Conjecture on integer solutions to the equation $ (ab + 1) \mid (a^{2}+b^{2})$ Inspired by the egregious problem in IMO 1988, I simulated the integer solutions to the equation
$$ (ab + 1) \mid (a^{2} + b^{2}) \tag{*}$$
for $1 \leq a, b \leq 3000$ and conjectured that every solution arises as an adjacent pair of numbers in the following sequence
$$ a_{0} = 0, \quad a_{1} = m, \quad \text{and} \quad a_{n+2} = ma_{n+1} - a_{n} $$
for some $m \in \Bbb{Z}$. Indeed, you can check that any pair $(a, b) = (a_{n}, a_{n+1})$ gives rise to a solution to $\text{(*)}$ with
$$ \frac{a_{n+1}^{2} + a_{n}^{2}}{a_{n+1}a_{n} + 1} = m^{2}. $$
I was unable to prove this conjecture as I'm ham-handed at number theory. Can you help me prove or disprove this?
 A: This problem has solve it six years ago in china,can you konw chinese Chinese language?
and this problem full solution on this book(page 457):you can download :
http://ishare.iask.sina.com.cn/f/22755033.html?sudaref=www.baidu.com&retcode=0
because this problem solution is very ugly,so  I only post This problem results
we have $(a,b)=(a_{k},b_{k})$,and

$$a_{k}=\dfrac{d}{\sqrt{d^4-4}}\left(\left(\dfrac{d^2+\sqrt{d^4-4}}{2}\right)^k- \left(\dfrac{d^2-\sqrt{d^4-4}}{2}\right)^k\right)$$
  and
  $$b_{k}=a_{k+1},k=1,2,\cdots,b_{k}\ge a_{k}$$
  where $d^2=m=gcd(a,b)$

A: 
I simulated the integer solutions to the equation $ (ab + 1) \mid (a^{2} + b^{2})$ for $1 \leq a, b \leq 3000$ and conjectured that every solution arises as an adjacent pair of numbers in the following sequence
$$ a_{0} = 0, \quad a_{1} = m, \quad \text{and} \quad a_{n+2} = ma_{n+1} - a_{n} $$
for some $m \in \Bbb{Z}$. 

Yes, that is another way to state what happens in the proof of IMO 1988/6 that inspired your calculations.   
The proof describes an algorithm that reduces an integer solution $(a,b)$ with $0 < a \leq b$ to a smaller solution. Repeating the reduction many times, one arrives at $a=0$ and $b=m$ for some integer $m$. Doing the reduction process in reverse is the same as your recurrence relation (for constant integer $m$), and will reproduce all solutions $(a,b)$ with that value of $m$.  
Taking the union over all $m$ gives your conjecture (for positive $m$).  
