To find $x$, I need to somehow relate $\cos (\arctan (x))$ and $x$. One way to do that is to use a right triangle with angle $\arctan (x)$. Then $\cos (\arctan (x))$ is the adjacent side over the hypotenuse, and $x$ is the opposite side over the adjacent side. So I can draw a right triangle with angle $\arctan (x)$, adjacent side $1$, and opposite side $x$. Then the hypotenuse is $\sqrt{1 + x^2}$ by the Pythagorean theorem. So I have $\cos (\arctan (x)) = \frac{1}{\sqrt{1 + x^2}}$. But I also have $\cos (\arctan (x)) = x$. So I can equate them and get $x = \frac{1}{\sqrt{1 + x^2}}$. Squaring both sides, I get $x^2 = \frac{1}{1 + x^2}$. Multiplying both sides by $1 + x^2$, I get $x^4 + x^2 = 1$. This is a quadratic equation in $x^2$, so I can use the quadratic formula to find $x^2$. I get $x^2 = \frac{-1 \pm \sqrt{1 + 4}}{2}$. But $x^2$ has to be positive, so I only take the positive root. I get $x^2 = \frac{-1 + \sqrt{5}}{2}$. This is the value of $x^2$. 
There exists a positive real number $x$ such that $ \cos (\arctan (x)) = x $. Find the value of $x^2$.