SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $f:\mathbb{Z}\to \mathbb{Z}$ satisfy the given functional equation. Putting $n=0$ in this equation, we get $$f(2m+f(m)+f(m)f(0))=m\forall m\in \mathbb{Z};$$ therefore, $f$ is surjective, so there exists $u\in \mathbb{Z}$ such that $f(u)=-1$. With $m=u$, the given equation gives us $$f(2u-1-f(n))=-n+u.$$ Now, if $a,b$ are some integers such that $f(a)=f(b)$, then $$u-a=f(2u-1-f(a))=f(2u-1-f(b))=u-b,$$ which implies that $a=b$. Hence, $f$ is also injective. Next, putting $n=u$, we have $$f(2m+f(m)-f(m))=uf(m)+m\iff f(2m)=uf(m)+m(\ast )$$ for all $m\in \mathbb{Z}$. In ( $\ast$ ), letting $m=0$, we see that $f(0)=uf(0)$, so $u=1$ or $f(0)=0$. If $f(0)=0$, then with $m=u$, (*) would imply $f(2u)=-u+u=0=f(0)$, i.e. $2u=0\iff u=0$ (since $f$ is injective), and thus $f(0)=-1$, a contradiction! Hence, $u=1\iff f(1)=-1$ and ( $\ast$ ) becomes $$f(2m)=f(m)+m$$ for all $m\in \mathbb{Z}$. Here, letting $m=1$, we obtain $f(2)=0$. In the given functional equation, putting $m=n=0$, we have $$f\left(f(0)+f(0{)}^2\right)=0\iff f(0)+f(0{)}^2=2\iff f(0)=1\text{ or }f(0)=-2.$$ If $f(0)=1$, then it would follow from the given equation with $m=0,n=2$ that $f(1)=2$, a contradiction (because $f(1)=-1$ )! Hence, $f(0)=-2$. In the functional equation, putting $n=0$, we get $$f(2m-f(m))=m\forall m\in \mathbb{Z}$$ Using this, we see that if $f(m)=m-2$ then $f(m+2)=m$; but $f(0)=-2,f(1)=-1$, we can easily prove by induction that $f(n)=n-2$ for all $n\ge 0$. In the given equation, letting $m=1$, we obtain $$f(2-1-f(n))=-n+1\iff f(1-f(n))=1-n\forall n\in \mathbb{Z}.$$ Now replacing $n$ by $n+3$ and letting $n\ge -3$, we have $f(1-f(n+3))=-n-2\iff f(1-(n+1))=-n-2\iff f(-n)=-n-2$. So $f(n)=n-2$ for all $n\in \mathbb{Z}$. It is easy to check that this function satisfies the given condition.