Honsberger's Identity

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Theorem

Let $F_k$ be the $k$th Fibonacci number.


Then:

$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$


Negative Indices

Let $n \in \Z_{< 0}$ be a negative integer.

Let $F_n$ be the $n$th Fibonacci number (as extended to negative integers).

Then Honsberger's Identity:

$F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

continues to hold, whether $m$ or $n$ are positive or negative.


Proof 1

From the initial definition of Fibonacci numbers, we have:

$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$\forall m \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$


Basis for the Induction

$\map P 1$ is the case:

\(\ds F_{m + 1}\) \(=\) \(\ds F_{m - 1} + F_m\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds F_{m - 1} \times 1 + F_m \times 1\)
\(\ds \) \(=\) \(\ds F_{m - 1} F_1 + F_m F_2\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds F_{m - 1} F_n + F_m F_{n + 1}\) for $n = 1$

and so $\map P 1$ is seen to hold.


$\map P 2$ is the case:

\(\ds F_{m + 2}\) \(=\) \(\ds F_{m + 1} + F_m\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds F_{m - 1} + F_m + F_m\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds F_{m - 1} \times 1 + F_m \times 2\)
\(\ds \) \(=\) \(\ds F_{m - 1} F_2 + F_m F_3\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds F_{m - 1} F_n + F_m F_{n + 1}\) for $n = 2$

and so $\map P 2$ is seen to hold.


This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ and $\map P {k - 1}$ are true, where $k > 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$F_{m + k} = F_{m - 1} F_k + F_m F_{k + 1}$

and:

$F_{m + k - 1} = F_{m - 1} F_{k - 1} + F_m F_k$


from which it is to be shown:

$F_{m + k + 1} = F_{m - 1} F_{k + 1} + F_m F_{k + 2}$


Induction Step

This is our induction step:

\(\ds F_{m + k + 1}\) \(=\) \(\ds F_{m + k} + F_{m + k - 1}\) Definition of Fibonacci Number
\(\ds \) \(=\) \(\ds F_{m - 1} F_k + F_m F_{k + 1} + F_{m - 1} F_{k - 1} + F_m F_k\) Induction Hypothesis
\(\ds \) \(=\) \(\ds F_{m - 1} \paren {F_k + F_{k - 1} } + F_m \paren {F_{k + 1} + F_k}\)
\(\ds \) \(=\) \(\ds F_{m - 1} F_{k + 1} + F_m F_{k + 2}\) Definition of Fibonacci Number

So $\map P k \land \map P {k - 1} \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall m, n \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$

$\blacksquare$


Proof 2

\(\ds \) \(\) \(\ds F_{m - 1} F_n + F_m F_{n + 1}\)
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m - 1} - \hat \phi^{m - 1} } {\sqrt 5} \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} + \dfrac {\phi^m - \hat \phi^m} {\sqrt 5} \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\sqrt 5}\) Euler-Binet Formula
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n - 1} - \phi^{m - 1} \hat \phi^n - \phi^n \hat \phi^{m - 1} + \hat \phi^{m + n - 1} + \phi^{m + n + 1} - \phi^m \hat \phi^{n + 1} - \phi^{n + 1} \hat \phi^m + \hat \phi^{m + n + 1} } 5\)
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n - 1} \paren {1 + \phi^2} + \hat \phi^{m + n - 1} \paren {1 + \hat \phi^2} - \phi^{m - 1} \hat \phi^n \paren {1 + \phi \hat \phi} -\phi^n \hat \phi^{m - 1} \paren {1 + \phi \hat \phi} } 5\)
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n - 1} \paren {1 + \phi^2} + \hat \phi^{m + n - 1} \paren {1 + \hat \phi^2} } 5\) as $\phi \hat \phi = -1$
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n - 1} \paren {2 + \phi} + \hat \phi^{m + n - 1} \paren {2 + \hat \phi} } 5\) as both $\phi$ and $\hat \phi$ satisfy $x^2 = x + 1$
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n - 1} \paren {2 + \dfrac {1 + \sqrt 5} 2} + \hat \phi^{m + n - 1} \paren {2 + \dfrac {1 - \sqrt 5} 2} } 5\) Definition of $\phi$ and $\hat \phi$
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n - 1} \paren {\dfrac {5 + \sqrt 5} 2} + \hat \phi^{m + n - 1} \paren {\dfrac{ 5 - \sqrt 5} 2} } 5\)
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n - 1} \paren {\dfrac {1 + \sqrt 5} 2} - \hat \phi^{m + n - 1} \paren {\dfrac {1 - \sqrt 5} 2} } {\sqrt 5}\) dividing numerator and denominator by $\sqrt 5$
\(\ds \) \(=\) \(\ds \dfrac {\phi^{m + n} - \hat \phi^{m + n} } {\sqrt 5}\) Definition of $\phi$ and $\hat \phi$
\(\ds \) \(=\) \(\ds F_{m + n}\) Euler-Binet Formula

$\blacksquare$


Source of Name

This entry was named for Ross Honsberger.


Sources