let $F_0 = 1$, $F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$ the fibonacci sequence. is easy to prove, with induction, that $y_n = F_{2n}$

Peter wrote: Prove that the sequence $\{y_{n}\}_{n \ge 1}$ defined by \[y_{0}=1, \; y_{n+1}= \frac{1}{2}\left( 3y_{n}+\sqrt{5y_{n}^{2}-4}\right)\] consists only of integers. I will propose here a different way: without proving any reltion between $y_{n}$ and $F_{n}$ and without using induction . So, we have that $2y_{n+1}-3y_{n}=\sqrt{5y_{n}^{2}-1}$, squaring both sides we get $4y_{n+1}^{2}+9y_{n}^{2}-12y_{n+1}y_{n}=5y_{n}^{2}-4$, which gives as $y_{n+1}^{2}+y_{n}^{2}=3y_{n+1}y_{n}-1$ and also $y_{n}^{2}+y_{n-1}^{2}=3y_{n}y_{n-1}-1$, subtracting from each other we get $y_{n+1}^{2}-y_{n-1}^{2}=3y_{n}(y_{n+1}-y_{n-1})$, therefore \[y_{n+1}=3y_{n}-y_{n-1}\] and because $y_{1}$ and $y_{2}$ are integers, then $y_{n}$ is always an integer.

Peter wrote: The sequence $ \{a_{n}\}_{n \ge 1}$ is defined by \[ a_{1}=1, \; a_{n+1}=2a_{n}+\sqrt{3a_{n}^{2}+1}. \] Show that $ a_{n}$ is an integer for every $ n$. Peter wrote: Prove that the sequence $ \{y_{n}\}_{n \ge 1}$ defined by \[ y_{0}=1, \; y_{n+1}= \frac{1}{2}\left( 3y_{n}+\sqrt{5y_{n}^{2}-4}\right) \] consists only of integers. One may establish the following result. Theorem. Let $ u$, $ v$ and $ \lambda$ be given integers with $ 2v \geq \lambda u$. Suppose that the sequence $ \{x_{n}\}_{n \ge 1}$ satisfies the recurrence relation \[ x_{n+1}= \frac{{\lambda}x_{n}+\sqrt{ \left({\lambda}^{2}-4 \right){x_{n}}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right) }}{2}, \;\; n \in \mathbb{N}\] and has the initial condition $ x_{1}=u$. Then, we have $ x_{2}=v$. Also, $ \{x_{n}\}_{n \ge 1}$ is the sequence of integers. Taking $ u=1$ and $ v=\lambda=2m$, we get the result also given by Tiks (See M6): Corollary. Suppose that the sequence $ \{x_{n}\}_{n \ge 1}$ satisfies the recurrence relation \[ x_{n+1}= m x_{n}+\sqrt{ \left({m}^{2}-1 \right){x_{n}}^{2}+1 \right) }, \;\; n \in \mathbb{N}\] and has the initial condition $ x_{1}=1$. Then, we have $ x_{2}=2m$. Also, $ \{x_{n}\}_{n \ge 1}$ is the sequence of integers. Proof of theorem. First, we give an inductive proof of the fact that the sequence $ \{x_{n}\}_{n \ge 1}$ of real numbers is well-defined. The initial term is defined by $ x_{1}=u$. Since \[ \left({\lambda}^{2}-4 \right){u}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right) = \left( 2v-\lambda u \right)^{2}\geq 0, \] we find that \[ x_{2}= \frac{{\lambda}u+\sqrt{ \left({\lambda}^{2}-4 \right){u}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right) }}{2}= \frac{{\lambda}u+\vert 2v-\lambda u \vert }{2}=v \] is well-defined. Now, we fix $ n \geq 2$ and suppose that $ x_{k}$ is well-defined for all $ 1 \leq k \leq n$. To show that $ x_{n+1}$ is well-defined, we need to show that \[ \left({\lambda}^{2}-4 \right){x_{n}}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right) \] is nonnegative. By the inductive hypothesis, we obtain \[ x_{n}= \frac{{\lambda}x_{n-1}+\sqrt{ \left({\lambda}^{2}-4 \right){x_{n-1}}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right) }}{2}, \] This means that $ x_{n}$ is a root of the quadratic equation \[ t^{2}-\lambda x_{n-1}t+{x_{n-1}}^{2}-\left( u^{2}-\lambda uv+v^{2}\right) =0 \] so that \[ {x_{n}}^{2}-\lambda x_{n-1}x_{n}+{x_{n-1}}^{2}-\left( u^{2}-\lambda uv+v^{2}\right) =0 \] or \[ u^{2}-\lambda uv+v^{2}={x_{n}}^{2}-\lambda x_{n-1}x_{n}+{x_{n-1}}^{2}. \] It therefore follows that \[ \left({\lambda}^{2}-4 \right){x_{n}}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right)=\left({\lambda}^{2}-4 \right){x_{n}}^{2}+4 \left({x_{n}}^{2}-\lambda x_{n}x_{n-1}+{x_{n-1}}^{2}\right) \] or \[ \left({\lambda}^{2}-4 \right){x_{n}}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right)= \left( \lambda x_{n}-2 x_{n-1}\right)^{2}\geq 0. \] Now, we prove the main result. We claim that \[ x_{n+2}= \lambda x_{n+1}-x_{n}\;\; or \;\; x_{n+2}= x_{n}\] for each $ n \in \mathbb{N}$. Using this claim, one may immediately give an inductive proof of the fact that $ x_{n}$ is integer for all $ n \in \mathbb{N}$. To establish the claim, we begin with the recurrence relation \[ x_{n+1}= \frac{{\lambda}x_{n}+\sqrt{ \left({\lambda}^{2}-4 \right){x_{n}}^{2}+4 \left( u^{2}-\lambda uv+v^{2}\right) }}{2}. \] We find that $ x_{n+1}$ is a root of the quadratic equation \[ t^{2}-\lambda x_{n}t+{x_{n}}^{2}-\left( u^{2}-\lambda uv+v^{2}\right) =0 \] so that \[ {x_{n+1}}^{2}-\lambda x_{n}x_{n+1}+{x_{n}}^{2}= \left( u^{2}-\lambda uv+v^{2}\right) \] This means that $ {x_{n+1}}^{2}-\lambda x_{n}x_{n+1}+{x_{n}}^{2}$ is constant. It follows that, for all $ n \in \mathbb{N}$, \[ {x_{n+2}}^{2}-\lambda x_{n+1}x_{n+2}+{x_{n+1}}^{2}={x_{n+1}}^{2}-\lambda x_{n}x_{n+1}+{x_{n}}^{2}\] so that \[ {x_{n+2}}^{2}-{x_{n}}^{2}-\lambda x_{n+1}x_{n+2}+\lambda x_{n}x_{n+1}=0 \] or \[ \left( x_{n+2}-x_{n}\right) \left( x_{n+2}-\lambda x_{n+1}+x_{n}\right)=0. \] This implies that, for each $ n \in \mathbb{N}$, one of the following relations holds: \[ x_{n+2}= x_{n}\;\; or \;\; x_{n+2}= \lambda x_{n+1}-x_{n}. \] This completes the proof.