Solving an equation for an unknown How can I solve the following equation for $q$? I'm totally stuck. I have done everything up to this point though.
$$\left(q + \sqrt{q^2-1}\right)^{2(N+1)} = 1,$$ where N is a natural number.
Attempt: Using DeMoivre's formula for the roots of unity, we have that for $$z^M = 1,$$the $M^{th}$ roots of unity are given by $$z=\exp\left(\frac{2ik\pi}{M}\right).$$ So $$q+\sqrt{q^2-1}=\cos\left(\frac{\pi k}{N+1}\right) + i\sin\left(\frac{\pi k}{N+1}\right), \ k=1,2,3, \ldots, 2N+1$$
What next?
 A: q = 1 or -1 are the only solutions. Observe that: (q + (q^2 - 1)^1/2)^2 = 1/(q - (q^2 - 1)^1/2)^2. So if q > 1 ==> LHS > 1. Contradiction. If q < -1 ==> multiply both sides with conjugate q - (q^2 - 1)^1/2 and get ((q - (q^2 - 1)^1/2)^2)^(N+1) > 1 since:
q - (q^2 - 1)^2 < - 1
A: Is it possible to take the 2(N+1)th root so that the left side becomes $ q + \sqrt {{q^2} - 1}  =  \pm 1 $ then you can do $ \sqrt {{q^2} - 1}  = 1 - q $  square both sides and get  $ {q^2} - 1 = 1 - 2q + {q^{2\,\,\,}}$ (note: only positive 1 is taken first )   then find q = 1 of course for negative one get also q=-1.
A: If $q>1$, then we have clearly no solution.
If $q\in [-1,1]$, then there is a $\vartheta\in (0,2\pi)$, such that
$$
\cos\vartheta=q
$$
and 
$$
\sqrt{q^2-1}=\pm i\sqrt{1-q^2}= i\sin \vartheta.
$$
(The minus sign is absorbed in the choice of $\vartheta$.) Then your equation looks like
$$
(\cos\vartheta+i\sin\vartheta)^{2N+2}=1.
$$
So in this case
$$
q=\cos\left(\frac{2k\pi}{2N+2}\right),\,\,\,k=0,1,\ldots,2N+1.
$$
Finally, if $q<-1$, then we simply want to study the cases
$$
q+\sqrt{q^2-1}=1\quad \text{and}\quad q+\sqrt{q^2-1}=-1,
$$
which are dealt with easily.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
$\ds{\pars{q + \root{q^{2} - 1}}^{2\pars{N + 1}} = 1:\ {\large q = ?}}$

Set $\ds{q \equiv \cosh\pars{t}}$. Then,
  \begin{align}
1&=\bracks{\cosh\pars{t} + \root{\cosh^{2}\pars{t} - 1}}^{2\pars{N + 1}}
=\bracks{\cosh\pars{t} + \sinh\pars{t}}^{2\pars{N + 1}}=\expo{2\pars{N + 1}t}
\end{align}

$$
1 = \expo{2\pars{N + 1}t}\quad\imp\quad 2\pars{N + 1}t_{n}=2n\pi\ic
\quad\imp\quad t_{n} = {n \over N + 1}\,\pi \ic\,;\qquad n \in {\mathbb Z}
$$

\begin{align}
q_{n} = \cosh\pars{t_{n}} =
{\expo{n\pi\,\ic/\pars{N + 1}} + \expo{-n\pi\,\ic/\pars{N + 1}} \over 2}
=\cos\pars{{n \over N + 1}\,\pi}
\end{align}

$$\color{#00f}{\large%
q_{n} = \cos\pars{{n \over N + 1}\,\pi}\,,\qquad n \in {\mathbb Z}}
$$
