A strange divisibility
Posted: Mon May 15, 2017 1:53 am
Determine all pairs of positive integers $(x,y)$ such that \[ \dfrac{x^2}{2xy^2-y^3+1} \] is a positive integer.
$\frac {x^2}{2xy^2-y^3+1}=k \in \mathbb{N}$
If $y=1$ , we can easily find $x=2k$ . Now, let $y>1$. Then, $\frac {x^2}{2xy^2-y^3+1}=k$ gives us the quadratic equation $x^2-2ky^2x+k(y^3-1)=0$ . Here the discriminant $X=4k^2y^4-4ky^3+4k$ is a perfect square . Notice that $(2ky^2-y-1)^2 < X < (2ky^2-y+1)^2$ . So, $X= (2ky^2-y)^2$ and $y^2=4k$. This implies $y=2n$ is an even. Then, solving the quadratic equation, we find $x=n$ and $x=8n^4-n$ .
All the solutions are $(x,y)=(2k,1),(n,2n),(8n^4-n,2n)$
If $y=1$ , we can easily find $x=2k$ . Now, let $y>1$. Then, $\frac {x^2}{2xy^2-y^3+1}=k$ gives us the quadratic equation $x^2-2ky^2x+k(y^3-1)=0$ . Here the discriminant $X=4k^2y^4-4ky^3+4k$ is a perfect square . Notice that $(2ky^2-y-1)^2 < X < (2ky^2-y+1)^2$ . So, $X= (2ky^2-y)^2$ and $y^2=4k$. This implies $y=2n$ is an even. Then, solving the quadratic equation, we find $x=n$ and $x=8n^4-n$ .
All the solutions are $(x,y)=(2k,1),(n,2n),(8n^4-n,2n)$