Mock AIME 3 Pre 2005 Problems/Problem 4

Problem

and are complex numbers such that

$\begin{align*}\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\end{align*}$ (Error compiling LaTeX. Unknown error_msg)

Compute .

Solution

We let (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 . 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 . Thus, applying Vieta's formulas backwards, and 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 (the power sums). Then from , we have the recursion . It follows that .

See Also

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