I got that no function existed. $ m = n = 0$ gives $ f(f(0)) = f(0)$, then $ m = 0, n = f(0)$ gives $ f(0) = 0$. $ m = 0$ gives $ f(f(n)) =-n$. Suppose that $ f(1) = a$. Then $ f(a) =-1$, $ f(-1) =-a$, and $ f(-a) = 1$. We can see here that $ a$ can't be $ 0$, $ 1$, or $ -1$. But now let $ m = 1, n =-a$. We get $ f(2) = 2a$. We can induct to show that $ f(n) = na$ for all $ n\in\mathbb{Z}$. But now consider $ n = a$. We get $ f(a) = a^{2}\neq-1$, contradiction.

MellowMelon wrote: I got that no function existed. That is indeed correct.

Let $P(m,n)$ be $f(m+f(n))=f(m)-n$.It is easy to show that $f$ is bijective. $P(0,0)\Rightarrow f(f(0))=f(0)$.From injectivity,$f(0)=0$. $P(0,n)\Rightarrow f(f(n))=-n$.Then $f(m+f(n))=f(m)+f(f(n))$.From surjectivity,$f(m+n)=f(m)+f(n)(\forall m,n\in \mathbb{Z})$.This is Cauchy equation.Hence ∃$c\in \mathbb{Z}$ s.t. $f(m)=cm$.Substituting this into $P(m,n)$,$c^2n=-n$ which implies $c^2=-1$.This is absurd.Therefore there is no such $f$.$\blacksquare$

Let $P(m,n)$ denote the given assertion. $P(0,0)\implies f(f(0))=f(0) \text{ and } P(0,f(0))\implies f(0)=0.$ $P(0,n)\implies f(f(n))=-n.$ So $f$ is surjective, let $f(z)=y.$ $P(x,z)\implies f(x+y)=f(x)+f(y).$ So the well known solution is $f(x)=kx,$ checking yields a contradiction. So no function exists.