<s>[INST] 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?  
 [/INST] Supposing that
$$ \frac{a^2 + b^2}{ab + 1} = k$$ then 
$$ a^2 - a(kb) + (b^2 - k) = 0 $$ 
So using quadratic formula gives: $$ a = \frac{kb \pm \sqrt{k^2b^2 -4(b^2 -k)}}{2}$$
The solutions are when $k = b^2 $ and thus $a= kb = b^3$
So $$ b = 1 , a = 1 , k= 1$$
$$b = 2 , a = 8, k = 4$$
$$ b= 3, a= 27, k = 9 $$ and so on....
There are other ways of getting more solutions than these such as b = 8, a = 30 and k = 4 where k is not $b^2$ or $a = b^3$ and of course the zero one. I have not yet found a way to find more solutions than this.
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