2006 IMO Problems/Problem 4
Determine all pairs of integers such that
If is a solution then obviously and is a solution too. For we get the two solutions and .
Now let be a solution with ; without loss of generality confine attention to . The equation rewritten as shows that the factors and are even, exactly one of them divisible by . Hence and one of these factors is divisible by but not by . So Plugging this into the original equation we obtain or, equivalently Therefore For this yields , i.e. , which fails to satisfy . For equation gives us implying . Hence ; on the other hand cannot be by . Because is odd, we obtain , leading to . From we get . These values indeed satisfy the given equation. Recall that then is also good. Thus we have the complete list of solutions : , , , .