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Let $ x $ and $ y $ be two positive integers. Prove that

$$(xy+1) | (x^2+y^2)\;\implies \frac{x^2+y^2}{xy+1}\;\text{ is a perfect square}$$

I assumed there exists $ k\in \Bbb N $ such that $$x^2+y^2-k(xy+1)=0$$ or $$x^2-kyx+y^2-k=0$$ i said that the discriminant $$\Delta=k^2y^2+4k-4y^2$$ must satisfy $$\sqrt{\Delta}\in \Bbb N$$ but it seems to complicate . Thanks in advance.

hamam_Abdallah
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  • Does this answer your question? [Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer](https://math.stackexchange.com/questions/28438/alternative-proof-that-a2b2-ab1-is-a-square-when-its-an-integer) – Sebastian Monnet Jul 19 '20 at 14:06
  • Thank yo so much. Appreciate your time. – hamam_Abdallah Jul 19 '20 at 14:08
  • Even [Wikipedia](https://en.wikipedia.org/wiki/Vieta_jumping) has an article devoted to this problem and the technique used to solve it. – Arthur Jul 19 '20 at 14:14
  • Ah yeah this was the famous question 6. The method used to solve it is called "Vieta Jumping" - it's a form of contradiction. The solutions to the equation are connected in some way, and you show that, using the way the solutions are connected - that the expression must be a perfect square (otherwise you contradict some minimality condition). It's super genious – Riemann'sPointyNose Jul 19 '20 at 14:36

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