How many natural numbers $x, y$ are possible if $(x - y)^2 = \frac{4xy}{(x + y - 1)}$. How many natural numbers $x$, $y$ are possible if $(x - y)^2 = \frac{4xy}{x + y - 1}$.
Does this system has infinite solutions which can be generalized for some integer $k \geq 2?$
$(x - y)^2(x + y) = (x + y)^2 ;$
since $(x + y)$ can not be $0$ ;
$(x - y)^2 = (x + y);$ 
$x^2 - x(2y + 1) + y^2 - y = 0$;
For $x$ to be integer, discriminant($D$) should be perfect square;
$D = 8y + 1;$
$y = k(k + 1)/2$;
$(x, y) = (\frac{k(k + 1)}{2},\frac{k(k - 1)}{2})$ or vice versa;
infinite possibilities

It is easy to verify that $(x, y) = (\frac{k(k + 1)}{2},\frac{k(k - 1)}{2})$ is solution of $(x - y)^2 = \frac{4xy}{x + y - 1}$.
we substitute $x$ and $y$ in  $(x - y)^2 = \frac{4xy}{x + y - 1}$.
$$\left(\frac{k(k+1)}{2}-\frac{k(k-1)}{2}\right)^2 = \frac{4.[k(k-1)/2][k(k+1)/2]}{[k(k-1)/2]+[k(k+1)/2]-1}$$ 
after simplification we get $$k^2=k^2$$
$$LHS=RHS$$
therefore it has infinitely many solution for each values of $k$.
