4
$\begingroup$

Is it true that for all square-free $n$, and for all $k>1$, we have $SO_k(\mathbb{Q})\subsetneq SO_k(\mathbb{Q(\sqrt{n})})$?

So far I have only discovered this: if $n$ has no prime factors $\equiv -1 \;\bmod 4$, then there exist $x,y\in \mathbb{Z}^+$ such that $(\dfrac{x}{\sqrt{n}})^2+(\dfrac{y}{\sqrt{n}})^2=1$, from which a rotation is easily constructed. The other cases seem more obscure.

$\endgroup$
2
$\begingroup$

If $n$ is square-free, there is a nontrivial solution $(x,y) \in \mathbb{Z}^2$ of Pell's equation $x^2 - ny^2 = 1$. Then $$ \Bigl( \frac{1}{x} \Bigr)^2 + \Bigl( \frac{y}{x} \sqrt{n} \Bigr)^2 = 1. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.