This is based on an earlier question here.
Given a positive integer $n$, is this method for finding all positive integers $x$ and $y$ with $x \le y$ such that
$\dfrac1{\sqrt{n}} =\dfrac1{\sqrt{x}}+\dfrac1{\sqrt{y}} $
correct?
All variables here are positive integers,
My solution:
For each representation $n=ku^2v^2 $ with $(u, v)=1, u \le v $ we have $x=u^2z, y=v^2z, $ with $z=k(u+v)^2 $.
Check:
$\begin{array}\\ \dfrac1{\sqrt{x}}+\dfrac1{\sqrt{y}} &=\dfrac1{\sqrt{u^2z}}+\dfrac1{\sqrt{v^2z}}\\ &=\dfrac1{u\sqrt{z}}+\dfrac1{v\sqrt{z}}\\ &=\dfrac1{\sqrt{z}}\left(\dfrac1{u}+\dfrac1{v}\right)\\ &=\dfrac1{\sqrt{k(u+v)^2}}\left(\dfrac{u+v}{uv}\right)\\ &=\dfrac1{\sqrt{k}uv}\\ &=\dfrac1{\sqrt{ku^2v^2}}\\ &=\dfrac1{\sqrt{n}}\\ \end{array} $
My derivation.
$\dfrac1{\sqrt{n}} =\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}} $.
$\sqrt{n} =\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}} $.
$\sqrt{n}(\sqrt{x}+\sqrt{y}) =\sqrt{xy} $
Squaring,
$n(x+y+2\sqrt{xy}) =xy $ so $\sqrt{xy} =\dfrac{xy-n(x+y)}{2n} $.
Therefore $xy$ is a square since the square root of an integer is rational if and only if the integer is the square of an integer.
If $(x, y)=z$, we must have $x=u^2z, y=v^2z$ with $(u, v)=1, u\le v$.
So $n(u^2z+v^2z+2uvz) =u^2v^2z^2 $ or $n(u^2+v^2+2uv) =u^2v^2z $ or $n(u+v)^2 =u^2v^2z $.
Since $(u, v)=1$, $(uv, u+v)=1$ so that, for some $k$, $z=k(u+v)^2 $ so $n =ku^2v^2 $.
A solution is always $k=n, u=v=1, z=4n, x=4n, y=4n$.
The example that inspired this:
If $n=20=2^25$, $(k, u, v) = (20, 1, 1), (5, 1, 2) $. For the latter, $z=5(1+2)^2=45, x=45, y=4\cdot 45=180 $.
Another example:
If $n=100=2^25^2 $ then $(k, u, v) =(100, 1, 1), (25, 1, 2), (4, 1, 5), (1, 2, 5) $ for which, respectively, $(z, x, y) =(400, 400, 400), (225, 225, 900), (144, 144, 3600), (49, 196, 1225) $.
We begin by rearranging terms to remove the surds from the divisors. \begin{align} \dfrac1{\sqrt{n}}&=\dfrac1{\sqrt{x}}+\dfrac1{\sqrt{y}} \quad 1\le x\le y\tag{1}\\ \sqrt{xy}&=\sqrt{nx}+\sqrt{ny}\tag{2}\\ xy&=n (x + 2\sqrt{x y} + y)\tag{3}\\ \dfrac{xy}{n}&=\big(\sqrt{x}+\sqrt{y}\big)^2\tag{4}\\ \end{align}
It is clear that $\,n|xy.\quad$ It is useful when $\,xy\,$ is a perfect square, when $\,x\,$ or $\,y\,$ are squares themselves, when $\,y=4x,\,$ and when $\,x=y=4n.\quad$ A number of these solutions are shown below. $$(n,x,y)\in \big\{(1,4,4),(2,8,8),(3,12,12),(4,9,36),(4,16,16),(5,20,20),(6,24,24),(7,28,28),(8,18,72),(8,32,32),(9,16,144),(9,36,36),(10,40,40),(11,44,44),(12,27,108),(12,48,48) \cdots (28,112,112),\cdots\big\}$$