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.

Published by ayushkhaitan3437

Hello! My name is Ayush Khaitan, and I'm a graduate student in Mathematics.

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