rijul,i think that for $m=n=0$ we get $f(0)=0$ then $f(n)(f(n)+f(-n))=0$ ,$n\in{Z}$, change $ n\rightarrow{-n}$, gives $(f(n)+f(-n))=0$ for $m=2,n=1$ $(f(3)-1)(f(2)-1)=2$ gives $4$ cases $1.f(2)=2, f(3)=3$, $2.f(2)=3, f(3)=2$, $3.f(2)=0, f(3)=-1$, $4.f(2)=-1, f(3)=0$ for case 1. $f(n)=n$, $n\in{Z}$ for case 2. $f(4)=9$, $8f(5)=20$ rejected for case 3. $f(n)=0$ for $n=2k$ ,$f(n)=1$ for $n=4k+1$, $f(n)=-1$ for $n=4k+3$ for case 4. $f(n)=0$ for $n=3k$ ,$f(n)=1$ for $n=3k+1$, $f(n)=-1$ for $n=3k+2$