Does there exist a field where all even degree equations have solutions but not all odd degree equations?

Question

Does there exist a field $F$, preferably a subfield of the complex numbers, such that all even degree polynomial equations have solutions in $F$, but such that for every positive odd integer $k$ greater than or equal to $3$, there exist $k$-th degree polynomial equations which do not have solutions in $F$? Basically, such a field $F$, if it exists, would be an "anti"-real-closed field. In a real closed field, every odd degree polynomial equation has solutions, but for every positive even integer $k$, there exist $k$-th degree polynomial equations with no solutions.

Answer 1

No. Suppose that every polynomial over $F$ of even degree has a root in $F$, and pick $p \in F[x]$ of odd degree. Then, $\deg (p^2)$ is even, so $p(\alpha)^2 = 0$ for some $\alpha \in F$. But that means that $p(\alpha) = 0$, so in particular the arbitrary polynomial $p$ of odd degree has a root in $F$.

As @Ron Kaminsky remarked in the comments, there's an informally dual argument: Instead consider the polynomial $q(x) := p(x^2)$. It has even degree $2 \deg p$ so has a root $\alpha \in F$, that is, $0 = q(\alpha) = p(\alpha^2)$, hence $\alpha^2 \in F$ is a root of $p$.

Answer 2

While I can't guarantee that such a field exists, I can guarantee you a field such that nothing other than $0,1,2$ or $2^n$ degree polynomials irreducible over $\mathbb{Q}$ have solutions: the field of constructible numbers. I think that'll be as good as you can get.