Does anyone have a solution? Thanks!

I couldn't make too much progress. Especially since, if you look at my last line I got stumped since I reach a contradiction. I don't know what I did wrong in my thinking, but maybe this approach can be used to find a solution. My initial intent was to use a generating function to find an explicit form for $a_{n}$ and then construct from that the necessary $\{ x_{n} \}$ and $\{ y_{n} \} .$ Hopefully rchokler who also appears to be doing the problem can give a solution.

As a start, we can try solving the recursive equation for a closed form, but I am not sure if that would be the way to go. Characteristic equation: $\lambda^2-6\lambda+1=0$. Roots: $\lambda=3\pm2\sqrt{2}$. Formula for the $\{a_n\}$ sequence: $\frac{4-\sqrt{2}}{4}(3+2\sqrt{2})^{n+1}+\frac{4+\sqrt{2}}{4}(3-2\sqrt{2})^{n+1}$ Another possibility is to prove that every element $a_n$ is a factor of a number of the $y^2+7$ form, since that would establish the existence of the appropriate $y_n$'s sequence. Once we know that the desired sequence of $y_n$'s exists, then the existence of the corresponding $x_n$'s is trivial. That is just a tip for now.

Here are my findings. Nevermind the solution to the recursive equation. Instead, I found by inspection that the sequences will satisfy $x_n-y_n=a_{n-1}$ We know that $a_n=\frac{y_n^2+7}{x_n-y_n}$ so $y_n^2+7=a_na_{n-1}$ We now need to prove that $a_na_{n-1}-7$ is a perfect square for all $n\geq 1$. This would mean that the following two sequences will work: $y_0=1$, $y_1=9$, $y_n-6y_{n-1}+y_{n-2}=0$. $x_n=a_{n-1}+y_n$

I claim that setting $y_0=1$, $y_1=9$, and $y_n=6y_{n-1}-y_{n-2}$ and $x_n=y_n+a_{n-1}$ gives a valid construction. It suffices to prove that $y_n=\sqrt{a_na_{n-1}-7}$, since then we will have $\frac{y^2_n+7}{x_n-y_n}=\frac{a_na_{n-1}}{a_{n-1}}=a_n$ as desired. Characteristic equations give that $a_n=\frac{4-\sqrt{2}}{4} (3+2\sqrt{2})^{n+1} + \frac{4+\sqrt{2}}{4}(3-2\sqrt{2})^{n+1}$ Now we have $$y_n=\sqrt{a_na_{n-1}-7} = \sqrt{(\frac{4-\sqrt{2}}{4} (3+2\sqrt{2})^{n+1} + \frac{4+\sqrt{2}}{4}(3-2\sqrt{2})^{n+1})(\frac{4-\sqrt{2}}{4} (3+2\sqrt{2})^n + \frac{4+\sqrt{2}}{4}(3-2\sqrt{2})^n)-7}$$$$= \sqrt{\frac{1}{8}((3-\sqrt{2})^2(3-2\sqrt{2})^{2n}+(3+\sqrt{2})^2(3+2\sqrt{2})^{2n}-14} = \frac{3\sqrt{2}+2}{4}(3+2\sqrt{2})^n-\frac{3\sqrt{2}-2}{4}(3-2\sqrt{2})^n$$ Clearly, this satisfies $y_n=6y_{n-1}-y_{n-2}$, and $y_0=1$, $y_1=9$. Therefore, we have $y_n=\sqrt{a_na_{n-1}-7}$, as desired. Uhh for the motivation of the solution itself - I read the above post and mindlessly decided to bash using characteristic equations Well actually in the beginning I thought that I should prove $\left( \frac{-7}{a_n} \right)=1$ (Jacobi Symbol) but then I saw that $a_0a_1 - 7$ is a square so I thought I should prove the stronger statement that $a_na_{n-1}-7$ is a square. For the recursion setup for $y_n$ - I based $\sqrt{a_na_{n-1}-7}$ for $n=0,1,2,3$ which was enough guidance for me to believe that the above claim was really true.

rkm0959 wrote: I claim that setting $y_0=1$, $y_1=9$, and $y_n=6y_{n-1}-y_{n-2}$ and $x_n=y_n+a_{n-1}$ gives a valid construction. It suffices to prove that $y_n=\sqrt{a_na_{n-1}-7}$, since then we will have $\frac{y^2_n+7}{x_n-y_n}=\frac{a_na_{n-1}}{a_{n-1}}=a_n$ as desired. Characteristic equations give that $a_n=\frac{4-\sqrt{2}}{4} (3+2\sqrt{2})^{n+1} + \frac{4+\sqrt{2}}{4}(3-2\sqrt{2})^{n+1}$ Now we have $$y_n=\sqrt{a_na_{n-1}-7} = \sqrt{(\frac{4-\sqrt{2}}{4} (3+2\sqrt{2})^{n+1} + \frac{4+\sqrt{2}}{4}(3-2\sqrt{2})^{n+1})(\frac{4-\sqrt{2}}{4} (3+2\sqrt{2})^n + \frac{4+\sqrt{2}}{4}(3-2\sqrt{2})^n)-7}$$$$= \sqrt{\frac{1}{8}((3-\sqrt{2})^2(3-2\sqrt{2})^{2n}+(3+\sqrt{2})^2(3+2\sqrt{2})^{2n}-14} = \frac{3\sqrt{2}+2}{4}(3+2\sqrt{2})^n-\frac{3\sqrt{2}-2}{4}(3-2\sqrt{2})^n$$ Clearly, this satisfies $y_n=6y_{n-1}-y_{n-2}$, and $y_0=1$, $y_1=9$. Therefore, we have $y_n=\sqrt{a_na_{n-1}-7}$, as desired. May I ask what the motivation of your solution is?

My solution aims at Pell's equation $$x^2-2y^2=7$$We will choose $\{y_n\}$: $y_0=5$, $y_1=31$, and $y_{n}=6y_{n-1}-y_{n-2}$ It is easy to prove that $2a_{n}^2-7=y_{n}^2$ for all $n \ge 0$. Therefore we can choose $\{x_n\}$ so that $x_{n}=y_{n}+2a_{n}$ and the solution follows.