Problem (Algebra):
Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying

$$
f(f(m)+n)+f(m)=f(n)+f(3 m)+2014
\tag{1}
$$

for all integers $m$ and $n$.

Solution:
Let $f$ be a function satisfying (1). Set $C=1007$ and define the function $g: \mathbb{Z} \rightarrow \mathbb{Z}$ by $g(m)=f(3 m)-f(m)+2 C$ for all $m \in \mathbb{Z}$; in particular, $g(0)=2 C$. Now (1) rewrites as

$$
f(f(m)+n)=g(m)+f(n)
$$

for all $m, n \in \mathbb{Z}$. By induction in both directions it follows that

$$
f(t f(m)+n)=\operatorname{tg}(m)+f(n)
\tag{2}
$$

holds for all $m, n, t \in \mathbb{Z}$. Applying this, for any $r \in \mathbb{Z}$, to the triples $(r, 0, f(0))$ and $(0,0, f(r))$ in place of $(m, n, t)$ we obtain

$$
f(0) g(r)=f(f(r) f(0))-f(0)=f(r) g(0) \text {. }
$$

Now if $f(0)$ vanished, then $g(0)=2 C>0$ would entail that $f$ vanishes identically, contrary to (1). Thus $f(0) \neq 0$ and the previous equation yields $g(r)=\alpha f(r)$, where $\alpha=\frac{g(0)}{f(0)}$ is some nonzero constant.

So the definition of $g$ reveals $f(3 m)=(1+\alpha) f(m)-2 C$, i.e.,

$$
f(3 m)-\beta=(1+\alpha)(f(m)-\beta)
\tag{3}
$$

for all $m \in \mathbb{Z}$, where $\beta=\frac{2 C}{\alpha}$. By induction on $k$ this implies

$$
f\left(3^{k} m\right)-\beta=(1+\alpha)^{k}(f(m)-\beta)
\tag{4}
$$

for all integers $k \geqslant 0$ and $m$.

Since $3 \nmid 2014$, there exists by (1) some value $d=f(a)$ attained by $f$ that is not divisible by 3 . Now by (2) we have $f(n+t d)=f(n)+t g(a)=f(n)+\alpha \cdot t f(a)$, i.e.,

$$
f(n+t d)=f(n)+\alpha \cdot t d
\tag{5}
$$

for all $n, t \in \mathbb{Z}$.

Let us fix any positive integer $k$ with $d \mid\left(3^{k}-1\right)$, which is possible, since $\operatorname{gcd}(3, d)=1$. E.g., by the Euler-Fermat theorem, we may take $k=\varphi(|d|)$. Now for each $m \in \mathbb{Z}$ we get

$$
f\left(3^{k} m\right)=f(m)+\alpha\left(3^{k}-1\right) m
$$

from (5), which in view of (4) yields $\left((1+\alpha)^{k}-1\right)(f(m)-\beta)=\alpha\left(3^{k}-1\right) m$. Since $\alpha \neq 0$, the right hand side does not vanish for $m \neq 0$, wherefore the first factor on the left hand side cannot vanish either. It follows that

$$
f(m)=\frac{\alpha\left(3^{k}-1\right)}{(1+\alpha)^{k}-1} \cdot m+\beta
$$



So $f$ is a linear function, say $f(m)=A m+\beta$ for all $m \in \mathbb{Z}$ with some constant $A \in \mathbb{Q}$. Plugging this into (1) one obtains $\left(A^{2}-2 A\right) m+(A \beta-2 C)=0$ for all $m$, which is equivalent to the conjunction of

$$
A^{2}=2 A \quad \text { and } \quad A \beta=2 C .
\tag{6}
$$

The first equation is equivalent to $A \in\{0,2\}$, and as $C \neq 0$ the second one gives

$$
A=2 \quad \text { and } \quad \beta=C .
\tag{7}
$$

This shows that $f$ is indeed the function mentioned in the answer and as the numbers found in (7) do indeed satisfy the equations (6) this function is indeed as desired.

Final answer: $f(n) = 2 n+1007$