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?  

For given positive integers a and b, and applying simple division of $(a^2 + b^2)$ by $(ab + 1)$ we have quotient as a/b and remainder as $(b^2 - a/b)$.
For the result to be a perfect division, the remainder must be zero. Therefore $(b^2 - a/b) = 0$ which gives the relation $a=b^3$.
Applying $a=b^3$ in the equation $x=(a^2 + b^2)/(ab + 1)$, we have $x=b^2$ which is a perfect square of an integer.
All the number pairs $(a,b)$ of the form $(b^3,b)$ are solutions of the above equation and the number which is a perfect square is of the form $x = b^2$ applicable for all $b$ belonging to a set of positive integers.
