IMO Team Selection Test 3, June 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 $$f(-f(x)-f(y))=a(-ax-b-ay-b)+b=-a^{2}x-a^{2}y-2ab+b.$$ 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 $f(x)=x-1$.

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