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Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer

Whenever $1+ab$ divides $a^2+ b^2$, I need to prove that $\frac{a^2+b^2}{1+ab}$ will be a perfect square.

This is the last problem in Exercise 1.3 of number theory book by Ivan Niven. The chapter only has dealt with GCD and LCM concepts so far, along with a few division algorithms. However using the abovesaid principles I have failed to reach any solution.

So any help please?

Bhargav
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  • didnt quite get the result in the search i made before posting this q, thanks anyways to the mods and chandrasekhar – Bhargav Dec 25 '11 at 16:59
  • New and better solution without using vieta jumping method here http://math.stackexchange.com/questions/28438/alternative-proof-that-a2b2-ab1-is-a-square-when-its-an-integer/646382#646382 – MathGod Jan 23 '14 at 06:55

1 Answers1

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Right this is a IMO $1988$ problem. Check out this link.

  • New and better solution without using vieta jumping method here http://math.stackexchange.com/questions/28438/alternative-proof-that-a2b2-ab1-is-a-square-when-its-an-integer/646382#646382 – MathGod Jan 23 '14 at 06:54