Where I made a mistake during factoring $x^6+x^5+x^4+x^3+x^2+x+1$?

Question

In order to factor the expression, due to symmetry of coefficients if $r_1,r_2,r_3$ are zeros of $x^6+x^5+x^4+x^3+x^2+x+1$ then $\frac{1}{r_1}, \frac{1}{r_2} , \frac{1}{r_3}$ are also zeros. So we can rewrite:

$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2-(r_1+\frac{1}{r_1})x+1)+(x^2-(r_2+\frac{1}{r_2})x+1)(x^2-(r_3+\frac{1}{r_3})x+1)$$

And it can be written as:$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)(x^2+cx+1)$$

$$=x^6+(a+b+c)x^5+(3+ab+ac+bc)x^4+(abc+a+b+c)x^3+(3+ab+ac+bc)x^2+(a+b+c)x+1$$

In orther to find$a,b,c$ I should solve the system :

${ \begin{cases}{a+b+c=1} \\ {ab+ac+bc=-2} \\ {abc=0}\end{cases} }$

from the equation $abc=0$ We have $a=0$ (Since our derivation was symmetric in $a,b,c$ it doesn't matter to put $a=0$ or $b=0$ or $c=0$).

${ \begin{cases}{b+c=1} \\ {bc=-2}\end{cases} }$

and from that we have $b=2$ , $c=-1$ , $a=0$

So I get :

$$x^6+x^5+x^4+x^3+x^2+x+1=(x^2+1)(x^2+2x+1)(x^2-x+1)=(x^2+1)(x+1)^2(x^2-x+1)$$

But my answer is wrong because if we multiply the equation by $x-1$ We should obtain $x^7-1$. But:

$$(x-1)(x^2+1)(x+1)^2(x^2-x+1)=(x-1)(x+1)(x^2+1)(x+1)(x^2-x+1)=(x^4-1)(x^3+1) \ne x^7-1$$

I checked my answer several times but I couldn't find my mistake.

Answer 1

Noe that $$ (x^2+1)(x+1)^2(x^2-x+1)= x^6 + x^5 + x^4 + 2x^3 + x^2 + x + 1, $$ so you forgot a factor $2$, as remarked above. The polynomial $$ \Phi_7(x)=1+x+x^2+x^3+x^4+x^5+x^6 $$ is irreducible over the integers and over the rational numbers by the Gauss lemma and Eisenstein's criterion. Therefore such a factorization cannot hold.

Answer 2

To factor over $\mathbb{R}$, note that: $$ \begin{align} x^6+\cdots+1 &=\frac{1}{x-1}\left(x^7-1\right) \end{align} $$ where $x^7-1$ has its roots being the complex seventh roots of unity. One such root is of course $1$, while the other six pair up with their complex conjugate: $$ \begin{align} &x^6+\cdots+1\\ &=\frac{1}{x-1}(x-1)\left(x-e^{2\pi i/7}\right)\left(x-e^{-2\pi i/7}\right)\left(x-e^{4\pi i/7}\right)\left(x-e^{-4\pi i/7}\right)\left(x-e^{6\pi i/7}\right)\left(x-e^{-6\pi i/7}\right)\\ &=\left(x^2-2\cos(2\pi/7)x+1\right)\left(x^2-2\cos(4\pi/7)x+1\right)\left(x^2-2\cos(6\pi/7)x+1\right) \end{align} $$ This is the full factorization over $\mathbb{R}$.

Any factorization over $\mathbb{Q}$ would have to be composed of polynomials where complex conjugate roots are paired. So a factorization over $\mathbb{Q}$ is either trivial (the original polynomial is irreducible) or with all other factorings under consideration, at least one of those quadratic polynomials would be present, and one of $\cos(2\pi/7)$, $\cos(4\pi/7)$, $\cos(6\pi/7)$ is rational. But the only way that the cosine of a rational multiple of $\pi$ can be a rational number is if said cosine is $0$, $\pm\frac12$, or $\pm1$. (This last fact not trivial, just something I've memorized. A proof is accessible though.) And of course those three cosine values are not in $\{0,\pm\frac12,\pm1\}$. Therefore the original polynomial must be irreducible.