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

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The discussion revolves around the math puzzle a^2 + b^2 = k(ab + 1), where a and b are positive integers. The main question is whether k can only be a fraction or a perfect square, with the initial assertion being that k is always a fraction. It is clarified that the equation should be expressed as k = (a^2 + b^2) / (1 + ab) and the task is to show that if k is an integer, it must be a perfect square. The proof and its historical context are referenced in a Wikipedia article on Vieta jumping. The thread emphasizes the mathematical implications of the equation and its connection to a specific problem from the 1988 International Mathematical Olympiad.
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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.
 

Thread 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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