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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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    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 $y$.2012-07-12

1 Answers 1

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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 $k$.