Square roots

$\sqrt{20} = 2\sqrt{5} \longrightarrow \frac{1}{2\sqrt{5}} = \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{a\sqrt{5}} + \frac{1}{b\sqrt{5}}$ . Then it's sufficient to find ordered pairs of positive integers $(a, b)$ such that $\frac{1}{2} = \frac{1}{a} + \frac{1}{b}$ and the only solutions are $\begin{cases} (a,b) = (4,4) \\ (a,b) = (3,6) \\ (a,b) = (6,3) \end{cases}$ . Therefore, the only solutions are $\begin{cases} (x,y) = (80,80) \\ (x,y) = (45,180) \\ (x,y) = (180,45) \end{cases}$ Important Note.- If $\frac{1}{2\sqrt{5}} = \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = \frac{1}{a\sqrt{c}} + \frac{1}{b\sqrt{d}}$ where $a, b$ are positive integers numbers and $c,d$ are free-square positive integers numbers, then squaring this equation, we get $\frac{1}{20} = \frac{1}{a^2c} + \frac{2}{ab\sqrt{cd}} + \frac {1}{b^2d}$ and then LHS is a rational number and this implies that $c = d$ , because if $c \neq d$ then RHS is an irrational number, and this is a contradiction... Therefore, $c = d$ and then $c = d = 5$ due to Fundamental Arithmetic Theorem and squaring again...