Find all positive integers x,y satisfying $ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}$ Find all positive integers $x$,$y$ satisfying $ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}$
$$ \frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{\sqrt{20}}\\
\frac{1}{\sqrt{x}} +\frac{1}{\sqrt{y}} =\frac{1}{2\sqrt{5}}$$
By hit and trial I found one pair value of $x$ & $y$ i.e $(80, 80)$
But is there any other way to solve this tricky question and find all possible value of $x$ and $y$.

After squaring we get
$$
\frac{2}{\sqrt{xy}}=\frac{1}{20}-\frac{1}{x}-\frac{1}{y}
$$
which implies $xy=n^2$ is a square. Thus we can rewrite the original equation as
$$
\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{n}=\dfrac{1}{\sqrt{20}}
$$
or, as well,
$$
\left(\frac{1}{x}+\frac{1}{n}\right)\sqrt{x}=\frac{1}{10}\sqrt{5}
$$
Since the right-hand side belongs to $\mathbb{Q}(\sqrt{5})$, the left-hand side should as well, so we conclude that $x=5X$, $y=5Y$ and so $n=5m$.
$$
\frac{1}{\sqrt{X}}+\frac{\sqrt{X}}{m}=\frac{1}{2}
\qquad\text{also written as}\qquad
\left(\frac{1}{X}+\frac{1}{m}\right)\sqrt{X}=\frac{1}{2}
$$
Now this implies $X$ is a square.
If $x\ge y$, then $20<x\le 80$, so $4<X\le16$. This yields only $X=9$ or $X=16$, which correspond to $x=45,y=180$ or $x=80,y=80$.
