There are some positive integer pairs(x,y) that satisfy $\frac{1}{ \sqrt{x}}+\frac{1}{ \sqrt{y}}=\frac{1}{ \sqrt{20}}$
How many different possible values of the product x and y are there?
I found one value when $x=y$
$\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{x}}=\frac{1}{\sqrt{20}}$
which simplifies to
$\frac{2\sqrt{x}}{x}=\frac{1}{\sqrt{20}}$
which gives $x=y=80$
therefore one possible value of the product is $80^2$
but I don't know how to prove this is the only solution or if there are more
suggestions, help, and solutions would all be appreciated.
taken from the 2019 IWYMIC/SAIMC(Question 7)
https://chiuchang.org/imc/wp-content/uploads/sites/2/2019/08/SAIMC-2019_Keystage-3_Individual_Final.x17381.pdf
Best Answer
First, we have $\frac{1}{\sqrt x} - \frac{1}{\sqrt{20}} = -\frac{1}{\sqrt y}$, so $\frac{20 + x - 4\sqrt{5 x}}{20x} = \frac{1}{y}$. As $x$ and $y$ are rational, so is $\sqrt{5x}$, thus $x = 5\cdot a^2$ for some integer $a$. Similarly $y = 5\cdot b^2$. Substituting it into original equation, we get $$\frac{1}{a} + \frac{1}{b} = \frac{1}{2}$$ $$2a + 2b = ab$$ $$(a - 2)(b - 2) = 4$$
As $4 = 1 \cdot 4 = 2 \cdot 2 = 4 \cdot 1$ are the only decompositions of $4$ to product of positive integers, we have variants $a - 2 = 1, b - 2 = 4$, $a - 2 = 2, b - 2 = 2$, $a - 2 = 4, b - 2 = 1$. They correspond to $x = 45, y = 180$, $x = y = 80$ and $x = 180, y = 45$.