Generate recurrence relation for a hyperbolic equation I want to generate the recurrence relation for the integer solution of the equation 3m² - 2n² = 1 ; I'm a tenth grader, and so it took me pretty long to at first get out of the normal Quadratic Formula and stuff and realize that there exists things like Cayley-Hamilton Theorem, but unfortunately, I'm inexperienced enough to even try that !! I used Diophantine solving methods, manipulating it but end up messing with the equation ... With reference to Ryan's Generating Integer solution from Hyperbolic Equation , I'd appreciate if one can explain Will Jagy's solution and how he arrived at the recurrence xn+2 =  6xn+1 - xn
 A: The equation is the same as
$$(3m)^2-6n^2=3$$
which in effect is the same as
$$r^2-6n^2=3$$
as each solution of this has $3\mid r$.
That's the same as
$$(r+n\sqrt6)(r-n\sqrt6)=3.$$
Now
$$(5+2\sqrt6)(5-2\sqrt6)=1.$$
If
$$r'+n'\sqrt6=(5+2\sqrt6)(r+n\sqrt6)$$
then
$$r'-n'\sqrt6=(5-2\sqrt6)(r-n\sqrt6)$$
which leads to
$$r'^2-6n'^2=(r'+n'\sqrt6)(r'-n'\sqrt6)=(5+2\sqrt6)(r+n\sqrt6)
(5-2\sqrt6)(r-n\sqrt6)=3.$$
Then $r'=5r+12n$ and $n'=2r+5n$. If we let $m'=r'/3=5m+4n$
then $n'=6m+5n$ and
$$3m'^2-2n'^2=1.$$
So we can start with $(m_0,n_0)=(1,1)$ and then
recursively define $$(m_{k+1},n_{k+1})=(5m_k+4n_k,6m_k+5n_k).$$
Then
$$m_{k+2}=5m_{k+1}+4n_{k+1}=5m_{k+1}+24m_k+20n_k=10m_{k+1}-m_k$$
and similarly for $n_k$.
A: I would expect you to do better with Conway's topograph method. Everything involved here can be done without actual matrices, including the recursion. 
I guess I will put some books, including a very good new one, CONWAY and HATCHER. There is also (I bought a copy) a well illustrated new one by WEISSMAN 
This time it is 
$$ x_{n+2} = 10 x_{n+1} - x_n \; \; , $$
$$ y_{n+2} = 10 y_{n+1} - y_n \; \; . $$
These come from the combined one-step rule
$$ (x_{n+1}, y_{n+1}) = (5 x_n + 4 y_n \; , \;  6 x_n + 5 y_n)  $$
which you get pretty quickly with Conway. Here is a fairly basic version of the topograph for $3x^2 - 2 y^2,$ showing several locations with $3 x^2 - y^2 = 1$ labelled with the appropriate $\left( \begin{array}{c} x \\ y \end{array} \right)$ in green

Then the separation into two rules can be done without matrices, I wrote that out at http://math.stackexchange.com/questions/1786760/recursive-system-solution/1786850#1786850
Mon Apr  2 10:34:50 PDT 2018
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 x:  881   y:  1079
 x:  8721   y:  10681
Mon Apr  2 10:34:53 PDT 2018

