South African Mathematical Olympiad, Third Round

Suppose $f$ satisfies the condition.
By taking $(a, b, c) = (0, 0, 0)$, we get $3f(0) + 3f(0)^3 = f(0)^3$, so that either $f(0) = 0$, or $3 = -2f(0)^2$. The latter is not possible in $\mathbb{Z}$, so we must have $f(0) = 0$.
By taking $(a, b, c) = (n, -n, 0)$, we get $f(n^3) + f(-n^3) = 0$, resulting in
$$f(-n^3) = -f(n^3) \text{ for all } n \in \mathbb{Z}. \qquad (1)$$
By taking $(a, b, c) = (n, 0, 0)$, we get
$$f(n^3) = f(n)^3 \text{ for all } n \in \mathbb{Z}. \qquad (2)$$
By combining (1) and (2), we see that, for any $n \in \mathbb{Z}$,
$$f(-n)^3 = f((-n)^3) = f(-n^3) = -f(n^3) = -f(n)^3 = (-f(n))^3,$$
so that $f(-n) = -f(n)$, i.e., $f$ is an odd function.
Now take $(a, b, c) = (k, 1-k, 0)$, so that
$$f(k)^3 + f(1-k)^3 + 3f(k)f(1-k)f(1) = f(1)^3 \text{ for all } k \in \mathbb{Z}. \qquad (3)$$
Also, from (2), $f(1) = f(1)^3$, so that $f(1) \in \{-1, 0, 1\}$.
First, if $f(1) = 0$, then from (3), $f(k) = -f(1-k) = f(k-1)$ for all $k \in \mathbb{Z}$, so that $f(n) = 0$ for all $n \in \mathbb{Z}$ (using induction and the fact that $f$ is odd).
Second, if $f(1) = 1$, then from (3), $f(k)^3 + f(1-k)^3 + 3f(k)f(1-k) = 1$, for all $k \in \mathbb{Z}$, i.e., the Diophantine equation $X^3 + Y^3 + 3XY = 1$ is satisfied by $(X, Y) = (f(k), f(1-k))$. This equation can be rewritten as $(X + Y - 1)(X^2 - XY + Y^2 + X + Y + 1) = 0$. Note that $X^2 - XY + Y^2 + X + Y + 1 = 0$ is only solvable in $\mathbb{Z}$ if $X = Y = -1$ (otherwise the discriminant is negative when considered as a quadratic in $X$). So there are two options here:

1. $f(k) = 1 - f(1-k) = 1 + f(k-1)$ for all $k \in \mathbb{Z}$. Then it follows immediately by induction and the fact that $f$ is odd that $f(n) = n$ for all $n \in \mathbb{Z}$.
2. There is some $k_0 \in \mathbb{Z}$ such that $f(k_0) = f(1-k_0) = -1$. Then $f(-k_0) = 1 = 1 - f(1+k_0)$ implies that $f(1+k_0) = 0$. Hence, by using $(a, b, c) = (1+k_0, 1-k_0, -1)$ in the functional equation, we get that $0 - 1 - 1 + 3f(2)(-1)(1) = 1$, so that $f(2) = -1$. Thus $k_0 = 2$ is the smallest positive value of $k_0$ with the property that $f(k_0) = -1 = f(1-k_0)$. We have therefore established the base case of the proof by induction that $(f(3k), f(3k+1), f(3k+2)) = (0, 1, -1)$ for all $k \ge 0$. Assume now that this statement is true for some $k \ge 0$. Then, from $f(-3k-2) = 1 = 1 - f(1+3k+2)$, we find that $f(3+3k) = f(3(k+1)) = 0$. Using $(a, b, c) = (3k, 1, 3)$ in the functional equation (and recalling that $f(3) = f(1+2) = f(1+k_0) = 0$), we get $0 + 1 + 0 + 3(0)(1)(0) = f(3k+4)^3$, so that $f(3k+4) = f(3(k+1)+1) = 1$. Similarly, with $(a, b, c) = (3k, 2, 3)$, we see that $f(3k+5) = f(3(k+1)+2) = -1$, and the induction is complete. It follows now from the fact that $f$ is odd that $(f(3k), f(3k+1), f(3k+2)) = (0, 1, -1)$ for all $k \in \mathbb{Z}$, i.e.,
$$f(n) = \begin{cases} 0 & \text{if } n \equiv 0 \pmod{3} \\ 1 & \text{if } n \equiv 1 \pmod{3}, & \text{for all } n \in \mathbb{Z}. \\ -1 & \text{if } n \equiv 2 \pmod{3} \end{cases} \quad (4)$$
Finally, the observation that whenever $f$ satisfies the equation then $-f$ also satisfies the equation, takes care of the case $f(1) = -1$.
It is straightforward to check that the three functions $f(n) = 0$, $f(n) = n$ and $f(n) = -n$ (for all $n \in \mathbb{Z}$) satisfy the given functional equation. The function (4) requires a little more effort (and time) to check, but offers no difficulty. Its negative then follows automatically. We conclude that there are five functions that solve the equation, as described above.