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

1. If a sequence is geometric, then its terms are of the form:

${\displaystyle X_0,\hspace{0.5em}{X}_1={\mathrm{rX}}_0,\hspace{0.5em}{X}_2=r^2{X}_0,\hspace{0.5em}...X_n=r^n{X}_0.}$

Now, I have to find r, if the sequence is Fibonacci-type. Using the definition of a geometric sequence, I obtain:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{r^{n+2}{X}_0}& =r^{n+1}{X}_0+r^n{X}_0\\ \multicolumn{1}{c}{r^{n+2}}& =r^{n+1}+r^n\end{array}.}$

I see that $r$ must be different from 0, so I could divide both sides by $r$ which leads to the quadratic equation:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{r^2}& =r+1\\ \multicolumn{1}{c}{r^2-r-1}& =0\\ \multicolumn{1}{c}{r_{1,2}}& =\frac{1\pm \sqrt{5}}{2}.\end{array}}$

So

${\displaystyle \frac{X_1}{X_0}=\frac{1+\sqrt{5}}{2}\hspace{0.5em}\mathrm{or}\hspace{0.5em}\frac{1-\sqrt{5}}{2}=\varphi \hspace{0.5em}\mathrm{or}\hspace{0.5em}\frac{-1}{\varphi }.}$

which completes the proof of (1).

2. If the ratio of the first two terms is given by

${\displaystyle X_1=\frac{1+\sqrt{5}}{2}{X}_0}$

then, using the recursive formula for the sequence, I obtain:

${\displaystyle X_2=X_0+X_1=X_0+\frac{1+\sqrt{5}}{2}{X}_0=\frac{3+\sqrt{5}}{2}.}$

I observe that:

${\displaystyle \frac{3+\sqrt{5}}{2}=\frac{6+2\sqrt{5}}{4}=\frac{5+1+2\sqrt{5}}{4}=(\frac{1+\sqrt{5}}{2}{)}^2.}$

So this gives $X_2={\varphi }^2{X}_0=\varphi X_1.$ and I have shown $1+\varphi ={\varphi }^2$.

I calculate $X_3$: $X_3=X_2+X_1={\varphi }^2{X}_0+\varphi X_0=\varphi (\varphi +1)X_0={\varphi }^3{X}_0.$

But this is not enough, I have to utilise induction to show the sequence is geometric, that is $X_n={\varphi }^n{X}_0$ for all $n$. In the general case, I have, using the recurrence formula for the Fibonacci sequence:

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{X_{n+2}}& =X_{n+1}+X_n\\ \multicolumn{1}{c}{}& ={\varphi }^{n+1}{X}_0+{\varphi }^n{X}_0\\ \multicolumn{1}{c}{}& ={\varphi }^n(\varphi +1)X_0\\ \multicolumn{1}{c}{}& ={\varphi }^{n+2}{X}_0\end{array}.}$

So by the axiom of induction, as I have shown $X_n={\varphi }^n{X}_0$ is true for $n=1$ and $2$, it is true for all $n$ and so the sequence is geometric.

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