Mock AIME 3 Pre 2005 Problems/Problem 4

Problem

$\zeta_1, \zeta_2,$ and $\zeta_3$ are complex numbers such that

$\zeta_1+\zeta_2+\zeta_3=1$ $\zeta_1^2+\zeta_2^2+\zeta_3^2=3$ $\zeta_1^3+\zeta_2^3+\zeta_3^3=7$

Compute $\zeta_1^{7} + \zeta_2^{7} + \zeta_3^{7}$ .

Solution

We let $e_1 = \zeta_1 + \zeta_2 + \zeta_3,\ e_2 = \zeta_1\zeta_2 + \zeta_2\zeta_3 + \zeta_3\zeta_1,\ e_3 = \zeta_1\zeta_2\zeta_3$ (the elementary symmetric sums). Then, we can rewrite the above equations as

$\begin{align*}\zeta_1+\zeta_2+\zeta_3&=e_1 = 1\\ \zeta_1^2+\zeta_2^2+\zeta_3^2&= e_1^2 - 2e_2 = 3$ (Error compiling LaTeX. Unknown error_msg)

from where it follows that $e_2 = -1$ . The third equation can be factored as

$\begin{align*}7 =\zeta_1^3+\zeta_2^3+\zeta_3^3 &= (\zeta_1+\zeta_2+\zeta_3)(\zeta_1^2+\zeta_2^2+\zeta_3^2-\zeta_1\zeta_2-\zeta_2\zeta_3 -\zeta_3\zeta_1)+3\zeta_1\zeta_2\zeta_3\\ &= e_1^3 - 3e_1e_2 + 3e_3,$ (Error compiling LaTeX. Unknown error_msg)

from where it follows that $e_3 = 1$ . Thus, applying Vieta's formulas backwards, $\zeta_1, \zeta_2,$ and $\zeta_3$ are the roots of the polynomial

$\begin{align}x^3 - x^2 - x - 1 = 0 \Longleftrightarrow x^3 = x^2 + x + 1\end{align}$ (Error compiling LaTeX. Unknown error_msg)

Let $s_n = \zeta_1^n + \zeta_2^n + \zeta_3^n$ (the power sums). Then from $(1)$ , we have the recursion $s_{n+3} = s_{n+2} + s_{n+1} + s_n$ . It follows that $s_4 = 7 + 3 + 1 = 11, s_5 = 21, s_6 = 39, s_7 = \boxed{071}$ .

Mock AIME 3 Pre 2005 (Problems, Source)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

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Mock AIME 3 Pre 2005 Problems/Problem 4

Problem

Solution

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