Mathematical Reflections 2017, Issue 2 - Problem J407

Problem:
Solve in positive real numbers the equation $$\sqrt{x^4-4x}+\dfrac{1}{x^2}=1.$$

Proposed by Titu Andreescu, University of Texas at Dallas, USA

First solution:
The equation has a real solution if and only if $x \leq -1$ or $x \geq \sqrt[3]{4}$. We have $$x^4-4x=\left(1-\dfrac{1}{x^2}\right)^2 \iff x^4-4x+\dfrac{4}{x^2}=\left(1-\dfrac{1}{x^2}\right)^2+\dfrac{4}{x^2},$$ i.e.
$$\left(x^2-\dfrac{2}{x}\right)^2=\left(1+\dfrac{1}{x}\right)^2$$
So, $$x^2-\dfrac{2}{x}=\pm \left(1+\dfrac{1}{x^2}\right).$$
Observe that $x^2-\dfrac{2}{x}=-1-\dfrac{1}{x^2}$ gives $x^4+(x-1)^2=0$, which has no real solutions. So,
$$x^2-\dfrac{2}{x}=1+\dfrac{1}{x^2} \iff x(x^3-1)=x^2+x+1 \iff x(x-1)=1.$$ So, we get $x^2-x-1=0$, which yields $x=\dfrac{1+\sqrt{5}}{2}$.

Second solution:
The given equation is equivalent to $$x\sqrt{x^4-4x}=x-\dfrac{1}{x}.$$ Squaring both sides, we get $x^6-4x^3=x^2-2+\dfrac{1}{x^2}$. Hence, $$(x^3-2)^2=\left(x+\dfrac{1}{x}\right)^2.$$ Since $x^3>4$, we have $x^3-2=x+\dfrac{1}{x}$. So, $x^4-2x=x^2+1$, which gives $x^4=(x+1)^2$, i.e. $x^2=x+1$. So, $x=\dfrac{1+\sqrt{5}}{2}$.