Indefinite binary quadratic forms, integer coefficients and discriminant not a square, possess an automorphism group; taking the Hessian matrix $H,$ an automorphism element is an integer matrix $P$ such that $P^T H P = H.$ The oriented part ($P$ has positive determinant) is infinite cyclic.
For the form $A x^2 + B xy + C y^2,$ with discriminant $\Delta= B^2 - 4 AC$ positive but not a square, Hessian
$$
H =
\left(
\begin{array}{cc}
2A & B \\
B & 2 C
\end{array}
\right),
$$
all positive determinant $P$ are given by integer solutions to $\tau^2 - \Delta \sigma^2 = 4,$ with
$$
P =
\left(
\begin{array}{cc}
\frac{\tau - B \sigma}{2} & - C \sigma \\
A \sigma & \frac{\tau + B \sigma}{2}
\end{array}
\right)
$$
For the case of $x^2 + Bxy+y^2$ with $B^2 > 4,$ the automorphisms take on the familiar Vieta appearance, and the negative determinant ones can be taken to be interchanging the variables. With a fixed target $T,$ any expression $x^2 + B xy + y^2 = T$ can be transported, by automorphisms, to a region satisfying desired inequalities. These desired solutions can be thought of as representative points in a group orbit. Siegel's description of counting solutions comes down to counting the representative solutions, that is the number of orbits, as the number of literal representations of a fixed target number is infinite.
Probably worth pointing out that finding a generator for the (oriented part) of the automorphism group requires solving $\tau^2 - \Delta \sigma^2 = 4,$ where $\Delta > 0$ is not a square. This would be a bit much. However, for the special case $x^2 + B xy + y^2,$ the discriminant is $\Delta = B^2 - 4,$ so that $B^2 - \Delta 1^2 = 4.$ There is also the single-variable "jumping" argument, done extremely well in Hurwitz 1907
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LEMMA 1
Given integers $$ M \geq m > 0, $$ along with positive integers $x,y$ with
$$ x^2 - Mxy + y^2 = m. $$
Then $m$ is a square.
PROOF.
First note that we cannot have integers $xy < 0$ with $ x^2 - Mxy + y^2 = m, $ since then
$ x^2 - Mxy + y^2 \geq 1 + M + 1 = M + 2 > m.$ If we have a solution with $x > 0$ and $xy \leq 0,$ it follows that $y=0.$
This is the Vieta jumping part, with some extra care about inequalities. We begin with
$$ y > x $$ and
$$ y < Mx. $$ We have
$$ x^2 - Mxy + y^2 > 0, $$
$$ x^2 > Mxy - y^2 = y(Mx - y) > x(Mx-y), $$
$$ x > Mx - y > 0. $$
That is, the jump
$$ (x,y) \mapsto (Mx - y,x) $$
takes us from one ordered solution to another ordered solution while strictly decreasing $x+y.$
Within a finite number of such jumps we violate the conditions we were preserving; we reach a solution $(x,y)$ with $y \geq Mx,$ that is
$x > 0$ but $Mx-y \leq 0.$ Since $(Mx - y,x) $ is another solution we know that $Mx-y = 0.$ Therefore $x^2 = m$ and $m$ is a square.
......
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This one takes a little more work.
LEMMA 2
Given integers $$ m > 0, \; \; M \geq m+3, $$ there are no integers $x,y$ with
$$ x^2 - Mxy + y^2 = -m. $$
The contrapositive of lemma 2 is that when there are solutions, $M \leq m+2.$ The bound is sharp, achieved at $x=y=1 \; .$ As a side note, if $M \leq 2,$ then $x^2 - M xy + y^2$ is positive or positive semi-definite. So, the contrapositive gives the other example at the wikipedia article, that if $xy$ divides $x^2 + y^2 + 1$ and $x,y >0,$ then actually $x^2 + y^2 + 1 = 3xy.$
Compare the contrapositive with Lemma 3 below...
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This one took a whole bunch more work. I also needed some help from Gerry Myerson.
LEMMA 3
Given integers $$ m > 0, \; \; M > 0, $$ such that there are integers $x,y$ with
$$ x^2 - Mxy + y^2 = -Mm, $$ then
$$ M \leq (m+1)^2 + 1 \; . $$
The bound is sharp, achieved with $x=1, \; y=m+1 \; .$
