Proving $x^2 + x + 1$ is a factor of $P_n (x)=(x+1)^{2n+1} + x^{n+2}$.

Question

Further says to consider $P_n(\omega)$ and $P_n(\omega^2)$ wherein $\omega$ is a cube root of unity. $\omega \not=1$.

Found this from a examination paper with no solutions. I understand it relates to roots of unity, but I'm unsure how I can bring that into play. Any explanation on how to approach these types of questions is appreciated thank you.

Answer 1

Generally speaking , a polynomial $P \in \mathbb{C}[t]$ of degree $n$ divides some other element of $\mathbb{C}[t]$ say Q of degree $\ge n$, if and only if

$$Q(\xi_j)=0$$ where $\{\xi_{j}: j=1, \cdots ,n\}$ is the zero set of $P$

Therefore , in this case we've $q(x)=(x-\omega)(x-\omega^2)$ and $q(x) \mid P_n(x)$, iff

$P_n (\omega^j)=0$ for $j=1,2$

Now $$P_n(\omega)=(\omega +1)^{2n+1}+(\omega)^{n+2}=-\omega^{4n+2}+\omega^{n+2}=0$$

as $4n+2=n+2$ in $\mathbb{Z}_3$( or just as we're working in $\mathbb{Z}_{3}$)

Similarly $$P(\omega^2)=(\omega^2 +1)^{2n+1}+(\omega^2)^{n+2}=-\omega^{2n+1}+\omega^{2n+4}=0$$

Again because we're working with the exponents in $\mathbb{Z}_{3}$

Conclusion: $q(x) \mid P_n(x)$

Answer 2

Let $\omega$ be any root of $x^2+x+1$. We will show that $\omega$ is a root of $(x+1)^{2n+1} + x^{n+2}$.

First note that we can rearrange $x^2+x+1$ to $x+1 = -x^2$ so we know that $(\omega+1)^{2n+1} = (-\omega^2)^{2n+1} = -\omega^{4n+2}$.

Thus $(\omega+1)^{2n+1} + \omega^{n+2} = -\omega^{4n+2} + \omega^{n+2} = -\omega^{n+2}(\omega^{3n} - 1).$

So we just need to show that $\omega^{3n} - 1 = 0$.

This can been shown via the identity $u^3-1 = (u-1)(u^2+u+1)$ with $u = \omega^n$.

Answer 3

Induction is really helpful here, did you try it? lets check the case $n=0$: $$(x+1)^{2\cdot0+1}+x^{0+2} = x+1+x^2$$ which is of course multiplier of $x^2+x+1$.

Assume, that your claim is true for every $k\leq n$, and lets prove the next step ($n+1$):

we need to prove that $x^2+x+1$ is factor of $(x+1)^{2(n+1)+1}+x^{n+1+2}$ and this can be proved easily:

$$\begin{align}(x+1)^{2n+3}+x^{n+3}&=(x+1)^2\cdot (x+1)^{2n+1}+x\cdot x^{n+2}\\ &=(x^2+2x+1)\cdot(x+1)^{2n+1}+x\cdot x^{n+2}\\ &=(x^2+x+1)\cdot(x+1)^{2n+1}+x\cdot(x+1)^{2n+1}+x\cdot x^{n+2}\\ &=(x^2+x+1)\cdot(x+1)^{2n+1}+x\cdot((x+1)^{2n+1}+x^{n+2}) \end{align}$$

where both terms are multipliers of $x^2+x+1$

Answer 4

Since others have shown some methods, here is an idiosyncratic approach - equivalent to others, but interesting in structure.

Suppose $x^2+x+1=0$ then $x+1=-x^2$ and $x^3=-x^2-x=x+1-x=1$

Then $P_n(x)=\left(-x^2\right)^{2n+1}+x^{n+2}=-x^{4n+2}+x^{n+2}=-x^{n+2}+x^{n+2}=0$

and this shows that $x^2+x+1$ is a factor. Technically this is most economically expressed by saying that we are working modulo (the ideal generated by) $x^2+x+1$. An elementary explanation would use a version of the factor theorem.

Versions of this method occasionally appear on the site, advocated by me and one or two others. In assignments and exams methods known to tutors and examiners are recommended - but for computation, I have always found this approach convenient.