Number of solutions to $x^2+ax=y^2+by$

Question

I'm trying to find the number of positive integer solutions, in terms of $a$ and $b$, to the equation:

$$x^2+ax=y^2+by$$

I have tried many approaches but couldn't seem to get an answer. Is this possible?

EDIT:

I should make clear that $x$ and $y$ are variables whose only restrictions are that they must be positive integers, and $a$ and $b$ are positive integer constants.

I am looking to find the number of unique $x,y$ pairs given the constants $a$ and $b$.

Answer 1

$$ (2x+2y+a+b)(2x-2y+a - b)= a^2 - b^2 $$ Given all ways to factor $a^2 - b^2 = UV$ with $U \equiv V \equiv a+b \pmod 2 \; ...$

It is necessary to include pairs with $U,V$ negative as well as positive.

Answer 2

Above equation shown below:

$x^2+ax=y^2+by$ -------$(1)$

Equation $(1)$ has parametric solution & is given below:

$x=10-23k+9k^2$

$y=5-19k+18k^2$

$a=8-23k+11k^2$

$b=31-19k-8k^2$

For $k=3$, we get $(x,y,a,b)= (11,55,19, (-49))$