How to factorize :$f(x)=x^n+x+1 \ \ \ \ \ \ : n=3k+2 ,k\in \mathbb{N}$

Question

How to factorize :

$$f(x)=x^n+x+1 \ \ \ \ \ \ : n=3k+2 ,k\in \mathbb{N}$$

And :

$$g(x)=x^n+x-1 \ \ \ \ \ \ : n=3k+2 ,k\in2m-1 \ \ \ , \ \ m\in\mathbb{N}$$

My try :

$$f(x)=x^n+x+1=x^{3k+2}+x+1$$

$$=(x^{3k+2}+x^{3k+1}+x^{3k})-(x^{3k+2}+x^{3k+1}+x^{3k})+(x+1)$$

Now what ?

@MichaelBurr According to wolfram, there does seem to be a nice factorisation which follows a pattern. — mrnovice, Jun 02 '17 at 14:22
You can also take a look at Bill Dubuque's answer here. The third cyclotomic polynomial comes up surprisingly often. — Jyrki Lahtonen, Jun 02 '17 at 14:41

Michael Rozenberg · Accepted Answer · 2017-06-02T14:29:30.290

3

$$x^{3k+2}+x+1=x^{3k+2}-x^2+x^2+x+1=x^2((x^3)^k-1)+x^2+x+1$$ and use $x^3-1=(x-1)(x^2+x+1)$.

For $n=6m-1$ we have $$x^n+x-1=x^n+x^2-x^2+x-1=x^2(x^{n-2}+1)-(x^2-x+1)$$ and use $x^3+1=(x+1)(x^2-x+1).$

edited Jun 02 '17 at 14:29

answered Jun 02 '17 at 14:23

Michael Rozenberg

score 2 · Answer 2 · answered Jun 02 '17 at 14:24

2

Hint: evaluate $f(x)$ at $\omega$, where $\omega=e^{2\pi i/3}$, so $\omega^3=1$, and $\omega^2 + \omega + 1 = 0$.

answered Jun 02 '17 at 14:24

B. Mehta

2 Answers2