If $f(f(x))=x^2-x+1$, what is $f(0)$? 
Suppose that $f\colon\mathbb{R}\to\mathbb{R}$ without any further restriction. If $f(f(x))=x^2-x+1$, how can one find $f(0)$?

Thanks in advance.
 A: $f(f(0))=f(f(1))=1$. Apply $f$ once again: $f(f(f(0)))=f(f(f(1)))=f(1)=f(0)^2-f(0)+1=f(1)^2-f(1)+1$.
That leads to $f(1)=1$, hence $f(0)^2-f(0)=0$ and $f(0)$ can only be $0$ or $1$.
But $f(0)=0$ leads to $f(f(0))=0$, contra $f(f(0))=1$, so $\color{red}{f(0)=1}$.
A: If such a function $f$ exists, then $f(0) = 1$, but such a function $f$ does not exist. See the paper:


*

*When is $f(f(x)) = az^2 +bz+c$?
R. E. Rice, B. Schweizer and A. Sklar
The American Mathematical Monthly
Vol. 87, No. 4 (Apr., 1980), pp. 252-263
(Link to PDF not behind the JSTOR paywall)


Edit: Such a function does not exist in $\mathbb{C}$! Or in any algebraically closed field of characteristic zero. But you can have such a function in the reals. See the epilogue of the paper (page 262).
A: I think it is more easy from what it seems. You know that 
$$f(f(0))=1,$$
then
$$f(1)=f(f(f(0)))=f(0)^2-f(0)+1,\;(*)$$
but also
$$f(f(1))=1^2-1+1=1,$$
so
$$f(1)=f(f(f(1)))=f(1)^2-f(1)-1$$
and you can calculate $f(1)$ from here, and then substitute its value in $(*)$
A: We have that 
$$f(f(f(x)))=f(x)^2-f(x)+1.$$ Since $f(f(0))=1$ we get that $$f(1)=f(0)^2-f(0)+1.$$ That is
$$f(0)^2-f(0)+1-f(1)=0.$$ Repeating the process, since $f(f(1))=1$ we get that $$f(1)=f(1)^2-f(1)+1.$$ That is
$$(f(1)-1)^2=f(1)^2-2f(1)+1=0.$$ Thus, $f(1)=1.$ So it is $$f(0)^2=f(0),$$ from where $f(0)=0$ or $f(0)=1.$
If $f(0)=0$ then $f(f(0))=f(0)=0$ which contradicts $f(f(0))=1.$ So, it must be $f(0)=1.$
