Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$ 
Find infinitely many pairs of integers $a$ and $b$ with $1 < a < b$, so that $ab$ exactly divides $a^2 +b^2 −1$. What are the possible values of $$\frac{x^2+y^2-1}{xy}$$?

I have discovered infinitely many pairs. They are $(n, n+1)$. This solution works for all $n$. For all of them the value of $$\frac{x^2+y^2-1}{xy}=2$$
But I want to know if there are any other such solutions. Also what are all the possible values. I have discovered 2. There may be many such values. Are they finite? Can we prove their finite or infinite existence? 
Any help would be appreciated.
 A: Note that the usual problem, and quite difficult if you do not know what to expect, is slightly different, https://en.wikipedia.org/wiki/Vieta_jumping#Example_2
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Define integer $k \geq 3$ as
$$k = \frac{x^2+y^2-1}{xy}$$
We are solving
$$ x^2 - kxy + y^2 = 1. $$
The discriminant of the quadratic form is $\Delta = k^2 - 4$ which is positive but not a square. As a result, if we can solve $\tau^2 - \Delta \sigma^2$ we can construct the generator of the (oriented) automorphism group of the form. Now, $k^2 - \Delta = 4,$ so $\tau = \pm k, \sigma = \pm 1.$
Given a form (positive nonsquare discriminant) of coefficients $\langle A,B,C\rangle$ we can take the generator (in $SL_2 \mathbb Z$) as
$$
\left(
\begin{array}{rr}
\frac{\tau - B \sigma}{2} && -C \sigma \\
A \sigma && \frac{\tau + B \sigma}{2}
\end{array}
\right)
$$
With $\langle A,B,C\rangle = \langle 1,-k,1\rangle$ we get
$$  U = 
\left(
\begin{array}{rr}
k && -1 \\
1 && 0
\end{array}
\right)
$$
and will stick with this one, as convenient for keeping $x \geq y \geq 0.$
The transformation from one solution $(x,y)$ to the next is matrix multiplication with the column vector $(x,y)^T$ on the right. That is,
$$ \color{blue}{ (x,y) \mapsto (kx-y, x)}. $$
The first few solutions with $k \geq 3$ are
$$ (1,0), $$
$$ (k,1), $$
$$ (k^2 - 1,k), $$
$$ (k^3 - 2k,k^2 - 1), $$
$$ (k^4 - 3 k^2 + 1,k^3 - 2k). $$
Oh, as far as actual division, $(1,0)$ is not legal in the original fraction. Life is like that. 
Let's try $k=3$ in the fraction:
$(3,1)$ gives $9/3 = 3.$ $(8,3)$ gives $72/24 = 3.$ 
The final thing is Cayley-Hamilton. The matrix I named $U$ above satisfies $U^2 - k U + I = 0,$ or $U^2 = kU - I.$ As a result, both $x$ and $y$ satisfy linear recurrences,
$$ x_{j+2} = k x_{j+1} - x_j, $$ 
$$ y_{j+2} = k y_{j+1} - y_j. $$
Again with $k=3, $we get $x$ in
$$ 1, 3, 8, 21, 55, 144,  $$
Note that  these are every second Fibonacci number.
With $k=4, $we get $x$ in
$$ 1, 4, 15, 56, 209, 780,  $$
With $k=5, $we get $x$ in
$$ 1, 5, 24, 115, 551, 2640,  $$
With your $k=2$ you have $x^2 - 2xy + y^2 = 1,$ or $(x-y)^2 = 1,$ or $x-y = \pm 1.$
A: Every $n$ works.
$$\forall n\in\mathbb{Z}^+. \,\,\,\,n\cdot n \cdot (n^2-1) = n^2+ (n^2-1)^2 -1$$
So,
$$\forall n\in\mathbb{Z}^+.\,\,\,\,\ \frac{n^2 + (n^2-1)^2 -1}{n \cdot (n^2 -1)} = n$$
