Find $a+b$

Similar solution as @Joshua Lowrance 's

Given that $2x-1 = \sqrt 5\implies x = \dfrac {1+\sqrt 5}2 = \varphi$ , the gold ratio . Then $x$ satisfies $x^2 - x - 1 = 0$ or

$\begin{aligned} x^2 & = x + 1 = F_1x + F_0 & \small \blue{\text{where }F_n \text{ is the }n\text{th Fibonacci number.}} \\ x^3 & = x^2 + x = 2x + 1 = F_2x + F_1 \\ x^4 & = 2x^2 + x = 3x + 2 = F_3x + F_2 \\ \cdots & = \cdots \\ \implies x^{17} & = F_{17}x + F_{16} \end{aligned}$

Therefore $a+b = F_{17} + F_{16} = F_{18} = \boxed{2584}$ .

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Reference: Fibonacci number

$2x-1=\sqrt{5}=>x=\phi=\frac{1+\sqrt{5}}{2}$

$\phi^{17}=1597\phi+987$

$1597+987=\boxed{2584}$