plagueis wrote: Let $\mathbb{Z}$ be the set of integers. Find all functions $f: \mathbb{Z}\rightarrow \mathbb{Z}$ such that, for any two integers $m, n$, $$f(m^2)+f(mf(n))=f(m+n)f(m).$$ The only solutions that work are $\boxed{f\equiv 0},\boxed{f\equiv 2},\boxed{f\equiv \text{id}}$.These clearly work so now we prove these are the only. Plugging $(0,n)\implies f(0)(f(n)-2)=0\implies f(0)=0$ or $f(n)=2$.

PS-I think I have flawed in the case $f(1)=0$

That's correct, the induction can only be done backwards. Anyway, you can cover all the integers by choosing some sufficiently small $n$ depending on $m$, and plug it in the original equation to get $f(m)^2=f(m^2)=0$ for all integers $m$.

Claim : $ f(m) = 0 , f(m) = 2 \ and f(m) = m \forall m \in \mathbb{Z} $ are the only solutions. Let $ P(m,n) : f(m^2) + f(mf(n)) = f(m+n)f(m) \forall m,n \in \mathbb{Z} \dots (1) $ $$ P(0, m ) : 2f(0) = f(m)f(0) \implies f(m) = 2 \ or \ f(0) = 0 $$Now observe $f(m) = 2$ is indeed a solution so now we go to the next case. Case 1 : $ f(0)=0$ $$ P(m,0) : f(m^2) = f(m)^2 \implies f((-m)^2) = f(m^2) = f(-m)^2$$$$\implies f(m)^2 = f(-m)^2 \implies f(m) = \pm f(-m) \implies f \ is \ either \ odd \ or \ even $$$$P(1,0) : f(1) = f(1)^2 \implies f(1) = 0 \ or \ f(1)= 1 $$ Case 1.1 : $ f(1) = 0 $ $$ P(m,1) : f(m)^2 = f(m+1)f(m) \implies f(m) = 0 \ or \ f(m) = f(m+1) $$Indeed $f(m) = 0 \forall m \in \mathbb{Z} $ is a solution. Now when $ f(m+1) = f(m) \implies f(m) = 0 \forall m \in \mathbb{Z}^+ $( by simple induction) And as $f$ is either odd or even we observe that $f(m) = 0 \forall m \in \mathbb{Z}^+ \cup \mathbb{Z}^- $ and as $f(0)=0$ we get - $$f(m) = 0 \forall m \in \mathbb{Z} $$ Case 1.2 : $ f(1) = 1 $ $$ P(m,1) : f(m)^2 +f(m) = f(m+1)f(m) \implies f(m)= 0 \ or \ f(m+1) = f(m) + 1 $$$$ f(m+1) = f(m) + 1 \implies f(m) = m \forall m \in \mathbb{Z}^+ ( \ by \ simple \ induction ) $$Now when $f$ is even $ \implies f(m) = | m | \forall m \in \mathbb{Z} $ plugging it back in $eqn(1)$ we observe it isn't a solution. When $f$ is odd $ \implies f(m) = m \forall m \in \mathbb{Z} $ which indeed is a solution. All solutions of $eqn(1)$ are $$ \boxed{f(m) = 2 \forall m \in \mathbb{Z} }$$$$ \boxed{ f(m) = 0 \forall m \in \mathbb{Z} } $$$$ \boxed{ f(m) = m \forall m \in \mathbb{Z} }$$

The only solutions are $f(x)=0$, $f(x)=2$ and $f(x)=x$. Let $P(m,n)$ be the given assertion. From $P(0,n)$ we have that $2f(0)=f(0)f(n)$. Thus we have two cases. The first case is when $f(0) \neq 0$, and we get that $f(n)=2$., for any $n$. The second case is when $f(0)=0$. From $P(m,-m)$ we get that $f(m^2)+f(mf(-m))=0$. From $P(m,0)$ we get that $f(m)^2=f(m^2)=f(-m)^2$. Thus we have that $f(1)^2=f(1)$ Thus we have two cases. The first case is when $f(1)=1$. From $P(m,1)$, we have that $f(m)\left( f(m)+1 \right)=f(m)f(m+1)$, which by induction implies that $f(m)=m$, for every positive integer $n$. Now choose $n \geq 0$ and $m \leq 0$, but choose in such a fashion that $m+n \geq 0$. Thus we have that $m^2+f(mn)=(m+n)f(m)$ But now set $m = -1$, to get that $1+f(-n)=(n-1)f(-1)$. From $P(1,-1)$ we get that $f(1)+f(f(-1))=0 \implies f(f(-1))=-f(1)=-1$. But notice that $f(-1)^2=1$: Now if $f(-1)=1$, we would get that $f(1)=-1$, which is impossible, thus $f(-1)=-1$. By induction on the upper relation we get that $f(-n)=-n$, for any positive $n$. Thus $f(x)=x$, for all integers $x$. The second case is when $f(1)=0$, this easily implies that $f(f(n))=0$ and $f(m)^2=f(m+1)f(m)$, which easily implies, by induction, that $f(x)=0$.