Lucas Number 2n in terms of Square of Lucas Number n

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Theorem

Let $L_n$ denote the $n$th Lucas number.


Then:

$L_{2 n} = {L_n}^2 + 2 \left({-1}\right)^n$


Proof

Let:

$\phi = \dfrac {1 + \sqrt 5} 2$
$\hat \phi = \dfrac {1 - \sqrt 5} 2$


Note that we have:

\(\ds \phi \hat \phi\) \(=\) \(\ds \dfrac {1 + \sqrt 5} 2 \dfrac {1 - \sqrt 5} 2\)
\(\ds \) \(=\) \(\ds \dfrac {1 - 5} 4\) Difference of Two Squares
\(\ds \) \(=\) \(\ds -1\)


Then:

\(\ds L_{2 n}\) \(=\) \(\ds \phi^{2 n} + \hat \phi^{2 n}\) Closed Form for Lucas Numbers
\(\ds \) \(=\) \(\ds \phi^{2 n} + 2 \phi^n \hat \phi^n + \hat \phi^{2 n} - 2 \phi^n \hat \phi^n\)
\(\ds \) \(=\) \(\ds \paren {\phi^n + \hat \phi^n}^2 - 2 \paren {-1}^n\) simplifying, and from above: $\phi \hat \phi = -1$
\(\ds \) \(=\) \(\ds {L_n}^2 - 2 \paren {-1}^n\) Closed Form for Lucas Numbers

Hence the result.

$\blacksquare$


Sources