prove polynomial division for any natural number

Question

Show that for any natural numbers $a$, $b$, $c~$ we have $~x^2 + x + 1|x^{3a+2} + x^{3b+1} + x^{3c}$.

Any hints on what to use?

Answer 1

We have $$x^3-1=(x-1)\left(x^2+x+1\right)$$ This means that $$\begin{align}x^3-1&\equiv0\pmod{x^2+x+1}\\x^3&\equiv1\pmod{x^2+x+1}\end{align}$$ Now, substitute it $$\begin{align}x^{3a+2}+x^{3b+1}+x^{3c}&\equiv x^2\left(x^3\right)^a+x\left(x^3\right)^b+\left(x^3\right)^a\\&\equiv x^2\cdot1^a+x\cdot1^b+1^c\\&\equiv x^2+x+1\\&\equiv0\pmod{x^2+x+1}\end{align}$$

Answer 2

Hint : $x^2+x+1$ is a factor of $x^3-1$ , so we have $x^3\equiv 1\ (\ mod\ x^2+x+1\ )$

Answer 3

$x^2+x+1$ splits in $\mathbf C$ as $(x-j)(x-\bar j)$, so it is enough to prove $j$ is a root of $p(x)=x^{3a+2} + x^{3b+1} + x^{3c}$. Since $j^3=1$,we have $$p(j)=j^2+j+1=0.$$