Equation involving fields Let $ K $ be a field and $f:K-\left \{ 0 \right \}\rightarrow K-\left \{ 0 \right \}$ a function with $f(f(x))=x^{-1},\forall x\in K-\left \{ 0 \right \}$ and $ f(1)\neq 1.$
Knowing that $ f^{2}(x)-f(x)+1=0 $ has an unique solution in $K-\left \{ 0 \right \} $, determine $ f(2)$.
I haven't found something useful yet.

Suppose that $a$ is a root of the polynomial $x^2-x+1$.
Then $(x-a)$ divides the polynomial $x^2-x+1$.
This means that $(x-a)(x-a)=x^2-x+1$ (because the root is unique).
it follows that $x^2-2a+1=x^2-x+1$ and so $2a=1$ and $a^2=1$.
we conclude that $1=1^2=(2a)^2=4a^2=4$.
So the field must have characteristic $3$, and so $2=-1$.
Let $z=f(1)$. Notice $z^{-1}=f(f(z))=f(f(f(1))=f(1)=z$. It follows that $z^2=1$. The only two roots of the polynomial $x^2-1$ are $1$ and $-1$. We conclude $z=-1$.
So we have $f(1)=-1$, and since $1=f(f(1))=f(-1)$ we have $1=f(-1)=f(2)$.
