# 1985 AHSME Problems/Problem 23

## Problem

If $$x = \frac{-1+i\sqrt{3}}{2} \qquad\text{and}\qquad y=\frac{-1-i\sqrt{3}}{2},$$ where $i^2 = -1$, then which of the following is not correct?

$\mathrm{(A)\ } x^5+y^5 = -1 \qquad \mathrm{(B) \ }x^7+y^7 = -1 \qquad \mathrm{(C) \ } x^9+y^9 = -1 \qquad$

$\mathrm{(D) \ } x^{11}+y^{11} = -1 \qquad \mathrm{(E) \ }x^{13}+y^{13} = -1$

## Solution 1

We can write \begin{align*}&x = \cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right) = e^{\frac{2\pi}{3}i}, \text{ and} \\ &y = \cos\left(-\frac{2\pi}{3}\right)+i\sin\left(-\frac{2\pi}{3}\right) = e^{-\frac{2\pi}{3}i},\end{align*} which gives \begin{align*}&x^k = e^{\frac{2\pi k}{3}i} = \cos\left(\frac{2\pi k}{3}\right)+i\sin\left(\frac{2\pi k}{3}\right), \text{ and} \\ &y^k = e^{-\frac{2\pi k}{3}i} = \cos\left(-\frac{2\pi k}{3}\right)+i\sin\left(-\frac{2\pi k}{3}\right) = \cos\left(\frac{2\pi k}{3}\right)-i\sin\left(\frac{2\pi k}{3}\right),\end{align*} using the fact that $\cos$ is an even function and $\sin$ is an odd function.

Accordingly, $$x^k+y^k = 2\cos\left(\frac{2\pi k}{3}\right),$$ and upon substituting the values $k = 5,7,9,11,13$ from the answer choices, we find that $x^k+y^k = -1$ for all such values except $k = 9$, where $x^9+y^9 = 2\cos(6\pi) = 2 \neq -1$. Thus the answer is $\boxed{\text{(C)} \ x^9+y^9 = -1}$.

## Solution 2

Notice that $x+y = -1$ and $xy = 1$, so \begin{align*}1 &=(x+y)^2 \\ &= x^2+2xy+y^2 \\ &=x^2+y^2+2,\end{align*} and hence $x^2+y^2=-1$. Similarly, \begin{align*}x^3+y^3 &= (x+y)^3-3x^2y-3xy^2 \\ &= \left(-1\right)^3-3xy(x+y) \\ &= -1-3(1)\left(-1\right) \\ &=2,\end{align*} and \begin{align*}x^5+y^5 &= (x+y)^5-5x^4y-5xy^4-10x^3y^2-10x^2y^3 \\ &= \left(-1\right)^5-5xy\left(x^3+y^3+2xy\left(x+y\right)\right) \\ &= -1-5(1)\left(2+2\left(1\right)\left(-1\right)\right) \\ &=-1.\end{align*}

Now let $z_n = x^{2n+1}+y^{2n+1}$. Then, using the results $x^2+y^2 = -1$ and $xy = 1$ from above, we obtain \begin{align*}-z_n &= \left(x^{2n+1}+y^{2n+1}\right)\left(x^2+y^2\right) \\ &= x^{2n+3}+y^{2n+3}+x^2y^{2n+1}+x^{2n+1}y^2 \\ &= x^{2n+3}+y^{2n+3}+\left(xy\right)^2\left(x^{2n-1}+y^{2n-1}\right) \\ &= x^{2n+3}+y^{2n+3}+x^{2n-1}+y^{2n-1} \\ &= z_{n+1}+z_{n-1}.\end{align*}

Again from above, $z_1 = 2$ and $z_2 = -1$, so \begin{align*}1 &= -z_2 \\ &= z_3+z_1 \\ &= x^7+y^7+2,\end{align*} giving $z_3 = x^7+y^7 = -1$. Similarly, \begin{align*}1 &= -z_3 \\ &= z_4+z_2 \\ &= x^9+y^9-1,\end{align*} giving $z_4 = x^9+y^9 = 2 \neq -1$, meaning that the answer must be $\text{(C)}$. To confirm this, we further note that \begin{align*}-2 &= -z_4 \\ &= z_5+z_3 \\ &= x^{11}+y^{11}-1,\end{align*} giving $z_5 = x^{11}+y^{11} = -1$, and finally \begin{align*}1 &= -z_5 \\ &= z_6+z_4 \\ &= x^{13}+y^{13}+2,\end{align*} giving $x^{13}+y^{13} = -1$, which shows that the only false statement is indeed $\boxed{\text{(C)} \ x^9+y^9 = -1}$.