Show that $\mathbb{Q}(i)(\sqrt{1+2i})\neq \mathbb{Q}(i)(\sqrt{1-2i})$. Hint: $\mathbb{Q}(i)(\sqrt{1+2i})=\mathbb{Q}(i)(\sqrt{1-2i})$ would imply $(1+2i)/(1-2i)$ is square in $\mathbb{Q}(i)$. Why would that be the case? (I can work the rest out).
Okay, edited. Suppose $\mathbb{Q}(i)(\sqrt{1+2i})= \mathbb{Q}(i)(\sqrt{1-2i})$, then $\sqrt{1-2i} \in \mathbb{Q}(i)(\sqrt{1+2i})$, therefore $\sqrt{1-2i} = a +b\sqrt{1+2i}$, with $a,b \in \mathbb{Q}(i)$, so $1-2i = a^2 + b^2(1+2i)+ 2ab\sqrt{1+2i}$. If both $a,b \neq 0$, we get $\sqrt{1+2i} \in \mathbb{Q}(i)$, which is false. If $b=0$ then $\sqrt{1-2i} \in \mathbb{Q}(i)$ which can't be the case either, so $a=0$ and $\sqrt{1-2i} = b\sqrt{1+2i}$ thus $(1+2i)/(1-2i)= 1/b^2$, a square in $\mathbb{Q}(i)$.