# Under the which condition, factorisation of $a_1^n+a_2^n+\cdots+a_n^n-na_1a_2a_3...a_n ?$ is possible?

Under the which condition, factorisation of the polynomial $$a_1^n+a_2^n+\cdots+a_n^n-na_1a_2a_3...a_n ?$$ is possible?

I know possible cases:

$$a^2+b^2-2ab=(a-b)^2$$

and

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$

There are $$2$$ things I'm interested in here. For which number $$n$$ is factorization possible? For which number $$n$$ it is not possible?

What I'm interested in here is general factorization possible? If not, is there any proof?

Based on the comments, I understand that general factorization is not possible.However, it is still unknown whether factorization is possible when $$n > 3$$.

• PARI/GP screens the case $n=4$ as irreducible. Currently, I check the case $n=5$ Oct 23, 2022 at 9:38
• For $n=5$ it is also irreducible. Oct 23, 2022 at 9:41
• @DietrichBurde Thank you , the case $n=5$ is already time-consuming in PARI/GP. Maybe , $n=3$ is the last reducible case ? Oct 23, 2022 at 9:43
• For $n=4$ see also this post, ad for $n=5$ this post, or this post. Oct 23, 2022 at 11:59
• Galois theory is rather for solvability , so I do not think it helps much here. I am not even aware of galois groups for multivariate polynomials, although this can probably be done. Oct 23, 2022 at 12:33

## 3 Answers

$$\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\QQ}{\mathbb Q}\newcommand{\ZZ}{\mathbb Z}$$The polynomial is irreducible over $$\mathbb C$$ for all $$n>3$$. This has been confirmed for $$n=4$$ and $$n=5$$ by some of the comments, so we may henceforth assume $$n>5$$. The proof outline is as follows:

1. Reduce the question of reducibility over $$\CC$$ to reducibility over $$\FF_p$$ for particular primes $$p$$. (Note that we use reducibility over $$\FF_p$$, not its algebraic closure.)
2. Show that reducibility over $$\FF_p$$ implies reducibility of $$x^n-t$$ for many values of $$t\in\FF_p$$, by substituting particular values and showing that we can recover irreducibility of the polynomial of interest.
3. Show that, in enough cases, these values of $$t$$ comprise all of $$\FF_p$$.

Lemma 1. Suppose $$f\in\ZZ[x_1,\dots,x_n]$$ is homogeneous and reducible over $$\CC$$, and let $$g\in\ZZ[y]$$ have no double roots in $$\CC$$. There exist infinitely many primes $$p$$ for which $$f\in\FF_p[x_1,\dots,x_n]$$ is reducible and $$g\in\FF_p[y]$$ splits completely.

Proof. We first need to express the idea that "$$f$$ is reducible" as a bunch of polynomial equations, and this is where we use the homogeneity of $$f$$. Since factors of homogeneous polynomials are homogeneous, $$f$$ is reducible if and only if can be written as the product $$f_1f_2$$ for some polynomials $$f_1$$ and $$f_2$$ of degrees $$r$$ and $$s$$, respectively. Suppose $$f$$ is reducible over $$\CC$$ with these degrees. Let variables $$w_1,\dots,w_M$$ and $$z_1,\dots,z_N$$ represent the coefficients of arbitrary degree $$r$$ and $$s$$, respectively, homogeneous polynomials. Letting $$x^{\alpha_1},\dots,x^{\alpha_M}$$ and $$x^{\beta_1},\dots,x^{\beta_N}$$ be the corresponding monomials (each represents the product $$x_1^{a_1}\cdots x_n^{a_n}$$ for some nonnegative integers $$a_1,\dots,a_n$$), a solution to $$\left(\sum_{i=1}^M w_ix^{\alpha_i}\right)\left(\sum_{j=1}^N z_jx^{\beta_j}\right)=f(x_1,\dots,x_n)$$ with $$w_1,\dots,w_M,z_1,\dots,z_N\in k$$ for a field $$k$$ implies reducibility of $$f$$ over that field (and we know that such a solution exists with $$k=\CC$$). Expanding and equating coefficients, we get some polynomials $$h_1,\dots,h_K\in\ZZ[w,z]$$ so that $$f$$ is reducible if these polynomials have a common zero.

