Problem 474

Starting with the given relationship between $x$ and $y$ and manipulating algebraically, we obtain successively $\frac{1}{x+y}=\frac{1}{x}-\frac{1}{y}$ $xy=(x+y)y-(x+y)x$ $xy=xy+y^{2}-x^{2}-xy$ $x^{2}+xy-y^{2}=0$ $\frac{x^{2}}{y^{2}}+\frac{x}{y}-1=0$ where $t=\frac{x}{y}$. Since $x>0$ and $y>0$, then $t>0$. Using the quadratic formula $t=\frac{-1 \pm \sqrt{1^{2}-4(1)(-1)}}{2}=\frac{-1 \pm \sqrt{5}}{2}$. Since $t>0$, then $\frac{x}{y}=t=\frac{\sqrt{5}-1}{2}$. Therefore, $\$\left(\frac{x}{y}+\frac{y}{x}\right)^{2}=\left(\frac{\sqrt{5}-1}{2}+\frac{2}{\sqrt{5}-1}\right)^{2}=\left(\frac{\sqrt{5}-1}{2}+\frac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}\right)^{2}=\left(\frac{\sqrt{5}-1}{2}+\frac{\sqrt{5}+1}{2}\right)^{2}=(\sqrt{5})^{2}=5$