<s>[INST] What is the algebraic intuition behind Vieta jumping in IMO1988 Problem 6? Problem 6 of the 1988 International Mathematical Olympiad notoriously asked:

Let $a$ and $b$ be positive integers and $k=\frac{a^2+b^2}{1+ab}$. Show that if $k$ is an integer then $k$ is a perfect square.

The usual way to show this involves a technique called Vieta jumping. See Wikipedia or this MSE post.
I can follow the Vieta jumping proof, but it seems a bit strained to me. You play around with equations that magically work out at the end. I don't see how anyone could have come up with that problem using that proof.
Is there a natural or canonical way to see the answer to the problem, maybe using (abstract) algebra or more powerful tools? In addition, how can someone come up with a problem like this?
 [/INST] I don't have a strong math background, I think my solution is more suitable for the Math olympiad contest as a young student.
Problem Definition:
Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^2+b^2$. Show that
 $\frac{a^2+b^2}{ab+1}$
is the square of an integer.
Proof:
Let $ t = \frac{a^2+b^2}{ab+1}$.
W.L.O.G we assume $1 \le a \le b$.
If $t=1$, we are done.
If $a=b$, we have  $2 b^2 = t (b^2+1) \ge 2 (b^2+1)$, contradiction.
If $a = 1$, we have $1+b^2 = t(b+1)$. 
As $t \equiv 1 (mod~b)$, we have  $t \ge 1+b$, contradiction.
Therefore, the major focus is on $t>1$ and $1 < a < b$.
Since $a^2 \equiv t ~(mod~b)$,  we define $k$ such that $a^2 = t + k b$, hence $k$ is an integer and $k < a$. 
Note that 
$ t = \frac{a^2+b^2}{ab+1} < \frac{a}{b} + \frac{b}{a}  \le 1 + b/2 \le b $.
If $k<0$, we have $a^2 = t+kb \le t - b \le 0$, contradiction.
If $k=0$, we have $t = a^2$, done.
Hence, the only remaining case is $0<k<a$.
In this case, we claim that 
$$ t = \frac{k^2+a^2}{k a+1} $$
Substitute $t$, we have  $ a^2+b^2 = (a^2 - k b)(ab+1)$, hence 
$$ b = (a^2 - k b)a - k = a t - k$$
As a result, $k+b = at$ and $kb = a^2-t$, that is, $k$ and $b$ are the solution of the equation
$$ x^2 - at x + (a^2-t) = 0 $$
One solution is $x = b$, the other solution is $x = k$, therefore the claim follows.
Consequently, whenever a feasible pair $(a,b)$ exists, we create another feasible pair $(k,a)$. 
By replacing $(a,b)$ by $(k,a)$ the pair become smaller as  $0<k<a<b$.
Because all the other case will result in $t$ is a square number, hence by repeating such replacement, we will finally get into the case where $t$ is a square number.
\qed
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