<s>[INST] Simplify $7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$ Let $\varphi=\frac{1+\sqrt5}2$ (the golden ratio).
How can I simplify the following expression?
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$$
 [/INST] The answer of  @Vladimir Reshetnikov: is hard to improve on, so we'll try to answer some of the questions posed in the comments.
If $\phi=\frac{1+ \sqrt{5}}{2}$ is the positive root of $x^2 - x-1=0$ then 
\begin{eqnarray}
\tan(2 \arctan(\phi^n)) = \frac{2\phi^n}{ 1- \phi^{2n}} = \begin{cases} -\frac{2}{L_{n}} \text{ if $n$ odd} \\
-\frac{2}{F_{n}\sqrt{5}} \text{ if $n$ even } 
\end{cases}
\end{eqnarray}
where $(F_n)$ is the Fibonacci sequence and $(L_n)$ is the Lucas sequence. 
Indeed, for every $v$ we have
$$(1+ iv)^2 = 1- v^2 + 2 i v= -v\cdot ( v -\frac{1}{v} - 2 i)$$
Now consider $v = \phi^n$. We get 
$$(1+ i \phi^n)^2 = - \phi^n\cdot ( \phi^n - \phi^{-n} - 2 i)$$
Note that $\phi$, $-\frac{1}{\phi}$ are the roots of $x^2 - x -1=0$. Therefore, if $n$ odd then $\phi^n - \phi^{-n}= \phi^n + \left(-\frac{1}{\phi}\right)^n=L_n$, while if $n$ even then $\phi^n - \phi^{-n} = \phi^n - \left(-\frac{1}{\phi}\right)^n= F_n \sqrt{5}$ .
Assume $n$ is odd. In order to write $2\arctan(\phi^n)$ as a combination of several $\arctan$'s of rational numbers, we decompose the number $L_n - 2 i$ in the Gaussian integers. It appears that for $n >5$ the number $L_n - 2 i$ keeps involving some distinct Gaussian primes from the previous ones. 
Assume now $n$ is even. We need to decompose the number $2 + F_{n} \sqrt{-5}$ into some products of elements. Note that in the ring $\mathbb{Z}[\sqrt{-5}]$ the decomposition into irreducibles is not unique, so we have some problems. 
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