Golden fibs

We have to thank Andrei Lazanu of School No. 205, Bucharest, Romania for this solution.

Given a general Fibonacci sequence $X_n$, that is a sequence satisfying the Fibonnaci relation: $X_{n+2}= X_{n+1}+ X_n$, I have to prove the following (where $\phi$ is the golden ratio):

1. If the sequence is geometric, then the ratio of the first two terms is given by $X_1:X_0= \phi$ or $X_0:X_1=-\phi$

2. If the ratio of the first two terms is $X_1:X_0= \phi$ or $X_0:X_1=-\phi$ then the sequence is geometric.

1. If a sequence is geometric, then its terms are of the form: $$X_0,X_1=r{X}_0,X_2=r^2{X}_0,...X_n=r^n{X}_0.$$

Now, I have to find r, if the sequence is Fibonacci-type. Using the definition of a geometric sequence, I obtain: $$\begin{array}{rl}{r}^{n+2}{X}_0& =r^{n+1}{X}_0+r^n{X}_0\\ r^{n+2}& =r^{n+1}+r^n\end{array}.$$ I see that $r$ must be different from 0, so I could divide both sides by $r$ which leads to the quadratic equation: $$\begin{array}{rl}{r}^2& =r+1\\ r^2-r-1& =0\\ r_{1,2}& =\frac{1\pm \sqrt{5}}{2}.\end{array}$$ So $$\frac{X_1}{X_0}=\frac{1+\sqrt{5}}{2}\mathrm{o}\mathrm{r}\frac{1-\sqrt{5}}{2}=\varphi \mathrm{o}\mathrm{r}\frac{-1}{\varphi }.$$ which completes the proof of (1).

2. If the ratio of the first two terms is given by $$X_1=\frac{1+\sqrt{5}}{2}{X}_0$$ then, using the recursive formula for the sequence, I obtain: $$X_2=X_0+X_1=X_0+\frac{1+\sqrt{5}}{2}{X}_0=\frac{3+\sqrt{5}}{2}.$$ I observe that: $$\frac{3+\sqrt{5}}{2}=\frac{6+2\sqrt{5}}{4}=\frac{5+1+2\sqrt{5}}{4}=(\frac{1+\sqrt{5}}{2}{)}^2.$$ So this gives $X_2=\phi^2 X_0=\phi X_1.$ and I have shown $1+\phi = \phi^2$.

I calculate $X_3$: $X_3 = X_2 + X_1 = \phi^2 X_0 + \phi X_0 = \phi(\phi+1)X_0=\phi^3 X_0.$

But this is not enough, I have to utilise induction to show the sequence is geometric, that is $X_n=\phi^nX_0$ for all $n$. In the general case, I have, using the recurrence formula for the Fibonacci sequence: $$\begin{array}{rl}{X}_{n+2}& =X_{n+1}+X_n\\ & ={\varphi }^{n+1}{X}_0+{\varphi }^n{X}_0\\ & ={\varphi }^n(\varphi +1)X_0\\ & ={\varphi }^{n+2}{X}_0\end{array}.$$ So by the axiom of induction, as I have shown $X_n=\phi^nX_0$ is true for $n=1$ and $2$, it is true for all $n$ and so the sequence is geometric.

A similar proof works when $X_0=-\phi X_1$.