Find $f(x)$ from $f(f(x))=x^2 - x + 1$

Question

I have this: $$ f\colon \mathbb{R} \to \mathbb{R}, $$ $$ f(f(x)) = x^2 - x + 1 $$ I need to show that $f(1) = 1$ and I need to show that $g(x) = x^2 - xf(x) + 1$ is not an one-to-one fuction.

I know how to solve the second problem but I have no idea how to find $f(x)$. Any help is appreciated

Answer 1

For the first part, take:

$\forall x \in \mathbb{R}: f(x)^2-f(x)+1 = (f\circ f)(f(x)) = f \circ (f\circ f)(x) = f(x^2-x+1)$. Take $x=1$ and we get : $f(1)^2-f(1)+1 = f(1)$. Now factoring gives us $(f(1)-1)^2 = 0$, so $f(1) = 1$.

Answer 2

Hint for the first part, use:

$$f(x^2-x+1) = f(f(f(x))) = f(x)^2 - f(x) + 1$$