Functional Equation / Vietnam TST 2014

Problem (Vietnam Team Selection Test 2014)

Find all function $f:\mathbb{Z}\to \mathbb{Z}$ and such that :

$$f(2m+f(m)+f(m)f(n))=nf(m)+m,\forall m,n\in \mathbb{Z}(1)$$

Solution :

It is easy to get $f$ is a injective.

Let $f(0)=a$. In $(1)$, put $m=0$ :

$$f(a+af(n))=an,\forall n\in \mathbb{Z}(2)$$

Consider first the special case when $a=0$ or $f(0)=0$. Put $n=0$ in $(1)$, we get :

$$f(2m+f(m))=m,\forall m\in \mathbb{Z}$$

Which mean that $f$ is surjective. Suppose $b$ be a integer such that $f(b)=-1$. In $(1)$ put $m=n=b$, $f(2b)=0$.

Implies $f(2b)=0=f(0)$. From the injection of $f$, we deduce $b=0$ or $f(0)=-1$. Which is a contracditon !

We just consider $a\ne 0$. We put $n=1$ in $(2)$ :

$$f(a+af(1))=a$$

Or $f(a+af(1))=f(0)$. Implies $a=0$ or $f(1)=-1$. However, we just consider $f(1)=-1$.

In $(1)$, put $n=1$ :

$$f(2m)=f(m)+m,\forall m\in \mathbb{Z}(3)$$

From $(3)$ and $f(1)=-1$, we have $a=f(0)=-2$. We deduce from $(2)$ :

 $$f(-2-2f(n))=-2n,\forall n\in \mathbb{Z}(4)$$

Put $m=1$ in $(1)$ :

$$f(1-f(n))=-n+1,\forall n\in \mathbb{Z}(5)$$

We replace $n$ by $1-f(n)$ in $(4)$ and use the equation $(5)$ :

$$f(-2-2f(1-f(n)))=-2(1-f(n)),\forall n\in \mathbb{Z}$$

$$f(2n-4)=2f(n)-2,\forall n\in \mathbb{Z}(6)$$

We deduce from $(3)$ and $(6)$ that :

$$f(n-2)+n=2f(n),\forall n\in \mathbb{Z}(7)$$

Using $(7)$ and $f(0)=-2,f(1)=-1$ and induction, we have a solution for this problem :

$$\boxed{f(n)=n-2,\forall n\in \mathbb{Z}}$$