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.
 A: 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:


*

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

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

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

A: If you manipulate the expression, we get
$$
\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{20}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{20}}{\sqrt{20x}}.
$$
Therefore, $y=\frac{20x}{x-2\sqrt{20x}+20}.$  In order for there to be any hope of $y$ being an integer, $\sqrt{20x}=2\sqrt{5x}$ must be an integer.  Therefore, the only options that need to be considered are when $x$ is five times a square.  Moreover, since $(80,80)$ is a solution, you only need to check $x$ values up to $80$.  
In other words, the only values for $x$ to check are $x=5,20,45$. Of these, $x=5$ and $x=20$ are too large, so we are only left with considering $x=45$.
When $x=45$, $y=\frac{20\cdot 45}{45-2\sqrt{20\cdot 45}+20}=180.$
Therefore, $(45,180)$, $(80,80)$, and $(180,45)$ are the pairs.
