Let $P(m,n)$ the assertion of the given F.E. $P(0,0)$ $$f(0)=0$$$P(m,0)$ $$f(m)=m^2$$And we can verify that this works, hence $\boxed{f(n)=n^2}$ is the only function that works, thus we are done

MathLuis wrote: Let $P(m,n)$ the assertion of the given F.E. $P(0,0)$ $$f(0)=0$$$P(m,0)$ $$f(m)=m^2$$And we can verify that this works, hence $\boxed{f(n)=n^2}$ is the only function that works, thus we are done sorry the problem now is fixed .

The equation rewrites as: $$f(m+n) + f(m)f(n) = n^2 f(m) + m^2 f(n) + (m+n)^2 -m^2 n^2, \quad \forall m,n.$$Define $g:\mathbb{Z}\to\mathbb{Z}$ as $g(n) = f(n)-n^2$. Rewriting the equation above in terms of $g$ implies that solving it for $f$ is equivalent to solving the functional equation below for $g$: $$g(m+n) + g(m) g(n) = 0, \quad \forall m,n.$$Set $n=0$. We get $$g(m)\left(g(0)+1\right)=0, \quad \forall m.$$ Case I: $g(0) \neq -1$ Then $g \equiv 0$ hence $f(n) = n^2, \forall n$. Case II: $g(0) = -1$ - Setting $n = -m$ gives: $g(m)g(-m)=1$. Hence for any $m$, $g(m)=1$ or $-1$ i.e. $g(m)^2=1$. - Setting $m=n$ therefore implies: $g(2m) = -g(m)^2 = -1$. - Setting $m=2t, n=1$ gives $g(2t+1)=g(1)$. Thus $g(m)=-1$ for $m$ even and $g(m)=g(1)$ for $m$ odd where $g(1) = 1$ or $-1$. Hence $f(m)=m^2-1$ for $m$ even and $f(m)=m^2+g(1)$ for $m$ odd where $g(1) = 1$ or $-1$.

MathLuis wrote: Let $P(m,n)$ the assertion of the given F.E. $P(0,0)$ $$f(0)=0$$$P(m,0)$ $$f(m)=m^2$$And we can verify that this works, hence $\boxed{f(n)=n^2}$ is the only function that works, thus we are done what about case when f(0)=-1, because when we use P(0;0) we have f(0)(f(0)+1)=0

aidar_nasirov wrote: MathLuis wrote: Let $P(m,n)$ the assertion of the given F.E. $P(0,0)$ $$f(0)=0$$$P(m,0)$ $$f(m)=m^2$$And we can verify that this works, hence $\boxed{f(n)=n^2}$ is the only function that works, thus we are done what about case when f(0)=-1, because when we use P(0;0) we have f(0)(f(0)+1)=0 That was my sol to un-edited post so it doesnt work now as the AoPSer who posted this gave a different version at the beggining