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2
votes
Diophantine approximation on spheres
dodd is right. Every point over $\mathbb{Z}[1/2]$ must have coordinates in $\frac{1}{2} \mathbb{Z}$, since by clearing denominators we get four squares of integers, not all even, summing to a power of …
3
votes
Accepted
Diophantine approximation on spheres
Here is a proof that $S(\mathbb{Z}[\frac{1}{p}])$ lies dense in $S^3$ for all primes $p \equiv 1 \pmod{4}$.
Since this is a wholly algebraic/arithmetical question, it is easier to switch to algebro-ge …