$ m = n = 0$ gives $ f(f(0)) = f(0)$. $ n = 0$ gives $ f(m+f(0)) = f(m)$, but $ n = f(0)$ also gives $ f(m+f(0)) = f(m)+f(0)$. From the previous two statements we conclude that $ f(0) = 0$. Then $ m = 0$ gives $ f(f(n)) = n$. Suppose $ f(k) = 0$ for some $ k$. Then $ n = k$ gives $ f(m) = f(m)+k$, so $ k = 0$. Now put $ m = f(1), n =-1$ in to get $ f(f(1)+f(-1)) = 0$, so $ f(1) =-f(-1)$. We now prove by induction that $ f(k) = kf(1)$ for all positive integers $ k$. Base case $ k = 1$ is trivial. Suppose it is true for $ k$. Then $ m = k, n = f(1)$ gives $ f(k+1) = (k+1)f(1)$, completing the induction. We also prove by induction that $ f(-k) =-kf(1)$ for all positive integers $ k$. Base case $ k = 1$ has already been established. Suppose it is true for $ k$. Then $ m =-k, n = f(-1)$ gives $ f(-k-1) =-kf(1)+f(-1) = (-k-1)f(1)$, completing the induction. So we have $ f(k) = kf(1)$ for all integers $ k$. Let $ k = f(1)$, and we get $ f(1)^{2}= f(f(1)) = 1$. So $ f(1) = 1$ or $ f(1) =-1$. We see that $ f(k) = k$ and $ f(k) =-k$ both work, and they are the only solutions.

$n=-f(m)$ gives $f(m+f(-f(m)))=0$. Substituting $n$ by $m+f(-f(m))$ gives $f(m)=f(m)+m+f(-f(m))$, so $f(-f(m))=-m$. Substitute $n$ by $-f(n)$, we get $f(m-n)=f(m)-f(n)$. Letting $m-n=x,n=y$ gives Cauchy equation.

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 the Cauchy equation.So ∃$c\in \mathbb{Z}$ s.t. $f(m)=cm$.Substituting this into $P(m,n)$, we get $c^2n=n$.Then $c=\pm 1$, so $\boxed{f(m)=m, f(m)=-m}$.It is easy to check that both solutions satisfy the condition.$\blacksquare$

First by fixing m and moving n we get that f is a covering function . If f(a)=0 then n=a , we get that f(0)=0 . Now we get f(f(n))=n (1). If f(b)=1 then n=b , we get that f(m+1)=f(m)+b , so f(m)=mb (2). By using (1),(2) we understand b=1 or -1 , From this we get f(m)=m , f(m)=-m

Let $P(m,n)$ denote the given assertion. $P(m,0)$ forces $f(0)=0.$ Then $P(0,n)$ gives $f$ is an involution. Now $P(m,f(n))$ gives $f$ is additive. Hence $f(x)=kx.$ And we get $f(x)=x$ and $f(x)=-x$ only, which work.