We know these polynomials have a common zero in $$\CC$$; let one such zero be $$(a_1',\dots,a_M',b_1',\dots,b_N')$$. Then the ideal $$I$$ generated by $$h_1,\dots,h_K$$ in $$\CC[w,z]$$ is contained in the maximal ideal $$(w_1-a_1',\dots,w_M-a_M',z_1-b_1',\dots,z_N-b_N')$$ and is thus proper; in particular, there exist no polynomials $$j_1,\dots,j_K\in\CC[w,z]$$ for which $$h_1j_1+\cdots+h_Kj_K=1$$. This means there exist no such polynomials in $$\overline\QQ[w,z]$$, and as a result the ideal $$(h_1,\dots,h_K)\subset\overline\QQ[w,z]$$ is proper. This means it is contained in some maximal ideal $$\mathfrak m$$ of $$\overline\QQ[w,z]$$; by Hilbert's Nullstellensatz such an ideal must be $$(w_1-a_1,\dots,w_M-a_M,z_1-b_1,\dots,z_N-b_N)$$ for some $$(a_1,\dots,a_M,b_1,\dots,b_N)\in\overline\QQ^{M+N}$$. This means that $$f$$ is reducible over $$\overline\QQ$$.

We now convert this factorization into one over $$\FF_p$$ for infinitely many primes $$p$$. Let $$L$$ be the number field generated by the union of $$\{a_1,\dots,a_M,b_1,\dots,b_N\}$$ and the set of roots $$\{v_1,\dots,v_k\}$$ of $$g$$ in $$\overline\QQ$$. By the primitive element theorem we can write $$L=\QQ(\alpha)$$ for some $$\alpha\in L$$, whence there exist polynomials $$p_1,\dots,p_M,q_1,\dots,q_N,r_1,\dots,r_k\in\QQ[x]$$ for which $$p_i(\alpha)=a_i$$, $$q_i(\alpha)=b_i$$, and $$r_i(\alpha)=v_i$$. Let $$t\in\ZZ[x]$$ be the minimal polynomial of $$\alpha$$. Note that, since $$L\cong \QQ(x)/(t)$$, $$t(x)\mid h_j\big(p_1(x),p_2(x),\dots,p_M(x),q_1(x),\dots,q_N(x)\big)$$ for every $$1\leq j\leq K$$ and $$t(x)\mid g(r_i(x))$$ for every $$1\leq i\leq k$$ (both of these statements hold in $$\ZZ[x]$$). By an elementary result of Schur there exist infinitely many primes modulo which $$t$$ has a root. Take any such prime $$\ell$$; we claim that $$f$$ is reducible and and that $$g$$ splits completely modulo all but finitely many such primes. Indeed, let $$\beta\in\FF_\ell$$ be a root of $$f$$ in $$\FF_\ell$$, and define $$a_i^{(\ell)}=p_i(\beta)$$, $$b_j^{(\ell)}=q_j(\beta)$$, and $$v_j^{(\ell)}=r_i(\beta)$$. We have for each $$1\leq j\leq K$$ that $$h_j\big(a_1^{(\ell)},\dots,a_M^{(\ell)},b_1^{(\ell)},\dots,b_N^{(\ell)}\big)=h_j\big(p_1(\beta),\dots,p_M(\beta),q_1(\beta),\dots,q_N(\beta)\big)=t(\beta)u_j(\beta)=0$$ for some polynomial $$u_j\in\ZZ[x]$$, and for each $$1\leq i\leq k$$ that $$g(v_i^{(\ell)})=g(r_i(\beta))=t(\beta)v_i(\beta)=0$$ for some $$v_i\in\ZZ[x]$$. This implies by the definition of the $$h_j$$ that, over $$\FF_\ell$$, $$\left(\sum_{i=1}^M a_i^{(\ell)}x^{\alpha_i}\right)\left(\sum_{j=1}^N b_j^{(\ell)}x^{\beta_j}\right)=f(x_1,\dots,x_n),$$ so $$f$$ is reducible over $$\FF_\ell$$. Also, the values $$v_i^{(\ell)}$$ are each roots of $$g$$; since the $$v_i$$ are distinct, the polynomials $$r_i$$ are also distinct, and so modulo only finitely many primes does $$r_i(\beta)=r_j(\beta)$$ (such primes divide the resultant of $$t$$ and $$r_i-r_j$$, which is a constant since $$t$$ is irreducible). This means that $$g$$ has exactly the roots $$v_i^{(\ell)}$$ for large enough $$\ell$$, as desired. $$\square$$

Corollary 2. If $$f=x_1^n+\cdots+x_n^n-nx_1\cdots x_n$$ is reducible over $$\CC$$, there exist infinitely many primes $$p\equiv 1\pmod n$$ for which $$f$$ is reducible over $$\FF_p$$.

