<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] 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.
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