TITLE: Is the Pythagorean closure of $\mathbb Q$ equal to the field of constructible numbers?
QUESTION [15 upvotes]: A Pythagorean field is one in which every sum of two squares is again a square. $\mathbb Q$ is not Pythagorean, which is easy to see. I have read a theorem online which says that every field has a unique (up to isomorphism) Pythagorean closure. I haven't found the proof so I thought I should start with the most familiar field, $\mathbb Q$.  
I was thinking if it would be possible to somehow imagine or characterize the Pythagorean closure $\mathbb P$ of $\mathbb Q.$ I know that in the field $\mathbb E$ of all constructible numbers, every positive number is a square because it is possible to construct square roots of already constructed numbers (by drawing a certain right triangle and its altitude). So $\mathbb E$ must be Pythagorean. $(1)$ But is it equal to $\mathbb P?$ If it's not, then what is the $\mathbb P$-dimension of $\mathbb E$?
Surely, for any $q_1,q_2,\ldots,q_n\in \mathbb Q,$ we must have $\sqrt{q_1^2+q_2^2+\cdots + q_n^2}\in \mathbb P$ by a simple induction. $(2)$ Is this all we have to adjoin to $\mathbb Q$ to obtain $\mathbb P?$ It looks like it is but I haven't done anything with infinite extensions and I don't know how to handle this.

REPLY [4 votes]: I treat the concept of Pythagorean closure in these notes.  In fact, I speak briefly about a class of fields being "closable" and give the Pythagorean closure as an example of that.  
Let $F$ be a field.  Here are two key (but easy) observations:
(CC1) Any algebraically closed field extension of $F$ is Pythagorean.
(CC2) If $\{K_i\}_{i \in I}$ is a family of Pythagorean field extensions of $F$ -- all contained in some large field $L$ -- then also $K = \bigcap_{i \in I} K_i$ is a Pythagorean field extension of $F$.  
It follows that inside any algebraic closure $\overline{F}$ of $F$ there is a literally unique Pythagorean closure: the intersection of all Pythagorean algebraic extensions of $F$.  (When $F = \mathbb{Q}$, Cam McLeman has given a spot-on description of it.)