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1988 IMO Question Six - mklabgr - 09-09-2026
1988 IMO Question Six — Solving the Hardest Problem on the Hardest Test
Author: Ethan Horsfall
Published: January 5, 2020, in Cantor’s Paradise
The article explains the celebrated Problem 6 of the 1988 International Mathematical Olympiad, often regarded as one of the hardest IMO problems ever posed. The problem asks: if positive integers $a,b$ satisfy $\frac{a^2+b^2}{ab+1}=x$ for some integer $x$, prove that $x$ must be a perfect square.
The proof uses elementary algebra but develops a powerful descent argument. Writing the equation as $a^2+b^2=x(ab+1)$ and treating it as a quadratic in one variable gives $b^2-xab+(a^2-x)=0$. If $b$ is one integer root, Vieta’s formulas provide another integer root, allowing one solution to be replaced by a smaller related solution. The argument first establishes bounds such as $a,b\geq\sqrt{x}$, and then repeatedly applies this root-switching process.
If both variables always remained strictly larger than $\sqrt{x}$, the descent would continue indefinitely, contradicting the existence of a smallest positive solution. Therefore eventually one obtains $a=\sqrt{x}$ or $b=\sqrt{x}$. Since $a$ and $b$ are integers, $\sqrt{x}\in\mathbb Z$, and consequently $x$ is a perfect square. This ingenious method became a classic example of Vieta jumping, combining quadratic equations, Vieta’s formulas and infinite descent to solve an apparently very difficult Diophantine problem using only elementary mathematics.
Key takeaways: The central equation is $\frac{a^2+b^2}{ab+1}=x$; interpreting it as a quadratic creates a second integral solution through Vieta’s formulas; repeatedly moving to a smaller solution produces an infinite-descent argument; and the process ultimately forces $a=\sqrt{x}$ or $b=\sqrt{x}$, proving that $x$ is a perfect square.
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