<s>[INST] Understanding a proof that, if $xy$ divides $x^2+y^2+1$ for positive integers $x$ and $y$, then $x^2+y^2+1=3xy$ This is a Worked Example from Brilliant.org's entry on Vieta Root Jumping.

Let $x$ and $y$ be positive integers such that $xy$ divides $x^2+y^2+1$. Prove that $$x^2+y^2+1=3xy$$

The solution proposes that $x^2+y^2+1=k(xy)$ where $k$ is an integer.
It claims that there exists a minimum solution $(x,y)$ that has the minimum value of $x+y$.
So, they use $t$ to replace $x$ to show that $t^2-kty+y^2+1=0$
Then $t_1=x$ is one solution. By vieta formula,$t_1+t_2=ky$
Then $t_2=ky-x=\frac{x^2+y^2+1}{x}-x=\frac{y^2+1}{x}$
which implies $t_2\lt y$ then $t_1+t_2\lt x+y$.
So, the minimum condition only exists when $x=y$
I am ok as far, but after that it says, $x^2$ divided by $2x^2+1$, $x^2$divided by $1$.
So $k=3$.

But, why they can get $k=3$? $k=3$ only when $x$ and $y$ be the minimum solution. Why $k$ cannot be multiple of $3$? 

 [/INST] Suppose there is a larger $k$. Then we can apply Viete's root jump and 'descend' the roots infinitely, which is impossible since there is no infinite descending chain in the naturals. This is the crux of the whole article you linked to.
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