The positive solution of the quadratic equation \( x^{2}-x-1=0 \) is denoted by \( \phi \), and commonly known as the golden section or golden number. Solve the equation \[ x \cos \left(\frac{1}{2} \arctan 2\right)=\sqrt{\phi}, \quad x \in \mathbb{R} . \] Give the answer in the form \( \sqrt[n]{m} \), where \( m \) and \( n \) are positive integers.

To solve the equation \( x \cos \left( \frac{1}{2} \arctan 2 \right) = \sqrt{\phi} \), we can start by finding \( \cos \left( \frac{1}{2} \arctan 2 \right) \). 1. Let \( \theta = \arctan 2 \). This means \( \tan \theta = 2 \), which gives us the opposite and adjacent sides of a right triangle: opposite = 2 and adjacent = 1. Therefore, we can find the hypotenuse using the Pythagorean theorem: \[ \text{hypotenuse} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}. \] 2. Now we can find \( \sin \theta \) and \( \cos \theta \): \[ \sin \theta = \frac{2}{\sqrt{5}}, \quad \cos \theta = \frac{1}{\sqrt{5}}. \] 3. Using the half-angle identity for cosine: \[ \cos \left( \frac{\theta}{2} \right) = \sqrt{\frac{1 + \cos \theta}{2}} = \sqrt{\frac{1 + \frac{1}{\sqrt{5}}}{2}} = \sqrt{\frac{\sqrt{5} + 1}{2\sqrt{5}}}. \] 4. Simplifying: \[ \cos \left( \frac{\theta}{2} \right) = \sqrt{\frac{(\sqrt{5} + 1)}{2\sqrt{5}}}. \] To find \( x \), substitute back into the original equation: 5. The equation now reads: \[ x \cdot \sqrt{\frac{\sqrt{5} + 1}{2\sqrt{5}}} = \sqrt{\phi}. \] 6. Recall that the golden ratio \( \phi = \frac{1 + \sqrt{5}}{2} \). Hence \( \sqrt{\phi} = \sqrt{\frac{1 + \sqrt{5}}{2}} \). 7. Rearranging gives: \[ x = \frac{\sqrt{\phi}}{\cos \left( \frac{1}{2} \arctan 2 \right)} = \frac{\sqrt{\frac{1 + \sqrt{5}}{2}}}{\sqrt{\frac{\sqrt{5} + 1}{2\sqrt{5}}}}. \] 8. Multiplying through by the conjugate, the x simplifies as follows: \[ x = \sqrt{\frac{1 + \sqrt{5}}{2}} \cdot \sqrt{\frac{2\sqrt{5}}{\sqrt{5} + 1}} = \sqrt{\frac{(1 + \sqrt{5}) \cdot 2\sqrt{5}}{2(\sqrt{5} + 1)}} = \sqrt{5} \cdot 1 = \sqrt{5}. \] Now simplify: \[ x = \sqrt{5} \cdot \sqrt{\frac{1 + \sqrt{5}}{2}}/\sqrt{\frac{\sqrt{5} + 1}{2}} = \sqrt{\frac{5(1 + \sqrt{5})}{\sqrt{5} + 1}}. \] Putting everything together, we need a numerical output in the form \( \sqrt[n]{m} \). Evaluating gives, \[ x = \sqrt[4]{5}. \] So \( m = 5 \) and \( n = 4 \). Thus, the answer is: \[ \boxed{\sqrt[4]{5}}. \]