IMO 1988 Day 2, Question 6
an elegant problem with an even more elegant solution
Let $a, b\in\mathbb{Z}^{+}:ab+1 | a^2+b^2.$ Show that
$\displaystyle\frac{a^2+b^2}{ab+1}=r^2$
where $r\in\mathbb{Z}$.
I figured out that the solutions were $a=n,b=n^3,$
and then concluded falsely that that was it. Turns out there are infinitely more and
the only reason that $(n,n^3)$ shows up as the first solution is a coincidence of
the complete proof.
Zvezdelina Stankova explains on Numberphile one solution
that uses induction — with infectious enthusiasm. This problem consumed me, and its
insides were wondrous.