Peter wrote: The sequence $\{y_{n}\}_{n \ge 1}$ is defined by \[y_{1}=y_{2}=1,\;\; y_{n+2}= (4k-5)y_{n+1}-y_{n}+4-2k.\] Determine all integers $k$ such that each term of this sequence is a perfect square. Let us assume that $k\geq 4$ is an integer such that all the members of the sequence $y_{n}$ are squares. We have that $y_{3}=2k-2$ then $k=2t^{2}+1$ for some $t\geq 2$. From the recurrence we obtain that $y_{5}=256t^{6}-96t^{4}+8t^{2}+1$. On the other hand from $t\geq2$ follows that \[(16t^{3}-3t-1)^{2}<y_{5}=256t^{6}-96t^{4}+8t^{2}+1<(16t^{3}-3t)^{2}\] therefore $y_{5}$ can't be a square $\Rightarrow k\leq 3$. When $k=1$ then the sequence only consist of $0$s and $1$s. The case $k=2$ is impossible because $y_{3}=2k-2$ should be a perfect square. For $k=3$ we get that $y_{n+2}=7y_{n+1}-y_{n}-2$ and we will prove that for all $n\geq 1$ $y_{n}$ is a square. Denote $a_{n}=y_{n}-\frac{2}{5}$ then $a_{n+2}=7a_{n+1}-a_{n}$ and $a_{1}=a_{2}=\frac{3}{5}$. Solving the equation ${\lambda}^{2}-7{\lambda}+1=0$ we find that \[{\lambda}_{1}=\frac{7+3\sqrt{5}}{2},{\lambda}_{2}=\frac{7-3\sqrt{5}}{2}.\] Consequently, $a_{n}=c_{1}{{\lambda}_{1}}^{n}+c_{2}{{\lambda}_{2}}^{n}$ where $c_{1}$ and $c_{2}$ are constants. From $a_{1}=a_{2}=\frac{3}{5}$ we can show that $c_{1}=\frac{9-4\sqrt{5}}{5}=(\frac{\sqrt{5}-2}{\sqrt{5}})^{2}$ and $c_{2}=\frac{9+4\sqrt{5}}{5}=(\frac{\sqrt{5}+2}{\sqrt{5}})^{2}$. Thus, \[a_{n}=((\frac{\sqrt{5}-2}{\sqrt{5}})(\frac{3+\sqrt{5}}{2})^{n})^{2}+((\frac{\sqrt{5}+2}{\sqrt{5}})(\frac{3-\sqrt{5}}{2})^{n})^{2}\] and \[y_{n}=((\frac{\sqrt{5}-2}{\sqrt{5}})(\frac{3+\sqrt{5}}{2})^{n}+(\frac{\sqrt{5}+2}{\sqrt{5}})(\frac{3-\sqrt{5}}{2})^{n})^{2}=x_{n}^{2}\] Now it is easy to show that $x_{n+2}=3x_{n+1}-x_{n}$ for all $n\geq 1$ and $x_{1}=x_{2}=1$. Therefore $x_{n}\in{N}$ for all $n\geq{1}$ and $y_{n}=x_{n}^{2}$. Answer: $k=1$ and $k=3$.