IMO Shortlist

We are tasked with determining all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that the following functional equation holds for all integers $x, y$:

$$f(x - f(y)) = f(f(x)) - f(y) - 1.$$

To solve this problem, we will analyze the equation by substituting various values initially to find a pattern or constraints on $f$.

### Step 1: Simplification with Substitutions

1. **Substituting $x = f(y)$:**

$$f(0) = f(f(f(y))) - f(y) - 1.$$

Let $c = f(0)$, then the equation becomes:

$$c = f(f(f(y))) - f(y) - 1. \tag{1}$$

2. **Substituting $y = 0$:**

$$f(x - f(0)) = f(f(x)) - c - 1.$$

Replacing $c$ gives:

$$f(x - c) = f(f(x)) - c - 1. \tag{2}$$

### Step 2: Exploring Constant and Linear Solutions

1. **Case: $f(x) = -1$ for all $x$**

For this to hold, the original condition becomes:

$$-1 = (-1) - (-1) - 1,$$

which simplifies to:

$$-1 = -1,$$

thus, valid for all $x$.

2. **Case: $f(x) = x + 1$ for all $x$**

For the substitution:

$$f(x - (y + 1)) = (x + 1 + 1) - (y + 1) - 1,$$

which simplifies to:

$$f(x - y - 1) = x + 1 - y - 1,$$

$$f(x - y - 1) = x - y = (x-y) + 1 - 1 = f(x-y),$$

This verifies the functional equation is satisfied for all integers $x, y$.

### Conclusion

From both cases, we verify that there are two possible solutions for $f$:

- $f(x) = -1$ for all $x \in \mathbb{Z}$.
- $f(x) = x + 1$ for all $x \in \mathbb{Z}$.

Therefore, the functions that satisfy the given equation are:
$$\boxed{f(x) = -1 \text{ for all } x \in \mathbb{Z} \text{ or } f(x) = x + 1 \text{ for all } x \in \mathbb{Z}}.$$