Proof. Let $$g=\Phi_n(y)$$ be the $$n$$th cyclotomic polynomial, which splits completely in $$\FF_p$$ if and only if $$n\mid p-1$$ (i.e. if and only if there are exactly $$n$$ $$n$$th roots of unity in $$\FF_p^\times$$). Applying Lemma 1 to $$f$$ and $$g$$ gives the result.

Let $$p\equiv 1\pmod n$$ be a prime, and let $$B\subset\FF_p^\times$$ be the unique subgroup of index $$n$$, i.e. $$B=\{x^n: x\in\FF_p^\times\}$$. Let $$A=\{0\}\cup B\subset\FF_p$$. Given a set $$S$$ inside an abelian group and a positive integer $$m$$, let $$mS=\{s_1+s_2+\cdots+s_m\colon s_1,\dots,s_m\in S\}$$. We will need the following somewhat technical result, mostly related to additive combinatorics.

Lemma 3. Let $$n>5$$. For all sufficiently large $$p\equiv 1\pmod n$$, $$(n-2)A=\FF_p$$.

Proof. Note that, since $$A$$ is permuted by multiplication by elements of $$B$$, $$mA$$ is as well for any $$m\geq 1$$, and so $$mA$$ consists of $$0$$ and some multiplicative cosets of $$B$$. In particular, $$|mA|\equiv 1\pmod{|B|}$$. By the Cauchy-Davenport theorem , we have $$|2A|\geq 2\big(|A|-1\big)+1=2|B|+1$$ with equality if and only if $$A$$ forms an arithmetic progression $$\{a+rd\colon r\in\{0,1,\dots,|A|-1\}\}$$ in $$\FF_p$$.

Suppose first that $$A$$ does not form an arithmetic progression. Then $$|2A|>2|B|+1$$. This implies $$|2A|\geq 3|B|+1$$, which means $$|4A|=|2(2A)|\geq \max(2\big(|2A|-1\big),p)\geq \max(6|B|+1,p).$$ This implies, using repeated applications of Cauchy--Davenport, \begin{align*} |(n-2)A| &=\big|4A+\underbrace{A+A+\cdots+A}_{n-6\text{ copies}}\big|\\ &\geq \max\big(|4A|+(n-6)(|A|-1),p\big)\\ &\geq \max(6|B|+1+(n-6)|B|,p\big)=\max(n|B|+1,p)=p. \end{align*} This means that $$(n-2)A=\FF_p$$.

Now, suppose $$A$$ does form an arithmetic progression; we will show $$p=n+1$$. Suppose not; such a progression must be of the form $$\{-sd,-(s-1)d,\dots,-d,0,d,2d,\dots,rd\}$$ for some nonnegative integers $$r$$ and $$s$$ with $$r+s=|B|$$, since $$0\in A$$. If $$r>1$$, then $$d\in B$$ implies that $$B=\{-s,-(s-1),\dots,-1,1,2,\dots,r\}$$ since $$x\in B$$ if and only if $$xd\in B$$. Since $$B$$ is a subgroup, $$2r$$ must be in $$B$$, which means $$p\mid 2r-j$$ for some $$j\in\{-s,\dots,r\}$$, and so $$p$$ has a multiple in the interval $$[r,2r+s]$$. In particular, $$p<2r+s\leq 2(r+s)=2|B|$$, a contradiction since $$|B|=(p-1)/n$$.

On the other hand, if $$r\leq 1$$, then $$s>0$$ (since $$p>n+1$$), and $$-d\in B$$, implying $$B\subset \{-1,1,2,\dots,s\}$$. The same argument now applies; $$2s\in B$$, meaning that $$p$$ is at most $$2|B|-1$$, again a contradiction. $$\square$$

We now use our lemmas to prove the desired result. Suppose that $$f=x_1^n+\cdots+x_n^n-nx_1\cdots x_n$$ is reducible over $$\CC$$, and let $$p\equiv 1\pmod n$$ be a large prime for which $$f$$ is reducible over $$\FF_p$$, guaranteed to exist by Corollary 2.

