How do you solve $k(a^2-b^2)=2(ax-by)$?

Question

let $a,b,c,d,x,y,k$ be all non-zero positive integers >1. If $a^2-b^2 \neq0$,how do you find all the pairs $(x,y)$ such that $k(a^2-b^2)=2(ax-by)$. I have found so far only solutions where $\gcd(x,y)>1$. Are there any other ones if $\gcd(x,y)=1$? if so, how do i find them?

Answer 1

For the equation: $$q(a^2-b^2)=2(ax-by)$$

If parameter is specified $q$ - the solution seems cumbersome, but if you find you only need: $x,y$

Then the solution has the form:

$$x=\frac{q(a+b)+bk}{2}$$

$$y=\frac{q(a+b)+ak}{2}$$

$k$ - asked us integer corresponding parity.

Answer 2

$$q(a^2-b^2)=2(ax-by)$$ Note that $$x_0=y_0=\frac{q(a+b)}{2}$$ is a special solution of this linear Diophantine equation.
Then all the parametric solutions can be written as $$x=x_0+2bt$$ and $$y=y_0+2at$$ where $t$ is a parameter.