TITLE: Can you tie up these Laurent sequences?
QUESTION [9 upvotes]: 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.

REPLY [10 votes]: 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 by Joshua Alman, Cesar Cuenca and Jiaoyang Huang).

REPLY [7 votes]: As to Question 2, the sequence $(x_n^2)$ solves the same recursive relation as $(y_n)$, with the same initial values, therefore they coincide.