By Lemma 3, there exist $$a_1,\dots,a_{n-2}\in A$$ for which $$t=-(a_1+\cdots+a_{n-2})$$ is a primitive root. For each $$1\leq i\leq n-2$$, let $$b_i\in\FF_p$$ be so that $$b_i^n=a_i$$. Define $$f_1(x)=f(x,0,b_1,\dots,b_{n-2})=x^n+\sum_{i=0}^{n-2}b_i^n-0=x^n-t.$$ We first claim that $$f_1$$ is irreducible. Indeed, if $$x^n-t$$ has a degree $$d$$ divisor, then it has roots in $$\FF_{p^d}$$, so it suffices to show that $$x^n-t$$ has no roots in $$\FF_{p^d}$$ for any $$1\leq d. If there is such a root, then $$t$$ is in the subgroup of index $$n$$ of $$\FF_{p^d}^\times$$, and so this subgroup contains the subgroup $$\FF_p^\times$$ of index $$\frac{p^d-1}{p-1}$$ in $$\FF_{p^d}^\times$$. This implies $$n\mid\frac{p^d-1}{p-1}$$. Since $$p\equiv 1\pmod n$$, $$\frac{p^d-1}{p-1}=\sum_{i=0}^{d-1}p^i\equiv \sum_{i=0}^{d-1}1=d\pmod{n},$$ so $$d$$ must be at least $$n$$, as desired.

Now, suppose that $$f=gh$$ with $$g,h\in\FF_p[x_1,\dots,x_n]$$; we'll show that one of them is constant, contradicting the reducibility of $$f$$ over $$\FF_p$$. Defining $$g_1(x)=g(x,0,b_1,\dots,b_{n-2})\text{ and }h_1(x)=h(x,0,b_1,\dots,b_{n-2}),$$ we see that $$f_1=g_1h_1$$, so either $$g_1$$ or $$h_1$$ is constant; without loss of generality let it be $$g_1$$. Then $$h_1$$ contains a scalar multiple of $$x^n$$ as a monomial, and so $$h$$ contains a scalar multiple of $$x_1^n$$ a monomial. Since factors of homogeneous polynomials are homogeneous, $$\deg h=n$$ implies $$\deg g=0$$, which means $$g$$ is constant. $$\square$$

Trying my hand at this. Using a few elementary ideas from algebraic geometry.

For starters assume that the polynomial $$F(x_1,x_2,\ldots,x_n)=x_1^n+x_2^n+\cdots+x_n^n-nx_1x_2\cdots x_n$$ factors as a product of two polynomials $$F=HG$$, $$H,G\in\Bbb{C}[x_1,x_2,\ldots,x_n]$$. It follows immediately that $$H$$ and $$G$$ must also be homogeneous polynomials. Hence the zero loci $$V(F)$$, $$V(G)$$ and $$V(H)$$ of $$F,G,H$$ respectively are all codimension one algebraic subsets of the projective space $$\Bbb{P}^{n-1}(\Bbb{C})$$. By the properties of projective spaces the intersection $$V(G)\cap V(H)$$ has at least one component of codimension two. Furthermore, every point $$P\in V(G)\cap V(H)$$ is a singularity of $$V(F)$$. That is, $$F$$ as well as its gradient $$\nabla F=(\partial_{x_1}F,\ldots,\partial_{x_n}F)$$ vanish at $$P$$. This follows immediately from the product rule of (partial) differentiation: $$\partial_{x_i}F=(\partial_{x_i}G)H+G(\partial_{x_i}H)$$ implying that $$\partial_{x_i}F(P)=0$$ whenever $$G(P)=0=H(P)$$.

Assume that $$n\ge4$$. Then any codimension two component has dimension $$(n-1)-2\ge1$$ (the projective space itself has dimension $$n-1$$). So if we can show that the singularities of $$V(F)$$ are isolated points, we have shown that $$F$$ is irreducible as long as $$n\ge4$$. This is the plan of attack.

So assume that $$P=[a_1:a_2:\cdots : a_n]$$ is a singularity of $$V(F)$$. In addition to the equation $$F(P)=0$$ we thus have the equations $$\partial_{x_i}F(P)=na_i^{n-1}-na_1\cdots\hat{a_i}\cdots a_n=0\qquad(*)$$ for all $$i$$. I adopted the convention that the hat marks the variable missing from the product.

