Fix an integer $k\geq3$. Define the two families of sequences $\{x\_n\}$ and $\{y\_n\}$ according to the rules:
$$x\_n=\frac{x\_{n-1}^2+x\_{n-2}^2+\cdots+x\_{n-k+1}^2}{x\_{n-k}} \qquad n\geq k$$
and
$$y\_n=\frac{(y\_{n-1}+y\_{n-2}+\cdots+y\_{n-k+1})^2}{y\_{n-k}} \qquad n\geq k$$
with initial conditions $x\_j=y\_j=1$ for $j=0,1,\dots,k-1$.



> 
> **QUESTIONS.**
> 
> 
> (1) It seems that both $x\_n$ and $y\_n$ are always positive integers.
> 
> 
> (2) It also appears true that $y\_n=x\_n^2$ for all $n$.
> 
> 
> 


Any ideas of a proof? Of course, if (2) holds and $x\_n\in\mathbb{N}$ then $y\_n\in\mathbb{N}$.


**REMARK.** Special cases are also appreciated, say for $k=3$ etc.



Can you tie up these Laurent sequences?
Suppose we know that $y\_j=x\_j^2$ for $j=n-1, \ldots, n-k$. Then
$$x\_n^2=\left(\frac{x\_{n-1}^2+x\_{n-2}^2+\cdots+x\_{n-k+1}^2}{x\_{n-k}} \right)^2=\frac{(y\_{n-1}+y\_{n-2}+\cdots+y\_{n-k+1})^2}{y\_{n-k}} =y\_n.$$


If all $y\_n$ are integers then from requrrent relation follows that they are squares. It means that all $x\_n$ are also integers. So $\{y\_n\}$ are integers iff $\{x\_n\}$ are integers.


Integrality of $\{x\_n\}$ is a special case of more general result (see case (1) in Theorem 3.9 from the article [Laurent Phenomenon Sequences](https://arxiv.org/pdf/1309.0751v2.pdf) by Joshua Alman, Cesar Cuenca and Jiaoyang Huang).


