Conjecture:
The set of unique roots of $$f_k(x)=\sum_{j=1}^k x^j-x^{-j} \;,\;\; x \not=0$$ is given by $e^{i \pi \phi_k}$, where $$\frac{1}{2}\phi_k=\{0, \frac{1}{2}, \underbrace{\frac{\pm1}{k+1},\frac{\pm1}{k},\frac{\pm2}{k+1},\frac{\pm2}{k},\frac{\pm3}{k+1},...}_\text{$k-1$ elements} \}$$
Q: If true, how can this be proven?
I unfortunately don't have much experience with proving stuff (I'm a physicist), so any insights would greatly appreciated!
This may very well be some lemma, but I thought it might be useful since Mathematica wasn't able to solve this kind of equation for $k>12$.
Note that the roots of $f_k$ are equivalent to the roots of $f_{k,n}=\sum_{j=1}^k x^{n+j}-x^{n-j}$. Also, if $x \not = 1$ $f_k=\frac{x^{-k}(x^{k}-1)(x^{k+1}-1)}{x-1}$.
The process through which I reached this conjecture (btw, sorry if "conjecture" sounds pretentious, but I didn't know what else to call it) is described below.
Just to get an overview, here's one way of writing out $f_k=0$:
\begin{array}{c|c} k & \\ \hline 1 & x^2-1=0\\ 2 & x^4+x^3-x-1=0 \\ 3 & x^6+x^5+x^4-x^2-x-1=0 \\ \vdots &\vdots \end{array}
I noticed a high degree of regularity when plotting the solutions.
The solutions to the first 12 $k$ are shown below, plotted in the complex plane.
$\hskip1.4in$ 
(I have no idea why Mathematica doesn't recognize $-1$ as a solution to $f_1$, but oh well)
I now organized the solutions such that one might see a pattern:
If one looks at the list of the roots for $k\in \{1,...,12\}$, they can all be written as a phase only, $e^{i \pi \phi}$, where $\phi$ is some rational number. For instance, the roots of $f_2$ are $\{1,e^{\pm \frac{2}{3} i \pi},-1\}$ so $\phi_2=\{0,\pm \frac{2}{3},1 \}$. I'm assuming that this pattern continues indefinitely.
All $0<\phi<1$ appears with both a positive and a negative sign, and all sets of solutions include $\phi=\{0,1\}$. The non-negative and non-trivial ($0$ and $1$) values for $\phi_{k}$ for $k\in \{1,...,12\}$ are listed here (written as reduced fractions, going from smallest to largest values):
\begin{array}{c|c} k & \phi_k \\ \hline 2 & \frac{2}{3}\\ 3 & \frac{1}{2} \;\; \frac{2}{3}\\ 4 & \frac{2}{5} \;\; \frac{1}{2} \;\; \frac{4}{5}\\ 5 & \frac{1}{3} \;\; \frac{2}{5} \;\; \frac{2}{3} \;\; \frac{4}{5}\\ 6 & \frac{2}{7} \;\; \frac{1}{3} \;\; \frac{4}{7} \;\; \frac{2}{3} \;\; \frac{6}{7}\\ 7 & \frac{1}{4} \;\; \frac{2}{7} \;\; \frac{1}{2} \;\; \frac{4}{7} \;\; \frac{3}{4} \;\; \frac{6}{7}\\ 8 & \frac{2}{9} \;\; \frac{1}{4} \;\; \frac{4}{9} \;\; \frac{1}{2} \;\; \frac{2}{3} \;\; \frac{3}{4} \;\; \frac{8}{9}\\ 9 & \frac{1}{5} \;\; \frac{2}{9} \;\; \frac{2}{5} \;\; \frac{4}{9} \;\; \frac{3}{5} \;\; \frac{2}{3} \;\; \frac{4}{5} \;\; \frac{8}{9} \\ 10 &\frac{2}{11} \;\; \frac{1}{5} \;\; \frac{4}{11} \;\; \frac{2}{5} \;\; \frac{6}{11} \;\; \frac{3}{5} \;\; \frac{8}{11} \;\; \frac{4}{5} \;\; \frac{10}{11}\\ 11 & \frac{1}{6} \;\; \frac{2}{11} \;\; \frac{1}{3} \;\; \frac{4}{11} \;\; \frac{1}{2} \;\; \frac{6}{11} \;\; \frac{2}{3} \;\; \frac{8}{11} \;\; \frac{5}{6} \;\; \frac{10}{11}\\ 12 & \frac{2}{13} \;\; \frac{1}{6} \;\; \frac{4}{13} \;\; \frac{1}{3} \;\; \frac{6}{13} \;\; \frac{1}{2} \;\; \frac{8}{13} \;\; \frac{2}{3} \;\; \frac{10}{13} \;\; \frac{5}{6} \;\; \frac{12}{13} \end{array}
I now rewrote this (by extending by $2$ here and there) to get
\begin{array}{c|c} k & \phi_k \\ \hline 2 & \frac{2}{3}\\ 3 & \frac{2}{4} \;\; \frac{2}{3}\\ 4 & \frac{2}{5} \;\; \frac{2}{4} \;\; \frac{4}{5}\\ 5 & \frac{2}{6} \;\; \frac{2}{5} \;\; \frac{4}{6} \;\; \frac{4}{5}\\ 6 & \frac{2}{7} \;\; \frac{2}{6} \;\; \frac{4}{7} \;\; \frac{4}{6} \;\; \frac{6}{7}\\ 7 & \frac{2}{8} \;\; \frac{2}{7} \;\; \frac{4}{8} \;\; \frac{4}{7} \;\; \frac{6}{8} \;\; \frac{6}{7}\\ 8 & \frac{2}{9} \;\; \frac{2}{8} \;\; \frac{4}{9} \;\; \frac{4}{8} \;\; \frac{6}{9} \;\; \frac{6}{8} \;\; \frac{8}{9}\\ \vdots & \vdots \end{array}
which clearly reveals the pattern given above.
Here are the first 100 sets of solutions:
And 500:
(1)
You can pretty much make this equivalent to finding the roots of a polynomial (with highest coefficient 1). Pick e.g. n=k, then you're dealing with a polynomial which is a simple function after all.
In that case you can prove it directly. For the $s=2k$ roots you have found write: $(x-x_1)(x-x_2)...(x-x_s)$ and show that this is the same polynomial as yours (by opening all the brackets and executing the multiplications). Of course when opening the brackets group the roots appropriately so that your calculations get simpler.
I think something along these lines would work.
(2)
Another identical way to do it:
If you take $x^k \cdot f_k(x)$ you get a polynomial of degree $2k$ and 0 is not a root of that polynomial $P(x)$. So the roots of that polynomial are the same as the roots of $f_k(x)$. Now just take the roots/numbers you have found and prove that P(x) is indeed zero at these roots. That would be sufficient since we know a polynomial of degree $2k$ has exactly $2k$ complex roots (some of them may be with frequency > 1 though but still the total count is $2k$).
(3)
Actually from this formula it's most obvious.
$f_k=\frac{x^{-k}(x^{k}-1)(x^{k+1}-1)}{x-1}$
From here you can conclude that the roots are the k-th roots of 1 and the (k+1)-th roots of 1 in the complex plane. So from this expression of the function, you almost have nothing to prove, it's all known formulas.
Roots of unity