66th Belarusian Mathematical Olympiad

a) $f(x)=-1$ or $f(x)=x+1$.

(Alternative solution by I. Voronovich) Setting $y=f(x)$ in the given equation $$f(x-f(y))=f(f(x))-f(y)-1,(1)$$ we obtain $f(x-f(f(x)))=-1$, i.e., there exists an integer $\lambda$ such that $f(\lambda )=-1$. Set $y=\lambda$ in (1), then $$f(f(x))=f(x+1).(2)$$ Substituting $f(x)+f(y)$ for $x$ in (1), we obtain $$f(f(x))=f(f(x)+f(y))-f(y)-1.$$ Due to symmetry $$f(f(y))=f(f(y)+f(x))-f(x)-1.$$ Therefore $f(f(x))-f(x)=f(f(y))-f(y)$, so $f(f(x))-f(x)=c$ or, in view of (2), $$f(x+1)-f(x)=c,c\in \mathbb{Z}(3)$$

If $c=0$, then $f(x+1)=f(x)$ for all integer $x$, so $f(x)$ is a constant function, thus from (1) it follows that $c=-1$, i.e., $f(x)=-1$ for all integer $x$.

If $c>0$ or $c<0$, then we conclude from (3) that $f(x)$ is increasing or decreasing, respectively. Anyway, $f$ is injective, so (2) gives $f(x)=x+1$.

Both the functions obviously satisfy the initial equation (1).

b) $f(x)=-2$ or $f(x)=x+2$.

(Solution by A. Asanau.) First, in the same way as in a) we can easily establish for any $f:\mathbb{Z}\to \mathbb{Z}$ satisfying the given equation $$f(x-f(y))=f(f(x))-f(y)-2(1)$$ the following equalities: $$f(x-f(f(x)))=-2,(2)$$ $$f(f(x))=f(x+2),(3)$$ $$f(x+2)-f(x)=c,(4)$$ where $c$ is a constant. In particular, (2) shows that $-2\in E(f)$, the range of $f$. Now we have two cases.

I. $c=0$. Let $f(0)=a$, $f(1)=b$. Then from (4) we have $f(x+2)=f(x)$, so $f(x)=a$ for all even $x$ and $f(x)=b$ for all odd $x$. Note that one of $a$, $b$ equals $-2$ in view of (2). We claim that $a=b=-2$. Suppose that $a\ne b$. Then $$f(x)=f(y)⟺x\equiv y(\mathrm{mod}2).$$ Therefore (3) implies $f(x)\equiv x+2\equiv x(\mathrm{mod}2)$. In particular, $f(0)\equiv 0(\mathrm{mod}2)$, whence $f(0)=a=-2$. Now we set $x=0$, $y=1$ in (1); then we get $f(-b)=f(-2)-b-2=-b-4$. But $f(-b)$ equals either $b$ or $-2$. In both cases we have $b=-2$, contrary to $a\ne b$. So $a=b=-2$, and $f(x)=-2$ for all $x$. It is easy to see that this function satisfies the initial equation.

II. $c\ne 0$. Let again $f(0)=a$, $f(1)=b$. Then from (4) we have $$f(2n)=cn+a,f(2n+1)=cn+b.(5)$$ In particular, (5) implies that, on the one hand, $f(x)\to \infty$ as $x\to \infty$, and, on the other hand, ${\lim }_{x\to \infty }\frac{f(x)}{x}=\frac{c}{2}$. Then from (3) we have $$\frac{f(f(x))}{f(x)}=\frac{f(x+2)}{f(x)}\to 1\text{ as }x\to \infty ,$$ but $\frac{f(f(x))}{f(x)}\to \frac{c}{2}$. Hence, $c/2=1$, i.e. $c=2$. Then (5) becomes $$\begin{array}{rlr}f(x)& =x+a& \text{for even }x,\\ f(x)& =x+d& \text{for odd }x(\text{where }d=b-1).\end{array}(6)$$ 1) Now, if $f$ is an injective function, then (3) immediately implies $f(x)=x+2$, and this function satisfies (1).

2) Suppose that $f$ is not injective. But from (6) it easily follows that $f$ is injective on the set of even numbers and also injective on the set of odd numbers. Thus we see that some $f(2n)$ and $f(2m+1)$ are equal. From (6) it follows that $2n-2m-1=d-a$, i.e., $a$ and $d$ are of different parity. Then (6) implies that all values of $f$ are of the same parity. Since $-2\in E(f)$, all the values of $f(x)$ are even. Then from (3) we have $f(x+2)=f(f(x))=f(x)+a$. Thus (4) gives $a=c=2$. In particular, $d$ is odd. So, (6) becomes $$\begin{array}{rlr}f(x)& =x+2& \text{for even }x,\\ f(x)& =x+d& \text{for odd }x.\end{array}$$ One can verify that for any odd $d$ this function is a solution of (1).