<s>[INST] Prove the inequality $\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|<\frac{1}{mn}$ 
Prove that the inequality $$\left|\frac{m}{n}-\frac{1+\sqrt{5}}{2}\right|<\frac{1}{mn}$$ holds for positive integers $m, n$ if and only if $m$ and $n$ where $m > n$ are two successive terms of the Fibonacci sequence.

I thought about using the explicit formula for the Fibonacci sequence, which is $$F_n = \dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right),$$ but this seems to get computational. Is there an easier way to think about this?
 [/INST] $\varphi = \frac{1+\sqrt{5}}{2}$ is the root of $x^2-x-1=0$. In fact $x^2-x-1=\left(x - \frac{1+\sqrt{5}}{2}\right) \cdot \left(x - \frac{1-\sqrt{5}}{2}\right)$. Taking $x=\frac{F_{n+1}}{F_n}$, we have:
$$\left|\left(\frac{F_{n+1}}{F_n}\right)^2-\frac{F_{n+1}}{F_n}-1\right|=\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left|\frac{F_{n+1}}{F_n} - \frac{1-\sqrt{5}}{2}\right| \Leftrightarrow $$
$$\left|\frac{F_{n+1}^2-F_{n+1}F_n-F_n^2}{F_n^2}\right|=\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left|\frac{F_{n+1}}{F_n} + \frac{\sqrt{5}-1}{2}\right| \Leftrightarrow $$
Using Will's hint (a particular case of this):
$$\frac{1}{F_n^2}=\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left(\frac{F_{n+1}}{F_n} + \frac{\sqrt{5}-1}{2}\right) \Rightarrow$$
$$\frac{1}{F_n^2}>\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| \cdot \frac{F_{n+1}}{F_n}$$
or
$$\left|\frac{F_{n+1}}{F_n} - \frac{1+\sqrt{5}}{2}\right| < \frac{1}{F_n \cdot F_{n+1}}$$
Now, let's assume $\left|\frac{m}{n} - \frac{1+\sqrt{5}}{2}\right| < \frac{1}{mn}, m>n$ then, using the same polynomial:
$$\left|\left(\frac{m}{n}\right)^2-\frac{m}{n}-1\right|=\left|\frac{m}{n} - \frac{1+\sqrt{5}}{2}\right| \cdot \left|\frac{m}{n} - \frac{1-\sqrt{5}}{2}\right| < $$
$$\frac{1}{mn}\cdot \left(\frac{m}{n} + \frac{\sqrt{5}-1}{2}\right)$$
or
$$\left|m^2-mn-n^2\right|< \frac{n}{m} \cdot \left(\frac{m}{n} + \frac{\sqrt{5}-1}{2}\right)<\frac{n}{m} \cdot \left(\frac{m}{n} + 1\right)<2$$
which means $\left|m^2-mn-n^2\right|=0$ or $\left|m^2-mn-n^2\right|=1$. The first one, being related to  $x^2-x-1=0$, has no integer solutions other than $m=n=0$ which is not $m>n$. Fibonacci numbers satisfy the latter. 
Are Fibonacci the only numbers satisfying $\left|m^2-mn-n^2\right|=1$? Well, $1$ and $0$ do and if we assume $m=n+q$ we have:
$$|m^2-mn-n^2|=|(n+q)^2 - (n+q)n - n^2|=|n^2+2nq+q^2 - n^2 - nq - n^2|=$$
$$|q^2+nq-n^2|=|n^2-nq-q^2|=1$$
clearly $n>q>0$, otherwise $q^2+nq-n^2 \geq n^2 + n^2 - n^2\geq 1$ (equality in fact is possible only for $n=q=1$). Next we take $n=q+q_1$ and so on, leading to $m>n>q>q_1>...>q_p > 0$, i.e. it's a finite process. $q_p=1$ otherwise it can be "split" further (e.g. $q_p=2$ leads to $q_{p+1}=1$ and $|2^2-2\cdot 1 -1^1|=1$). Eventually, this leads to Fibonacci numbers only.
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