IMO Shortlist 1988 problem 9
Dodao/la:
arhiva2. travnja 2012. Let
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
and
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
be two positive integers such that
![a \cdot b + 1](/media/m/8/1/9/819861a7e4bb22c90f07346e54d19776.png)
divides
![a^{2} + b^{2}](/media/m/6/d/3/6d3d14b2df344553404ea47d410902c5.png)
. Show that
![\frac {a^{2} + b^{2}}{a \cdot b + 1}](/media/m/4/c/4/4c4fb573cadf28fdad98bca01609efca.png)
is a perfect square.
%V0
Let $a$ and $b$ be two positive integers such that $a \cdot b + 1$ divides $a^{2} + b^{2}$. Show that $\frac {a^{2} + b^{2}}{a \cdot b + 1}$ is a perfect square.
Izvor: Međunarodna matematička olimpijada, shortlist 1988