As a follow-up to my question about dyadic rational boundary points of the Mandelbrot set I wondered about dyadic rational points of the unit circle, and thought about the extension to higher dimensions.
For the 1D case, $\left\{\pm 1\right\}$ is the whole space.
For the 2D case, I think it might be $\left\{(\pm 1, 0), (0, \pm 1)\right\}$: it is known that the rational points on the unit circle take the form (see comments to linked question) $$ \left( \frac{1-t^2}{1+t^2} , \frac{2t}{1+t^2} \right), t \in \mathbb{Q} $$ I have no complete proof, but taking $1 < t \in \mathbb{N}$ the fractions are not dyadic as the square root of Mersenne number is irrational. Taking dyadic $t = \frac{p}{2^q}$ in reduced form gives a fraction with denominator $2^{2q}+p^2$ that is indivisble by $2$ unless $t \in \{ 0, \pm 1 \}$.
Not sure about the 3D case, apart from the similar permutations of $(\pm 1, 0, 0)$.
For the 4D case, $\left\{\left( \pm\frac{1}{2} , \pm\frac{1}{2} , \pm\frac{1}{2} , \pm\frac{1}{2} \right)\right\}$ is another subset of points.
For higher dimensions, the lower dimensional cases are embedded with 0 in the new dimensions, but there may be other new cases.
Question: is there a classification of dyadic rational points on spheres in all dimensions?
This really about finding integer solutions of $$x_1^2+\cdots+x_n^2=4^m.\tag1$$ This corresponds to the dyadic point $(x_1/2^m,\ldots,x_m/2^m)$. To avoid trivialities and degeneracies, stick to solutions with $m\ge1$ and not all $x_i$ even, so at least one is odd. Also let's avoid cases where some $x_i=0$.
There are no such solutions when $n\in\{1,2,3\}$. This is impossible modulo $4$. For $n=4$ and $m=1$ we have $(\pm1,\pm1,\pm1,\pm1)$ but when $m\ge2$ then ($1$) is impossible modulo $8$.
Non-trivial solutions start multiplying for $n\ge5$. There are no longer obstructions modulo powers of $2$. We have $(3,2,1,1,1)$ for $n=5$ etc. I'm not sure about finiteness any more...