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The famous problem number 6 of the 1988 International Mathematical Olympiad is about showing that if $a,b$ are non-negative integers such that $\frac{a^2+b^2}{ab+1}$ is an integer, then it is a square number.

Given $A\subseteq \mathbb{N}$, let $\mu^+(A) = \lim\sup_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n+1}$ be the upper density of $A$.

Let $I$ be the set of positive integers $n$ such that there are positive integers $a,b$ with $n^2 = \frac{a^2+b^2}{ab+1}$.

(I write positive integers $a,b$ above, because $n^2 = \frac{0^2 + n^2}{0\cdot n + 1}$.)

What is the value of $\mu^+(I)$?

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From the solution for the problem (Vieta jumping), one can see that if there is one pair $(a,b)$ for which $\frac{a^2+b^2}{ab+1}=k$ then there are infinitely many. For instance, $(a,b)=(n^3, n)$ also works.

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  • $\begingroup$ This is not an answer. $\endgroup$
    – Fan Zheng
    Commented Aug 16, 2019 at 7:03
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    $\begingroup$ Yes it is. He explained how the trivial solution $(a, b) =(0, n) $, which was excluded in the question, gives rise to a solution in positive integers $(a, b)=(n^3,n)$ for any $n$. $\endgroup$
    – Achim Krause
    Commented Aug 16, 2019 at 7:42

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