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I think most of you may know the well-known problem:

"Let $x$ and $y$ be positive integers such that $xy + 1$ divides $x^{2} + y^{2}$. Show that $\frac {x^{2} + y^{2}}{xy + 1}$ is the perfect square."

I know that this problem is meant for using the 'Vieta jumping' method. But I wanted to solve this problem in another way.

First, using Wolfram Alpha, I wanted to make an equation.

$\frac{x^{2} + y^{2}}{xy + 1} = k$

and solved over $x$ (or $y$, it's same)

The result was $\frac {1}{2}$$\sqrt{4 k - 4 x^2 + k^2 x^2} + \frac {1}{2}$$kx$, $-\frac {1}{2}$$\sqrt{4 k - 4 x^2 + k^2 x^2} + \frac {1}{2}$$kx$ (I am first using Tex...)

I thought that if $y$ is an integer, "$4 k - 4 x^2 + k^2 x^2$ has to be a perfect square" (Even though it is a perfect square, because of $\frac {1}{2}$, it could be a rational number, not an integer. But at least $4 k - 4 x^2 + k^2 x^2$ has to be a perfect square).

I have tried many integers with $k$ and $x$, and concluded that $4 k - 4 x^2 + k^2 x^2$ can be a perfect square only when $k$ is a perfect square, which means the initial hypothesis was right. (For example, if $k$ is $4$, $12x^2 + 16$ can make a perfect square when $x$ is 2, 8, 30....)

So what I want to ask you is that "how to prove $4 k - 4 x^2 + k^2 x^2$ can be a perfect square only when $k$ is a perfect square".

The next thing I am curious about is, it is said that 11 of the IMO participants solved perfectly. I don't think all of them used the same 'Vieta jumping' method. What do you think of more "general ways" to solve this problem for other participants?

Thank you. (Sorry for the bad English)

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    $\begingroup$ 1988 IMO 6 has been discussed several times on math.stackexchange, probably also on art of problem solving website. Have you checked to see whether any non-Vieta solutions have been posted on those sites? $\endgroup$
    – Gerry Myerson
    Commented Jul 11, 2020 at 1:28
  • $\begingroup$ @GerryMyerson Thank you for the comment. Of course, I have seen several posts about non-Vieta solutions about this problem but I think I have never seen this kind of approach to this problem... or maybe my skills are not good enough to understand other solutions... $\endgroup$
    – SG Kwon
    Commented Jul 12, 2020 at 4:24
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    $\begingroup$ The way the last paragraph of your question is written, it seemed to me that you hadn't seen any non-Vieta solutions, or at any rate that you were asking to see more. But from your comment, it appears you aren't interested in other solutions, you're only interested in whether your approach leads to a solution. Is that right? Then it's not clear to me that reducing the problem to deciding whether $4k-4x^2+k^2x^2$ square implies $k$ square is making any progress toward a solution. $\endgroup$
    – Gerry Myerson
    Commented Jul 12, 2020 at 5:49

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