We are given the equation: \[ x^2 + \frac{1}{x^2} = 2 \]
Step 1: Identifying the Nature of \( x \) From the identity: \[ x^2 + \frac{1}{x^2} = 2 \Rightarrow (x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2} \] Setting \( y = x + \frac{1}{x} \), we can express: \[ y^2 = x^2 + 2 + \frac{1}{x^2} \] Substituting \( x^2 + \frac{1}{x^2} = 2 \): \[ y^2 = 2 + 2 = 4 \Rightarrow y = \pm 2 \] Thus, \( x + \frac{1}{x} = 2 \) or \( x + \frac{1}{x} = -2 \).
Step 2: Higher Powers of \( x \) Using the recurrence relation: \[ x^n + \frac{1}{x^n} = (x^{n-2} + \frac{1}{x^{n-2}})(x^2 + \frac{1}{x^2}) - (x^{n-4} + \frac{1}{x^{n-4}}) \] Since \( x + \frac{1}{x} = 2 \), we establish the sequence: \[ x^2 + \frac{1}{x^2} = 2 \] \[ x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = 2^2 - 2 = 2 \] Continuing similarly, we observe that: \[ x^8 + \frac{1}{x^8} = 2 \] \[ x^{16} + \frac{1}{x^{16}} = 2 \] \[ x^{32} + \frac{1}{x^{32}} = 2 \] \[ x^{64} + \frac{1}{x^{64}} = 2 \] \[ x^{128} + \frac{1}{x^{128}} = 2 \] \[ x^{256} + \frac{1}{x^{256}} = 2 \]
Final Answer: \[ \boxed{2} \]