$C$ has an extension of degree $p$ if and only if $p\neq2$. Where $C$ is the field of all constructible real numbers 
Let $C$ be the field of all constructible real numbers and let $p$ be a prime. Prove that $C$ has an extension of degree $p$ if and only if $p\neq2$.  

for $\Rightarrow$ direction:
For $C\subset E$ such that $[E:C]=p$, if $p=2$ then any element on $E$ will be constructible. Because constructible numbers are in quadratic extensions.  (Is that correct?)  
And for the $\Leftarrow$ direction,
I'm trying  to find a non constructible number so that it has a minimum polynomial in $C[x]$ with degree $p$
But I really don't see how to complete.
Appreciate your help
 A: It is not true that every constructible number is in a quadratic extension of $\mathbb Q$, what is true is that a number is constructible if and only if the extension $\mathbb Q(\alpha)/\mathbb Q$ can be refined into a sequence of quadratic extensions.
Your first direction: suppose $\alpha$ has degree $2$ over $C$. This means it is the root of a quadratic equation with coefficients in $C$, so it can be constructed from two construcible numbers, and hence is constructible itself.
Conversely, we want to produce an extension of odd prime degree over $C$. Let's just take $\sqrt[p]3$ which is a root of $x^p - 3$. That polynomial is irreducible by Eisenstein's criterion. If this polynomial remains irreducible in $C[x]$ we are done. Otherwise, it factors, and therefore factors in $L[x]$ for some $L\subseteq C$ which is finite over $\mathbb Q$. Since $L\subseteq C$ it has degree $2^t$ for some $t$. But then it is obvious that this cannot happen; look at the diamond consisting of $\mathbb Q,L, \mathbb Q(\sqrt[p]3),L(\sqrt[3]p)$. Since the two bottom degrees are coprime, they do not drop when they lift, and so $\alpha$ has degree $p$ over $L$ contradicting the assumption that its degree dropped.
