$\omega$ is a solution of $x^2+x+1=0$, find $\omega^{10}+\omega^5+3$

Question

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.

Answer 1

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

Answer 2

Hint:

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

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

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

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

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

Answer 3

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 $2$.

Answer 4

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

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

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

Answer 5

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

Answer 6

Alternatively: $$\begin{align}\omega^2+\omega+1&=0\\ \omega ^2&=-\omega-1\\ \omega^{10}&=-(\omega +1)^5=-\omega^5-5\omega^4-10\omega^3-10\omega^2-5\omega-1\\ \omega^{10}+\omega^5+3&=-5\omega^2(\omega^2+\omega+1)-5\omega(\omega^2+\omega+1)-1+3=2. \end{align}$$