In a field with characteristic other than 2, if we have two elements in our field, $a,b \in F$ such that $F(\sqrt{a})= F(\sqrt{b})$ and $[F(\sqrt{a}):F]=[F(\sqrt{b}):F]=2$
Then we can deduce that $ab \in F^2$
I am not sure how to do this. How do we know that $ab \in F^2$ ? Why is this true?
Note that since $F(\sqrt{a})=F(\sqrt{b})$ and both are of degree $2$, then we can express $\sqrt{a}$ as $\sqrt{a}=r+s\sqrt{b}$ for some $s,b\in{F}$. Squaring both sides we get $$a = r^2 + s^2b + 2rs\sqrt{b}.$$ That means that $2rs=0$, and since the characteristic is not $2$, then $rs=0$, so $r=0$ or $s=0$.
Can you take it from there?