Peter wrote: Let $\{x_{n}\}_{n\ge0}$ and $\{y_{n}\}_{n\ge0}$ be two sequences defined recursively as follows \[x_{0}=1, \; x_{1}=4, \; x_{n+2}=3 x_{n+1}-x_{n},\] \[y_{0}=1, \; y_{1}=2, \; y_{n+2}=3 y_{n+1}-y_{n}.\] Prove that ${x_{n}}^{2}-5{y_{n}}^{2}+4=0$ for all non-negative integers. Suppose that $a$, $b$ are two positive integers such that $a^{2}-5b^{2}+4=0$. Prove that there exists a non-negative integer $k$ such that $a=x_{k}$ and $b=y_{k}$. We have that $x_{n+2}=3x_{n+1}-x_{n}$ and $y_{n+2}=3y_{n+1}-y_{n}$ subtracting from each other we get that $x_{n+2}-y_{n+2}=3(x_{n+1}-y_{n+1})-(x_{n}-y_{n})$. Now by induction it is easy to show that $x_{n}\equiv y_{n}(mod 2)$ for all $n\in{N}$. Let us define the sequences of positive integer numbers $x_{n}'$ and $y_{n}'$ in the following way \[x_{n+1}'=\frac{5y_{n}+3x_{n}}{2},y_{n+1}'=\frac{3y_{n}+x_{n}}{2}\] for all $n\in{N_{0}}$. Then we can easily see that $x_{1}'=x_{1}=4$,$x_{2}'=x_{2}=11$ and $y_{1}'=y_{1}=2$,$y_{2}'=y_{2}=5$ and \[x'_{n+2}=3x_{n+1}'-x_{n}',y_{n+1}=3y_{n+1}'-y_{n}'\] consequently, $x_{n}'\equiv x_{n}$ and $y_{n}'\equiv y_{n}$. (a) We have proved that $x_{n+1}=\frac{5y_{n}+3x_{n}}{2}$ and $y_{n+1}=\frac{3y_{n}+x_{n}}{2}$, then let us notice that \[5y_{n+1}^{2}-x_{n+1}^{2}=5(\frac{3y_{n}+x_{n}}{2})^{2}-(\frac{5y_{n}+3x_{n}}{2})^{2}=5y_{n}^{2}-x_{n}^{2}\] then $5y_{n+1}^{2}-x_{n+1}^{2}=5y_{0}^{2}-x_{0}^{2}=4$ for all $n\in{N}$. (b) Let us assume that $(a_{0},b_{0})$ is a pair of positive integers so that $5b_{0}^{2}-a_{0}^{2}=4$ and $b_{0}>1$(if $b_{0}=1$ then $a_{0}=1$ and the claim is obvious). We have that $9b_{0}^{2}-a_{0}^{2}>5b_{0}^{2}-a_{0}^{2}=4>0$ then $3b_{0}>a_{0}$ and ,also, $a_{0}^{2}=5b_{0}^{2}-4>\frac{25}{9}b_{0}^{2}$(because $b_{0}\geq 2$) so $3a_{0}>5b_{0}$. Now define a new pair of positive integer numbers $(a_{1},b_{1})$ as $b_{1}=\frac{3b_{0}-a_{0}}{2},a_{1}=\frac{3a_{0}-5b_{0}}{2}$ and note that $5b_{1}^{2}-a_{1}^{2}=5b_{0}^{2}-a_{0}^{2}=4$ and $a_{1}+b_{1}=a_{0}-b_{0}<a_{0}+b_{0}$. Similarly, if $b_{1}>1$ we can define a new pair $(a_{2},b_{2})$ of positive integers for which $5b_{2}^{2}-a_{2}^{2}=4$ and $a_{2}+b_{2}<a_{1}+b_{1}$. Because $a_{i}+b_{i}\geq 2$ then the process of constructing new pairs is limited, which means that we will eventually get to a pair $(a_{k},b_{k})$ for which $b_{k}=1$ and so $a_{k}=1$. Finally note that because $b_{m+1}=\frac{3b_{m}-a_{m}}{2},a_{m+1}=\frac{3a_{m}-5b_{m}}{2}$ then $a_{m}=\frac{5b_{m+1}+3a_{m+1}}{2},b_{m}=\frac{3b_{m+1}+a_{m+1}}{2}$. Therefore, considering that $a_{k}=x_{0}$ and $b_{k}=y_{0}$ we conclude that $a_{t}=x_{k-t}$ and $b_{t}=y_{k-t}$ for all $t\leq k$, so $(a_{0},b_{0})=(x_{k},y_{k})$ as the problem claims.