Let $F$ be a field such that $\operatorname{Char} (F)\neq 2$. Let $E,E'$ be extension fields of $F$ , $[E:F]=[E':F]=2$.
Prove $E,E'$ are $\text{$F$-isomorphism} \iff F^{\times}\cap (E^\times)^2 = F^{\times}\cap (E'^\times)^2.$
My attempt:
Suppose $E=F(a)$.
$$a^2 \in (E^\times)^2 \implies a^2=x+ay, \quad x,y\in F$$
Any idea ? Appreciate any help.
"$\Rightarrow$" Let $\sigma$ be the isomorphism, if $u=\alpha^2\in F^\times\cap (E^\times)^2$, then $\sigma(u)=u=(\sigma(\alpha))^2\in F^\times\cap (E'^\times)^2$.
"$\Leftarrow$" Suppose $E=F(a)$ and $a^2=x+ay,x,y∈F$. Then $(a-1/2y)^2=x+1/4y^2\in F^\times\cap (E^\times)^2=F^\times\cap (E'^\times)^2$. So there is $b\in E'$, $(b-1/2y)^2=x+1/4y^2$. The element $b$ can not belong to $F$ since $a$ is not in $F$. Mapping $a$ to $b$ we get $E=F[a]\cong F[b]=E'$.