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

Question

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 $n > 3$.

Answer 1

$\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\QQ}{\mathbb Q}\newcommand{\ZZ}{\mathbb Z}$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 $\FF_p$.

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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.

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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. $\square$

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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 $1\leq d<n$. 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. $\square$

Answer 2

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

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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.

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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.

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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 $$x^3+y^3+z^3-3xyz=(x+y+z)(x+\omega y+\omega^2z)(x+\omega^2y+\omega z).$$

Answer 3

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.