Problem 6 of the 1988 International Math Olympiad reads:
Let a and b be positive integers such that a b + 1 divides . Show that is the square of an integer.
There’s a writeup of a solution at the Art of Problem Solving wiki, but I found it incredibly sketchy and difficult to read and the article was locked, so here’s what’s hopefully an easier-to-read exposition of the same solution:
In other words, we’d like to show that any solution in positive integers to the equation has c a perfect square.
First, suppose we have a solution with a = b. Then divides . Applying the Euclidean algorithm,
so we must have . Since a is positive, we must have a=1. In this case, so c = 1 which is a perfect square. So we may assume in what follows that .
Fix a positive integer c for which the equation has solutions in the positive integers, and let (a, b, c) be such a solution with the minimum possible value of a. Note that by symmetry we must have a < b, since otherwise (b, a, c) is a solution with a smaller first coordinate.
Rewrite the equation as
and notice that this says that b is a root of the quadratic equation
Let r denote the other root. Then b + r = c a and . It follows that is an integer, and that
So r < a. This must mean that , as otherwise the solution (r, b, c) to the original equation would contradict the minimality of a.
Since b + 1 is positive, we must have r + 1 positive as well. Since r is an integer, we must have r = 0, so
or in other words .