We have $f(f(0)) = f(f(1)) = 1$. Now, $f(x^2-x+1) = f(f(f(x))) = f(x)^2 - f(x) + 1$, so $f(0)^2 - f(0) + 1 = f(1)^2 - f(1) + 1 = f(1)$. So $f(1)^2 - 2f(1) + 1 = 0$, hence $\boxed{f(1)=1}$. It follows $f(0)(f(0)-1) = 0$. Assume $f(0) = 0$; then $1=f(f(0)) = f(0) = 0$, absurd. So $\boxed{f(0)=1}$.

It's not end of this problem. It's necessary to show that such function exists. But I heard that none of teams on Baltic Way showed this function and 10 out of 11 teams got 5/5 pts.

Swistak wrote: It's not end of this problem. It's necessary to show that such function exists. But I heard that none of teams on Baltic Way showed this function and 10 out of 11 teams got 5/5 pts. I dont think showing the existence is demanded (the problem statement supposed existence) But, if really required, it's not very difficult to build infinitely many solutions.

it suffices to notice the following conclusion: if $f(x_1)=f(x_2)$,then $x_1=x_2$ or $x_1+x_2=1$.

pco wrote: But, if really required, it's not very difficult to build infinitely many solutions. I was required thru pm to give some examples. Here is one construction : Let $g(x)=x^2-x+1$. Notice that $g(1-x)=g(x)$, $f(x)\ge x$ $\forall x$ and $g(x)$ is increasing over $[\frac 12, +\infty)$ 1) Defining $f(x)$ over $[\frac 12, 1)$ =================== Let $a\in(\frac 12,\frac 34)$ Let the sequence $a_n$ defined as $a_0=\frac 12$, $a_1=a$ and $a_{n+2}=g(a_n)$ $\forall n\ge 0$ $a_n$ is an increasing sequence whose limit is $1$ Let $h_0(x)$ be any bijection from $[a_0,a_1)\to [a_1,a_2)$ It's easy to build a sequence of bijections $h_n(x)$ from $[a_n,a_{n+1})\to[a_{n+1},a_{n+2})$ as $h_{n+1}(x)=g(h_n^{-1}(x))$ $\forall n\ge 0$ We can define $f(x)$ over $[\frac 12,1)$ as : $\forall x\in [a_n,a_{n+1})$ : $f(x)=h_n(x)$ Obviously $f(f(x))=x^2-x+1$ $\forall x\in [\frac 12, 1)$ 2) defining $f(x)$ for $x=1$ ================= Choose $f(1)=1$ Obviously $f(f(x))=x^2-x+1$ $\forall x\in [1,1])$ 3) defining $f(x)$ over $(1,+\infty)$ ==================== Let $b>1$ and $c\in(b,g(b))$ 3.1) Defining $f(x)$ over $[b,+\infty)$ Let the sequence $b_n$ defined as $b_0=b$, $b_1=c$ and $b_{n+2}=g(b_n)$ $\forall n\ge 0$ $b_n$ is an increasing sequence whose limit is $+\infty$ Let $k_0(x)$ be any bijection from $[b_0,b_1)\to [b_1,b_2)$ It's easy to build a sequence of bijections $k_n(x)$ from $[b_n,b_{n+1})\to[b_{n+1},a_{b+2})$ as $k_{n+1}(x)=g(k_n^{-1}(x))$ $\forall n\ge 0$ We can define $f(x)$ over $[b,+\infty)$ as : $\forall x\in [b_n,b_{n+1})$ : $f(x)=k_n(x)$ Obviously $f(f(x))=x^2-x+1$ $\forall x\in [b,+\infty)$ 3.2) Defining $f(x)$ over $(1,b)$ Let $g_{1+}$ be the restriction of $g(x)$ over $(1,+\infty)$. $g_{1+}$ is a bijection from $(1,+\infty)\to(1,+\infty)$ Let the sequence $c_n$ defined as $c_0=c$, $c_1=b$ and $c_{n+2}=g_{1+}^{-1}(c_n)$ $\forall n\ge 0$ $c_n$ is a decreasing sequence whose limit is $1$ It's easy to build a sequence of bijections $l_n(x)$ from $[c_{n+2},c_{n+1})\to[c_{n+1},c_n)$ as : $l_0(x)=k_0^{-1}(g(x))$ $l_{n+1}(x)=l_n^{-1}(g(x))$ $\forall n\ge 0$ We can define $f(x)$ over $(1,b)$ as : $\forall x\in [c_{n+2},c_{n+1})$ : $f(x)=l_n(x)$ Obviously $f(f(x))=x^2-x+1$ $\forall x\in (1,b)$ 4) Defining $f(x)$ over $(-\infty,\frac 12)$ ========================= Just choose $f(x)=f(1-x)$

PCO how do come with such solutions that seems nearly impossible? what is the motivation?

I have uploaded a video explaining this problem: Link: Video Solution

$f(f(f(x))=f(x)^2-f(x)+1=f(x^2-x+1)$ Setting $x=1$, we obtain $f(1)^2-2f(1)+1=0$, so $f(1)=1$. Setting $x=0$, we get $f(0)^2-f(0)+1=1$, so $f(0)\in\{0,1\}$ Set $x=0$ in the original equation; we get $f(f(0))=1$, so $f(0)$ cannot be zero. Thus $\boxed{f(0)=1}$.