1988 IMO Problems/Problem 6

Problem

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^{2} + b^{2}$ . Show that $\frac {a^{2} + b^{2}}{ab + 1}$ is the square of an integer.

Solution

Choose integers $a,b,k$ such that $a^2+b^2=k(ab+1)$ Now, for fixed $k$ , out of all pairs $(a,b)$ choose the one with the lowest value of $\min(a,b)$ . Label $b'=\min(a,b), a'=\max(a,b)$ . Thus, $a'^2-kb'a'+b'^2-k=0$ is a quadratic in $a'$ . Should there be another root, $c'$ , the root would satisfy: $b'c'\leq a'c'=b'^2-k<b'^2\implies c'<b'$ Thus, $c'$ isn't a positive integer (if it were, it would contradict the minimality condition). But $c'=kb'-a'$ , so $c'$ is an integer; hence, $c'\leq 0$ . In addition, $(a'+1)(c'+1)=a'c'+a'+c'+1=b'^2-k+b'k+1=b'^2+(b'-1)k+1\geq 1$ so that $c'>-1$ . We conclude that $c'=0$ so that $b'^2=k$ .

This construction works whenever there exists a solution $(a,b)$ for a fixed $k$ , hence $k$ is always a perfect square.

1988 IMO (Problems) • Resources
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1988 IMO Problems/Problem 6

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