$K \neq \sum K^2 \implies K$ admits an ordering

Question

$K$ is a field. $K^2 = \{a^2 | a \in K \}$, $\sum K^2$ is set of all sum of squares. I have to prove an implication:

$K \neq \sum K^2 \implies K$ admits an ordering

$K \neq \sum K^2$, so $-1 \notin \sum K^2$, so $K$ is formally real. It implies that $\sum K^2$ is pre-positive cone. I am not sure what to do now. I know that relation $a<b \iff b-a\in P$, where $P$ is a positive cone, satisfies conditions for being an order relation. I know also taht every positive cone of field is a pre-positive cone.

Answer 1

There is no sum of all squares.
Let C be the set of all finite sums of squares.
Show C is an order cone.
Let K be a maximal order cone with C subset K.
Show that K is a linear cone.

If you want to work with positive cones,then remove
0 from the numbers that are being squared.