IMO 1988, problem 6 In 1988, IMO presented a problem, to prove that $k$ must be a square if $a^2+b^2=k(1+ab)$, for positive integers $a$, $b$ and $k$.
I am wondering about the solutions, not obvious from the proof.  Beside the trivial solutions a or $b=0$ or 1 with $k=0$ or $1$, an obvious solution is $a=b^3$ so that the equation becomes $b^6+b^2=b^2(1+b^4)$ . Are there any other solutions?  

Not quite a proof to the original IMO problem, but there definately is a very easy way to compute all possible answers.
Also it demonstrates that here, Vieta jumping is basically just using
symmetry to jump between the two solutions of the good old quadratic formula.
Nothing complicated below, but I haven't seen this explanation anywhere else.
Suppose:
$$
{{a^2+b^2} \over {1+ab}} = c
$$
Then using the quadratic formula to solve b (and a little rewriting) gives:
$$
b = {ac \pm \sqrt {a^2 (c^2-4) + 4c} \over 2}
$$
Using this we can compute all answers for any given c. 
Eg: assume c = 4. Then we get:
$$
b = {a \cdot 4 \pm \sqrt {a^2 \cdot 12 + 16} \over 2}
$$
We know that a = 0 will always provide a solution. That gives us:
$$
a = 0 \Rightarrow b = {{0 \cdot 4\pm \sqrt {0 \cdot 12 + 16}} \over 2} = 2 \Rightarrow (a,b) = (0,2) 
$$
But because of the symmetry, if (0,2) is a solution, then (2,0) must
also be a solution. And since the quadratic formula has 2 solutions,
it will produce another solution.
$$
a = 2 \Rightarrow \space b = {{2 \cdot 4 \pm \sqrt {4 \cdot 12 + 16}} \over 2} = 0 \space or \space 8 \Rightarrow (a,b) = (2,8)
$$
$$
a = 8 \Rightarrow \space b = {{8 \cdot 4 \pm \sqrt {64 \cdot 12 + 16}} \over 2} = {32 \pm 28\over 2} = 2 \space or \space 30 \Rightarrow (a,b) = (8,30)
$$
$$
a = 30 \Rightarrow \space b = {{30 \cdot 4 \pm \sqrt {900 \cdot 12 + 16}} \over 2} = {120 \pm 104\over 2} = 8 \space or \space 112 \Rightarrow (a,b) = (30,112)
$$
Same can be done for c = 9, 16, 25, etc.
