Prove that $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$

Question

Prove that $x^{3m}+x^{3n+1}+x^{3p+2}$ is divisible by $x^2+x+1$ in ring $ {\displaystyle \mathbb {R}}$$[x]$ where $m, n, p \in {\displaystyle \mathbb {N}}$

I have tried to use factor theroem to write down $x^{3m}+x^{3n+1}+x^{3p+2}$ as a product of $x^2+x+1$ and something. But I cannot find the second factor.

Thanks in Advance.

Answer 1

Note: $$\frac{x^{3m}+x^{3n+1}+x^{3p+2}}{x^2+x+1}=\frac{x^{3m}-1+x^{3n+1}-x+x^{3p+2}-x^2+(x^2+x+1)}{x^2+x+1}=\\ \frac{(x^{3m}-1)+x(x^{3n}-1)+x^2(x^{3p}-1)+(x^2+x+1)}{x^2+x+1}.$$ And: $$x^{3k}-1 \equiv 0 \pmod{x^2+x+1}, k\in \mathbb N.$$

Answer 2

Hint : Because of $$x^3-1=(x-1)(x^2+x+1)$$ we have $$x^3\equiv 1\mod (x^2+x+1)$$

Answer 3

It is basicaly the same solution as above. Let $$p(x)= x^{3m}+x^{3n+1}+x^{3p+2} $$ If $a$ is a solution of $x^2+x+1=0$ then $a^3=1$ so $$p(a)= a^{3m}+a^{3n+1}+a^{3p+2} = (a^3)^m +(a^3)^na+(a^3)^pa^2 = 1+a+a^2 =0$$ The same holds for the second solution $b$ which is different from $a$. So $p(x)$ is divisible by $$(x-a)(x-b) = x^2+x+1$$

Answer 4

$$ x^{3m}+x^{3n+1}+x^{3p+2}=(x^2+x+1)x^{3p}+x^{3m}-x^{3p}+x\left(x^{3n}-x^{3p}\right) $$