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
$$f(x - f(y)) = f(f(x)) - f(y) - 1, \qquad (1)$$
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
$$f(f(x)) = f(x + 1). \qquad (2)$$
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),
$$f(x+1) - f(x) = c, \quad c \in \mathbb{Z} \qquad (3)$$

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
$$f(x - f(y)) = f(f(x)) - f(y) - 2 \quad (1)$$
the following equalities:
$$f(x - f(f(x))) = -2, \quad (2)$$
$$f(f(x)) = f(x + 2), \quad (3)$$
$$f(x + 2) - f(x) = c, \quad (4)$$
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 $a \neq b$. Then
$$f(x) = f(y) \iff x \equiv y \pmod{2}.$$
Therefore (3) implies $f(x) \equiv x + 2 \equiv x \pmod{2}$. In particular, $f(0) \equiv 0 \pmod{2}$, 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 $a \neq b$.
So $a = b = -2$, and $f(x) = -2$ for all $x$. It is easy to see that this function satisfies the initial equation.

II. $c \neq 0$. Let again $f(0) = a$, $f(1) = b$. Then from (4) we have
$$f(2n) = cn + a, \quad f(2n + 1) = cn + b. \quad (5)$$
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
$$\begin{aligned} f(x) &= x + a & \text{for even } x, \\ f(x) &= x + d & \text{for odd } x \quad (\text{where } d = b - 1). \end{aligned} \quad (6)$$
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
$$\begin{aligned} f(x) &= x + 2 & \text{for even } x, \\ f(x) &= x + d & \text{for odd } x. \end{aligned}$$
One can verify that for any odd $d$ this function is a solution of (1).