BMO 2017

If $y=1$, then $2x\mid x^2\iff x=2n$, $n\in \mathbb{N}$. So, the pairs $(x,y)=(2n,1)$, $n\in \mathbb{N}$ satisfy the required divisibility.

Let $y>1$ such that $x^2$ is divisible by $2x{y}^2-y^3+1$. There exists $m\in \mathbb{N}$ such that $$x^2=m(2x{y}^2-y^3+1),\text{ e.t. }{x}^2-2m{y}^{2}x+(m{y}^3-m)=0.$$ The discriminant of the last quadratic equation is equal to $\Delta =4m^2{y}^4-4m{y}^3+4m$. Denote $$A=4m(y^2-1)+(y-1{)}^2,B=4m(y^2+1)-(y+1{)}^2.$$ For $y>1$, $y\in \mathbb{N}$ and $m\in \mathbb{N}$ we have $$A>0,B=4m(y^2+1)-(y+1{)}^2>2(y^2+1)-(y+1{)}^2=(y-1{)}^2\ge 0\Rightarrow B>0.$$ We obtain the following estimations for the discriminant $\Delta$: $$\Delta +A=(2m{y}^2-y+1{)}^2\ge 0\Rightarrow \Delta <(2m{y}^2-y+1{)}^2;$$ $$\Delta -B=(2m{y}^2-y-1{)}^2\ge 0\Rightarrow \Delta >(2m{y}^2-y-1{)}^2.$$ Because the discriminant $\Delta$ must be a perfect square, we obtain the equalities: $$\Delta =4m^2{y}^4-4m{y}^3+4m=(2m{y}^2-y{)}^2\iff y^2=4m\Rightarrow y=2k,k\in \mathbb{N},m=k^2,k\in \mathbb{N}.$$ The equation $x^2-8k^{4}x+k(8k^4-k)=0$ has the solutions $x=k$ and $x=8k^4-k$, where $k\in \mathbb{N}$.

Finally, we obtain that all pairs of positive integers $(x,y)$, such that $x^2$ is divisible by $2x{y}^2-y^3+1$, are equal to $(x,y)\in \{(2k,1),(k,2k),(8k^4-k,2k)\mid k\in \mathbb{N}\}$.