Vieta Root Jumping - 1

Let $\frac{x^{2}+y^{2}+1}{xy} = k$ . Let, Among all pairs $(x,y)$ which satisfy the above equation let $(X,Y)$ be the solution which minimizes the sum $x+y$ . As the system is symmetrical, WLOG let $X>Y$ .

Now, Consider the equation $\frac{t^{2}+Y^{2}+1}{tY} = k$ .

Then,

$t^{2}-kYt+Y^{2}+ 1 = 0$ is a quadratic in $t$ . We know that $t_{1} = X$ is a solution to this quadratic.

So, by vieta's roots, the other root is $t_{2} = kY-X = \frac{Y^{2} +1}{X}$

But $t_{2}$ is also an integer. Also, $X>Y \geq 1$ .

So, $t_{2}= \frac{Y^{2} +1}{X} < X$

So, $t_{2} + Y < X+Y$ , which contradicts the minimality of $X+Y$ .

So, our assumption that $X>Y$ is false. So, $X=Y$ and thus $X^{2} | 2X^{2}+1$ .

And this forces that $X=1$ . And thus $\boxed{k=3}$ .