Is every field of characteristic zero whose nonzero elements are squares or the additive inverse of squares but not both a Euclidean ordered field? A Euclidean ordered field is an ordered field $F$ for which every positive element of $F$ is a square.
Now, call a field $F$ "quasi-Euclidean" if for any nonzero $x \in F$, exactly one of $x$ and $-x$ is a square. Then, this definition leads to the following question:

Is every "quasi-Euclidean" field of characteristic zero in fact a Euclidean ordered field?

If a "quasi-Euclidean" field $F$ is not a Euclidean ordered field, then there must be two nonzero elements $x, y \in F$ for which $x^2+y^2$ is not a square, and hence there must be a third nonzero element $z \in F$ for which $x^2+y^2+z^2=0$.
Note that there do exist "quasi-Euclidean" fields of nonzero characteristic, and of course, they cannot be ordered fields. For example, $\mathbb{F}_p$ (the Galois field of order $p$) is "quasi-Euclidean" for any prime $p \equiv 3 \pmod 4$.
 A: The existence of lots of counterexamples follows from the following lemma.
Lemma: Let $k$ be a field in which $-1$ is not a square and let $x\in k$ be such $-x$ is not a square.  Then $-1$ is not a square in $k(\sqrt{x})$.
Proof: If $x$ is a square then $k(\sqrt{x})=k$ so we are done, so let us assume $x$ is not a square.  If $(a+b\sqrt{x})^2=-1$ for $a,b\in k$ then we must have $2ab=0$ and $a^2+xb^2=-1$.  We cannot have $b=0$ since $-1$ is not a square in $k$, so we must have $a=0$ so $xb^2=-1$.  But then $-x=1/b^2$ is a square in $k$, a contradiction.
Using this lemma, you can start with any field in which $-1$ is not a square and repeatedly adjoin square roots to preserve this property.  In particular, you can iterate transfinitely to eventually get a field in which for all $x$, either $x$ or $-x$ is a square.  Since $-1$ is not a square, both options cannot be true at once if $x$ is nonzero.
For instance, you could start with $k=\mathbb{Q}$ and then adjoin a square root of $-2$ as the first step (or a square root of $-n$ for any $n>0$ that is not a square in $\mathbb{Q}$).  In the end you will get a quasi-Euclidean field in which $-2$ is a square.
