Whether $SO_k(\mathbb{Q})\subsetneq SO_k(\mathbb{Q(\sqrt{n})})$

Question

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.

Answer 1

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 $$ \Bigl( \frac{1}{x} \Bigr)^2 + \Bigl( \frac{y}{x} \sqrt{n} \Bigr)^2 = 1. $$