Case 1. We have $$a_i=0$$ for at least two indices $$i=i_1$$ and $$i=i_2$$, $$i_1\neq i_2$$. Then the product $$a_1a_2\cdots \hat{a_j}\cdots a_n$$ vanishes for all $$j=1,\ldots,n$$. The equation $$\partial_{x_j} F(P)=0$$ reads $$na_j^{n-1}=0$$, forcing all the coordinates to vanish, a case excluded by the properties of homogeneous coordinates.

Case 2. We have $$a_{i_0}=0$$ and $$a_j\neq0$$ for all $$j\neq i_0$$. In this case $$P$$ does not satisfy the equation $$\partial_{x_{i_0}}F(P)=0$$, so there are no singularities of this type.

Case 3. If all the homogeneous coordinates $$a_i$$ are non-zero, then for each pair of indices $$i,j$$ we get by multiplying the equations $$\partial_{x_i}F(P)=0$$ by $$x_i$$ that $$n a_i^n=na_1a_2\cdots a_n=na_j^n.$$ This implies that $$a_1^n=a_2^n=\cdots=a_n^n$$. Therefore the singularities are among the points with homogeneous coordinates $$[\zeta^{m_1}:\zeta^{m_2}:\cdots:\zeta^{m_n}],$$ where $$\zeta=e^{2\pi i/n}$$ is a primitive $$n$$th root of unity, and the integers $$m_i$$ are drawn from the set $$\{0,1,\ldots,n-1\}$$. In particular we see that there are at most $$n^{n-1}$$ singular points on $$V(F)$$ (we can always scale $$m_1=0$$).

The main claim follows from this.

It may be instructive to see that in the case $$n=3$$ the singularities are $$[1:1:1], [1:\omega:\omega^2], [1:\omega^2:\omega]$$ with the square of each coordinate being the product of the other two, all according to $$(*)$$. These are the points of pairwise intersections of the lines $$x+y+z=0$$, $$x+\omega y+\omega^2z=0$$, $$x+\omega^2y+\omega z=0$$ corresponding to the factorization $$x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+\omega^2z)(x+\omega^2y+\omega z).$$

• Thank you I appreciated. Does this happen all the time in math? A high-school level question cannot always be answered without higher mathematics?
– User
Oct 31, 2022 at 23:38
• LOL @User. Depends. More often than not I strive to find solutions requiring as little technology as possible. This time I could not think of one. It may be due to my relative inexperience, but when dealing with polynomials in several variables A) I don't know many other techniques, B) I have met such polynomials mostly in the context of (algebraic) geometry. C) Here I was also mildly amused by the contrast in the methodology to Carl's argument (+1), so I decided to see whether it works. I only spotted Macavity's comment later. That would lead to a simpler variation of this approach. Nov 1, 2022 at 4:38
• Ah, yes, I couldn't understand anything (mostly) when I read it. But mathematics is so beautiful that I wish I could understand it. I'm reading something I don't understand. But, honestly I still enjoy it. Thank you again. I'll put a bounty on the question again. Because both answers must be rewarded...
– User
Nov 1, 2022 at 9:11
• @User That is not necessarily a fruitful way of thinking about it. You see that I looked at factorizations with factors having complex coefficients. So did Vlad. If you specialize all but one of the variables, you end up with a polynomial in a single variable. And, because $\Bbb{C}$ is algebraically closed, only linear polynomials are irreducible there. In the rings of polynomials in two or more variables this is no longer an obstacle. That is the reason Vlad left two variables (and why Macavity left three variables, in the comment). Nov 5, 2022 at 7:22
• Irreducibility of univariate polynomials is a more interesting question only when we are interested in factors with coefficients in a "small" field (=a finite field or a finite extension of a suitable "ground field" such as $\Bbb{Q}$). Nov 5, 2022 at 7:26

I think the easiest to do it is with a specialization argument for $$n\geq 4$$. Namely it suffices to show that $$f(a_1,a_2,t_3,\ldots, t_n)$$ is irreducible over $$\mathbb{C}[a_1,a_2]$$ for a particular choice of the variables $$a_i=t_i\in \mathbb{C}$$ for $$i\geq 3$$.

Setting $$t_3=1$$ and $$t_4=\ldots=t_n=0$$ we should show that $$a_1^n+a_2^n+1$$ is irreducible over $$\mathbb{C}[a_1,a_2]$$. But this easy with Eisenstein's criterrion; namely $$a_2^n+1$$ splits into distinct linear factor and pick one of them.