[Math] IMO 1988, problem 6

Question

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?

Answer 1

This is a famous problem, here is one of the solutions that I like the most that I read it in a book previously, but later in a topic on here I realized the importance of the problem (The credit goes to T. Andreescu & R. Gelca If I remember it correctly, but I'm not sure, since 11 individuals in that year solved this problem and I'm not sure about their solutions):

Solution: Suppose that $\displaystyle \frac{a^2+b^2}{a.b+1}=x$ We want to prove that for every non-negative integer pairs $(\alpha,\beta)$ with the property that $\displaystyle \frac{\alpha^2+\beta^2}{\alpha\beta+1}=x$ and $\alpha \geq \beta$ the pair that minimizes $\alpha+\beta$ must imply $\beta=0$. If that happened, then $x=\alpha^2$. So, suppose that $(\alpha,\beta)$ is such a pair that minimizes $\alpha+\beta$ but $\beta>0$. then we can obtain the equation $y^2 - \beta xy + \beta^2 -x =0$ from $\displaystyle \frac{y^2+\beta^2}{y\beta+1}=x$. This equation has $\alpha$ as one of its roots and since the sum of the roots is $\beta x$, the other root must be $\beta x - \alpha$. Now if we prove that $0 \leq \beta x - \alpha < \alpha$ then we're done because this contradicts the minimality of $(\alpha,\beta)$

$x = \displaystyle \frac{\alpha^2+\beta^2}{\alpha\beta+1}<\frac{2\alpha^2}{\alpha \beta} = \frac{2 \alpha}{\beta} \implies \beta x-\alpha < \alpha$

and it's also possible to show that $\beta.x - \alpha \geq 0$ but honestly I don't remember that part of the proof and I leave it to you. That completes the proof.

(Also check that wikipedia link provided by pre-kidney).