<s>[INST] Diophantine equation $(x+y)(x+y+1) - kxy = 0$ The following came up in my solution to this question, but buried in the comments, so maybe it's worth a question of its own.  Consider the Diophantine equation
$$ (x+y)(x+y+1) - kxy = 0$$
For $k=5$ and $k=6$ the positive integer solutions are given by successive terms of OEIS sequences A032908 and A101265 respectively.
Are there positive integer solutions for any other positive integers $k$? 
I know there are solutions if you remove the positivity requirement, but I've been unable to find any with positive integers.
If my programming is correct, I've been able to rule out all $k \le 500$.
 [/INST] What I got so far:
First, solve for $x$:
$$
x=\frac{k y-2 y-1\pm\sqrt{k^2 y^2-4 k y^2-2 k y+1}}{2} 
$$
$x$ is only an integer, if the discriminant is a perfect square, so:
$$
k^2 y^2-4 k y^2-2 k y+1=z^2
$$
Solving this for $y$:
$$
y=\frac{k\pm\sqrt{k^2 z^2-4 k z^2+4 k}}{k^2-4 k}
$$
This can only happen if:
$$
k^2 z^2-4 k z^2+4 k=w^2\Rightarrow w^2-(k-4)kz^2=4k
$$
Which is a Pell equation.
If we have $w$ and $z$, then (if I made no mistakes):
$$
x=\frac{1}{4} \left(\frac{4-w}{k-4}-\frac{w}{k}+2 z\right)\qquad y=\frac{k-w}{(k-4) k}
$$
$$
x=\frac{1}{4} \left(\frac{w}{k}+\frac{w+4}{k-4}-2 z\right)\qquad y=\frac{k+w}{(k-4) k}
$$
Will give us the solutions
</s>