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 $y$.
 A: 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 $x=80,y=80$.
A: By symmetry, we can assume $x \ge y$ and then duplicate the solutions.  A simple way to to note that $20 \lt x \le 80$ to make the sum large enough and not too large.  That is only $60$ numbers to try.  A spreadsheet with copy down will make it easy, finding the pair (45,180).  
Another way to find it is to note $\frac 1{\sqrt{20}}=\frac 1{2\sqrt 5}$ and remember that $\frac 13+\frac 16=\frac 12$, again giving $(45,180)$ and $\frac 14+\frac 14=\frac 12$ giving $(80,80)$
The algebraic approach is a mess.
$$\frac 1{\sqrt x}+\frac 1{\sqrt y}=\frac 1{\sqrt {20}}\\
\frac 1{\sqrt x}=\frac 1{\sqrt {20}}-\frac 1{\sqrt y}\\
\frac 1x=\frac 1{20}-\frac 2{\sqrt{20y}}+\frac 1y\\
\frac 1x-\frac 1y-\frac 1{20}=\frac 1{\sqrt{20y}}\\
\frac 1{x^2}+\frac 1{y^2}+\frac 1{400}-\frac 2{xy}-\frac 1{10x}+\frac 1{10y}=\frac 1{20y}\\
400y^2+400x^2+x^2y^2-800xy-40xy^2+40x^2y=20x^2y$$
but at least we can note that $x^2y^2$ must have a factor of $20$, so either one has a factor $10$ or one has a factor $2$ and the other a factor $5$.  We can view this as a quadratic in one of the variables and feed it to the quadratic equation, but that will lead to testing cases in a more complicated way.
