Let $P(x,y): f (2m + f (m) + f (m)f (n)) = nf (m) + m$ Clearly $f(m)=m-2,m\in Z$ is a solution, i will prove that is the only one. $P(m,0): f(2m+f(m)+f(m)f(0))=m$ so $f$ is surjective. Let $c\in Z,f(c)=1$ $P(c,n): f(2c+1+f(n))=n+c$ so $f$ is $1-1$ $P(0,1):f(f(0)+f(1)f(0))=f(0)\Leftrightarrow f(0)(1+f(1))=0$ Case 1:$f(0)=0$ Let $k\in Z,f(k)=-1$ $P(k,k): f(2k)=0=f(0)\Leftrightarrow k=0\Rightarrow f(0)=-1$, contradiction. Case 2:$f(1)=-1$ $P(m,1): f(2m)=f(m)+m$ $P(1,1): f(2)=0$ $P(2,1):f(4)=2$ $P(1,-1):f(2-1-f(-1))=1+1=2=f(4)$ so $f(-1)=-3$ $P(m,-1): m-f(m)=f(2m+f(m)-3f(m))=f(2(m-f(m))=f(m-f(m))+m-f(m)$ $\Leftrightarrow f(m-f(m))=0=f(2)\Leftrightarrow m-f(m)=2\Leftrightarrow \boxed{f(m)=m-2}$