Stuck on a proof involving the floor of a ratio of natural logarithms

So, I'm working on a proof of a personal problem that I solved years ago unrigorously, and am now trying to rigorously prove. The problem I have is as follows:

Given some $v\in\Bbb Z$, $v\ge2$, find a formula giving $m\in\Bbb Z$, $m>0$ such that

$$\frac{(2+\sqrt{5})^m-(2-\sqrt{5})^m}{\sqrt{5}}\le v<\frac{(2+\sqrt{5})^{m+1}-(2-\sqrt{5})^{m+1}}{\sqrt{5}}$$

Now, if - as I wholly suspect - the $(2-\sqrt{5})^m$ and $(2-\sqrt{5})^{m+1}$ terms can be ignored due to their lack of effect on the final result (as $|2-\sqrt{5}|<\frac{1}{2}$ and $\lim_{m\to\infty}|(2-\sqrt{5})^m|=0$), then we have:

$$m\le \frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}<m+1$$

which implies that:

$$m=\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\rfloor$$

which certainly seems to work empirically. However, the difficulty I'm having is in proving that that $(2-\sqrt{5})^m$ term can be dropped since we're taking the floor of the final result. Boiling it down, I need to prove that:

$$\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\rfloor=\lfloor\frac{\ln{(v\sqrt{5}+(2-\sqrt{5})^m)}}{\ln(2+\sqrt{5})}\rfloor$$

for all $v\in\Bbb Z$, $v\ge2$ and $m\in\Bbb Z$, $m>0$ (as above), and I'm having a heck of a time figuring out how best to go about it. Any pointers in the correct direction would be appreciated!

Let $$v = \frac{(2+\sqrt{5})^m-(2-\sqrt{5})^m}{\sqrt{5}}$$ with even $m$. Then $$(2+\sqrt{5})^{m-1}<(2+\sqrt{5})^m-1<v\sqrt{5}< (2+\sqrt{5})^m.$$ which implies $$m-1\le \frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}<m.$$ Therefore, $$\left\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\right\rfloor=m-1,$$ and your claim is false, unfortunately.
If, however, both inequalities are strict, $$F_m:=\frac{(2+\sqrt{5})^m-(2-\sqrt{5})^m}{\sqrt{5}}< v<\frac{(2+\sqrt{5})^{m+1}-(2-\sqrt{5})^{m+1}}{\sqrt{5}},$$ then (knowing that $F_m$ is integer) $F_m+1\le v\le F_{m+1}-1$. Therefore, $$(2+\sqrt5)^m < \sqrt{5}F_{m+1} + \sqrt{5}\le v\sqrt{5}\le F_{m+1}-\sqrt{5} <(2+\sqrt5)^{m+1},$$ whence $$\left\lfloor\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}\right\rfloor=m.$$
• Could you elaborate on how you got to $$(2+\sqrt{5})^m-1<v\sqrt{5}< (2+\sqrt{5})^m$$ by setting $$v = \frac{(2+\sqrt{5})^m-(2-\sqrt{5})^m}{\sqrt{5}}$$ I'm afraid I'm not certain I understand how you arrived thereat. – Jeremy Holland Sep 10 '15 at 13:10
• Could you elaborate on how you derived $$m-1\le \frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}<m$$ based on $$(2+\sqrt{5})^m-1<v\sqrt{5}< (2+\sqrt{5})^m$$ I'm afraid I'm not able to see how you got there. – Jeremy Holland Sep 14 '15 at 6:50
• I believe I understand that $$(2+\sqrt{5})^{m-1}<(2+\sqrt{5})^{m}-1\quad\forall\ m\in\Bbb Z,m>0$$ and so $$m-1<\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}<m.$$ Is it therefore logically correct to be able to make the looser assertion $$m-1\le\frac{\ln{(v\sqrt{5})}}{\ln(2+\sqrt{5})}<m?$$ – Jeremy Holland Sep 14 '15 at 14:25