, 2023

Solution 1

Starting with the given relationship between $x$ and $y$ and manipulating algebraically, we
obtain successively $$\begin{align*} \dfrac{1}{x+y} & = \dfrac{1}{x} - \dfrac{1}{y} \\ xy & = (x+y)y - (x+y)x \qquad \text{(multiplying by $xy(x+y)$)}\\ xy & = xy + y^2 - x^2 - xy \\ x^2 + xy - y^2 & = 0 \\ \dfrac{x^2}{y^2} + \dfrac{x}{y} - 1 & = 0 \qquad\text{(dividing by $y^2$ which is non-zero)}\\ t^2 + t - 1 & = 0\end{align*}$$ where $t = \dfrac{x}{y}$.

Since $x>0$ and $y>0$, then $t > 0$. Using the quadratic formula
$$t = \dfrac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2} = \dfrac{-1 \pm \sqrt{5}}{2}$$ Since $t>0$, then $\$\dfrac{x}{y}$ = t = $\text{5}$ -
1}{2}$.

Therefore, $$\begin{align*} \left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2 & = \left( \dfrac{\sqrt{5} - 1}{2} + \dfrac{2}{\sqrt{5} - 1} \right)^2 \\ & = \left( \dfrac{\sqrt{5} - 1}{2} + \dfrac{2(\sqrt{5}+1)}{(\sqrt{5} - 1)(\sqrt{5}+1)} \right)^2 \\ & = \left( \dfrac{\sqrt{5} - 1}{2} + \dfrac{2(\sqrt{5}+1)}{4} \right)^2 \\ & = \left( \dfrac{\sqrt{5} - 1}{2} + \dfrac{\sqrt{5}+1}{2} \right)^2 \\ & = (\sqrt{5})^2 \\ & = 5\end{align*}$$

Solution 2

Since $x,y > 0$, the following
equations are equivalent: $$\begin{align*} \dfrac{1}{x+y} & = \dfrac{1}{x} - \dfrac{1}{y} \\ 1 & = \dfrac{x+y}{x} - \dfrac{x+y}{y} \\ 1 & = \dfrac{x}{x} + \dfrac{y}{x} - \dfrac{x}{y} - \dfrac{y}{y} \\ 1 & = 1 + \dfrac{y}{x} - \dfrac{x}{y} - 1 \\ 1 & = \dfrac{y}{x} - \dfrac{x}{y} \\ -1 & = \dfrac{x}{y} - \dfrac{y}{x}\end{align*}$$ Therefore,
$$\begin{align*} \left(\dfrac{x}{y} + \dfrac{y}{x}\right)^2 & = \dfrac{x^2}{y^2} + 2\cdot\dfrac{x}{y} \cdot\dfrac{y}{x} + \dfrac{y^2}{x^2} \\ & = \dfrac{x^2}{y^2} + 2 + \dfrac{y^2}{x^2} \\ & = \dfrac{x^2}{y^2} - 2 + \dfrac{y^2}{x^2} + 4 \\ & = \dfrac{x^2}{y^2} - 2\cdot\dfrac{x}{y} \cdot\dfrac{y}{x} + \dfrac{y^2}{x^2} + 4 \\ & = \left(\dfrac{x}{y} - \dfrac{y}{x}\right)^2 + 4\\ & = (-1)^2 + 4 \\ & = 5\end{align*}$$