How to find all functions $f$ such that $f(f(m)+n)+f(m)=f(n)+f(3m)+2014$

Question

Find all functions $f: \mathbb{Z}\rightarrow\mathbb{Z}$ such that:

$$f(f(m)+n)+f(m)=f(n)+f(3m)+2014$$

for every $m, n \in \mathbb{Z}$.

I'm completely stuck after trying to plug different things into the above functional equation. I would like to know a complete solution to this problem please.

Answer 1

Much of what we see is just smoke and mirrors.

Any functional equation looks simple once it is solved. So I'd rather put the answer before the solution: $$f(x)=2x+1007$$ Now the boring part.

1. As noted by Samuel Adrian Antz, $$f(n+C)=f(n)+2014,\text{ where }C=f(0)\tag1$$ By applying this repeatedly, we may also find $f(n+2C),\;f(n+3C)$ and so on.
2. We have many things inside $f$ and very few things outside $f$, which is a sure sign that we need injectivity. Indeed, $f$ is injective. The proof is rather involved (read: on an entirely different level of "boring"), so feel free to skip directly to step 9.
3. Suppose there are some $k_1\ne k_2$ such that $f(k_1)=f(k_2)$. Then plug them instead of $m$: $$ f(f(k_1)+n)+f(k_1)=f(n)+\color{red}{f(3k_1)}+2014\\ f(f(k_2)+n)+f(k_2)=f(n)+\color{red}{f(3k_2)}+2014 $$ Everything except the red term in both equations is the same, so the red terms must also be the same. So if we have one pair of arguments giving the same values of $f$, then we may have another such pair by tripling them.
4. Now plug $k_1,\,k_2$ instead of $n$: $$ \color{red}{f(f(m)+k_1)}+f(m)=f(k_1)+f(3m)+2014\\ \color{red}{f(f(m)+k_2)}+f(m)=f(k_2)+f(3m)+2014 $$ By the same reasoning concerning the red term, now we may produce another such pair by shifting $k_1,\,k_2$ simultaneously by the value of $f$ anywhere, for example, by $C$ (which is its value at $0$).
5. Whatever our $C$ might be, there are some powers of $3$ such that $C$ divides $3^a-3^b$.
6. By tripling, produce the following identities: $$f(3^ak_1)=f(3^ak_2)\tag2$$ $$f(3^bk_1)=f(3^bk_2)\tag3$$
7. Now let's do some shifting. As I said before, $3^a-3^b$ is a multiple of $C$. So is $(3^a-3^b)k_1$. Well, let's shift (3) by that much: $$f(3^ak_1)=f(3^bk_2+(3^a-3^b)k_1)\tag4$$
8. By combining (2) and (4), we now have: $$f(3^ak_2)=f(3^bk_2+(3^a-3^b)k_1)$$ But wait, now the difference of arguments is $(3^a-3^b)(k_2-k_1)$, which is a multiple of $C$, so by repeated application of (1), they must differ by quite a number of $2014$'s, instead of being equal. A contradiction.
9. Now we know that from $f(k_1)=f(k_2)$ follows $k_1=k_2$, which immediately makes a lot of things a lot simpler. Plug $m=n$: $$f(f(n)+n) = f(3n)+2014 \stackrel{by\;(1)}{=\!=\!=} f(3n+C)\\ f(n)+n=3n+C\\ f(n)=2n+C$$
10. Now $f(C)=3C$, but also $f(C)=f(0+C)=f(0)+2014=C+2014$, so $C=1007$.

So it goes.