# IMO 1988/Problem 6

I spent some time thinking about the infamous IMO 1988/Problem 6 today:

For positive integers $a,b$, if $\frac{a^2+b^2}{ab+1}$ is an integer, prove that it is a square number.

After some initial false starts, I came up with this:

$a^2+b^2=(ab+1)(\frac{a}{b}-\frac{1}{b^2})+(b^2+\frac{1}{b^2})$. We need to eliminate the $\frac{1}{b^2}$, in order to have some hope of a quotient that is an integer. One way to do that is to have $(ab+1)\frac{1}{b^2}=(b^2+\frac{1}{b^2})$. This implies that $\frac{a}{b}=b^2$. We have thus proven that the quotient, which is $\frac{a}{b}$, is a square number.

Of course, this is not the complete solution. We might have $(ab+1)(\frac{a}{b^2}+l)=(b^2+\frac{1}{b^2})$, in which case the quotient could also be an integer. However, it turns out that the solution that I have written above is the only possible solution. Now we only need to justify that this is the only solution.