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

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==Problem==
 
==Problem==
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If <math> x=\frac{-1+i\sqrt{3}}{2} </math> and <math> y=\frac{-1-i\sqrt{3}}{2} </math>, where <math> i^2=-1 </math>, then which of the following is ''not'' correct?
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If <cmath>x = \frac{-1+i\sqrt{3}}{2} \qquad\text{and}\qquad y=\frac{-1-i\sqrt{3}}{2},</cmath> where <math>i^2 = -1</math>, then which of the following is not correct?
  
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<math> \mathrm{(A)\ } x^5+y^5=-1 \qquad \mathrm{(B) \ }x^7+y^7=-1 \qquad \mathrm{(C) \  } x^9+y^9=-1 \qquad </math>  
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<math> \mathrm{(A)\ } x^5+y^5 = -1 \qquad \mathrm{(B) \ }x^7+y^7 = -1 \qquad \mathrm{(C) \  } x^9+y^9 = -1 \qquad </math>  
  
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<math> \mathrm{(D) \  } x^{11}+y^{11}=-1 \qquad \mathrm{(E) \  }x^{13}+y^{13}=-1 </math>
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<math> \mathrm{(D) \  } x^{11}+y^{11} = -1 \qquad \mathrm{(E) \  }x^{13}+y^{13} = -1 </math>
  
 
==Solution 1==
 
==Solution 1==
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Notice that <math> x+y=-1 </math> and <math> xy=1 </math>. We have <math> 1=(x+y)^2=x^2+2xy+y^2=x^2+y^2+2\implies x^2+y^2=-1 </math>.  
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We can write <cmath>\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*}</cmath> which gives <cmath>\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*}</cmath> using the fact that <math>\cos</math> is an even function and <math>\sin</math> is an odd function.
  
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We also have <math> x^3+y^3=(x+y)^3-3x^2y-3xy^2=(-1)^3-3xy(x+y)=-1-3(1)(-1)=2 </math>.
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Accordingly, <cmath>x^k+y^k = 2\cos\left(\frac{2\pi k}{3}\right),</cmath> and upon substituting the values <math>k = 5,7,9,11,13</math> from the answer choices, we find that <math>x^k+y^k = -1</math> for all such values except <math>k = 9</math>, where <math>x^9+y^9 = 2\cos(6\pi) = 2 \neq -1</math>. Thus the answer is <math>\boxed{\text{(C)} \ x^9+y^9 = -1}</math>.
  
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Finally, <math> x^5+y^5=(x+y)^5-5x^4y-5xy^4-10x^3y^2-10x^2y^3 </math>
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==Solution 2==
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Notice that <math>x+y = -1</math> and <math>xy = 1</math>, so <cmath>\begin{align*}1 &=(x+y)^2 \\ &= x^2+2xy+y^2 \\ &=x^2+y^2+2,\end{align*}</cmath> and hence <math>x^2+y^2=-1</math>. Similarly, <cmath>\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*}</cmath> and <cmath>\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*}</cmath>
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<math> =(-1)^5-5xy(x^3+y^3+2xy(x+y))=-1-5(1)(2+2(1)(-1))=-1 </math>.
 
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Let <math> x^{2n+1}+y^{2n+1}=z </math>. We have
 
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<math> -z=(x^{2n+1}+y^{2n+1})(x^2+y^2) </math>
 
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<math> -z=x^{2n+3}+y^{2n+3}+x^2y^{2n+1}+x^{2n+1}y^2 </math>
 
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<math> -z=x^{2n+3}+y^{2n+3}+(xy)^2(x^{2n-1}+y^{2n-1}) </math>
 
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<math> -z=x^{2n+3}+y^{2n+3}+x^{2n-1}+y^{2n-1} </math>.
 
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For <math> n=2, z=-1 </math>, so
 
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<math> 1=x^7+y^7+x^3+y^3=x^7+y^7+2\implies x^7+y^7=-1 </math>.
 
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Therefore, for <math> n=3, z=-1 </math>, and
 
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<math> 1=x^9+y^9+x^5+y^5=x^9+y^9-1\implies x^9+y^9=2 </math>, so <math> x^9+y^9\not=-1, \boxed{\text{C}} </math>.
 
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As a check, we have <math> z=2 </math> for <math> n=4 </math>, and
 
  
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<math> -2=x^{11}+y^{11}+x^7+y^7=x^{11}+y^{11}-1\implies x^{11}+y^{11}=-1 </math>.
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Now let <math>z_n = x^{2n+1}+y^{2n+1}</math>. Then, using the results <math>x^2+y^2 = -1</math> and <math>xy = 1</math> from above, we obtain <cmath>\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*}</cmath>
  
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Again from above, <math>z_1 = 2</math> and <math>z_2 = -1</math>, so <cmath>\begin{align*}1 &= -z_2 \\ &= z_3+z_1 \\ &= x^7+y^7+2,\end{align*}</cmath> giving <math>z_3 = x^7+y^7 = -1</math>. Similarly, <cmath>\begin{align*}1 &= -z_3 \\ &= z_4+z_2 \\ &= x^9+y^9-1,\end{align*}</cmath> giving <math>z_4 = x^9+y^9 = 2 \neq -1</math>, meaning that the answer must be <math>\text{(C)}</math>. To confirm this, we further note that <cmath>\begin{align*}-2 &= -z_4 \\ &= z_5+z_3 \\ &= x^{11}+y^{11}-1,\end{align*}</cmath> giving <math>z_5 = x^{11}+y^{11} = -1</math>, and finally <cmath>\begin{align*}1 &= -z_5 \\ &= z_6+z_4 \\ &= x^{13}+y^{13}+2,\end{align*}</cmath> giving <math>x^{13}+y^{13} = -1</math>, which shows that the only false statement is indeed <math>\boxed{\text{(C)} \ x^9+y^9 = -1}</math>.
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Finally, for <math> n=5 </math> we have <math> z=-1 </math>, and
 
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<math> 1=x^{13}+y^{13}+x^9+y^9=x^{13}+y^{13}+2\implies x^{13}+y^{13}=-1 </math>, and this is true for all other answer choices.
 
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==Solution 2==
 
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Note that <math>x=\cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3})</math> and that $y=\cos(-\frac{2\pi}{3})+i\sin(-\frac{2\pi}{3})=\cos(\frac{2\pi}{3})-i\sin(\frac{2\pi}{3})
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=22|num-a=24}}
 
{{AHSME box|year=1985|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:47, 19 March 2024

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 $\boxed{\text{(C)} \ x^9+y^9 = -1}$.

See Also

1985 AHSME (Problems • Answer Key • Resources)
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Problem 22 		Followed by
Problem 24
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