Prove that in a finite field of characteristic $3$ $\exists x,y \in K^{*}$ such that $x^2+y^2\neq a^2, \forall a\in K$

Question

Prove that in a finite field of characteristic $3$ $\exists x,y \in K^{*}$ such that $x^2+y^2\neq a^2, \forall a\in K$.
I thought about using the fact that any element in a finite field is the sum of two squares, but I didn't get really far. I also considered the function $f: K \to K, f(x)=x^2$ and I studied its injectivity, but this also didn't help.

Answer 1

Assume the opposite. Then, an easy induction shows that every finite sum of squares in $K$ is a square. As $x \in K \longmapsto x^2$ is not injective, there is an element $x \in K$ that cannot be written as a sum of squares.

The set $F$ of the elements of $K$ that are sums of squares is clearly a subring (as $-1=2$ and $0$ is the empty sum) so is a proper subfield. So $k=|K|$ is a power of $f=|F|$, hence $f^2 \leq k$

But $F$ contains all the squares in $K$, so $\sqrt{k} \geq f \geq \frac{k}{2}$ so $k \leq 4$, so $K=\mathbb{F}_3$. But it is impossible, hence the conclusion.