SELECTION and TRAINING SESSION

(Solution by I. Voronovich.) Rewrite the given equation as $$f(f(m)+n)=f(n)-f(m)+f(3m)+c,(1)$$ where $c=2014$; all we need in this solution is that $c$ is even and coprime with 3. Let $f(0)=a$.

1) Setting $m=0$ we get $f(n+a)=f(n)+c$, then by standard induction to both sides $f(n+ka)=f(n)+kc$, $k\in \mathbb{Z}$; in particular, $f(ka)=a+kc$ and $f(a)=a+c$.

2) Set $m=n$ in (1), then $$f(n)+n=f(3n)+c=[\text{in view of 1}]=f(3n+a).$$

3) Setting $m=a$ in (1) we get $$f(n+f(a))=f(n)-f(a)+f(3a)+c,$$ or, in view of 1) $f(n+c)+c=f(n+c+a)=f(n)-c-a+a+3c+c$, or $f(n+c)=f(n)+2c$. Again, by easy induction $f(n+kc)=f(n)+2kc$.

4) Searching on injectivity: suppose that $f(s)=f(t)$. Then setting $m=s$ and $m=t$ in (1) we obtain $f(3s)=f(3t)$, and thus $f(3^{l}s)=f(3^{l}t)$ for all $l\in \mathbb{N}$. Since $(3;c)=1$, there exists an $l$ such that $3^l-1=nc$ for some $n\in \mathbb{N}$. Then in view of 3) we have $$f(3^{l}s)=f(3^{l}t)⟺f(s+(3^l-1)s)=f(t+(3^l-1)t),$$ or $f(s+nsc)=f(t+ntc)$, or $f(s)+2nsc=f(t)+2ntc$, $2nsc=2ntc$, so $s=t$. Thus $f$ is injective. Since in view of 2) $f(f(n)+n)=f(3n+a)$, we obtain $f(n)+n=3n+a$, $f(n)=2n+a$. Easy verification shows that the function is a solution of (1) if and only if $a=c/2$.