# 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 $$\zeta_1+\zeta_2+\zeta_3=e_1 = 1$$ $$\zeta_1^2+\zeta_2^2+\zeta_3^2= e_1^2 - 2e_2 = 3$$ from where it follows that $e_2 = -1$. The third equation can be factored as $$7 =\zeta_1^3+\zeta_2^3+\zeta_3^3 = (\zeta_1+\zeta_2+\zeta_3)(\zeta_1^2+\zeta_2^2+\zeta_3^2-\zeta_1\zeta_2-\zeta_2\zeta_3 -\zeta_3\zeta_1)+3\zeta_1\zeta_2\zeta_3\\ = e_1^3 - 3e_1e_2 + 3e_3,$$ 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 $$x^3 - x^2 - x - 1 = 0 \Longleftrightarrow x^3 = x^2 + x + 1$$ 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 $s_4 = 7 + 3 + 1 = 11, s_5 = 21, s_6 = 39, s_7 = \boxed{071}$.