Solve Odd Math Puzzle: a^2+b^2=k(ab+1)

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SUMMARY

The mathematical puzzle presented is defined by the equation a^2 + b^2 = k(ab + 1), where A and B are positive integers. The discussion confirms that k can only be a fraction or a perfect square, as derived from the expression k = (a^2 + b^2) / (ab + 1). This aligns with the problem #6 from the International Mathematical Olympiad (IMO) 1988, which states that if k is an integer, then k must be a perfect square. The proof and historical context of this problem can be found in the Wikipedia article on Vieta jumping.

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mathpleb
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This isn't for me, it's for a friend. I'm still teaching myself stuff before Trigonometry.

Anyways, he has a puzzle. a^2+b^2=k(ab+1).

A and B are given as positive integers.

Q: "Prove that K can only take on the value of fractions or squares."
 
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Isn't this clear, since $k = \frac{a^2 + b^2}{ab + 1}$ is always a fraction when $a$ and $b$ are positive integers? Or perhaps I am overlooking something?
Also, could it be you posted this accidentally in the "differential equations" section?

EDIT: I see it was already moved to the right section, thank you (Smile)
 
I suspect this is supposed to be problem #6 at IMO 1988, in which case it should read:

Let $\displaystyle k={{a^2+b^2}\over{1+ab}}.
$ Show that if $k$ is an integer then $k$ is a perfect square.​

The proof (along with its history) is given in the Wikipedia article on Vieta jumping.
 

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