KNOTT-4.1 — Fibonacci Power Linearization (Φⁿ = FₙΦ + Fₙ₋₁) & The Phibonacci Brick

Every integer power n ∈ Z of the Golden Ratio reduces to the linear form Φⁿ = Fₙ Φ + Fₙ₋₁ and (−φ)ⁿ = Fₙ₊₁ − Fₙ Φ = Fₙ₋₁ − Fₙ φ, while a rectangular Phibonacci brick of edges (1, φ, Φ) has unit volume 1, surface area 4Φ, and integer space diagonal 2.

Curated Source & Catalog Provenance

Catalog ID	KNOTT-4.1
Curated Source	Dr. Ron Knott, Phi’s Fascinating Figures (#powers)
Source Attribution	© 1996–2025 Dr. Ron Knott, University of Surrey — Explore Dr. Knott’s original exposition
Exact SymPy Validator	geometor.knott.phigits.construct_phi_powers_model() (0 sp.Float atoms)
GEOMETOR Concordance	geometor/divine, geometor/fundamentals

Constructive Synthesis & Exact Proof

Because \(\Phi^2 = \Phi + 1\), Dr. Ron Knott (KnottPropsPhi) emphasizes that \(\Phi\) is the unique positive proportion whose geometric power sequence \(1, \Phi, \Phi^2, \Phi^3, \dots\) is simultaneously an additive Fibonacci sequence (\(\Phi^{n+1} = \Phi^n + \Phi^{n-1}\)):

• Linearization of All Integer Powers: Induction over \(n \in \mathbb{Z}\) reduces every positive and negative power to \(\Phi^n = F_n \Phi + F_{n-1}\) and \((-\phi)^n = F_{n+1} - F_n \Phi\). Visually, the \(\Phi^2 \times \Phi^2\) square AEFH on A, E, F, H (with main diagonal AF) decomposes via internal cuts BJ and DG through C into unit square ABCD (\(1\)), \(\Phi \times \Phi\) square CGFJ (\(\Phi^2\)), and two \(1 \times \Phi\) golden rectangles (\(\Phi^4 = 3\Phi + 2 = 1 + 2\Phi + \Phi^2\)).

• The Unit-Volume Phibonacci Brick: Alongside the square, rectangle APQR (with edges PQ and RQ) displays the unfolded cross-section of Dr. Knott’s 3D Phibonacci brick of edges \((1, \phi, \Phi)\), whose volume is exact unit \(1 \cdot \phi \cdot \Phi = 1\), total surface area is \(2(\phi + \Phi + 1) = 4\Phi\), and 3D space diagonal has integer length \(\sqrt{1 + \phi^2 + \Phi^2} = \sqrt{4} = 2\).

Machine-Verified Exact Symbolic Invariants

Invariant	Exact Symbolic Value
PHI squared	\(\frac{\sqrt{5}}{2} + \frac{3}{2}\)
PHI to 5	\(\frac{11}{2} + \frac{5 \sqrt{5}}{2}\)
phi to 4	\(\frac{7}{2} - \frac{3 \sqrt{5}}{2}\)
brick volume	\(1\)
brick surface area	\(2 + 2 \sqrt{5}\)
brick space diagonal	\(2\)