How do I prove $F(a)=F(a^2)?$

You know that $[F(a):F]=[F(a):F(a^2)][F(a^2):F]$. The minimal polynomial of $a$ over $F(a^2)$, assuming that $a\notin F(a^2)$, is $x^2-a^2$, so we have $[F(a):F(a^2)]=2$. Now, what is the problem with that?

I am writing the converse part $F(a)\subset F(a^2).$ Enough to show that $a \in F(a^2).$ Let $a\notin F(a^2)$ then $ F(a^2)\subsetneq F(a). $ Since $a^2$ satisfies the polynomial $x^2-a^2\in F(a^2)$ so $[F(a):F(a^2)]=2$. Now $$ [F(a): F] = [F(a) : F(a^2)][F(a^2):F]=2[F(a^2):F].$$ which contradicts the fact that [F(a) : F] is odd. Thus $a \in F(a) $ and therefore $F(a) = F(a^2)$.