(0;0) (1;0) (2;3) ?

$x=0\Leftrightarrow y=0$. So we assume $x,y\neq 0$ Clearly $x,y>0$ as we have $y$ and $xy$ inside square root Squaring, $2x+2\sqrt{x^2-y}=xy$ So, $xy-2x\geq 0\implies y\geq 2$ Rearranging and squaring again, $4(x^2-y)=x^2y^2-4x^2y+4x^2$ Or, $x^2y^2-4x^2y+4y=0$ If $y\neq 0$, $x^2y-4x^2+4=0$ $y=\frac{4x^2-4}{x^2}$ So, $y\leq 4$ Hence we are left to check only for $y=2,3,4$ which can be done easily

Tato_ wrote: (0;0) (1;0) (2;3) ? $(1,0)$ doesnt work. So, $(x,y)=(0,0) (2,3)$