<s>[INST] How many ordered pairs $(x,y)$ are there such that ${1\over\sqrt x}+{1\over\sqrt y}={1\over\sqrt {20}}$, where both $x$ and $y$ are positive integers $${1\over \sqrt x}+{1\over \sqrt y}={1\over \sqrt {20}}$$
I could find one ordered pair that satisfies above equation, that is $(80,80)$. But the answer says that there are $3$ ordered pairs satisfying above equation.
 [/INST] We have the following equation:
$$\frac{1}{\sqrt x}+\frac{1}{\sqrt y}=\frac{1}{\sqrt {20}}$$
$\sqrt{20}=2\sqrt{5}$:
$$\frac{1}{\sqrt x}+\frac{1}{\sqrt y}=\frac{1}{2\sqrt 5}$$
Multiply both sides by $\sqrt 5$:
$$\frac 1 {\sqrt{\frac x 5}}+\frac 1 {\sqrt{\frac y 5}}=\frac 1 2$$
Basically, we want answers to $\frac 1 a+\frac 1 b=\frac 1 2$ where $a=\sqrt{\frac x 5}$ (meaning $x=5a^2$) and $b=\sqrt{\frac y 5}$ (meaning $y=5b^2$). In order to show $a$ and $b$ have to be integers, look at the beginning of @MichaelBurr's answer. Thus, we can manipulate this equation to get the following:
$$a=\frac{2b}{b-2}$$
Thus, $(b-2) \mid (2b)$. However, $b \equiv 2 \pmod{b-2}$, so $2b \equiv 4 \pmod{b-2}$. This means $(b-2) \mid 4$. Therefore, we have the following possibilities:


*

*$b-2=1 \implies a=6, b=3 \implies x=180, y=45$

*$b-2=2 \implies a=b=4 \implies x=y=80$

*$b-2=4 \implies a=3, b=6 \implies x=45, y=180$

</s>