Bernoulli's lemniscate with unit half-width is given by the polar equation $r=\sqrt{\cos 2\theta}$. Using this curve, in the first quadrant, we define the function $\operatorname{arcsl}$ according to this figure
(the author is Jacob Rus on Wikipedia) and the function $\operatorname{sl}$ as its inverse. The definition of the function $\operatorname{sl}$ can be extended to all real numbers (see here https://en.wikipedia.org/wiki/Lemniscate_elliptic_functions for more details).
Let $2\varpi$ be the perimeter of the lemniscate $r=\sqrt{\cos 2\theta}$. The n-division points of the lemniscate are the points dividing it to arcs of equal length. It can be proved that
The n-division points can be constructed using only straightedge and compass iff $n=2^r p_1\cdots p_m$ where $r$ is a non-negative integer and each $p_i$ is a distinct Fermat prime.
The problem of constructibility of the n-division points can be translated to the language of algebra as the problem of determining for what $n$ is the degree of the minimal polynomial of $\operatorname{sl}\frac{2\varpi k}{n}, (k\in\{1,\ldots,n\})$ in $\mathbb{Q}[x]$ a power of two. However, this uses the tacit assumption that the origin is a division point. (since for $k=n$ we get $\operatorname{sl}2\varpi=0$.)
If we drop this assumption, does the theorem still hold true? If it does, why?
In his book Galois theory, Cox ignores this subtlety. On p. 464, he writes
Here, the n-division points of the lemniscate are obtained as follows. Begin at the origin and follow the curve into the first quadrant, down into the fourth quadrant [...]
Note that "begin at the origin" is the case $k=n$. The question is what if one doesn't begin at the origin?