, 2024

Solution:
Summing the equations gives $a^{2} + b^{2} + c^{2} = 0$ and summing $a$ times the first equation and etc. gives $a^{3} + b^{3} + c^{3} = 0$. Let $a + b + c = k$. Then $a^{2} + b^{2} + c^{2} = 0$ means $ab + bc + ca = k^{2} / 2$, and $a^{3} + b^{3} + c^{3} = 0 \Longrightarrow -3abc = a^{3} + b^{3} + c^{3} - 3abc = (a + b + c)\left(a^{2} + b^{2} + c^{2} - ab - bc - ca\right) = -k^{3} / 2$, so $abc = k^{3} / 6$.
This means $a$, $b$, and $c$ are roots of the cubic

$$x^{3} - k x^{2} + \left(k^{2} / 2\right) x - \left(k^{3} / 6\right) = 0$$

Next, note that
$$\begin{aligned} a^{4} + b^{4} + c^{4} & = \sum_{\mathrm{cyc}} a\left(k a^{2} - \left(k^{2} / 2\right) a + \left(k^{3} / 6\right)\right) \\ & = \sum_{\mathrm{cyc}} k\left(k a^{2} - \left(k^{2} / 2\right) a + \left(k^{3} / 6\right)\right) - \left(k^{2} / 2\right) a^{2} + \left(k^{3} / 6\right) a \\ & = \sum_{\text{cyc}} \left(k^{2} / 2\right) a^{2} - \left(k^{3} / 3\right) a + \left(k^{4} / 6\right) \\ & = -k^{4} / 3 + k^{4} / 2 \\ & = k^{4} / 6 \end{aligned}$$

After this, there are two ways to extract the values of $k$.

- Summing squares of each equation gives
$$a^{4} + b^{4} + c^{4} = \sum_{\mathrm{cyc}} (a - b)^{2} = 2\left(a^{2} + b^{2} + c^{2}\right) - 2(ab + bc + ca) = -k^{2}$$

so
$$\frac{k^{4}}{6} = -k^{2} \Longrightarrow k = 0, \pm i \sqrt{6} .$$

- Summing $a^{2}$ times the first equation, etc. gives
$$a^{4} + b^{4} + c^{4} = \sum_{\mathrm{cyc}} a^{2}(b - c) = -(a - b)(b - c)(c - a) = -a^{2} b^{2} c^{2} = -\frac{k^{6}}{36}$$
so
$$\frac{k^{4}}{6} = -\frac{k^{6}}{36} \Longrightarrow k = 0, \pm i \sqrt{6} .$$

We can achieve $k = 0$ with $a = b = c = 0$. Letting $a$, $b$, and $c$ be the roots of $x^{3} - (i \sqrt{6}) x^{2} - 3x + (i \sqrt{6})$ will force one of $a^{2} = b - c$ and all other equalities or $a^{2} = c - b$ and all other equalities to hold, if the latter happens, swap $b$ and $c$. Finally, for these $(a, b, c)$, take $(-a, -c, -b)$ to get $-i \sqrt{6}$. Thus, all of these are achievable.