# Fibonacci Fashion

What have Fibonacci numbers to do with solutions of the quadratic
equation x^2 - x - 1 = 0 ?

## Problem

Given $F_n={1\over\sqrt5}(\alpha^n-\beta^n)$ where $\alpha$ and $\beta$ are solutions of the quadratic equation $x^2-x-1=0$ and $\alpha > \beta$ show that

(1)$\alpha\beta =-1$, $\alpha + \beta = 1$

(2)${ 1\over \alpha}+{1\over \alpha^2} = {1\over \beta} + {1\over \beta^2}=1$

(3)$F_1=F_2=1$ and $F_n + F_{n+1} = F_{n+2}$ and hence $F_n$ is the $n$th Fibonacci number and

(4) the sum $1 + F_1 + F_2 + \ldots F_n$ gives another Fibonacci number.

## Getting Started

Parts (2) and (3) of this problem use the results of previous parts.

In part (4) try some small values of $n$, look for a pattern and make a conjecture about the result you suspect might always be true, then prove your conjecture.

## Student Solutions

Here is another excellent solution from Andrei of Tudor Vianu National College, Bucharest, Romania.

We are given $F_n={1\over\sqrt5}(\alpha^n-\beta^n)$ where $\alpha$ and $\beta$ are solutions of the quadratic equation $x^2-x-1=0$ and $\alpha > \beta$.

(1) In the quadratic equation $ax^2+bx+c=0$ with roots $\alpha$ amd $\beta$, using Viete's relations for the sum and product of the roots, I obtain: $\alpha\beta =-b/a$, $\alpha + \beta = c/a$

In the particular case of the equation $x^2-x-1=0$, I have: $\alpha\beta =-1$, $\alpha + \beta = 1$

(2)$${1\over \alpha} + {1\over \alpha^2}={\alpha + 1\over \alpha^2}=1$$ as $\alpha$ satisfies $x^2=x+1$. Similarly for $\beta$.

(3) Here I shall prove that $F_1=1$, $F_2=1$ and $F_n + F_{n+1} = F_{n+2}$ and hence $F_n$ is the $n$th Fibonacci number. First, I calculate $\alpha$ and $\beta$: $\alpha={1+\sqrt5\over 2}$ and $\alpha={1+\sqrt5\over 2}$

$$\begin{eqnarray} F_1 &= {1\over \sqrt 5}\left( {1+\sqrt 5 \over 2} - {1-\sqrt 5 \over 2}\right) = 1 \\ F_2 &= {1\over \sqrt 5}(\alpha^2 - \beta^2) = {1\over \sqrt 5} (\alpha - \beta)(\alpha + \beta) = 1 \\ F_n +F_{n+1}- F_{n+2}&= {1\over \sqrt 5}[\alpha^n -\beta^n + \alpha^{n+1} -\beta^{n+1} - \alpha^{n+2} + \beta^{n+2}] \\ F_n + F_{n+1}- F_{n+2}&= {1\over \sqrt 5}[-\alpha^n(\alpha^2 - \alpha - 1) +\beta^n(\beta^2- \beta -1)]=0\end{eqnarray}$$

(4) I shall prove by induction the statement $P_n$ that $1 + F_1 + F_2 + \cdots F_n = F_{n+2}$.

I know that $F_1=1, F_2=1$ and by (3), $F_3=2, F_4=3, F_5=5, F_6=8,...$\\ For $n=1$, $P_1$ is $1+ F_1 = F_3$ which is evidently true as 1 + 1 = 2.

For $n=2$, $P_1$ is $1+ F_1 + F_2= F_4$ which is evidently true as 1 + 1 + 1 = 3.

For $n=3$, $P_1$ is $1+ F_1 + F_2 + F_3 = F_5$ which is evidently true as 1 + 1 + 1 + 2 = 5.

Now I assume that $P(k)$ is true for a fixed $k$ and I shall prove that $P(k+1)$ is also true, that is:

$$\begin{eqnarray} P_k &: 1 + F_1 + F_2 + \cdots F_k = F_{k+2} \\ P_{k+1}&: 1 + F_1 + F_2 + \cdots F_k + F_{k+1} = F_{k+3}\end{eqnarray}$$

$$\begin{eqnarray}1 + F_1 + F_2 + \cdots F_k + F_{k+1}&= (1 + F_1 + F_2 + \cdots F_k ) + F_{k+1}\\ &= F_{k+2} + F_{k+1} \\ &= F_{k+3}\end{eqnarray}$$

and hence the result is true for $n=k+1$. By the principle of mathematical induction the statement $P_n$ is true for all positive integer values of $n$ which completes the proof.

## Teachers' Resources

$F_n={1\over\sqrt5}(\alpha^n-\beta^n)$ where $\alpha$ and $\beta$ are solutions of the quadratic equation $x^2-x-1=0$ and $\alpha > \beta$ is the explicit formula for the $n$th Fibonacci number as given in this question.

The Golden Ratio is one of the roots of the quadratic equation and this explains the many connections between Fibonacci numbers and the Golden Ratio.

This problem complements the material in the article The Golden Ratio, Fibonacci Numbers and Continued Fractions

For a sequence of, mainly more elementary, problems on these
topics see

See this article if you want to know more about the method of proof by mathematical induction.