CCFU Proof 9 — Cassini Identity and H¹ Orbit

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CCFU Proof 9 — Cassini Identity and H¹ Orbit

05/16/2026

Given.
C₂: x(n+1) = x(n) + x(n-1)

Define (state-pair convention):
    Q(x,y) = y² − xy − x²

Claim.
Q(x_n, x_{n+1}) = (−1)^n · Q(x_0, x_1) for all n ≥ 0.

Proof by induction.

Base case (n=0): Q(x_0, x_1) = Q(x_0, x_1). ✓

Inductive step. Assume Q(x_n, x_{n+1}) = (−1)^n · Q(x_0, x_1). Then:

    Q(x_{n+1}, x_{n+2})
    = x_{n+2}² − x_{n+1}·x_{n+2} − x_{n+1}²
    = (x_{n+1}+x_n)² − x_{n+1}(x_{n+1}+x_n) − x_{n+1}²
    = x_{n+1}²+2x_{n+1}x_n+x_n² − x_{n+1}²−x_{n+1}x_n − x_{n+1}²
    = −x_{n+1}² + x_{n+1}x_n + x_n²
    = −(x_{n+1}² − x_{n+1}x_n − x_n²)
    = −Q(x_n, x_{n+1})
    = (−1)^{n+1} · Q(x_0, x_1)  ∎

Corollary (H¹ orbit).
For Fibonacci (x_0=0, x_1=1): Q(0,1) = 1, so Q(x_n, x_{n+1}) = (−1)^n.

For general initial conditions:
    Q(x_n, x_{n+1}) = (−1)^n · Q(x_0, x_1).

Under the change of variables:

    u = y − x/2
    v = (√5/2)·x

Then:

    u² − v² = (y−x/2)² − (5/4)x²
            = y² − xy + x²/4 − 5x²/4
            = y² − xy − x²
            = Q(x,y)

Therefore all C₂ states lie on u² − v² = (−1)^n · Q(x_0, x_1) in Minkowski(1,1).  ∎