$f(m) = 0$ and $f(m) = 2m$ My solution: $f(m-n + f(n)) = f(m) + f(n)$ $m = n = 0 \ yields \ f(f(0)) = 2f(0)$. $m = n \ yields \ f(f(m)) = 2f(m)$. Put $m = m+n-f(n)$ to obtain $f(m) = f(m+n-f(n)) + f(n)$. Setting $n=m$ yields $f(m) = f(2m-f(m)) + f(m)$ and $f(2m-f(m)) = 0$. Either $f(m) = 0$ for all m or $f$ is nonconstant and $2m - f(m) = b$ where b is a zero of $f$. Substituting $f(m) = 2m-b$ into the original equation, we conclude that $b=0$ and hence $f(m) = 2m$. I need to account for the possibility that $f$ has infinitely many zeros, thus, that $2m-f(m)$ is not constant.

i think this solution is not ok because you can't assume that $2m-f(m)=k$ has solution for every k if it is not costant.

sorry for my english

Showing injectivity was the key; I did not know how to do so. Could you please explain why $f$ cannot be periodic?

Because, if $f: \mathbb{Z}\to\mathbb{Z}$ is periodic, then its range is finite. Thus you can not get infinitely many different values in the sequence $f(m), f(f(m)), \ldots$. My solution does not use injectivity.

I did not read the previous solutions because they're boring. Here goes a simple one. If $f(n)=0$ for every $n$, then the equation is satisfied. Otherwise, $f(a)\neq 0$ for some $a$. Plugging $m=n=a$ yields to $f(f(a))=2f(a)$, and by induction $f^{(k)}(a)=2^{k-1}f(a)$, so $f$ has infinite image. $f(n)=0$ for some fixed $n$ implies $n=0$, otherwise $f(m-n)=f(m)$ for every $m$, and the image of $f$ would have cardinality at most $n$. Now plug $m=2n-f(n)$ to get $f(2n-f(n))=0$, yielding $2n-f(n)=0$ by the above argument. So $f(n)=2n$ for every $n$, which indeed satisfies the equation.

$m=n,f(f(n))=2f(n)$ Replace $n$ by $f(n)$ in the original equation,$f(m-f(n)+f(f(n)))=f(m)+f(f(n))$ $f(m+f(n))=f(m)+2f(n)$ Replace $m$ by $m+n$ in the original equation,$f(m+f(n))=f(m+n)+f(n)$ So $f(m+n)=f(m)+f(n)$ that's Cauchy equation. So $f(x)=cx$,then easy to know $c=0$ or 2.

Hi jin and MindFlyer, Next time please read the postings before yours. This could have saved you time writing the same stuff again Take it easy!

Azarus wrote: Find all functions $ f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that $ f(m - n + f(n)) = f(m) + f(n)$ for all $ m,n \in \mathbb{Z}$ I do not believe that the following solution has been posted yet:

$P(m,n)$ be the assertion of $f(m - n + f(n)) = f(m) + f(n)$ $P(0,0) \to f(f(0))=2f(0)$ $P(0,f(0)) \to f(f(0))=f(0)+2f(0)=2f(0) \to f(0)=0$ $P(0,n) \to f(f(n)-n)=f(n)$ Then $P(m,f(n)-n) \to f(m+n)=f(m)+f(n)$ It proves $f(x)=cx$ where $c \in \mathbb{Z}$ Plugging back we get $c=0$ or $c=2$

Let $P(m,n)$ be $f(m-n+f(n))=f(m)+f(n)$. $P(2n-f(n),n)\implies f(2n-f(n))=0$ Case1:$f\equiv 0$ This satisfies the condition. Case2:$\exists \alpha\in \mathbb Z$ s.t. $f(\alpha)\neq 0$ If $\exists n_0\in \mathbb Z$ s.t. $\beta =2n_0-f(n_0)\neq 0$. $P(m,\beta)\implies f(m-\beta)=f(m)$. This implies that $f$ is periodic.Thus $f$ is bounded.On the other hand, $P(m,m)\implies f(f(m))=2f(m)$.Then $f^{k}(\alpha)=2^{k-1}f(\alpha)$.Hence $|f|$ is unbounded which is absurd.Thus $\forall n\in \mathbb Z:f(n)=2n$ which satisfies the condition. From case1,2,the answer is $\boxed{f\equiv 0 \ \text{or} f(n)=2n(\forall n\in \mathbb Z)}\blacksquare$

IstekOlympiadTeam wrote: $P(m,n)$ be the assertion of $f(m - n + f(n)) = f(m) + f(n)$ $P(0,0) \to f(f(0))=2f(0)$ $P(0,f(0)) \to f(f(0))=f(0)+2f(0)=2f(0) \to f(0)=0$ $P(0,n) \to f(f(n)-n)=f(n)$ Then $P(m,f(n)-n) \to f(m+n)=f(m)+f(n)$ It proves $f(x)=cx$ where $c \in \mathbb{Z}$ Plugging back we get $c=0$ or $c=2$ I admire your solution, which doesn't use injectivity or periodicity. However, is there a way to prove injectivity directly (for $f(n)=2n$ )?

$P(0,0): f(f(0))=2f(0)$. $P(0,f(0)): f(f(0))=f(0)+2f(0)\implies f(0)=0$. $P(0,m): f(f(m)-m)=f(m)$. $P(m,f(n)-n): f(m+n)=f(m)+f(n)\implies f(x)=kx$ for some constant $k$. Plugging this back gives $km-kn+k^2n=km+kn\implies k^2n=2kn\implies k=\{0,2\}$. Thus, the only solutions are $\boxed{f\equiv0}$ and $\boxed{f(x)=2x\forall x\in\mathbb{R}}$, which work.

$P(n,n)\Rightarrow f(f(n))=2f(n)$ $P(m,f(n))\Rightarrow f(m+f(n))=f(m)+2f(n)$ $P(m+n,n)\Rightarrow f(m+n)=f(m)+f(n)$ so by Cauchy's FE $f(n)=cn$. Testing, we have $\boxed{f(n)=0}$ and $\boxed{f(n)=2n}$ as the only solutions.

$P(m+2n-f(n),n) : f(m+n) = f(m+2n-f(n)) + f(n) (1)$. $P(m,f(n)-n) : f(m+n) = f(m) + f(f(n)-n) (2)$. Now applying $m=n$ in $(1),(2)$ $\implies$ $f(3n-f(n)) = f(f(n) - n)$. Claim: $f$ is injective. Proof : Assume $f(a_1) = f(a_2)$. Now we have $f(z-a_1+f(a_1)) = f(z-a_2+f(a_2)$ for every $\ z\in \mathbb{Z}$ so if $k = z-a_1+f(a_1)$ then $f(b) = f(a_1 + (a_1-a_2))$ so $f$ is periodic. Now we have 2 cases, either $\forall x:$ $f(x) = 0$ or $\exists x: f(x)\neq 0$. first case is one of the answers, second case gives contradiction with $f$ being periodic so $z - a_1 + f(a_1) = z - a_2 + f(a_2)$ so $a_1 = a_2$. Now that $f$ is injective we have $f(3n-f(n)) = f(f(n) - n) \implies 2f(n) = 4n \implies f(n) = 2n$. Answers : $f(n) = 0,f(n) = 2n$.