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 $3$?
LEMMA
Given integers $$ m > 0, \; \; M > m+2, $$ there are no integers $x,y$ with $$ x^2 - Mxy + y^2 = -m. $$
PROOF
Calculus: $m+2 > \sqrt{4m+4},$ since $(m+2)^2 = m^2 + 4m + 4,$ while $\left( \sqrt{4m+4} \right)^2 = 4m + 4.$ Therefore also $$ M > \sqrt{4m+4} $$
We cannot have $xy < 0,$ as then $x^2 - M xy + y^2 \geq 2 + M > 0. $ It is also impossible to have $x=0$ or $y=0.$ From now on we take integers $x,y > 0.$
With $x^2 - Mxy + y^2 < 0,$ we get $0 < x^2 < Mxy - y^2 = y(Mx - y),$ so that $Mx - y > 0$ and $y < Mx.$ We also get $x < My.$
The point on the hyperbola $ x^2 - Mxy + y^2 = -m $ has both coordinates $x=y=t$ with $(2-M) t^2 = -m,$ $(M-2)t^2 = m,$ and $$ t^2 = \frac{m}{M-2}. $$ We demanded $M > m+2$ so $M-2 > m,$ therefore $t < 1.$ More important than first appears, that this point is inside the unit square.
We now begin to use the viewpoint of Hurwitz (1907). All elementary, but probably not familiar. We are going to find integer solutions that minimize $x+y.$ If $2 y > M x,$ then $y > Mx-y.$ Therefore, when Vieta jumping, the new solution given by $$ (x,y) \mapsto (Mx - y, x) $$ gives a smaller $x+y$ value. Or, if $2x > My,$ $$ (x,y) \mapsto (y, My - x) $$ gives a smaller $x+y$ value. We already established that we are guaranteed $My-x, Mx-y > 0.$
Therefore, if there are any integer solutions, the minimum of $x+y$ occurs under the Hurwitz conditions for a fundamental solution (Grundlösung), namely $$ 2y \leq Mx \; \; \; \; \mbox{AND} \; \; \; \; 2 x \leq My. $$ We now just fiddle with calculus type stuff, that along the hyperbola arc bounded by the Hurwitz inequalities, either $x < 1$ or $y < 1,$ so that there cannot be any integer lattice points along the arc. We have already shown that the middle point of the arc lies at $(t,t)$ with $t < 1.$ We just need to confirm that the boundary points also have either small $x$ or small $y.$ Given $y = Mx/2,$ with $$ x^2 - Mxy + y^2 = -m $$ becomes $$ x^2 - \frac{M^2}{2} x^2 + \frac{M^2}{4} x^2 = -m, $$ $$ x^2 \left( 1 - \frac{M^2}{4} \right) = -m $$ $$ x^2 = \frac{-m}{1 - \frac{M^2}{4}} = \frac{m}{ \frac{M^2}{4} - 1} = \frac{4m}{M^2 - 4}. $$ We already confirmed that $ M > \sqrt{4m+4}, $ so $M^2 > 4m+4$ and $M^2 - 4 > 4m.$ As a result, $ \frac{4m}{M^2 - 4} < 1.$ The intersection of the hyperbola with the Hurwitz boundary line $2y = Mx$ gives a point with $x < 1.$ Between this and the arc middle point, we always have $x < 1,$ so no integer points. Between the arc middle point and the other boundary point, we always have $y < 1.$ All together, there are no integer points in the bounded arc. There are no Hurwitz fundamental solutions. Therefore, there are no integer solutions at all.
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