Problem 6 of the 1988 International Math Olympiad reads:

Letaandbbe positive integers such thata b + 1divides . 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*.

Now

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 .