I have the following nonlinear recurrence
$$x_{n+1}=x_n+\frac{c}{x_{n-1}}$$
where $c$ is a constant - and if it makes things easier, let's assume it's a positive integer - with initial conditions $x_1=2$ and $x_2=3$. I'm asked to find a product of consecutive terms in the sequence $x_kx_{k+1}$, namely $x_{50}x_{51}$. How can I approach this?
One thing that came to mind was to rewrite as $(x_{n+1}-x_n)x_{n-1}=c$. Then $$\begin{cases} (x_{n+1}-x_n)x_{n-1}=c\\ (x_n-x_{n-1})x_{n-2}=c \end{cases}$$the idea being that I might be able to extract some explicit information about $x_nx_{n-1}$. Or perhaps to allow me to write $x_n$ in terms of products of consecutive terms.
Dividing one equation by the other gives $$\frac{(x_{n+1}-x_n)x_{n-1}}{(x_n-x_{n-1})x_{n-2}}=1\implies x_{n+1}x_{n-1}-x_nx_{n-1}=x_nx_{n-2}-x_{n-1}x_{n-2}$$ I'm thinking the next step would be to introduce products, like something along the lines of $$\begin{align*} x_{n+1}x_{n-1}-x_nx_{n-1}&=x_{n+1}x_{n-1}-x_{n+1}x_n+x_{n+1}x_n-x_nx_{n-1}\\ &=-x_{n+1}(x_n-x_{n-1})+x_{n+1}x_n-x_nx_{n-1} \end{align*}$$ but I'm not sure if there's any benefit to doing so.
Another manipulation that occurred to me would be to recursively expand the right side: $$\begin{align*} x_{n+1}&=\frac{c}{x_{n-1}}+x_n\\ &=c\left(\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}\right)+x_{n-1}\\ &=c\left(\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}+\frac{1}{x_{n-3}}\right)+x_{n-2} \end{align*}$$ though I don't really know what this would accomplish.