Difference between revisions of "1985 AHSME Problems/Problem 23"

Problem

If $x=\frac{-1+i\sqrt{3}}{2}$ and $y=\frac{-1-i\sqrt{3}}{2}$, 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

Notice that $x+y=-1$ and $xy=1$. We have $1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1$.

We also have $x^3+y^3=(x+y)^3-3x^2y-3xy^2=(-1)^3-3xy(x+y)=-1-3(1)(-1)=2$.

Finally, $x^5+y^5=(x+y)^5-5x^4y-5xy^4-10x^3y^2-10x^2y^3$ $=(-1)^5-5xy(x^3+y^3+2xy(x+y))=-1-5(1)(2+2(1)(-1))=-1$.

Let $x^{2n+1}+y^{2n+1}=z$. We have $-z=(x^{2n+1}+y^{2n+1})(x^2+y^2)$ $-z=x^{2n+3}+y^{2n+3}+x^2y^{2n+1}+x^{2n+1}y^2$ $-z=x^{2n+3}+y^{2n+3}+(xy)^2(x^{2n-1}+y^{2n-1})$ $-z=x^{2n+3}+y^{2n+3}+x^{2n-1}+y^{2n-1}$.

For $n=2, z=-1$, so $1=x^7+y^7+x^3+y^3=x^7+y^7+2\implies x^7+y^7=-1$.

Therefore, for $n=3, z=-1$, and $1=x^9+y^9+x^5+y^5=x^9+y^9-1\implies x^9+y^9=2$, so $x^9+y^9\not=-1, \boxed{\text{C}}$.

As a check, we have $z=2$ for $n=4$, and $-2=x^{11}+y^{11}+x^7+y^7=x^{11}+y^{11}-1\implies x^{11}+y^{11}=-1$.

Finally, for $n=5$ we have $z=-1$, and $1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1$, and this is true for all other answer choices.

Solution 2 (using polar complex numbers)

Note that $x=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})=e^{\frac{2\pi}{3}i}$ and that $y=\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3})=e^{-\frac{2\pi}{3}i}$.

Then $x^k=e^{\frac{2\pi k}{3}i}=\cos(\frac{2\pi k}{3})+i\sin(\frac{2\pi k}{3})$ and $y^k=e^{-\frac{2\pi k}{3}i}=\cos(-\frac{2\pi k}{3})+i\sin(-\frac{2\pi k}{3})=\cos(\frac{2\pi k}{3})-i\sin(\frac{2\pi k}{3})$.

Thus, $x^k+y^k=2\cos(\frac{2\pi k}{3})$. Testing $k = 5,7,9,11,13$ for choices A, B, C, D, and E, respectively, we find that for $k=9$, $x^9+y^9=2\cos(6\pi)=2\neq -1$. The answer is $\boxed{C}$.

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