Find $m\in \mathbb{R}$ so that $P(x)$ is divisible by $X^2+X+1$

Question

Let $P(X)=\left(X+1\right)^{2n+1}+\left(m-1\right)X^{n+2}$. Find $m\in \mathbb{R}$ so that $$X^2+X+1 | P(X),\forall n\in \mathbb{N}$$ I considered $y$ to be a root of $X^2+X+1$ and computed $P(y)$ but I'm getting $$y^{n-2}\left(y-m+1\right)=0$$ which means that $m$ is irrational. There must be a different approach. Got any other ideas?

Answer 1

Since $$X^3-1=(X-1)(X^2+X+1),$$ we obtain: $$P(X)\equiv -X^{4n+2}+(m-1)X^{n+2}=-X^{n+2}\left(X^{3n}-m+1\right)\equiv (m-2)X^{n+2},$$ which gives $m=2$.

Answer 2

Your method would have worked too: let $j$ be a root of $X^2+X+1$.

Since $j^2+j+1=0$ we can deduce the following:

1. $j\neq 0$
2. $j+1=-j^2$
3. $(j-1)(j^2+j+1)=0\text{, hence } j^3= 1$

Given this:$$P(j)=\left(-j^2\right)^{2n+1}+\left(m-1\right)j^{n+2}.$$ $$=-j^{4n+2}+\left(m-1\right)j^{n+2}$$ $$\ \ \ \ =-(j^3)^nj^{n+2}+\left(m-1\right)j^{n+2}$$ $$\!\!=-j^{n+2}+\left(m-1\right)j^{n+2}$$

Therefore $$P(j)=0\Longleftrightarrow j^{n+2}=(m-1)j^{n+2}$$ and since $j\neq 0$ you can divide both sides by $j^{n+2}$ and get $m=2$.