Prove that $L_n = \alpha^n +\beta^n$ for all integers $n\geq 0$ Let $\alpha =\left(\frac{1+\sqrt{5}}{2}\right)$ and $\beta = \left(\frac{1-\sqrt{5}}{2}\right)$. Prove that $L_n = \alpha^n +\beta^n$ for all integers $n\geq 0$ where $L_n$ denotes the Lucas numbers.
I managed to solve the base case:
$$L_2 = \left(\frac{1+\sqrt{5}}{2}\right)^2 + \left(\frac{1-\sqrt{5}}{2}\right)^2 = 3$$
I wasn't able to figure out how to proceed with the induction step and the rest of the proof. Any help is appreciated.

The crucial fact is that $\alpha$ and $\beta$ are the roots of $x^2=x+1$ and so $\alpha+\beta=1$. This makes induction work easily:
$L_0 = 2 = 1+1 = \alpha^0+\beta^0$
$L_1 = 1 = \alpha+\beta = \alpha^1+\beta^1$
$L_{n+2}=L_{n+1}+L_n
= \alpha^{n+1}+\beta^{n+1}+\alpha^n+\beta^n
= \alpha^{n+1}+\alpha^n+\beta^{n+1}+\beta^n
$
$\qquad
= \alpha^n(\alpha+1)+ \beta^n(\beta+1)
= \alpha^n\alpha^2+ \beta^n\beta^2
= \alpha^{n+2}+\beta^{n+2}
$
