Berkeley Math Circle Monthly Contest 1

Solution:
Given that
$$a + b + c = 0 \quad \text{and} \quad a^{n} + b^{n} + c^{n} = 0,$$
let
$$t = ab + bc + ca \quad \text{and} \quad u = abc.$$
Then $a$, $b$, and $c$ are the roots of the polynomial
$$f(z) = (z - a)(z - b)(z - c) = z^{3} + t z - u.$$
If $t = 0$, then $a$, $b$, and $c$ are the three cube roots of $u$ and thus all have the same absolute value; otherwise we can normalize, dividing $a$, $b$, and $c$ by a square root of $t$, so that $t = 1$. Now we have
$$f(z) = (z - a)(z - b)(z - c) = z^{3} + z - u.$$
Define $p_{k} = a^{k} + b^{k} + c^{k}$ for nonnegative integers $k$ (here we make the convention that $0^{0} = 1$). It is not hard to compute that $p_{0} = 3$, $p_{1} = 0$, $p_{2} = -2$, and, for $k \geq 0$,
$$p_{k+3} = u p_{k} - p_{k+1}.$$
This recursion allows us to think of each $p_{k}$ as a polynomial (with integer coefficients) in $u$. Our plan will be to prove that, for $k > 1$, these polynomials have only real roots. Then, given that $p_{n} = 0$, we can deduce that $f(z)$ is a real polynomial and therefore either has all real roots (which is impossible, since the sum of the 2nd powers of the roots is $-2$) or has a pair of complex conjugate roots which have the same absolute value.
The leading term of $p_{k}$ follows a pattern which is not difficult to verify by induction. It is:
$$\begin{array}{ll} 3 u^{i} & \text{if } k = 3i,\ i \geq 0 \\ \frac{k i}{2} u^{i-1} & \text{if } k = 3i + 1,\ i \geq 1 \\ -k u^{i} & \text{if } k = 3i + 2,\ i \geq 0 \end{array}$$
In particular, for $k \neq 1$, the leading coefficients of $p_{k}$ and $p_{k+3}$ have the same sign, and the degree of $p_{k+3}$ is one more than the degree of $p_{k}$. We will prove that the roots of $p_{k}$ alternate with the roots of $p_{k+3}$ on the number line, with no two coinciding, beginning and ending with a root of $p_{k+3}$. When $p_{k}$ is constant (that is, for $k = 0, 2$, or $4$) this statement is trivial, and we will use it as the base of an induction. For all other $k$, the induction hypothesis tells us there is exactly one root of $p_{k-3}$ between each pair of roots of $p_{k}$, and it suffices to prove the following statement:
$$\text{If } p_{k} = 0, \text{ then } p_{k+3} \text{ and } p_{k-3} \text{ have opposite signs.}$$
For brevity we will prove this only when $x = p_{k-3}$ is positive and $u$ is also positive; the proof readily generalizes when one or both are negative. (By the induction hypothesis $p_{k-3} \neq 0$, and if $u = 0$ then (1) gives $p_{k+3} = -p_{k+1} = p_{k-1} = -p_{k-3}$ for $k \geq 4$.) Assume for contradiction that $y = p_{k+3} \geq 0$. Using (1) repeatedly, we get
$$\begin{gathered} p_{k+1} = u p_{k} - p_{k+3} = -y \\ p_{k-2} = u p_{k-3} - p_{k} = u x \\ p_{k-1} = u p_{k-2} - p_{k+1} = u^{2} x + y \end{gathered}$$
Note that $p_{k-1}$, $p_{k-2}$, $p_{k-3}$ are all positive. Then, using the reverse recursion
$$p_{i} = \frac{p_{i+1} + p_{i+3}}{u}$$
we get that $p_{i} > 0$ for all $i < k$. Since $p_{2} = -2$, this is a contradiction.