As shown in the links above, the standard proof of Problem 6 of IMO 1988 is to consider a wider situation, and then clamp down on the situation of interest. Another proof in the same spirit is the topological proof, by Levinson and Purbhoo, of the Shapiro-Shapiro conjecture: They allow a certain Wronskian to have complex conjugate roots, and then later note that the result of interest follows from the case when all the roots are real. (Both proofs also proceed by leveraging polynomials.)

Special cases of Problem 6 of IMO 1988 might be easy to prove. Here is one:
It is impossible for k to be equal to 2 and for a to be equal to b.
Proof: We would have (a^2 + b^2)/(1 + ab) = 2. We would then have (a^2 + b^2)/2 = 1 + ab.
Applying the Arithmetic-Mean – Geometric-Mean Inequality to the quantities a^2 and b^2, we would then have (a^2 + b^2)/2 is less than or equal to ab. However, as is well-known, this inequality becomes an equality when the quantities in question are all the same. Since a = b, we then have (a^2 + b^2)/2 = ab, which contradicts the fact that (a^2 + b^2)/2 = 1 + ab. So, it cannot be the case that k = 2 and a = b. Q.E.D.
keywords: International Mathematics Olympiad, square, divisibility, Number Theory