Relation between Square of Fibonacci Number and Square of Lucas Number

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Theorem

Let $F_n$ denote the $n$th Fibonacci number.

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


Then:

$5 {F_n}^2 + 4 \paren {-1}^n = {L_n}^2$


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:

\(\ds 5 {F_n}^2\) \(=\) \(\ds 5 \paren {\dfrac {\phi^n - \hat \phi^n} {\sqrt 5} }^2\) Euler-Binet Formula
\(\ds \) \(=\) \(\ds \phi^{2 n} - 2 \phi^n \hat \phi^n + \hat \phi^{2 n}\) simplifying
\(\ds \) \(=\) \(\ds \phi^{2 n} + 2 \phi^n \hat \phi^n + \hat \phi^{2 n} - 4 \paren {\phi \hat \phi}^n\)
\(\ds \) \(=\) \(\ds \paren {\phi^n + \hat \phi^n}^2 - 4 \paren {-1}^n\) simplifying, and from above: $\phi \hat \phi = -1$
\(\ds \) \(=\) \(\ds {L_n}^2 - 4 \paren {-1}^n\) Closed Form for Lucas Numbers

Hence the result.

$\blacksquare$


Sources

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