# 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

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It will help us to help you if we know what $k$ is! –  Host-website-on-iPage Jul 12 '12 at 10:39
$(3,1)$ and $(-1,1)$ are two solutions. –  Host-website-on-iPage Jul 12 '12 at 10:41
I edited the post, see now. –  Bazinga Jul 12 '12 at 10:44
I don't see any problem with this, only thing is $k=\pm 1$ is not allowed. –  Aang Jul 12 '12 at 10:58
Maybe did not need the discriminant stuff. Let $x-y=k$. Then you found that $x+y=(x-y)^2=k^2$. Adding and subtracting, we can solve the two equation $x-y=k$, $x+y=k^2$ for $x$ and $y$. –  André Nicolas Jul 12 '12 at 11:14