Divisibility of a polynomial by another polynomial

[kalish1]

I have this question:

Find all numbers $n\geq 1$ for which the polynomial $x^{n+1}+x^n+1$ is divisible by $x^2-x+1$. How do I even begin?

**So far I have that $x^{n+1}+x^n+1 = x^{n-1}(x^2-x+1)+2x^n-x^{n-1}+1,$ and so the problem is equivalent to finding $n$ such that $2x^n-x^{n-1}+1$ is divisible by $x^2-x+1.$**

A solution that I found goes as follows (but I don't understand it!):

*Assume that $x^n+1=(x^m-x+1)Q(x).$ The polynomial $x^m-x+1$ has a real root in the interval $(0,1)$ but $x^n+1$ has no positive real roots. So, no such pairs $m,n$ exist.*

Any help?

This question has been crossposted here: Divisibility of a polynomial by another polynomial - Mathematics Stack Exchange

 

[jvicens]

To begin, we can rewrite the original polynomial as $x^{n+1}+x^n+1 = x^n(x+1)+1$. From this, we can see that the polynomial will only be divisible by $x^2-x+1$ if $x^n(x+1)+1$ is divisible by $x^2-x+1$.

Next, we can use polynomial division to rewrite $x^n(x+1)+1$ as $(x^2-x+1)Q(x)+R(x)$, where $Q(x)$ is the quotient and $R(x)$ is the remainder. If $x^2-x+1$ is indeed a factor, then the remainder $R(x)$ must be equal to 0.

Now, we can use the given information that $2x^n-x^{n-1}+1$ is divisible by $x^2-x+1$ to rewrite the remainder as $R(x) = 2x^n-x^{n-1}+1$. If we substitute this into our original equation, we get $(x^2-x+1)Q(x) + (2x^n-x^{n-1}+1) = x^n(x+1)+1$.

Since the remainder must equal 0, we can set the coefficients of like terms to equal each other. This gives us the equation $Q(x) = x^n$ and $2x^n-x^{n-1}+1 = 1$. Solving for $x^n$, we get $x^n = 0$ or $x^n = 1$.

If $x^n = 0$, then $x^n(x+1)+1 = 1$, which means that the remainder is not equal to 0. Therefore, $x^n$ cannot equal 0.

If $x^n = 1$, then $x^n(x+1)+1 = x+1$, which means that the remainder is not equal to 0. Therefore, $x^n$ cannot equal 1.

This leaves us with no possible solutions for $x^n$, which means that no such pairs $m,n$ exist and the original polynomial is not divisible by $x^2-x+1$ for any $n \geq 1$.