A field is quadratically closed if each of its elements is a square.
The field $\mathbb{F}_2$ with two elements is obviously quadratically closed.
However, testing some more finite fields with this property, I didn't find any more. Hence my question is:
Which finite fields $\mathbb{F}_{p^n}$ are quadratically closed and why?
Consider the squaring map from the multiplicative group of a finite field $F$ to itself. The kernel is $\{\pm1 \}$, i.e., it is trivial if and only if the characteristic of $F$ is $2$. Since this map is surjective if and only if it is injective, every element of $F$ is a square if and only if the characteristic of $F$ is $2$.
What about $\mathbb{F}_4$?
The multiplicative group of nonzero elements of a finite field is always cyclic. For odd $p$, that group has even order. This means there are always elements that aren't squares in the finite field.
Are you sure about your definition of quadratically closed field? I would have expected it to mean that every quadratic polynomial over the field has a root in the field. If the characteristic is not $2$, this is equivalent to the definition you gave (via the quadratic formula), but they are not equivalent in characteristic $2$.
$W^{\wedge}(F)$ is the Witt-Grothendieck ring of $F$, informally it can be thought of set of all formal differences of regular quadratic forms on $F$(this is not exactly right).
$F$ is quadratically closed $\Leftrightarrow$ $dim \colon W^{\wedge}(F) \to \mathbb{Z} $ is an isomorphism $ \Leftrightarrow \lbrace dim(q_{1})=dim(q_{2}) \Leftrightarrow q_{1}=q_{2} \rbrace$ .
Simply speaking,it just states for a given finite dimensional vector space $V$ over $F$ ,there should be exactly one quadratic form on $V$,upto isometry.
