, 2020

Substituting $x = y = 1$ yields $f(-2f(1)) = -1$.

Substituting $x = n$ and $y = 1$ yields $f(-f(n) - f(1)) = -n$.

Substituting $x = -f(n) - f(1)$ and $y = -2f(1)$ then yields
$$f(-f(-f(n) - f(1)) - f(-2f(1))) = 1 - (-f(n) - f(1)) - (-2f(1))$$
in which the left hand side expands as $f(-(-n) - (-1)) = f(n + 1)$ and the right hand as $1 + f(n) + f(1) + 2f(1) = f(n) + 3f(1) + 1$.

Writing $c = 3f(1) + 1$, we then get $f(n + 1) = f(n) + c$.

Applying induction in both directions, we see that $f(n + k) = f(n) + ck$ for all $k \in \mathbb{Z}$.

Substituting $n = 0$ then yields $f(k) = f(0) + ck$ for all $k \in \mathbb{Z}$, so $f$ is a linear function.

Now let $a, b \in \mathbb{Z}$ be such that $f(x) = ax + b$ for all $x \in \mathbb{Z}$. Then the left hand side of the functional equation evaluates as
$$f(-f(x) - f(y)) = a(-ax - b - ay - b) + b = -a^2x - a^2y - 2ab + b.$$
For all $x$ and $y$ this must be equal to $1 - x - y$. Therefore the coefficient for $x$ on both sides must be equal (for fixed $y$ both sides must give the same function in $x$), so $-a^2 = -1$, and therefore $a = 1$ or $a = -1$.

If $a = -1$, substituting $x = y = 0$ yields $2b + b = 1$, which contradicts $b$ being an integer.

If $a = 1$, substituting $x = y = 0$ yields $-2b + b = 1$, so $b = -1$.

Indeed, if $a = 1$ and $b = -1$, then the left hand side also expands to $1 - x - y$.

Therefore the only solution to the functional equation is the function $f(x) = x - 1$.

$\boxed{f(x) = x - 1}$