9. Suppose $a, b$, and $c$ are complex numbers satisfying
$$
\begin{aligned}
a^{2} & =b-c, \\
b^{2} & =c-a, \text { and } \\
c^{2} & =a-b .
\end{aligned}
$$

Compute all possible values of $a+b+c$.
Answer: $\quad 0, \pm i \sqrt{6}$
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 $a b+b c+c a=k^{2} / 2$, and
$$
a^{3}+b^{3}+c^{3}=0 \Longrightarrow-3 a b c=a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=-k^{3} / 2 \text {, }
$$
so $a b c=k^{3} / 6$.
This means $a, b$, and $c$ are roots of the cubic
$$
x^{3}-k x^{2}+\left(k^{2} / 2\right) x-\left(k^{3} / 6\right)=0
$$
for some $k$.
Next, note that
$$
\begin{aligned}
a^{4}+b^{4}+c^{4} & =\sum_{\text {cyc }} a\left(k a^{2}-\left(k^{2} / 2\right) a+\left(k^{3} / 6\right)\right) \\
& =\sum_{\text {cyc }} k\left(k a^{2}-\left(k^{2} / 2\right) a+\left(k^{3} / 6\right)\right)-\left(k^{2} / 2\right) a^{2}+\left(k^{3} / 6\right) a \\
& =\sum_{\text {cyc }}\left(k^{2} / 2\right) a^{2}-\left(k^{3} / 3\right) a+\left(k^{4} / 6\right) \\
& =-k^{4} / 3+k^{4} / 2 \\
& =k^{4} / 6 .
\end{aligned}
$$

After this, there are two ways to extract the values of $k$.
- Summing squares of each equation gives
$$
a^{4}+b^{4}+c^{4}=\sum_{\text {cyc }}(a-b)^{2}=2\left(a^{2}+b^{2}+c^{2}\right)-2(a b+b c+c a)=-k^{2},
$$
so
$$
\frac{k^{4}}{6}=-k^{2} \Longrightarrow k=0, \pm i \sqrt{6} .
$$
- Summing $a^{2}$ times the first equation, etc. gives
$$
a^{4}+b^{4}+c^{4}=\sum_{\text {cyc }} a^{2}(b-c)=-(a-b)(b-c)(c-a)=-a^{2} b^{2} c^{2}=-\frac{k^{6}}{36},
$$
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 $x^{3}-(i \sqrt{6}) x^{2}-3 x+(i \sqrt{6})$ 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.