All integer values of $\frac{a^2+b^2+1}{ab-1}$ Determine all possible values of $\frac{a^2+b^2+1}{ab-1}$ where $a,b$ are positive integers.
I am quite certain one should use a Vieta jumping argument, but I cannot complete it.
Let $\frac{a^2+b^2+1}{ab-1} = k \in \mathbb{Z}$ with $a+b$ being minimal. Clearly $k > 0$ and we have $a^2 - kab + b^2 + k + 1 = 0$. Without loss of generality $a\geq b$. We see that $x_1 = a$ and $x_2 = kb-a = \frac{b^2+k+1}{a}$ are the roots of $x^2 - kbx + b^2 + k + 1 = 0$. The expression $kb-a$ shows they are both integers and $\frac{b^2+k+1}{a}$ they are both positive. To reach a contradiction, it suffices to have $b + \frac{b^2+k+1}{a} < a + b$, i.e. $b^2 + k + 1 < a^2$. Hmm, if we can perhaps somehow assure $k\leq a$, we will just consider pairs $(a,b)$ with $a\neq b$ and $a+b$ minimal $-$ that's how we can rule out the case $a\neq b$ and the rest will be very easy. So any idea how to complete this argument, if it indeed works?
Any help appreciated!
 A: If the ratio is $n,$ then we have $x^2 - nxy + y^2 + (n+1) = 0,$ with both $x,y$ positive integers.
Now, if  $2y > nx,$ then we have $nx - y < y,$ which means that "jumping" the $y$ leads us to a smaller point in terms of $x+y.$ Same thing, if $2x > ny,$ jumping $x$ decreases $x+y.$ So, Hurwitz defined a "ground" solution ( Grundlösung) to be one with $2y \leq nx$ and $2x \leq ny.$ The picture becomes very simple, the part of the hyperbola arc between two slanted lines. 
If there are any integer points, Vieta Jumping takes us to an integer point on the hyperbolic arc with $x \geq \frac{2}{n} y$ and $y \geq \frac{2}{n} x.$ There are such points for $n=3,$ at point $(2,2),$ then $n=6,$ points $(1,2)$ and $(2,1).$ That is it. 
The solutions with $n=3$ are pairs $(x_{n+1}, x_n)$ from the sequence
$$ 2, 4, 10, 26, 68, 178, 466, ... $$
where $x_{n+2} = 3 x_{n+1} - x_n.$
The solutions with $n=6$ are pairs $(x_{n+2}, x_n)$ from the sequence
$$ 1, 1, 2, 4, 11, 23, 64, 134, 373, 781, 2174, 4552,...  $$
where $x_{n+4} = 6 x_{n+2} - x_n.$
Ummm for any $n \geq 7,$ the hyperbola crosses the line $y = 1$ with $1<x<2.$ Also crosses  the line $x = 1$ with $1<y<2.$






