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?
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5$$
 A: 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. 
A: As a preliminary result,
$$\frac{(1+i\phi)^7(1+i\phi^3)^2}{(1+i\phi^5)}=(-5+10i)(7+3\sqrt{5})\tag{1}$$
gives, switching to arguments,
$$ 7\arctan\phi+2\arctan\phi^3-\arctan\phi^5 = 3\pi-\arctan 2. \tag{2} $$
In the same way,
$$ 7\arctan\phi^{-1}+2\arctan\phi^{-3}-\arctan\phi^{-5} = \pi+\arctan 2. \tag{3} $$
In order to deal with the squared arctangent with Parseval-like techniques, it may be useful to recall that the Fourier transform of the arctangent function is given by $i\sqrt{\frac{\pi}{2}}\frac{e^{-|s|}}{s}$. Keep working.
A: We will use a well-known$^{[1]}$ formula for sum of arctangents:
$$\arctan u + \arctan v = \arctan \left( \frac{u+v}{1-uv} \right) \pmod \pi\tag1$$
The exact equality holds for $uv<1$, for other values there is additional term — an integer multiple of $\pi$ (it is easy to determine in each case).
Using this formula, we can establish the following identities:
$$\begin{align}
\arctan\varphi\phantom{^1}&=\frac\pi2-\frac12\,\arctan2,\tag2\\
\arctan\varphi^3&=\frac\pi4+\frac12\,\arctan2,\tag3\\
\arctan\varphi^5&=\pi-\frac32\,\arctan2.\tag4
\end{align}$$
Plugging these into the original expression in the question and expanding parentheses, we can see that all $\arctan2$ terms cancel, and we get the result
$$7\arctan^2\varphi+2\arctan^2\varphi^3-\arctan^2\varphi^5=\frac{7\pi^2}8.\tag5$$
