Tricky Problem

Simplify the equation and division by $w$ gives $\large w - \frac {1}{w} = 1 \Rightarrow (w - \frac {1}{w})^2 = 1 \Rightarrow w^2 + \frac {1}{w^2} = 3$

Squaring both sides of the equation and subtract both sides by $2$ repeatedly gives the following

$\large w^4 + \frac {1}{w^4} = 7, w^8 + \frac {1}{w^8} = 47, w^{16} + \frac {1}{w^{16}} = 2207$

Consider

$\large \left ( w^{16} + \frac {1}{w^{16}} \right ) \left ( w^4 + \frac {1}{w^{4}} \right ) = \left ( w^{20} + \frac {1}{w^{20}} \right ) + \left ( w^{12} + \frac {1}{w^{12}} \right )$ $\large \Rightarrow 2207 \cdot 7 = \left ( w^{20} + \frac {1}{w^{20}} \right ) + \left ( w^4 + \frac {1}{w^4} \right )^3 - 3 \left ( w^4 + \frac {1}{w^4} \right )$ $\large \Rightarrow w^{20} + \frac {1}{w^{20} } = 2207 \cdot 7 - 7^3 + 3 \cdot 7 = \boxed{15127}$

Given that: $(w+1)(w-1)=w\quad ...(1)$

$\Rightarrow w^2-w-1 =0 \quad \Rightarrow w = \frac{1\pm \sqrt{5}}{2}$

Of course, we can use a calculator and find $w^{20}+w^{-20}$ , but that is not really an algebraic solution. I am giving a more algebraic solution here.

Dividing (1) by $w: \quad \boxed{w - \dfrac{1}{w} = 1}$

$w-1=\dfrac{1}{w}\quad \Rightarrow \boxed{w+\dfrac{1}{w}} =2w-1=\boxed{\pm \sqrt{5}}$

$\boxed{w^2 - \dfrac{1}{w^2}} = \left( w - \dfrac{1}{w} \right) \left( w + \dfrac{1}{w} \right) = \boxed{\pm \sqrt{5}}$

$\left( w - \dfrac{1}{w}\right)^2 = 1^2\quad \Rightarrow w^2 -2 +\dfrac{1}{w^2} = 1\quad \Rightarrow \boxed{ w^2 + \dfrac{1}{w^2} = 3}$

$\left( w^2 - \dfrac{1}{w^2} \right) \left( w + \dfrac{1}{w} \right) = w^3 + w - \dfrac{1}{w} - \dfrac{1}{w^3} = (\pm \sqrt{5})(\pm \sqrt{5})$

$\Rightarrow \boxed{w^3 - \dfrac{1}{w^3} = 4}$

$\left( w^3 - \dfrac{1}{w^3} \right) \left( w^2 + \dfrac{1}{w^2} \right) = w^5 + w - \dfrac{1}{w} - \dfrac{1}{w^5} = (4)(3)$

$\Rightarrow \boxed{w^5 - \dfrac{1}{w^5} = 11}$

$\left( w^3 - \dfrac{1}{w^3} \right) \left( w^2 - \dfrac{1}{w^2} \right) = w^5 - w - \dfrac{1}{w} + \dfrac{1}{w^5} = (4)(\pm \sqrt{5})$

$\Rightarrow \boxed{w^5 + \dfrac{1}{w^5} = \pm 5\sqrt{5}}$

$\boxed{w^{10} - \dfrac{1}{w^{10}}} = \left( w^5 - \dfrac{1}{w^5} \right) \left( w^5 + \dfrac{1}{w^5} \right) = (11)(\pm 5\sqrt{5}) = \boxed{\pm 55\sqrt{5}}$

$\left( w^{10} - \dfrac{1}{w^{10}} \right) ^2 = w^{20} - 2 + \dfrac{1}{w^{20}} = (55^2)(5) = 15125$

$\Rightarrow w^{20} + \dfrac{1}{w^{20}} = 15125+2 =\boxed{15127}$

20th lucas number

If f(n)=w^n+1/w^n, as w-1/w=1,implies f(2)=1^2+2=3, f(4)=3^2-2=7 , f(4)*f(2)=f(6)-f(2) implies f(6)=18 Again f(6)* f(4)=f(10)+f(2) implies f(10)=123 -Finally, f(20)=f^2(10)-2