Golden ratio and floor function $\lfloor \phi ^2 n \rfloor - \lfloor \phi \lfloor \phi n \rfloor \rfloor = 1$ where $\phi = \frac{1+\sqrt{5}}{2}$ Prove that $\lfloor \phi ^2 n \rfloor - \lfloor \phi \lfloor \phi n \rfloor \rfloor = 1$ where $\phi = \frac{1+\sqrt{5}}{2}$ for all positive integers $n$
My thoughts: 
Tempted to take advantage of fact that $\phi-1 = 1/\phi$. 
So $ \frac{1}{\phi} \lfloor \phi ^2 n \rfloor = \frac{1}{\phi} \lfloor \phi \lfloor \phi n \rfloor \rfloor + \phi - 1  $ 
It looks like we are awfully close... 

Using $\phi^2 = \phi + 1$, we have
$$
\lfloor \phi^2 n \rfloor = \lfloor\phi n\rfloor + n,
$$
and, writing $\{x\}$ for the fractional part of a positive real number $x$, we have
$$
\lfloor \phi \lfloor \phi n \rfloor \rfloor = \lfloor \phi(\phi n - \{\phi n\})\rfloor = \lfloor \phi^2 n - \phi\{\phi n \}\rfloor = \lfloor \phi n - \phi \{\phi n\}\rfloor + n,
$$
so it suffices to show that
$$
\lfloor \phi n \rfloor = \lfloor \phi n - \phi \{\phi n\}\rfloor + 1.
$$
To make the argument more legible, define $x = \phi n$. Then we have to prove that
$$
\lfloor x \rfloor = \lfloor x - \phi\{x\}\rfloor + 1.
$$
But this is quite clear: since $1 < \phi < 2$, we subtract the fractional part of $x$ somewhere between once and twice from $x$. Strictly more than once tells us that the integral part the result must be at least 1 smaller, and strictly less than twice tells us that it cannot be 2 or more smaller.
