, 2015

Answer: there is only one function, $f(n) = 2n + 1007$

Solution: Let $f$ be a function satisfying the original equation, let $C = 1007$, and then define the function $g: Z \to Z$, for all integers $m, g(m) = f(3m) - f(m) + 2C$. Then we have $g(0) = 2C$, and the original equation can be rewritten as
$$f(f(m) + n) = g(m) + f(n)$$
holding for all $m, n \in Z$. Using induction we obtain
$$f(tf(m) + n) = tg(m) + f(n) \quad (1)$$
holding for all $m, n, t \in Z$. For any $r \in Z$, substituting $(r, 0, f(0))$ and $(0, 0, f(r))$ for $(m, n, t)$ we obtain:
$$f(0)g(r) = f(f(r)f(0)) - f(0) = f(r)g(0)$$
If $f(0) = 0$, then from $g(0) = 2C > 0$ we can obtain that for all $m, f(m) = 0$, which is a contradiction. Therefore $f(0) \neq 0$, and from the above equation we get $g(r) = \alpha f(r)$, where $\alpha = \frac{g(0)}{f(0)}$ is a nonzero constant.
Thus from the definition of $g$ we obtain $f(3m) = (1+\alpha)f(m) - 2C$, that is
$$f(3m) - \beta = (1 + \alpha)(f(m) - \beta) \quad (2)$$
holding for all $m \in Z$, where $\beta = \frac{2C}{\alpha}$. Using induction we obtain
$$f(3^k m) - \beta = (1 + \alpha)^k (f(m) - \beta) \quad (3)$$
holding for all integers $k \ge 0$ and $m$.
Since 3 does not divide 2014, from the original equation we can obtain that there exists $d = f(a)$ that is not divisible by 3. From (1)
we obtain $f(n + td) = f(n) + \alpha \cdot td$, that is
$$f(n + td) = f(n) + \alpha \cdot td \quad (4)$$
holding for all $n, t \in Z$.
Fix a positive integer $k$ such that $d|(3^k - 1)$; by Euler's theorem, we can take $k = \phi(|d|)$ so that this holds. Then from (4) we obtain that for all $m \in Z$
$$f(3^k m) = f(m) + \alpha(3^k - 1)m$$
Combining with (3) we can deduce that $((1 + \alpha)^k - 1)(f(m) - \beta) = \alpha(3^k - 1)m$. Since $\alpha \ne 0$, when $m \ne 0$ the right side is not zero, so the first term on the left side is also not zero, hence
$$f(m) = \frac{\alpha(3^k - 1)}{(1 + \alpha)^k - 1} \cdot m + \beta$$
So $f$ is a linear function; let $f(m) = Am + \beta$ for all $m \in Z$, where $A \in Q$ is a constant, and substituting into the original equation we get $(A^2 - 2A)m + (A\beta - 2C) = 0$ holding for all $m$, which is equivalent to
$$A^2 = 2A \text{ and } A\beta = 2C.$$
The first equality is equivalent to $A \in \{0, 2\}$, and from $C \ne 0$ we obtain
$$A = 2 \text{ and } \beta = C.$$
This tells us that $f$ is exactly the function in the answer, and this function does satisfy the original equation.