 3)}{k+1},}$$ while $x \equiv 0,1 \pmod {k+1}.$ Recall $x = \Sigma u_j, \; \; y = \Sigma u_j^3.$

• It's not too hard to show that $u_n = \frac{\left(\frac{1}{a}\right)^n-a^n}{\frac{1}{a}-a}$ where $a$ satisfy $a = \frac{1}{k-a}$ and from this the formula you put up is easily checked to be true. – Winther Jul 15 '16 at 23:41