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If $x$ is a complex number such that $x^2+x+1=0$, then the numerical value of $(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}$ is equal to?

A) 52 . B) 56 . C) 54. D)58 . E)None of these

Where is this question from? I'm pretty sure it comes from one high school math contest, does anyone one know which math contest and of course i can't solve, I have an answer key but I don't know the solution.

nar
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  • Hint: $x^2=-x-1$ and $\frac1x=-x-1$ – robjohn Jan 22 '19 at 03:25
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    Similar to what robjohn wrote, note that $x^2 + 1 = -x$, so if $x \neq 0$, which needs to be assumed, then dividing by $x$ gives $x + \frac{1}{x} = -1$. You can then take this to various powers to determine appropriate values for the other terms in your equation. – John Omielan Jan 22 '19 at 03:29
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    Welcome to MSE. robjohn and I have given you hints. Although you say you can't solve it, please give us a summary of what you've tried, so we can determine how to best help you. Thanks. – John Omielan Jan 22 '19 at 03:36
  • To answer the second question, this comes from mysmartmouth's MOCK AMC F 2006-2007 (roughly equivalent in difficulty to an AMC12 I believe) – user574848 Jan 22 '19 at 07:56

5 Answers5

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$x^2=-x-1.$

$x^2+1=-x\Rightarrow x+\frac{1}{x}=-1\Rightarrow \frac{1}{x}=-x-1=x^2$ $[$as $x \ne 0]$.

$x^3=1.$

$x^n+\frac{1}{x^n}=-1$ as $n = 3k+1,3k+2$

$x^n+\frac{1}{x^n}=2$ as $n = 3k$

So, here the answer is $18+4 \times9=54$

Offlaw
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Multiply both sides by $x-1\ne 0$: $$x^2+x+1=0 \Rightarrow (x-1)(x^2+x+1)=0 \Rightarrow x^3-1=0 \Rightarrow x^3=1\Rightarrow \\ x^{54}=1;x^{-54}=1 \quad (1)$$ Expand: $$(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}=\\ [x^2+x^4+\cdots+x^{54}]+[x^{-2}+x^{-4}+\cdots+x^{-54}]+2\cdot 27=\\ \frac{x^2(x^{54}-1)}{x^2-1}+\frac{x^{-2}(x^{-54}-1)}{x^2-1}+54\stackrel{(1)}=54.\\ $$

farruhota
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$x^3=1$

$x^2+1/x^2=-1$

We need $\sum_{r=0}^8\sum_{n=1}^3(x^{3r+n}+x^{-(3r+n)})^2$

$=\sum_{r=0}^8( (-1)^2+(-1)^2+2^2)$

$=9(6)$

lab bhattacharjee
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Too long for a comment.

Since you already received good explanations, I just focused on the more general problem of $$S_n=\sum_{k=1}^n \left(x^k+x^{-k}\right)^2 \qquad \text{where}\qquad x^2+x+1=0\implies x=i^{4/3}$$ These are the numbers which are congruent to $\{0, 1, 2\}$ modulo $6$ and there are several ways to write them. If you look here, you will find the nice $$S_n= 2 n-1+\frac{1}{\sqrt{3}}\sin \left(\frac{2 \pi }{3}n\right)+\cos \left(\frac{2 \pi }{3}n\right)$$

$$S_n=n-1+3 \left\lfloor \frac{n-1}{3}\right\rfloor $$

Claude Leibovici
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Thank you for all your answers, I just find the other solution to solve this, so I want to post it here.

Since $x^2+x+1=0$, so we can find

$x+\frac{1}{x}=-1$

and $x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2=(-1)^2-2=-1$. Let's keep going

$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=(-1)(-1)-(-1)=2$

$x^4+\frac{1}{x^4}=(x^2+\frac{1}{x^2})^2-2=(-1)^2-2=-1 $

$x^5+\frac{1}{x^5}=(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})-(x^2+\frac{1}{x^2})=(-1)(-1)-2=-1 $

$x^6+\frac{1}{x^6}=(x^5+\frac{1}{x^5})(x+\frac{1}{x})-(x^4+\frac{1}{x^4})=(-1)(-1)-(-1)=2$

......

So it has a period,-1,-1,2,-1,-1,2,-1,-1,-2.......until the last term.

The sum of the one period is $(-1)^2+(-1)^2+(2)^2=6$, and we have 27$\div$ 3=9, so the final value is equal to 6$\times$9=54.

nar
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