There exists a positive real number $x$ such that $ \cos (\arctan (x)) = x $. Find the value of $x^2$.
1. Let \( \theta = \arctan(x) \). By definition of the arctangent function, we have:
   \[
   \tan(\theta) = x
   \]
2. We need to find \( \cos(\theta) \) and set it equal to \( x \). Consider a right triangle where the angle \( \theta \) has an opposite side of length \( a \) and an adjacent side of length \( \frac{a}{x} \). The hypotenuse \( h \) of this triangle can be calculated using the Pythagorean theorem:
   \[
   h = \sqrt{\left(\frac{a}{x}\right)^2 + a^2} = \sqrt{\frac{a^2}{x^2} + a^2} = \sqrt{a^2 \left(\frac{1}{x^2} + 1\right)} = a \sqrt{\frac{1}{x^2} + 1}
   \]
3. The cosine of the angle \( \theta \) is given by:
   \[
   \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\frac{a}{x}}{a \sqrt{\frac{1}{x^2} + 1}} = \frac{\frac{a}{x}}{a \sqrt{\frac{1 + x^2}{x^2}}} = \frac{\frac{a}{x}}{\frac{a \sqrt{1 + x^2}}{x}} = \frac{1}{\sqrt{1 + x^2}}
   \]
4. We are given that \( \cos(\theta) = x \). Therefore, we have:
   \[
   \frac{1}{\sqrt{1 + x^2}} = x
   \]
5. To solve for \( x \), square both sides of the equation:
   \[
   \left(\frac{1}{\sqrt{1 + x^2}}\right)^2 = x^2 \implies \frac{1}{1 + x^2} = x^2
   \]
6. Rearrange the equation to form a quadratic equation:
   \[
   1 = x^2 (1 + x^2) \implies 1 = x^2 + x^4
   \]
7. Let \( y = x^2 \). Then the equation becomes:
   \[
   1 = y + y^2 \implies y^2 + y - 1 = 0
   \]
8. Solve the quadratic equation using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}
   \]
9. Since \( y = x^2 \) must be positive, we take the positive root:
   \[
   y = \frac{-1 + \sqrt{5}}{2}
   \]

The final answer is \( \boxed{\frac{\sqrt{5} - 1}{2}} \).