Equation $\sqrt{x}+\sqrt{y}=\sqrt{2013}$ in rationals Can we find all rational numbers $x,y$ such that $\sqrt{x}+\sqrt{y}=\sqrt{2013}$?
Certainly possible answers are $(2013,0)$ and $(0,2013)$.
If we square the equation, we get $x+y+2\sqrt{xy}=2013$, so $\sqrt{xy}$ must be rational.

if $x\neq 0$ then $x,\sqrt{xy},y$ make a geometric progression, so we can write $y=xq^2$ and $q$ is rational.
$x+y+2\sqrt{xy} = 2013$
$x + xq^2+2xq = x(q+1)^2=2013$
$x= \frac{2013}{(q+1)^2}, y=xq^2, q\in \mathbb Q^+\cup\{0\}$
if $x=0$ then $y = 2013$.
