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I am working on a scholarship exam practice assuming high school or pre-university math knowledge. I am stuck at the question below:

Let $\omega$ be a solution of the equation $x^2+x+1=0$. Then $\omega^{10}+\omega^5+3=.....$

My first question is how it would be possible since the discriminant of $x^2+x+1=0$ is less than $0$ so I am not sure how I can continue or start from here. The answer key provided is $2$. Please advise.

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$\omega$ is such that $\omega^2+\omega+1=0$ i.e. $\omega^2=-\omega-1$

Therefore $\omega^3=\omega \cdot \omega^2=\omega(-\omega-1)=-\omega^2-\omega=1$

$\omega^5=\omega^3 \cdot \omega^2=\omega^2=-\omega-1$

$\omega^{10}=(\omega^5)^2=(-\omega-1)^2=\omega^2+1+2\omega=\omega$

Hence $\omega^{10}+\omega^5+3=\omega-\omega-1+3=2$

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Dividing the polynomial $x^{10}+x^5+3$ by $x^2+x+1$ leaves a remainder of $2$. Plugging in $x=\omega$ yields your answer is $2$.

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  • $\begingroup$ +1 That's an extremely good approach for people not well versed with complex numbers. $\endgroup$ – AryanSonwatikar Jun 18 '19 at 9:45
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    $\begingroup$ Yes, I would go farther and say that the point of this problem is not to test one's understanding of complex numbers, but rather to test if one understands the relationship between roots of a polynomial and polynomial division. $\endgroup$ – pre-kidney Jun 18 '19 at 9:47
  • $\begingroup$ @pre-kidney is there a trick to see easily that the remainder is $2$? Or how would you recommend yo use division in this case? $\endgroup$ – Zacky Jun 18 '19 at 20:48
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    $\begingroup$ Yes, there is a trick @Zacky which Bill Dubuque wrote up extremely well here math.stackexchange.com/questions/3224765/… $\endgroup$ – pre-kidney Jun 19 '19 at 4:14
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Hint:

As $\omega$ is a solution of $x^2+x+1=0,$

$$\omega^2+\omega+1=0$$

$$\omega^3-1=(\omega-1)(\omega^2+\omega+1)=0$$

$$\omega^{10}=(\omega^3)^3\cdot\omega\text{ and }\omega^5=\omega^3\cdot\omega^2$$

$$\omega^2+\omega=?$$

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    $\begingroup$ How is $x^{10}=x$? $\endgroup$ – Trey Anupong Jun 18 '19 at 9:44
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    $\begingroup$ @TreyAnupong because $\omega^3 = 1$, then $\omega^{10} = \omega^9 \cdot \omega = \omega$ $\endgroup$ – dcolazin Jun 18 '19 at 9:45
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    $\begingroup$ @dcolazin $\omega$ is not $x$. This answer is confusing to a beginner, it should be $\omega$s. $\endgroup$ – Arnaud Mortier Jun 18 '19 at 10:00
  • $\begingroup$ @ArnaudMortier, Is it fine, now? $\endgroup$ – lab bhattacharjee Jun 19 '19 at 6:21
  • $\begingroup$ @labbhattacharjee Yes. $\endgroup$ – Arnaud Mortier Jun 19 '19 at 13:00
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$\omega^3=(\omega-1)(\omega^2+\omega+1)+1=1$.

$\omega^5=\omega^3\omega^2=\omega^2\ne1$ (as otherwise $\omega=-1-\omega^2=-2$, which is impossible)

$\omega^{10}+\omega^5+3=\dfrac{\omega^{15}-1}{\omega^5-1}+2=\dfrac{1-1}{\omega^5-1}+2=2$.

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  • $\begingroup$ You need to make sure that the denominator is not also $0$, otherwise this is nonsense. $\endgroup$ – Arnaud Mortier Jun 18 '19 at 10:01
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Alternatively: $$\begin{align}\omega^2+\omega+1&=0\\ \omega ^2&=-\omega-1\\ \omega^{10}&=-(\omega +1)^5=-\omega^5-5\omega^4-10\omega^3-10\omega^2-5\omega-1\\ \omega^{10}+\omega^5+3&=-5\omega^2(\omega^2+\omega+1)-5\omega(\omega^2+\omega+1)-1+3=2. \end{align}$$

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Since $$\omega^3-1=(\omega-1)(\omega^2+\omega+1)=0,$$ we obtain: $$\omega^{10}+\omega^5+3=\omega^{10}-\omega+\omega^5-\omega^2+\omega^2+\omega+3=$$ $$=\omega(\omega^9-1)+\omega^2(\omega^3-1)+(\omega^2+\omega+1)+2=2.$$

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    $\begingroup$ This doesn't bring anything more to the existing answers and is unreadable to a beginner. $\endgroup$ – Arnaud Mortier Jun 18 '19 at 10:03
  • $\begingroup$ @Arnaud Mortier I added something. See now. $\endgroup$ – Michael Rozenberg Jun 18 '19 at 10:09
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    $\begingroup$ This was mentioned in CRUDE. Why did you gold-badge reopen? Do you not understand how it is a special case of this?. We already have hundreds of (dupe) answers that show how to do these things. Why do you think we need more? $\endgroup$ – Bill Dubuque Jun 18 '19 at 19:30
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    $\begingroup$ I think the starting problem is not duplicate. $\endgroup$ – Michael Rozenberg Jun 18 '19 at 19:46
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    $\begingroup$ Please see this 8-year-old meta post on abstract duplicates. As someone who strives to bring order to motley inequalities, I would hope that you would be sympathetic to your peers who similarly strive to organize knowledge in other fields. $\endgroup$ – Bill Dubuque Jun 18 '19 at 19:53

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