§3.6a Solution
The complete set of solutions is given by
- For $k=1, f(x) \equiv C_{1} x+C_{2}(x \bmod 2)+C_{3}$.
- For $k=3, f(x) \equiv C_{1} x+C_{2} x^{2}$.
- For $k=9, f(x) \equiv C_{1} x+C_{2} x^{4}$.
- For all other $k$, only $f(x) \equiv C_{1} x$.

Here $C_{1}, C_{2}, C_{3}$ are arbitrary integers. We can check they work, so now we just want to show they are the only ones.

We will solve the functional equation for $f: \mathbb{Z} \rightarrow \mathbb{C}$, claiming that the above solution set remains the only one. If $k=1$, we can shift by constants to get $f(0)=0$; if $k \neq 1$ apply $a=b=c=0$ to get $f(0)=0$ anyways. Now note that $f(x) \equiv x$ is a solution, so we may shift by the identity to assume $f(-1)=f(1)$.
We will prove in this case, $f \equiv 0$ unless $k=1,3,9$.
Now plug in $(a, b, c)=(n+1,-n,-1)$ and $(a, b, c)=(1, n,-(n+1))$ gives
$$
\begin{aligned}
f(2 n+1)+f(-n+1)+f(-n-2) & =k(f(n+1)+f(-n)+f(-1)) \\
& =k(f(-n-1)+f(n)+f(1)) .
\end{aligned}
$$

The last two by induction imply $f$ is even. Now, by using this and $(a, b, c)=(n,-n, 0)$ we obtain
$$
\begin{aligned}
f(2 n+1)+f(n+2)+f(n-1) & =k(f(n+1)+f(n)+f(1)) \\
f(2 n)+2 f(n) & =k(2 f(n)+f(0)) \Longrightarrow f(2 n)=(2 k-2) f(n) .
\end{aligned}
$$

Thus $f$ is determined recursively by $f(1)$ (by induction). In particular, if $f(1)=0$ then $f(n) \equiv 0$ by induction.
Now, let us assume $f(1) \neq 0$, and hence by scaling $f(1)=1$. We can then compute:
$$
\begin{aligned}
f(1) & =1 \\
f(2) & =2 k-2 \\
f(3) & =k^{2} \\
f(4) & =4 k^{2}-8 k+4 \\
f(5) & =k^{3}-2 k^{2}+7 k-5 \\
f(6) & =2 k^{3}-2 k^{2} \\
f(7) & =4 k^{3}-6 k^{2}-4 k+7 \\
f(8) & =8 k^{3}-24 k^{2}+24 k-8 .
\end{aligned}
$$

Plug in $(a, b, c)=(5,-3,-2)$ and compute $f(8)+f(1)+f(7)=k(f(5)+f(3)+f(2))$, which simplifies to give
$$
k^{4}-13 k^{3}+39 k^{2}-27 k=0 \Longrightarrow k(k-1)(k-3)(k-9)=0
$$
so $k=1, k=3$, or $k=9$. In these cases it is easy to check by induction now that $f(n)=n \bmod 2, f(n)=n^{2}$, and $f(n)=n^{4}$.
Find all positive integers $k$ for which there exists a nonlinear function $f: \mathbb{Z} \rightarrow \mathbb{Z}$ which satisfies
$$
f(a)+f(b)+f(c)=\frac{f(a-b)+f(b-c)+f(c-a)}{k}
$$
for any integers $a, b, c$ with $a+b+c=0$.