How correct is my solution for IMO 1988 Problem 6?

Asked Sep 07 '21 at 13:18

Active Sep 07 '21 at 13:18

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Problem #6 at IMO 1988: Let a and b be positive integers such that $ab + 1$ divides $a^2 + b^2$. Prove that $\frac{a^2 + b^2}{ab + 1}$ is a perfect square.

Let this perfect square be $n$.

$a^2+b^2=n(ab+1)$
$a^2+b^2-abn-n=0$
$a^2-(bn)a+(b^2-n)=0$
$a=\frac{bn+\sqrt{b^2n^2-4(b^2-n)}}{2}$
$b^2n^2-4(b^2-n)$ must then be a square for $a$ to be an integer.
$4(b^2-n)=4b^2-4n$ is then a difference of two squares, and
$n$ must be a square as $4b^2$ is already a square.

asked Sep 07 '21 at 13:18

BobTheBuilding

4

If I understand correctly, your argument in the last line would conclude that since $5^2 - 4^2 = 6^2 - 27$, it should follow that $27$ is a perfect square... – Brian Moehring Sep 07 '21 at 13:37
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No, not good. Why must $n$ be a square if $4b^2$ is already a square? Compare with the beautiful solutions at the various duplicates. – Dietrich Burde Sep 07 '21 at 13:38
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Does this answer your question? Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer – Sep 07 '21 at 14:01
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For what it is worth, I explored the link in the comment of @JitendraSingh. I had trouble following some of the proofs. In that link, you can to to the answer by Amir Parvardi (current score = 33), and take the very 1st link in his answer. This takes you to an AOPS article that contains several proofs. In this AOPS article, the specific proof that I was able to follow was a descent argument by [james digol : may 18, 2010]. – user2661923 Sep 07 '21 at 16:04
Ah I see. I missed that, thank you! – BobTheBuilding Sep 07 '21 at 23:27

How correct is my solution for IMO 1988 Problem 6?

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