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
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $11$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $11$