Background. I thought up the problem of finding a regular $n$-sided polygon that has a diagonal with length $d_k$ such that the area of the polygon equals ${d_k}^2$. (Let $d_k$ denote the length of a diagonal that skips $k$ vertices, so that $d_0$ is the length of a side.) By doing some trigonometry I got the equation
$$8\sin^2\left(\frac{(k+1)\pi}{n}\right)=n\sin\left(\frac{2\pi}{n}\right).$$
I have checked some values of $n$ and it seems that $(4,0)$ and $(12,3)$ are the only (primitive and non-trivial) possible values for $(n,k)$.
Question. Are there other solutions?
One approach could be rewriting the equation using $\sin(x)=\frac i2(e^{-ix}-e^{ix})$. This gives $$4(\zeta_n^{k+1}+\zeta_n^{-k-1})-in(\zeta_n+\zeta_n^{-1})=8,$$ where $\zeta_n=e^{2\pi i/n}$.
Update. Except for $n=4$, we have $n=4p$ with $p\equiv3\pmod4$ prime. Furthermore, $\gcd(n,k+1)=4$.
See my answer below. Thanks to user dinoboy for the suggested technique.