The Pell equation $(2x+y)^2-5y^2=4$ has fundamental solution $(1,1)$. I need to find three more solutions, I know tha they are the Fibonacci numbers in ordered pairs $(2,3), (5,8),...$, but I don't know how to find them formally. Normally we would compute $x_i=(1+1*\sqrt5)^i$ when the equation the Pell equation is equal to $1$, but this is a little different, it is general (equal to $4$ and there is a $(2x+y)$ term).
Any solutions/hints are greatly appreciated.
0,2,4,6,..
. Then $y=abs(q)$ and $x=(p-y)/2$. Finally $(x,y)$=(1,0), (1,1), (2,3), (5,8), (13,21), (34,55), (89,144), (233,377), (610,987),..
$\endgroup$