<s>[INST] Lucas Numbers Proof $L_n = \alpha^n + \beta^n$ Proof by Induction:
Lucas numbers are recursively defined as:
$L_n = L_{n-1} + L_{n-2}$ where $L_1 = 1$ and $ L_2 = 3 $for $n \ge 3$
Show that:  $L_n = \alpha^n + \beta^n$  for $\alpha = {1+\sqrt{5}\over2}$ and $\beta = {1-\sqrt{5}\over2}$
Is this the correct approach?
base case:
n = 3
$L_3 = L_2 + L_1 = 4$
$L_3 = \alpha^3 + \beta^3 = 4$
inductive hypothesis:
Assume true: $L_k = L_{k-1} + L_{k-2} = \alpha^k + \beta^k$ for $k\ge3$
inductive step:
$L_{k+1} = L_k + L_{k-1} = \alpha^{k+1} + \beta^{k+1}$
$(\alpha^k + \beta^k) + L_{k-1} = \alpha^{k+1} + \beta^{k+1}$
$L_{k-1} = (\alpha^{k+1} - \alpha^k) + (\beta^{k+1} - \beta^k) = \alpha^{k-1} + \beta^{k-1}$
 [/INST] Note that for any $\alpha$, $\beta$, $k$, $$(\alpha^k + \beta^k)(\alpha + \beta) = \alpha^{k+1} + \alpha \beta^k + \alpha^k \beta + \beta^{k+1} = \alpha^{k+1} + \beta^{k+1} + \alpha \beta(\alpha^{k-1} + \beta^{k-1}).$$
Now, if $\alpha = (1+\sqrt{5})/2$, $\beta = (1-\sqrt{5})/2$, what are the sum $\alpha + \beta$ and product $\alpha\beta$, respectively?
This should suggest an explicit proof by induction.  For the inductive step, we claim that there exists at least one $n$ such that $L_k = \alpha^k + \beta^k$ for all $1 \le k \le n$.  Then in particular, $L_n = \alpha^n + \beta^n$ and $L_{n-1} = \alpha^{n-1} + \beta^{n-1}$.  But the identity we established above tells us that $L_{n+1} = L_n + L_{n-1} = \alpha^{n+1} + \beta^{n+1}$, completing the inductive step.
</s>