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If $x^2+x+1 = 0$ then find the value of $x^{1999}+x^{2000}$.

I first tried finding the solution of the given equation and then substituting it in the expression whose value we have to find but I wasn't able to simplify it.

In a different approach I moved the terms around a bit and arrived at $x^3 = 1$. But wouldn't that mean that $x = 1$ (which is clearly not possible since it wouldn't satisfy the given equation)? Any help would be appreciated.

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    $\begingroup$ $x^3=1$ has one real root and two complex roots. $x $ is not the real root so it must be one of the complex roots. Don't worry about it. You don't need to solve for x. Note x^3=1 so x^1998=(x^3)^666=1. So you have x^1999+x^2000=x+x^2=x^2+x+1-1=0-1=-1. $\endgroup$
    – fleablood
    Mar 26, 2017 at 5:30
  • $\begingroup$ @fleablood I still don't understand. Why do we have $x^3=1$? It doesn't satisfy $x^2+x+1=0$... $\endgroup$
    – Crescendo
    May 15, 2017 at 16:37
  • $\begingroup$ Yes it does. If $x = -\frac 12 \pm i\frac{\sqrt{3}}{2}$. Then both $x^3 = 1$. And $x^2 + x + 1 = 0$. Remember $x^3 = 1$ has THREE solutions. $x = 1$ is only one of them and it is the one that doesn't work. But the other two do. This is what is called a primative root of unity. And $x^n = 1$ will have $n$ solutions and some of them will satisisfy $x^{n-1} + .... + x + 1 = 0$ but $x =1$ will not, and those that do will be complex numbers..... $\endgroup$
    – fleablood
    May 15, 2017 at 16:49
  • $\begingroup$ The thing though is you don't have to actually solve $x^3 = 1$. You just have to know that there is a complex solution that satisfies $x^2 + x + 1=0$ and that whatever it is, $x^3 = 1$. $\endgroup$
    – fleablood
    May 15, 2017 at 16:53
  • $\begingroup$ @fleablood Ah okay thanks! $\endgroup$
    – Crescendo
    May 15, 2017 at 17:06

3 Answers 3

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Hint: As you have correctly observed, we can deduce that $x^3 = 1$. Now, note that $$ x^{1999} + x^{2000} = (x^{3})^{666}(x + x^2) $$

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    $\begingroup$ Note: We can only deduce $x^3 = 1\implies x = 1$ if we assume that $x$ is a real number (ignoring the possibility of non-zero characteristics). The equation $x^2 + x + 1 = 0$ has no real solutions, as you can check by looking at the discriminant. $\endgroup$
    – Ben Grossmann
    Mar 26, 2017 at 5:20
  • $\begingroup$ so is it that we can't calculate the cube root of a complex or an imaginary number? $\endgroup$
    – Aradhye Agarwal
    Mar 26, 2017 at 5:25
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    $\begingroup$ @AradhyeAgarwal it is the case that $x^3 = 1$ has three different solutions. One is $x = 1$. The other two are $$ x = -\frac 12 \pm \frac{\sqrt{3}}{2}i $$ these other two cube roots of $1$ are the complex solutions to $x^2 + x + 1 = 0$. $\endgroup$
    – Ben Grossmann
    Mar 26, 2017 at 5:27
  • $\begingroup$ Oh so it is always safe to take powers of a given equation but the roots should be taken with care as this can create additional solutions. Thanks! $\endgroup$
    – Aradhye Agarwal
    Mar 26, 2017 at 5:29
  • $\begingroup$ @AradhyeAgarwal that isn't exactly correct, but it's close enough that I'm afraid to create more confusion by correcting it. $\endgroup$
    – Ben Grossmann
    Mar 26, 2017 at 5:30
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$x^{1998}(x^2+x)=-x^{1998}=-(x^{3})^{666}=-1$

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There is nice relation here $$x^{2000}+x^{1999}=-x^{1998}=x^{1999}+x^{1997}$$ which is why $x^3=1$ So $$x^{1998}(x+x^2)=({x^{3}})^{666}(-1)=-1$$

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