69th Belarusian Mathematical Olympiad

Answer: $f(x)=x+1$. Let $A\subset \mathbb{Z}$ be the set of all integers $x$ such that $f(x)=x+1$. We prove that $A=\mathbb{Z}$.

Claim 1. $A\ne \varnothing$. Indeed, if $f(1)=a$ then $f(a)=f(f(1))=1\cdot f(1)-1^2+2=a+1$, so $a\in A$.

Claim 2. If $x_0\in A$ then $x_0+1\in A$. Indeed, $$f(x_0+1)=f(f(x_0))=x_{0}f(x_0)-x_0^2+2=x_0(x_0+1)-x_0^2+2=(x_0+1)+1,$$ therefore $x_0+1\in A$.

From claim 2 it follows that if $A\ne \mathbb{Z}$ then $A$ has its minimum element $b$ and $A=\{x\in \mathbb{Z}\mid x\ge b\}$. Suppose that $A\ne \mathbb{Z}$ and let $b\in A$ be the minimum element of $A$.

From the problem condition 2 we have $f(s)=b$ for some $s\in \mathbb{Z}$. Then $$b+1=f(b)=f(f(s))=sb-s^2+2\text{or}(s-1)(s+1-b)=0.$$ If $s=b-1$ then we have $f(b-1)=b=(b-1)+1$ which contradicts the minimality of $b$. So, $s=1$ and $f(1)=b$. If $b\le 1$ then $1\in A$ and we have $f(1)=2>b$ - a contradiction. Further, if $b=2$ then $f(1)=2=1+1$, so $1\in A$ - contrary to minimality of $b$. Hence, $b\ge 3$.

Now let $t\in \mathbb{Z}$ be such that $f(t)=1$. Then $$b=f(1)=f(f(t))=tf(t)-t^2+2=t-t^2+2\text{or}{t}^2-t+b-2=0,$$ which is impossible for $b\ge 3$.

The contradiction obtained shows that $A=\mathbb{Z}$, i.e. $f(x)=x+1$. It is easy to see that the function $f(x)=x+1$ is suitable.