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?
On
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
On
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.
On
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\pars{q + \root{q^{2} - 1}}^{2\pars{N + 1}} = 1:\ {\large q = ?}}$
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}
$$\color{#00f}{\large% q_{n} = \cos\pars{{n \over N + 1}\,\pi}\,,\qquad n \in {\mathbb Z}} $$
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.