I apologize for not having anything to show for an attempt. I've provided additional context, though, to keep it on topic. I hope it's enough.

The Question:

Factorise $$\begin{align}\, & 8wxyz \\ &+w^2yz+w^2xz+w^2xy \\ &+x^2yz+x^2wz+x^2wy \\ &+y^2xz+y^2wz+y^2wx \\ &+z^2xy+z^2wy+z^2wx \\ &+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \\ &+xyz+wyz+wxz+wxy=0\tag{I} \end{align}$$ over $\Bbb C$.


Here's a Q&A based on an answer from this meta question:

What are you studying?

A PhD in combinatorial group theory, first year.

What text is this drawn from, if any? If not, how did the question arise?

None. It arose as part of my research into the orders of abelianisations of certain cyclically presented groups.

What kind of approaches (to similar problems) are you familiar with?

Here's a related question about a system of equations generated by a smaller case:

The system $2XY+X^2+Y^2+X^2Y+Y+XY^2+X=0$, for distinct pairs $X, Y\in\{J, K, L\}$ for $n$th roots of unity $J, K, L$.

What kind of answer are you looking for? Basic approach, hint, explanation, something else?

A "simple" factorisation would suffice.

Is this question something you think should be able to answer? Why or why not?

No. I have little training in factorising polynomials in multiple variables.


The equation $(I)$ has $w=x=y=z=0$ as a solution.

Please help :)

NB: I $\color{red}{\text{suspect}}$ due to the context of the question that some of the roots are $n$th roots of unity. It would be weird if they weren't.

  • $\begingroup$ It is not quite clear what field this is over, and whether there are maybe additional constraints. Taking the real case at face value, suppose that $\,x=y=z=a \ne 0\,$ for example, then the equation reduces to the quadratic $\,(a + 3) w^2 + (3 a^2 + 14 a + 3) w + 3 a^2 + a=0\,$, which can certainly have non-zero $\,w\,$ roots of magnitude other than $1\,$. $\endgroup$ – dxiv Jul 15 '18 at 1:18
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    $\begingroup$ @dxiv I'll probably ask a separate question about a system of equations based on this, similar to the one I linked to in the question. $\endgroup$ – Shaun Jul 15 '18 at 1:40
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    $\begingroup$ Horner Form: $z (x (w (1 + w) + w x) + w x z) + y (x (w (1 + w) + w x) + y (w x + (w + (1 + w) x) z) + z (w (1 + w) + x (1 + w (8 + w) + (1 + w) x) + (w + (1 + w) x) z))$ $\endgroup$ – Mason Jul 15 '18 at 1:41
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    $\begingroup$ There are numerous professional tools for factoring polynomials. Have you tried turning any of them loose on this? $\endgroup$ – Steven Stadnicki Jul 15 '18 at 1:43
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    $\begingroup$ If it were 5wxyz instead of 8wxyz, there would be the factor 1+w+x+y+z $\endgroup$ – Empy2 Jul 15 '18 at 3:42

$$\begin{align}\, p_1(x, y, z, w)=\\ & 8wxyz \\ &+w^2yz+w^2xz+w^2xy \\ &+x^2yz+x^2wz+x^2wy \\ &+y^2xz+y^2wz+y^2wx \\ &+z^2xy+z^2wy+z^2wx \\ &+w^2xyz+wx^2yz+wxy^2z+wxyz^2 \\ &+xyz+wyz+wxz+wxy\tag{I} \end{align}$$

is a polynomial from $\mathbb{C}[x, y, z, w]=\mathbb{C}[x][y][z][w]$. So it can be seen as polynomial $p_2$ of the variable $w$ with coefficients from the integral domain $\mathbb{C}[x][y][z]$.

So we have $$\begin{align}\, p_2(w)=\\ &(yz+xz+xy +xyz)w^2\\ &+(x^2z+x^2y +y^2z+y^2x +z^2y+z^2x +x^2yz+xy^2z+xyz^2 +yz+xz+xy+8xyz)w\\ &+(x^2yz +y^2xz +z^2xy +xyz) \end{align}$$

The coefficients are from $\mathbb C[x,y,z]$. The constant term can be split into the irreducible factors $$x y z (x+y+z+1).$$

The coefficient of $w^2$ is a polynomial of $\mathbb C[x,y][z]$: $$(xy+x+y)z+xy$$ This is linear in $z$. So a factor can only be a common divisor of $xy+x+y$ and $xy$. But $$\gcd(xy+x+y,xy)=\gcd(x+y,xy)=\gcd(x^2,x+y)\\ \gcd(x,x+y)=\gcd(x,y)=1$$

So $$yz+xz+xy +xyz$$ is irreducible.

If $p_3(w)$ is a polynomial of degree $\gt 0$ that divides $p_2$, then $$p_2(w)=c_1w+c_0,$$ with $c_1, c_0 \in \mathbb C[x,y,z]$ and $$c_1 \mid (yz+xz+xy +xyz)$$ and $$c_0 \mid x^2yz +y^2xz +z^2xy +xyz.$$

So the possible values for $c_1$ are $$a, a(yz+xz+xy +xyz)$$ where $a \in \mathbb C$ and the possible values for $c_0$ are $$b, bx, bxy, bxyz,b(1+x+y+z), b(1+x+y+z)x, b(1+x+y+z)xy,b(1+x+y+z)xyz$$ where $b \in \mathbb C.$ Other divisors like $b y$ we will ignore here for symmetry reasons.

So if we check $$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$ we multiply $(2)$ by $$(b(1+x+y+z))^2$$ and get $$a(yz+xz+xy +xyz)w=b(1+x+y+z)$$

$$\begin{align}\, \\ (yz+xz+xy +xyz)(b(1+x+y+z))^2\\ +(x^2z+x^2y +y^2z+y^2x +z^2y+z^2x +x^2yz+xy^2z+xyz^2 +yz+xz+xy+8xyz)\\a(yz+xz+xy +xyz)b(1+x+y+z)\\ +(x^2yz +y^2xz +z^2xy +xyz)(a(yz+xz+xy +xyz))^2=0 \tag{3} \end{align}$$ and we have to check if there are appropriate $a$ and $b$ such that this equality holds. This task is simpler if we substitute $x,y,z$ by numbers and solve the resulting equations. I did this using a CAS and found out that all possible combinations result in $$a=b=0$$ So there are no proper factors.


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