TITLE: Diophantine equation $(x+y)(x+y+1) - kxy = 0$
QUESTION [13 upvotes]: 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 $k \le 500$.

REPLY [2 votes]: For the equation:
$$(x+y)(x+y+1)=kxy$$
Use the standard approach of using Pell equations.  If you use solutions of this equation.
$$p^2-k(k-4)s^2=1$$
Decisions can be recorded.
$$x=-(p^2+2kps+k(k-4)s^2)(p^2+2(k-2)ps+k(k-4)s^2)$$
$$y=-4ps(p^2+2kps+k(k-4)s^2)$$
If this equation Pell.
$$p^2-k(k-4)s^2=\pm1$$
Then decisions can be recorded.
$$x=\pm2(p-ks)s$$
Or.
$$x=\mp(p^2+2(k-1)ps+k(k-2)s^2)$$
For this.
$$y=\mp2(p+ks)s$$
Number  $p,s - $  can have any sign.
An interesting case of when.  $k=4$
The formulas can be seen that while the $s$ can be anything. And  $p=\pm1$