A Golden Equation

Given that

$\begin{aligned} \phi^{n-3} + k \phi^n & = \phi^{n+3} & \small \blue{\text{Multiply both sides by }\phi^{3-n}} \\ 1 + k \phi^3 & = \phi^6 & \small \blue{\text{Note that }\phi^n = F_n\phi + F_{n-1} \text{, where}} \\ 1 + k (F_3\phi + F_2) & = F_6 \phi + F_5 & \small \blue{F_n \text{ is the }n\text{th Fibonacci number.}} \\ 1 + 2k \phi + k & = 8 \phi + 5 \\ \implies k & = \boxed 4 \end{aligned}$

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Reference: Fibonacci number

$\phi^{n - 3} + k\phi^n = \phi^{n + 3}$

$\phi^{-3} + k = \phi^{3}$

$1 + k \phi^{3} = \phi^{6}$

$k = \frac{\phi^{6} - 1}{\phi^{3}}$

$k = \frac{9 + 4\sqrt{5} - 1}{2 + \sqrt{5}}$

$k = \frac{8 + 4\sqrt{5}}{2 + \sqrt{5}}$

$k = \frac{4(2 + \sqrt{5})}{2 + \sqrt{5}}$

$k = \boxed{4}$