Mathematical Reflections 2016, Issue 2 - Problem J369

Problem:
Solve the equation $$\sqrt{1+\dfrac{1}{x+1}}+\dfrac{1}{\sqrt{x+1}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Clearly, $x>0$. The given equation can be written as $$\dfrac{\sqrt{x+2}}{\sqrt{x+1}}+\dfrac{1}{\sqrt{x+1}}=\dfrac{x+1}{\sqrt{x}},$$ i.e. $$\sqrt{x^2+2x}+\sqrt{x}=(x+1)\sqrt{x+1}.$$ The last equation can be written as $$\dfrac{\sqrt{x}}{\sqrt{x+2}-1}=\sqrt{x+1},$$ which gives $$\sqrt{x}+\sqrt{x+1}=\sqrt{(x+2)(x+1)}.$$ Squaring both sides and reordering, we get $$x^2+x+1=2\sqrt{x^2+x},$$ i.e. $$(\sqrt{x^2+x}-1)^2=0.$$ The last equation is equivalent to $x^2+x-1=0$, which solved for $x>0$ gives $x=\dfrac{-1+\sqrt{5}}{2}$.