Closed Form for Lucas Numbers

Theorem
The Lucas numbers have a closed-form solution:
 * $L_n = \phi^n + \left({1 - \phi}\right)^n = \dfrac {\left({1 + \sqrt 5}\right)} {2^n} + \dfrac {\left({1 - \sqrt 5}\right)} {2^n}$

where $\phi$ is the golden mean.

Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
 * $L_n = \phi^n + \hat \phi^n$

Proof
Proof by induction:

For all $n \in \N$, let $P(n)$ be the proposition:
 * $L_n = \phi^n + \left({1 - \phi}\right)^n = \dfrac {\left({1 + \sqrt 5}\right)} {2^n} + \dfrac {\left({1 - \sqrt 5}\right)} {2^n} = \phi^n + \hat \phi^n$

Basis for the Induction
$P(0)$ is true, as this just says:
 * $\phi^0 + \hat \phi^0 = 1 + 1 = 2 = L_0$

$P(1)$ is the case:

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({j}\right)$ is true for all $0 \le j \le k + 1$, then it logically follows that $P \left({k + 2}\right)$ is true.

So this is our induction hypothesis:
 * $\forall 0 \le j \le k + 1: L_j = \phi^j + \hat \phi^j$

Then we need to show:
 * $L_{k + 2} = \phi^{k + 2} + \hat \phi^{k + 2}$

Induction Step
This is our induction step:

We observe that we have the following two identities:
 * $\phi^2 = \left({\dfrac {1 + \sqrt 5} 2}\right)^2 = \dfrac 1 4 \left({6 + 2 \sqrt 5}\right) = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$
 * $\hat \phi^2 = \left({\dfrac {1 - \sqrt 5} 2}\right)^2 = \dfrac 1 4 \left({6 - 2 \sqrt 5}\right) = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$

This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$.

Thus:

The result follows by the Second Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: L_n = \phi^n + \hat \phi^n$